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Unformatted text preview: .3) as n (5.2.5) n jk jk (I ; 2t L1 )U +1 2 = (I + 2t L2 )U n = (5.2.6a) n jk jk (I ; 2t L2 )U +1 = (I + 2t L1 )U +1 2 : (5.2.6b) When applied to the heat conduction equation with centered spatial di erences, this scheme is also identical to the ADI scheme (5.1.5). Let us verify that (5.2.5) and (5.2.6) are equivalent. Thus, operate on (5.2.6b) with I ; tL1 =2 to obtain n n jk jk = (I ; 2t L1 )(I ; 2t L2 )U +1 = (I ; 2t L1 )(I + 2t L1 )U +1 2 : n n jk jk = 5.2. Operator Splitting 15 The operators on the right may be interchanged to obtain (I ; 2t L1 )(I ; 2t L2 )U +1 = (I + 2t L1 )(I ; 2t L1 )U +1 2 : n n jk jk = Using (5.2.6a) yields (5.2.5). Example 5.2.3. D'Yakonov (cf. 4], Section 2.12) introduced the following scheme for solving (5.2.5) ^ (I ; t L1 )U +1 = (I + t L1 )(I + t L2 )U 2 2 2 n n jk jk ^ (I ; 2t L2 )U +1 = U +1: n (5.2.7b) n jk (5.2.7a) jk This scheme has the same order of accuracy and characteristics as the Peaceman-Rachford ADI scheme (5.2.6). Example 5.2.4. Douglas and Rachford 2] developed an alternative scheme for (5.2.1) using backward-di erence approximations. Thus, consider integrating (5.2.1) for a time step by the backward Euler method to obtain (I ; tL1 ; tL2 )u +1 = u + O( t)2: n n jk jk Let us rewrite this as (I ; tL1 ; tL2 + t2 L1L2 )u +1 = (I + t2 L1L2)u + n t2 1 2 (u LL +1 ; u n jk jk n n jk jk ) + O( t)3: As in Example 5.2.2, we may show that the next-to-last term on the right is O( t3) and, hence, may be neglected. Also neglecting the temporal discretization error term, discretizing the operators L1 and L2, and factoring the left side gives (I ; tL1 )(I ; tL2 )U +1 = (I + t2 L1 L2 )U : n n jk jk Douglas and Rachford 2] factored this as ^ (I ; tL1 )U +1 = (I + tL2 )U n n jk jk (5.2.8a) 16 Multi-Dimensional Parabolic Problemss ^ (I ; tL2 )U +1 = U +1 ; tL2 U n n jk (5.2.8b) n jk jk Assuming that the discrete spatial operators are second-order accurate, the local discretization error is O( t) + O( x2) + O( y2). This is lower order than the O( t2) local discretization error that would be obtained from the ADI factorization (5.2.6) however, backward di erencing gives greater stability than (5.2.6) which may be useful for nonlinear problems. Example 5.2.5. The previous examples suggest a simplicity that is not always present. Boundary conditions must be treated very carefully since the intermediate solutions or ^ U +1 2 or U +1 need not be consistent approximations of u(x (n + 1=2) t) or u(x (n + 1) t). Yanenko 8] presents a good example of the complications that can arise at boundaries. Strikwerda 7], Section 7.3, suggests using a combination of U and U +1 to...
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## This document was uploaded on 03/16/2014 for the course CSCI 6840 at Rensselaer Polytechnic Institute.
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# Career Definitions – MATHEMATICS
## ACTUARIAL SCIENCE (DEG/PG)
MATHEMATICS – It is a field dealing with statistical, mathematical and financial calculation involving probability of payments, contingencies in pension and insurance palms such as insurance against losses arising from death, disability, sickness and unemployment and determine the proper basis and methods for valuing liabilities and maintaining permanent financial stability of insurance & pension organisation.
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## ALGEBRA (M.PHIL)
Branch of ma-thematic dealing with properties of numbers and qualities by means of letters and other general symbols, system of this based on given axioms.
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## ANALYSIS (M.PHIL)
Branch of ma-thematic analysis concerned with properties of whole calculations of fractions.
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## APPLIED MATHEMATICS (DEG/PG/PhD)
Those mathematical topic of principal concerns in the resolution of problems in the physical and life sciences, in engineering and technology, in economics, linguistics, phonetics, and music, and indeed in all human endeavor involving problems that are amenable to mathematical analysis and solution.
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## BIO-STATISTICS (PG)
The branch of statistics that deals with relating to life, vital statistics.
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The use of fundamental principles of arithmetic in solving mathematical problems that occur in day- to- day business activities.
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## CONTINUUM MECHANICS (M.PHIL)
Mechanism of a continuous thing, quantity, or substance, a continuous series of elements passing into each other.
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## MATHEMATICAL ECONOMICS (DEG)
A branch of economics that seeks to verify economic theory and measure economic relationships by statistical and mathematical methods chiefly for the purpose of forecasting future events and choosing desirable policies.
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## MATHEMATICAL SCIENCE (PG)
The deductive study of shape, quantity, and dependence, the two main areas are applied mathematics and pure mathematics, the former erasing from the study of physical phenomena, the latter involving in the intrinsic study of mathematical structures.
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## MATHEMATICAL STATISTICS(PG)
Refers to using mathematical techniques for designing and improving statistical methods to obtain and interpret numerical information.
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## MATHEMATICS(PURE) (DEG/PG/PG DIP/PhD)
The deductive study of shape, quantity, and dependence. The two main areas of mathematics are applied mathematics and pure mathematics, the former arising from the study of physical phenomena. The latter involving the intrinsic study of mathematical structures.
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## NUMERICAL ANALYSIS(M PHIL)
The branch of mathematics dealing with the development and use of numerical methods for solving problems.
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## STATISTICS(DEG/PG DIP/PG)
The science dealing with the collection, analysis, interpretation, and presentation of masses of numerical data.
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## STOCHASTIC(M PHIL)
In econometric s, a formula that describes the relationship between two or more variables and that is tested with actual data. Making statistical allowance for error.
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after12thwhat | 836 | 3,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-38 | latest | en | 0.869207 |
https://slideplayer.com/slide/6839409/ | 1,631,820,276,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053717.37/warc/CC-MAIN-20210916174455-20210916204455-00130.warc.gz | 587,084,557 | 18,578 | # HW # 21- HOLT p. 68 # even Warm up
## Presentation on theme: "HW # 21- HOLT p. 68 # even Warm up"— Presentation transcript:
HW # 21- HOLT p. 68 # 19-52 even Warm up
Week 6, Day One HW # HOLT p. 68 # even Warm up Write each decimal as a fraction in simplest form. A. 8.75 B 0.27 –0.625 Write 13/6 as a decimal.
Warm Up Response A. 8.75 75 100 = 8 8.75 5 is in the hundredths place, so write hundredths as the denominator. = 8 3 4 Simplify by dividing by the greatest common divisor. B ,000 = 5 is in the ten-thousandths place. 0.2625 = 21 80 Simplify by dividing by the greatest common divisor. 27 100 C. 0.27 D. –0.625 5 / 8 E. Write as a decimal. 13/ 6 2.16 0.325
Homework Check Guided Practice # 1,11,17 0.625 11) ¾ 17) 3 21/100
Goals for Today Writing Repeating Decimals as Fractions
Chapter 1 quiz – pass back (≤20 pts see me at PT) Factor Trees II worksheet- compare with your neighbor DO class work: p. 68 # 1-18 AND 27, 37, 45 Clean out your binder and put your MMC project in your spiral (if you have not done this already).
To write a terminating decimal as a fraction, identify the place value of the digit farthest to the right. Then write all of the digits after the decimal point as the numerator with the place value as the denominator.
7 is in the hundredths place, so write hundredths as the denominator.
For Example Write each decimal as a fraction in simplest form. A. 5.37 7 is in the hundredths place, so write hundredths as the denominator. 37 100 = 5 5.37 B 2 is in the thousandths place, so write thousandths as the denominator. = 0.622 = Simplify by dividing by the greatest common divisor.
A fraction is in reduced, or simplest, form when the numerator and the denominator have no common divisor other than 1. Remember!
5 is in the hundredths place, so write hundredths as the denominator.
Example 2 Write each decimal as a fraction in simplest form. A. 8.75 5 is in the hundredths place, so write hundredths as the denominator. 75 100 = 8 8.75 = 8 3 4 Simplify by dividing by the greatest common divisor. B ,000 = 5 is in the ten-thousandths place. 0.2625 = 21 80 Simplify by dividing by the greatest common divisor.
Example: Writing Repeating Decimals as Fractions
_ Write 0.4 as a fraction in simplest form. x = … Let x represent the number. Multiply both sides by 10 because 1 digit repeats. 10x = 10( …) 10x = … Subtract x from both sides to eliminate the repeating part. Since x = …, use … for x on the right side of the equation. -x = … 9x = 4 9x = 4 Since x is multiplied by 9, divide both sides by 9. x =
Write 0.36 as a fraction in simplest form.
Example 2 __ Write 0.36 as a fraction in simplest form. x = … Let x represent the number. Multiply both sides by 100 because 2 digits repeat. 100x = 100( …) 100x = … Subtract x from both sides to eliminate the repeating part. Since x = …, use … for x on the right side of the equation. -x = … 99x = 36 99x = Since x is multiplied by 99, divide both sides by 99. x = = Write in simplest form.
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Similar presentations | 908 | 3,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-39 | latest | en | 0.888917 |
http://stackoverflow.com/questions/11009818/how-to-get-list-of-integer-from-string/11010028 | 1,406,086,001,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997872261.34/warc/CC-MAIN-20140722025752-00168-ip-10-33-131-23.ec2.internal.warc.gz | 147,487,602 | 18,230 | # How to get list of Integer from String
my string contains Integer separated by space:
String number = "1 2 3 4 5 "
How I can get list of Integer from this string ?
-
Have search for java-examples.com/java-string-split-example – Preet Sangha Jun 13 '12 at 7:10
You can use a Scanner to read the string one integer at a time.
Scanner scanner = new Scanner(number);
List<Integer> list = new ArrayList<Integer>();
while (scanner.hasNextInt()) {
}
-
+1 you should mention hasNextInt() and nextInt() :) – Eng.Fouad Jun 13 '12 at 7:13
@Eng.Fouad - I was getting to that. :) – Ted Hopp Jun 13 '12 at 7:14
hm nice example thx a lot – hudi Jun 13 '12 at 7:22
Just one question is this method faster or slower then method mention in other answers ? – hudi Jun 13 '12 at 7:26
All these examples will run in O(n) time. So the answer is there is no real performance difference. Edit: Actually, splitting the strings first like (suggested in another post), and then adding them to a collection will contain 2 x O(n) loops. This solution would be better – John Snow Jun 13 '12 at 7:28
split it with space, get an array then convert it to list.
-
when I convert array to list with method asList I get array of string not Integer – hudi Jun 13 '12 at 7:11
@hudi you can convert those string to integer using 'Integer.parseInt()' method – Chandra Sekhar Jun 13 '12 at 7:15
I dont like this method becaue when in this string isnt number then occurs exceptions and I looking for some faster method to convert this not one by one – hudi Jun 13 '12 at 7:16
String number = "1 2 3 4 5";
String[] s = number.split("\\s+");
List<Integer> myList = new List<Integer>();
for(int index = 0 ; index<5 ; index++)
-
ArrayList<Integer> lst = new ArrayList<Integer>();
for (String field : number.split(" +"))
-
Firstly,using split() method to make the String into String array.
Secondly,using getInteger() method to convert String to Integer.
-
String number="1 2 3 4 5";
List<Integer> l=new ArrayList<Integer>();
String[] ss=number.split(" ");
for(int i=0;i<ss.length;i++)
{
}
System.out.println(l);
-
Simple solution just using arrays:
// variables
String nums = "1 2 3 4 5";
// can split by whitespace to store into an array/lits (I used array for preference) - still string
String[] num_arr = nums.split(" ");
int[] nums_iArr = new int[num_arr.length];
// loop over num_arr, converting element at i to an int and add to int array
for (int i = 0; i < num_arr.length; i++) {
int num_int = Integer.parseInt(num_arr[i])
nums_iArr[i] = num_int
}
That pretty much covers it. If you wanted to output them, to console for instance:
// for each loop to output
for (int i : nums_iArr) {
System.out.println(i);
}
-
too much code for this easy task – hudi Jun 13 '12 at 7:27
Okay, simple modification: First for loop: Sub out the int num_int for nums_iArr = Integer.parseInt(num_arr[i]) Drops a line of code. Is that enough or would you suggest something more? – Healsgood Jun 13 '12 at 7:32
You can split it and afterwards iterate it converting it into number like:
String[] strings = "1 2 3".split("\\ ");
int[] ints = new int[strings.length];
for (int i = 0; i < strings.length; i++) {
ints[i] = Integer.parseInt(strings[i]);
}
System.out.println(Arrays.toString(ints));
-
You first split your string using regex and then iterate through the array converting every value into desired type.
String[] literalNumbers = [number.split(" ");][1]
int[] numbers = new int[literalNumbers.length];
for(i = 0; i < literalNumbers.length; i++) {
numbers[i] = Integer.valueOf(literalNumbers[i]).intValue();
}
-
I needed a more general method for retrieving the list of integers from a string so I wrote my own method. I'm not sure if it's better than all the above because I haven't checked them. Here it is:
public static List<Integer> getAllIntegerNumbersAfterKeyFromString(
String text, String key) throws Exception {
text = text.substring(text.indexOf(key) + key.length());
List<Integer> listOfIntegers = new ArrayList<Integer>();
String intNumber = "";
char[] characters = text.toCharArray();
boolean foundAtLeastOneInteger = false;
for (char ch : characters) {
if (Character.isDigit(ch)) {
intNumber += ch;
} else {
if (intNumber != "") {
foundAtLeastOneInteger = true;
intNumber = "";
}
}
}
if (!foundAtLeastOneInteger)
throw new Exception(
"No matching integer was found in the provided string!");
return listOfIntegers;
}
The @key parameter is not compulsory. It can be removed if you delete the first line of the method:
text = text.substring(text.indexOf(key) + key.length());
or you can just feed it with "".
- | 1,191 | 4,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-23 | latest | en | 0.715339 |
https://stats.libretexts.org/Bookshelves/Applied_Statistics/Introductory_Business_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.00%3A_Introduction_to_the_Central_Limit_Theorem | 1,695,694,761,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510130.53/warc/CC-MAIN-20230926011608-20230926041608-00226.warc.gz | 603,670,323 | 29,318 | # 7.0: Introduction to the Central Limit Theorem
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the Central Limit Theorem.
The Central Limit Theorem is one of the most powerful and useful ideas in all of statistics. The Central Limit Theorem is a theorem which means that it is NOT a theory or just somebody's idea of the way things work. As a theorem it ranks with the Pythagorean Theorem, or the theorem that tells us that the sum of the angles of a triangle must add to 180. These are facts of the ways of the world rigorously demonstrated with mathematical precision and logic. As we will see this powerful theorem will determine just what we can, and cannot say, in inferential statistics. The Central Limit Theorem is concerned with drawing finite samples of size $$n$$ from a population with a known mean, $$\mu$$, and a known standard deviation, $$\sigma$$. The conclusion is that if we collect samples of size $$n$$ with a "large enough $$n$$," calculate each sample's mean, and create a histogram (distribution) of those means, then the resulting distribution will tend to have an approximate normal distribution.
The astounding result is that it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact is that the distribution of sample means tend to follow the normal distribution.
The size of the sample, $$n$$, that is required in order to be "large enough" depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means. Sampling is done randomly and with replacement in the theoretical model.
This page titled 7.0: Introduction to the Central Limit Theorem is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 848 | 3,225 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-40 | latest | en | 0.668348 |
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# Answer to Question #87672 in Financial Math for Udeh George
Question #87672
P and Q are partners in a venture,P contributed $20,000 for nine months and Q contributed$50,000 for one year:find each person's share of profit of $6,300? 1 Expert's answer 2019-04-08T09:01:33-0400 If P contributed$20,000 for nine months and Q contributed $50,000 for one year, then P contributed 20,000*9 =$180,000 in total and Q contributed 50,000*12 = $600,000 in total. Their total contribution is$780,000, so P contributed 180,000/780,000 = 0.231 or 23.1% and Q contributed 600,000/780,000 = 0.769 or 76.9%.
Each person's share of profit of $6,300 is: P will receive 0.231*6,300 =$1,453.85,
Q will receive 6,300 - 1,453.85 = \$4, 846.15.
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https://www.shaalaa.com/question-bank-solutions/a-wheel-making-revolutions-about-its-axis-uniform-angular-acceleration-starting-rest-it-reaches-100-rev-sec-4-seconds-find-angular-acceleration-equations-rotational-motion_67156 | 1,618,854,853,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038916163.70/warc/CC-MAIN-20210419173508-20210419203508-00139.warc.gz | 1,076,991,633 | 9,462 | # A Wheel is Making Revolutions About Its Axis with Uniform Angular Acceleration. Starting from Rest, It Reaches 100 Rev/Sec in 4 Seconds. Find the Angular Acceleration. - Physics
Sum
A wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest, it reaches 100 rev/sec in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds.
#### Solution
Given
= 4s
Initial angular velocity = $\omega_0 = 0$
Final angular velocity = $\omega = 100 \text{ rev/s}$
$\omega = \omega_0 + \alpha t$
$\alpha = \frac{\omega}{t}$
$\alpha = \frac{100}{4}\text{ rev/ s}^2 = 25\text{ rev/ s}^2$
Now, we have
$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$
$\Rightarrow \theta = \frac{1}{2} \times 25 \times 16$
$= 200^\circ$
$\Rightarrow \theta = 200 \times 2\pi\text{ radians}$
$= 400\pi\text{ radians}$
Concept: Equations of Rotational Motion
Is there an error in this question or solution?
#### APPEARS IN
HC Verma Class 11, 12 Concepts of Physics 1
Chapter 10 Rotational Mechanics
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2-3 example5 - Student Grady Simonton Instructor Shawn...
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Unformatted text preview: Student: Grady Simonton Instructor: Shawn Parvini Date: 2/18/10 Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 - 16 weeks Book: Triola: Elementary Statistics, 11e Time: 11:06 ANI The table below shows the frequency distribution of home voltage measurements taken on 50 consecutive days. Use the frequency distribution to construct a histogram. Does the result appear to be a normal distribution? Home Voltage Class Frequency 123.3— 123.4 12 123.5— 123.6 11 123.7— 124.8 13 123.9— 124.0 12 124.1 — 124.2 2 A histogram is a graph consisting of bars of equal width drawn adjacent to each other (without gaps). The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to the frequency values. Use the given frequency distribution to construct the histogram. The correct graph is shown below. 20— _L ‘t‘ Frequency 3 U1 0 123.75 X Values 123.35 124.15 When graphed, a normal distribution has a "bell" shape. One characteristic of the bell shape is that the frequencies increase to a maximum and then decrease. Another characteristic is symmetry, with the left half of the graph roughly a mirror image of the right half. Examine the histogram. N ‘P _\ U1 I 0 Frequency 001 123.75 124.15 K Values 123.35 The histogram is not approximately symmetric because the far left side of the histogram is high, at 12, while the far right of the histogram is low, at 2. There is no obvious maximum. Use this information to determine if the frequency distribution appears to be normal. Page 1 ...
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{[ snackBarMessage ]} | 483 | 1,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-17 | latest | en | 0.797884 |
https://math.stackexchange.com/questions/2546640/how-many-times-one-has-to-throw-a-dice-to-obtain-a-multiplicity-of-6 | 1,621,002,205,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989526.42/warc/CC-MAIN-20210514121902-20210514151902-00580.warc.gz | 402,409,648 | 38,304 | # how many times one has to throw a dice to obtain a multiplicity of 6
I throw a dice until sum of results is divided by 6. What's an expectation value of number of throws?
I thought that it will be a sum: $\sum_{k=1}^{\infty} k \cdot \left(\frac{6^k - 6^{k-1} - 6^{k-2}\cdot5 - ... -6\cdot5^{k-2}}{6^k}\right)$ but the sum in the brackets is bigger than 1 for large $k$ so I had to made a mistake in my way of thinking.
HINT: The probability of the sum being $k$ modulo $6$ after any throw is $\frac 16$, so in particular you have $\frac 16$ probability that the sum will be divisible by $6$ after the $k$-th throw.
• so is it a sum: $\sum_{k=1}^\infty k* (\frac{5}{6})^{k-1} \frac{1}{6}$ ? – Filip Parker Dec 1 '17 at 22:13
• @FilipParker Your writing is a bit messy, but I guess you meant: $$\frac 16 \sum_{k=1}^{\infty} k\left(\frac 56 \right)^{k-1}$$ If yes, then it's true. Now use the geometric progression formula to find the actual value. – Stefan4024 Dec 1 '17 at 22:18 | 341 | 982 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-21 | latest | en | 0.865066 |
https://www.jackalopejobs.com/how-to-calculate-salary-from-hourly-rate/ | 1,716,442,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00260.warc.gz | 754,337,852 | 17,129 | # How to Calculate Salary From Hourly Rate?
The hourly rate is how much money is paid per hour. You can calculate your own rate if you know how many hours you usually work. If you don’t, there are calculators that will help you calculate it. These include an hourly wage calculator and a salary calculator.
An hourly rate is calculated by multiplying the rate of pay by the number of hours worked. For example, if your salary is \$100, your hourly rate is \$14. Similarly, if you work 120 hours, your hourly rate will be \$1680.
There are a few things to consider when calculating an hourly rate. First, you need to determine your weekly and annual average hours. This means that you’ll have to know how many days you work each week, how many days you work each month, and how many days you work each year.
A standard working week for a full-time employee is 40 hours. Depending on your work schedule, you might find yourself working less or more than that in a given month.
## How Can I Calculate Salary?
If you have an hourly wage, you can calculate how much you make by dividing your hourly rate by the number of hours you work in a week. This is a quick way to get an idea of how much your annual salary would be.
Calculating your salary from an hourly rate is not hard. All you need is a standard workweek, like 40 hours.
Hourly rates can be different depending on the type of job you have. You may be paid hourly, or you may be paid in the form of a semi-monthly or monthly payment. Some companies pay you for cell phone usage, company cars, and other extras.
The number of weeks worked is also important. For example, a full-time employee working a 40-hour workweek can take two weeks off each year. However, if you work more than four weeks a month, you might be paid every other month or every year.
In addition to the number of weeks you work, you should consider the number of hours you work per week. An employee who works a full-time 45-hours-per-week can be paid a salary of \$42,000.
## What is the Easiest Way to Calculate Salary?
If you work for a company that pays a salary, you’ll need to convert it into an hourly wage. The easiest way to do this is to use an hourly to salary calculator. You can use the calculator to determine the hourly rate for a salaried employee, or even the average monthly salary for someone working the same number of hours each week.
READ ALSO: How Many Weeks in a Salary Year?
To calculate the hourly rate, you first need to know how many hours you usually work. This can be done by tracking time for a month or two. For a full-time employee, the standard workweek is 40 hours. Using this as your approximation, divide your weekly salary by 40 to get your hourly rate.
Next, multiply the weekly rate by the number of paid hours you have each month. For instance, if you have a salary of \$14 per hour, your salary is \$1680.
After you have your annual and monthly income, subtract taxes, bonuses and other deductions to get your net pay. Your net pay is your max after all deductions.
## How is Salary Calculated in the Philippines?
Whether you are a new Filipino employee or a seasoned Pinoy worker, you might want to know how is salary calculated in the Philippines. There are many rules and regulations that govern this process. Some companies outsource the payroll process to third party service providers. Using an internal process is also viable. However, it requires more resources and can be a costly endeavor.
The Philippines has a unique set of employment laws that benefit both the employer and the employee. One of these is the overtime pay system. As long as the regular hours of work do not exceed eight in a day, an employee is entitled to receive overtime.
This payment is generally computed by multiplying the employee’s normal hourly rate by 1.25%. For every extra hour worked in a week, the employee is entitled to a time and a quarter of the regular hourly wage.
Another important factor to consider when computing overtime pay in the Philippines is the cost of living allowance (COLA). This allowance is paid to all minimum wage earners.
## How Do You Calculate Monthly Salary?
If you’re looking for the best way to calculate your monthly salary from hourly rate, you’ve come to the right place. This article will guide you through the steps required to make the conversion.
First, you need to find out how many hours you work. The average full time employee works about 40 hours per week. If your schedule isn’t set up that way, you can use a different number. For instance, if you usually work nine hours a day, you can use eight. However, you may need to add some overtime if you’re working a lot of overtime.
READ ALSO: How Long Do Salary Negotiations Take?
Once you know how many hours you’re working, you can calculate your salary. This may sound complicated, but it’s actually quite simple.
You can do this by using a calculator. There are several different types of salary calculators to choose from. They all include different features. Some of them will show you the exact amounts you’re owed for your paycheck. Other will show you the gross amount you’ll earn after you’ve taken deductions for taxes and pre-tax contributions.
## What is the Best Salary Calculator?
If you’re planning to switch careers or you’re interested in negotiating a new contract, a salary calculator can help you get a good idea of what you should be earning. A pay calculator can compare your current salary with other salaries in your industry, by your job title, and by your location.
There are a variety of salary calculators available on the Internet. Some are free to use and others cost a fee. Before you make a decision, you should take a look at the options and decide which one is right for you.
NerdWallet’s salary calculator is an easy-to-use tool that provides you with an estimate of your earnings. It also calculates the taxes you’ll owe, how much you can save, and how much you can afford to live on. The site is also an excellent resource for comparing cost of living, calculating transportation expenses, and reviewing the costs of education and other forms of quality of life.
Glassdoor’s Salary Calculator offers a number of features, including anonymous salaries and the ability to view salary information by industry or company. The tool is also designed with salary equity in mind, meaning it reflects what other users with similar jobs have been paid.
## What is the Formula to Calculate a Day Salary?
If you’re paid on a salary basis, you may need to know how to calculate a day’s salary from hourly rate. Getting an estimate is especially important during tax season.
There are many factors that can determine your pay. Some of them are the amount of time you work, the type of job you do, and your geographic location. Your hourly rate will be higher if you work a lot of overtime, and lower if you have low overtime needs.
When you’re calculating your annual salary, you should also take into account your benefits. These will help you stay healthy and motivate you to perform at your best.
READ ALSO: How to Do Salary Negotiation?
You can also consider the number of vacation days you have each year. For example, if you’re given nine vacation days each month, you should multiply the number of weeks you’re off each year by two. This will give you an average monthly income.
Aside from paying you for your time, your employer may pay you for your cell phone, a computer, and other benefits. These can add to your overall earnings.
## How Salary is Calculated For 30 And 31 Days?
Usually, employees receive their pay in a monthly, bi-monthly or semi-monthly cycle. The total salary is divided by the number of calendar days in a month, and then minus Sundays. Some organizations add holidays to their pay days.
The calculation of a per day wage is a more complex process. It is based on the average number of hours worked during a period. For example, if an employee works 30 hours a week at a base rate of \$15, he or she can expect to earn \$950. However, the annual rate is much larger.
In practice, however, calculating a per day wage based on the calendar day may not be the most accurate way to do it. One method is to multiply the total pay by the number of days in a month, which is easier than trying to calculate the total pay for each day.
The other option is to calculate the overall pay for a month and then divide it by the number of calendar days in a period. This calculation is a little more complex than the aforementioned method, but can be worth it in the end. | 1,820 | 8,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-22 | latest | en | 0.961311 |
http://www.multiplication.com/learn/learn-fact/2/x/7 | 1,490,737,190,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189903.83/warc/CC-MAIN-20170322212949-00155-ip-10-233-31-227.ec2.internal.warc.gz | 604,681,722 | 11,714 | # Learn a Fact: 2 x 7
Follow steps 1-6 to master this fact.
You're learning
2
x7
14
7
x2
14
2
X 6
12
2
X 8
16
## This picture and story will help you remember 2 x 7 = 14
2 (Shoe) x 7 (Surfin) = 14 (Four Kings)
2x7 | 97 | 225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-13 | longest | en | 0.858511 |
https://beta.geogebra.org/m/tV8BnPEh | 1,675,348,390,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500028.12/warc/CC-MAIN-20230202133541-20230202163541-00524.warc.gz | 144,607,049 | 10,360 | # Interior Angles: Triangles
## The interior angles of a triangle sum to two right angles
The objective of this activity is to understand WHY the interior angles of a triangle sum to two right angles
## Make a Construction
Create a triangle as the intersection of three lines: AB, BC, AC (plot three points on the plane and connect them with *lines* --not segments). Create a line parallel to AB through point C. Label points on each of the lines extending beyond the triangle, like this:
## Measure Angles
Please measure angles using the angle measurement tool. Please measure only the interior angles of ABC. Modify the color of each angle so that they are all different. | 142 | 679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-06 | latest | en | 0.882074 |
https://www.geographercorner.com/2022/03/climatology-125-mcqs-mock-test-1-for-nta-ugc-net.html | 1,721,764,414,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00200.warc.gz | 692,944,562 | 40,133 | # Climatology 125 MCQS Mock Test 1 FOR NTA UGC NET
Climatology 125 MCQS Mock Test 1 FOR NTA UGC NET
Climatology 125 MCQS Mock Test 1 NTA UGC NET, Climatology 125 MCQS Mock Test 1, NTA UGC NET PYQS GEOGRAPHY, Climatology 125 MCQS Mock Test 1, GEOGRAPHY PYQS UGC NET, NTA UGC NET GEOGRAPHY PAPER WISE PYQS, PYQS FOR UGC NET GEOGRAPHY,
Climatology 125 Questions Mock Test for NTA UGC NET & States SET/SLET Exam, This is the Part 1 of this series.
To attempt this Mock test, click on the below link
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Created on By netset corner
CLIMATOLOGY 125 MCQS PART 1
CLIMATOLOGY 125 MCQS PART 1
• This Test Contains 25 Questions
• Total Questions is 125
• All Questions is Compulsory
• No Negative Marking
• Specific Time given for this quiz
• Quiz / Test will start after clicking on Start Button
• After Completing Quiz, can see result in Percentages (%)
• Attempt Carefully, so lets start, best wishes
Dear Aspirants,You Need to Login this page by Your Name & Email Address to Access the Mock Test / Quiz.
1 / 25
Q.1. How much percentage of water vapour contains in the atmosphere
(A) 0-2 percent
(B) 0-4 percent
(C) 0-5 percent
(D) 0-10 percent
2 / 25
Q.2. In which of the following layer of atmosphere all weather phenomena occurs?
(A) Troposphere
(B) Stratosphere
(C) Mesosphere
(D) Thermosphere
3 / 25
Q.3. The Environmental Lapse Rate or Normal Lapse Rate (NLR) in the atmosphere is
(A) 5.60c per 1000 meter
(B) 6.50c per 1000 meter
(C) 6.000c per 1000 meter
(D) 6.650c per 1000 meter
4 / 25
Q.4. Which layer reflects the medium and high frequency radio waves back to the earth?
(A) Stratosphere
(B) Mesosphere
(C) lonosphere
(D) Exosphere
5 / 25
Q.5. Which of the following is the uppermost layer of the atmosphere
(A) Thermosphere
(B) Exosphere
(C) Ionosphere
(D) Mesosphere
6 / 25
Q.6. Match List-I with List-II from the codes given below
List-I List-II A. E-Layer 1. Southern Light B. F1 and F2 Layer 2. Kennelly Heaviside layer C. Ozone Layer 3. Northern Light D. Aurora Borealis 4. Stratosphere E. Aurora Australis 5. Appleton Layer
Codes:
A B C D E A. 2 3 5 4 1 B. 3 1 4 5 2 C. 2 5 4 1 3 D. 2 5 4 3 1
7 / 25
Q.7. Ozone gas is concentrated between
(A) 5 to 20 Km.
(B) 15 to 35 Km.
(C) 35 to 50 Km.
(D) 10 to 20 Km.
8 / 25
Q.8. The Ozone layer protects the earth from
(A) Cosmic Rays
(B) X-rays
(C) Infrared Rays
(D) Ultra Violet Rays
9 / 25
Q.9. The time taken by sun rays, to the earth is
(A) 20 minutes 8 second
(B) 8 minutes 20 second
(C) 9 minutes 35 second
(D) 10 minutes 20 second
10 / 25
Q.10. Match List-I with List-ll and from the using codes given below
List-I List-II A. Perihelion 1. Pyranometer B. Apehelion 2. 51 percent C. Heat Budget of the Earth 3. 4th July D. Isolation Measurement Instrument 4. 3rd January
Codes:
A B C D A. 1 2 3 4 B. 4 3 1 2 C. 4 3 2 1 D. 3 4 2 1
11 / 25
Q.11. The portion of incident radiation energy reflect back from the surface is called
(A) Heat Budget
(B) Photons
(C) Absorption
(D) Albedo
12 / 25
Q.12. The transfer of Heat through the molecular of matter in any body is called
(A) Convection
(B) Conduction
13 / 25
Q.13. The hottest place on the earth is considerable to be
(A) Quetta
(C) Al-Azizia
(D) Jaisalmer
14 / 25
Q.14. The transfer of heat energy through the movement of a mass of substance from one place to another is called
(A) Convection
(B) Conduction
15 / 25
Q.15. The temperature increases with increasing altitude due to certain conditions is called
(A) Normal Lapse Rate
(B) Inversion of Temperature
(C) Temperature Variation
16 / 25
Q.16. Read out the following conditions:
1. Cloudy Sky
2. Cold Dry Air
3. Strong Winds
4. Long Winter Nights
Which of the above conditions promote inversion of temperature?
(A) 1, 2 and 4
(B) 2, 3 and 4
(C) 1 and 4
(D) 2 and 4
17 / 25
Q.17. The coldest place on the earth surface is
(A) Ven Cover
(B) Verkhoyansk
(C) Vostok
(D) Norway
18 / 25
Q.18. Which of the following is known as the country of mind night sun?
(A) France
(B) Norway
(C) Netherlands
(D) Sweden
19 / 25
Q.19. What is Kinetic Energy?
(A) It is the energy of motion
(B) It is thermal energy
(C) It is latent energy
(D) It is the stored form of energy
20 / 25
Q.20. The standard air pressure at the sea level is
(A) 1002 MB
(B) 1025.13 MB
(C) 1013.25 MB
(D) 1008 MB
21 / 25
Q.21. The dividing line between day and night is called the
(A) Prime Meridian
(B) Standard Meridian
(C) International Date Line
(D) Circle of Illumination
22 / 25
Q.22. The average rate of albedo on the Earth is
(A) 25 percent
(B) 30 percent
(C) 35 percent
(D) 40 percent
23 / 25
Q.23. Which one of the following weather instruments is kept outside the Stevenson screen?
(A) Barometer
(B) Anemometer
(C) Hydrograph
(D) Maximum and minimum thermometers
24 / 25
Q.24. Which of the following belt is called “Doldrums”?
(A) Equatorial low pressure belt
(B) Sub-tropical high pressure belt
(C) Sub-polar low pressure belt
(D) Polar high pressure belt
25 / 25
Q.25. The Horse latitude refers to
(A) 25 degree – 35 degree latitude
(B) 20 degree – 35 degree latitude
(C) 30 degree - 40 degree latitude
(D) 30 degree - 35 degree latitude
The average score is 70%
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##### Click Here to Download the PDF OF Climatology 125 MCQS Mock Test 1
Syllabus:
UNIT- I
Climatology
Composition and Structure of Atmosphere; Insolation, Heat Budget of Earth, Temperature, Pressure and Winds, Atmospheric Circulation (air-masses, fronts and upper air circulation, cyclones and anticyclones (tropical and temperate), Climatic Classification of Koppen & Thornthwaite, ENSO Events (El Nino, La Nina and Southern Oscillations), Meteorological Hazards and Disasters (Cyclones, Thunderstorms, Tornadoes, Hailstorms, Heat and Cold waves Drought and Cloudburst , Glacial Lake Outburst (GLOF), Climate Change: Evidences and Causes of Climatic Change in the past, Human impact on Global Climate. | 1,881 | 5,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.66179 |
http://www.cnczone.com/forums/mechanical_calculations_engineering_design/51832-less_feed_requires_power.html | 1,369,025,560,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698289937/warc/CC-MAIN-20130516095809-00093-ip-10-60-113-184.ec2.internal.warc.gz | 379,772,008 | 17,717 | # Thread: Less feed requires more power??
1. ## Less feed requires more power??
I was trying to determine how much HP I would need for a given cut. And wondered something. Why is it according to Machinery's Handbook the higher your feed/revolution. The less power is needed?? I saw I needed 4HP for the cuts I was looking at. But figured I could use a smaller feed and reduce the needed power. Why is it, it holds opposite??
I was using the formula
Pc = KpCQW
2. Pc = KpCQW
Kp = .90 for rolled aluminum
C = 1 (For .3mm/rev)**
Q = 2.91096 Q = V/60ƒd
V = 286.512
ƒ = .3048
d = 2mm
W = 1
Pc = 2.61986 Kw or about 3HP
I figured that if I just didn't take as deep of a cut per revolution. I would be able to reduce the HP needed. Right? Or at least I thought. The problem is if I take .15mm/rev C now = 1.2. If you redo the equation. you're now needing 3.14348 Kw. Why do I need more power to take a smaller cut??
3. Originally Posted by Noodles87
.... Why is it according to Machinery's Handbook the higher your feed/revolution. The less power is needed?? ....
I do not have Machinery's Handbook handy so I cannot find out what material they we dealing with. However, I can tell you I know from experience both turning and milling different steel alloys that you can observe the power requirement falling sometimes when the feed is increased.
When machining steel if the conditions are such that the chip peeling off the work remains fairly cool, and by this I mean below a temperature at which the metal starts to discolor, then the steel retains practically 100% of its normal cold strength. However, if the conditions cause very significant heating of the metal coming off so that the chips are oxidized a brown or blue color, or even come off dull red, then the steel has lost a very large portion of its cold strength. I do not have a source handy for how much strength is lost but I think most simple carbon steels lose more than 60% of their cold strength at a temperature a little bit below dull red. You know that when bending steel by heating it you can take it up in temperature without much change in strength then at a certain point it bends very easily. In the machine the force needed to bend the chip and peel it off decreases in a similar manner so less power is needed.
4. Originally Posted by Noodles87
I was trying to determine how much HP I would need for a given cut. And wondered something. Why is it according to Machinery's Handbook the higher your feed/revolution. The less power is needed?? I saw I needed 4HP for the cuts I was looking at. But figured I could use a smaller feed and reduce the needed power. Why is it, it holds opposite??
I was using the formula
Pc = KpCQW
The power per cubic foot does indeed go down, and there's a bunch of places where this is documented. It's called the specific power requirement, i.e. power per volume. 1.5 hp per in^3 in steel, for example, is a specific power rating.
Most of the force in cutting is used to shear the metal (see shear plane/tool angle to cutting force studies), not to bend it. The less area you spend in shear while removing the same volume of material, the less power per cubic inch is required. Basic strength of materials. The force required for bending does go up, but nowhere near enough to equal the power gained by the shear force decreasing.
• Alright. Thanks a lot. I guess that makes sense.
If I were to use the "Optimum" speed and feed for turning aluminum. According to their feed and speed chart for carbide tooling. A speed of 2820 f/m and 36(0.001in)/revolution. That roughly translates to 10k rpms turning 25mm diameter. If I want to turn at 3500 rpm. Do I simply cross multiply and divde to get me my feed? Or is there another way in calculating it?
Example
X 3500
___ = _____
36 10000
X would equal my feed for 3500 r/m. Or is it not linear?
• Cutting speed in feet per minute.
Code:
```RPM = (cs * pi )/(12 * d)
or
RPM = cs * 3.82/d
2820 * 3.82 / .25 = 43089.6 RPM```
As you can see, 2820 is a rather ambitious speed to try and get a 1/4" endmill to run.
Code:
```IPM = RPM * Edges * Load
43089.6 * 3 * .001 = 129.27 IPM```
However, most machines can feed at 130 IPM.
It's rare for most machine shops to be able to spin an endmill over 10k. You don't need to be anywhere near the optimal value to get a good result.
• Sorry, I'm talking about turning.
• Originally Posted by Noodles87
Sorry, I'm talking about turning.
The only thing that changes in turning is the feed rate, because most people don't use balanced/triple/quad toolholders. Flutes goes to 1.
Also, I noticed you gave the diameter in mm and for some reason I read it as .25.
2820 * 3.82 / .984 = 10498
So yeah, you were right.
10498 * 1 * .001 = 10 ipm
Lathes just take the .001" though, you don't need IPM unless you want to program that way. Personally, .001" is kind of light. Depending on the tool you are using, you might not want to drop below .005" per rev. Sharp HSS, carbide, or a ground insert will do .001". A coated insert will not.
• Really? All turning is done with .001in per revolution? According to my handbook the "optimum" turning is "36" which I assumed was 36 times 0.001. There's also an "average" numbers given for turning 6160 aluminum. Which calls for a higher RPM with a slower feed. Which doesnt make sense to me either. I figured this would be a linear thing with feed/speed charts.
• In traditional machining (non-HSM), roughing/stock removal is best accomplished with a lower operating speed to control the temerature while using a high linear feed to remove as much material as possible. For finishing, you can allow the tool to get a little hotter since you won't be there long, which produces a better surface finish. The low feed does the same; the lower the feed, the smaller the ridges in your part are, the better the finish is. I don't know what Machinery's handbook is smoking by saying 36, because my copy is at work and I can't look it up right now. The ideal feed depends on the geometry of the tool and what you're trying to accomplish, not necessarily the work material.
You misunderstood what I meant by saying "the lathe takes .001 directly." What I mean is that lathes are traditionally programmed in feed per revolution, not feed per minute as a milling machine is. The lathe takes .001", .005", .010" etc values directly when in the appropriate mode, and to program in IPM (10, 20, etc) is unusual.
For some turning tasks, I've fed as much as .070" per revolution into the material for roughing a lot of stock down. For finishing, somewhere between .002" and .005" is satisfactory depending on tool geometry and nose radius.
• I thought I undrestood you right. The numbers you were giving, were per revolution. If it was .001 feed rate with an RPM of 1000. Your Z axis(or whichever) would be moving an 1 IPM respectively.
I'm just trying to figure out how much machine power I need to do the roughing cuts. Finish isn't a worry(for roughing) as there will be finishing passes. I wanted a max cutting speed of 3500 so I would be able to run up near 5000 on the smaller diamters. But maybe I'm looking into that too much?
I wanted to first know the machining power needed. To be able to find the forces acting on things and build my machine to handle those forces. At 3500 rpm I would be very happy to be able to cut at .2mm/revolution with a 2mm depth. Sure...I could settle for a depth of 1mm I guess. I could see for a feed of .1mm/revolution too. But if I dont have to, why should I, you know?
• How big (horsepower/swing) is your lathe, what are you trying to cut?
Aluminum is about .33 hp/in^3, steel is around 1.5 hp/in^3.
• Page 1 of 2 12 Last | 1,958 | 7,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2013-20 | latest | en | 0.963822 |
https://cs.stackexchange.com/questions/143226/algorithm-that-will-find-the-minimum-number-of-steps-to-get-from-state-j-to-st | 1,643,343,841,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305341.76/warc/CC-MAIN-20220128013529-20220128043529-00414.warc.gz | 251,023,501 | 33,838 | # Algorithm that will find the minimum number of steps to get from state $j$ to state $i$
Consider an adjacency matrix $$A$$ with elements $$[A]_{ij}=1$$ if one can reach state $$i$$ from state $$j$$ in one timestep, and $$0$$ otherwise. The matrix $$[A^k]_{ij}$$ represents the number of paths that lead from state $$j$$ to $$i$$ in $$k$$ timesteps. Derive an algorithm that will find the minimum number of step to get from state $$j$$ to state $$i$$.
Here is my attempt:
The algorithm takes as input:
• $$A$$ - Adjacency matrix
• $$j$$ and $$i$$ - the vertices for which we want to find the minimum path inbetween
Then it calculates in a while loop for which value of $$k$$, $$[A^k]_{ij}\neq 0$$ and returns that value of $$k$$. This way we get the minimum value of $$k$$ for which there exists a path.
Here is the pseudocode:
(Is this the correct way to do it and would there be a faster way?)
shortestPath(A,j,i):
if [A]_{ij}=1:
return 1
else:
f=0; k=1
while f=0:
f = [A^k]_{ij}
k = k + 1
return k
It can be improved further by using binary search on $$k$$, but you have to be careful with it since you want to multiply a small number of matrices (and computing $$A^r$$ for an arbitrary $$r$$, can take a lot of time using a simple multiplication algorithm) | 361 | 1,287 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2022-05 | latest | en | 0.834206 |
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3.4 Pp
1. 1. <ul><li>Bell Ringer: </li></ul><ul><li>Graph the system of inequalities from problem # 36 on your homework last night. </li></ul><ul><li>2. Is (-1,1) a solution of the system? </li></ul><ul><li>Why or why not? </li></ul>
2. 2. Solve linear programming problems. Objective
3. 3. optimization linear programming constraint feasible region New Vocabulary
4. 4. Check It Out! Example 1 Graph the feasible region for the following constraints. (Hint: Find the vertices) x ≥ 0 y ≥ 1.5 2.5 x + 5 y ≤ 20 3 x + 2 y ≤ 12
5. 5. Yum’s Bakery bakes two breads, A and B . One batch of A uses 5 pounds of oats and 3 pounds of flour. One batch of B uses 2 pounds of oats and 3 pounds of flour. The company has 180 pounds of oats and 135 pounds of flour available. Write the constraints for the problem and graph the feasible region. Example 1: Graphing a Feasible Region
6. 6. Graph the feasible region.
7. 7. The feasible region is a quadrilateral with vertices at (0, 0), (36, 0), (30, 15), and (0, 45). Check A point in the feasible region, such as (10, 10), satisfies all of the constraints.
8. 8. Why would we want to find a feasible region? objective function:
9. 11. Yum’s Bakery wants to maximize its profits from bread sales. One batch of A yields a profit of \$40. One batch of B yields a profit of \$30. Use the profit information and the data from Example 1 to find how many batches of each bread the bakery should bake. Example 2: Solving Linear Programming Problems
10. 12. Example 2 Continued Step 1 Let P = the profit from the bread. Write the objective function: P = 40 x + 30 y Step 2 Recall the constraints and the graph from Example 1. x ≥ 0 y ≥ 0 5 x + 2 y ≤ 180 3 x + 3 y ≤ 135
11. 13. Example 2 Continued Step 3 Evaluate the objective function at the vertices of the feasible region. Yum’s Bakery should make 30 batches of bread A and 15 batches of bread B to maximize the amount of profit. ( x , y ) 40 x + 30 y P(\$) (0, 0) 40(0) + 30(0) (0, 45) (30, 15) (36, 0)
12. 14. Ticket out the Door Use your own words to define Vertex Principle of Linear Programming. | 690 | 2,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.800396 |
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# MAT 222 WK 5 ASSIGNMENT
Read the following instructions to complete this assignment:
1. We define the following functions:
F(x) = 2x+5 g(x) =x2-3 h(x) = 7-x
3
· Compute (f – h) (4).
· Evaluate the following two compositions:
A: (f o g) (x)
B: (h o g) (x)
· Graph the g(x) function and transform it so that the graph is moved 6 units to the right and 7 units down.
· Find the inverse functions:
C: f -1 (x)
D: h-1 (x)
2. Write a two- to three-page paper (not including the title page) that is formatted in APA style and according to the Math Writing Guide. Format your math work as shown in the example, and be concise in your reasoning. In the body of your essay, do the following:
1. Demonstrate your solution to the above problems, making sure to include all mathematical work as well as explanations for each step. | 287 | 1,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-39 | latest | en | 0.881991 |
https://www.physicsforums.com/threads/fluid-mechanics-pitot-static-tube-manometer-system.342634/ | 1,527,048,476,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044534-00459.warc.gz | 812,487,884 | 15,248 | # Homework Help: Fluid Mechanics: pitot static tube/manometer system
1. Oct 3, 2009
### ywkim880801
1. The problem statement, all variables and given/known data
A water-filled manometer is connected to a Pitot-static tube to measure a nominal airspeed of 50 ft/s. It is assumed that a change in the manometer reading of 0.002 in. can be detected. What is the minimum deviation from the 50 ft/s airspeed that can be detected by this system?
2. Relevant equations
velocity measurement for a pitot-static tube
V=$$\sqrt{}2(p3-p4)/\rho$$ | 143 | 538 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-22 | latest | en | 0.821512 |
https://overunity.com/14699/coils-as-momentum/printpage/ | 1,585,622,536,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370499280.44/warc/CC-MAIN-20200331003537-20200331033537-00415.warc.gz | 597,922,031 | 5,935 | # Free Energy | searching for free energy and discussing free energy
## Solid States Devices => Tesla Technologgy => Topic started by: d3x0r on June 07, 2014, 09:14:40 AM
Title: Coils as momentum
Post by: d3x0r on June 07, 2014, 09:14:40 AM
I wish I hadn't gotten myself banned from physicsforums.com ; they seemed fairly knowledgable; only took 2 days too....
So the theory is, if I have two infinite capacitances on the ends of a coil (or 1 even and just make a LC loop standalone), if I could get a current to flow in the coil, then that current will flow forever, until the voltage of the capacitors is high enough to oppose the flow.. . but if there is no differential to start with, and the capacitance never fills up, the coil would keep conducting. True or false?
if I take a coil and use an external magnet to induce a current in a single direction and stop the magnet in place, this should allow the induced current to flow itself... ?
So I did; I took a 2000 turn 3.3mH air core coil and a 6" magnet and a few capacitors.
The first cap is 9.9nF (12kV?)
the second cap is 100nF (400V?)
the third cap is 6800uF (40V?)
the voltage rating doesn't really matter, I'm not going to exceed a few volts with a hand generator like this.
If it's the voltage difference only that stops the current flow, then a similar motion will generate a comparable voltage; unless voltage is somehow a exponential value of resistance such that 2V is resistance 4x, and 3V is resistance 40x.
(compiling accompanying video; will edit with update)
just musing... Tesla said 'there's no limit to the amount of power you can store in a capacitor'; that can be true if you know the momentum.
wish a I had some of those farad caps to test more.
Current * Inductance : Velocity * Mass
So, I was wondering, how fast does a coil's current increase for a certain voltage X applied to it? I guess it becomes a more complex question because of the resistance of the coil also?
------------------
So in the experiment I was able to charge 9.9nF to voltage 2.2V
I was able to charge 109.9nF to voltage 2V
I was able to charg 1000109nF to voltage 2V (1000uF + )
I was able to charge 6800109nF to voltage 1V (6800uF ) (I think here the resistance of the wire is a total cap on the current that can flow)
In 200ms; my motion from top to bottom is apparently about 200ms.... give or take a little but the same peeks in the middle.
While I was measuring the 200ms time, I inserted a resistor, my first was 10ohm with no effect, my secnod was 100ohm with no effect, but I left it in. (no effect on 9.9nF charge rate or level) but there was a notable effect at 6.8mF the voltage only charged to .25V and was very subdued... which leads me to go back to - the resistance of the coil is going to limit how much it can charge/conduct....
A higher inductance coil allows greater charge... that mobious coil is way better than the solenoid I ran... but maybe if there was a pancake instead? (I don't have a high inductance pancake), the large mobius is about 3.3mH also...
(I should do some math on total joules stored in the caps)
2V 9.2nF
Energy E = 18.4n Joules
2V 108nF
Energy E = 216n Joules
1V 6.8mF
Energy E = 3.4m Joules
3.4m Joules/sec = 3.4mW ; that's pretty straight forward... so these are jouls in 200ms or *5 of the above = watts....
9.2nF 0.092mW
108nF 1.080mW
6.8mF 17mW
Title: Re: Coils as momentum
Post by: d3x0r on June 07, 2014, 10:24:12 AM
In a simulator... a huge capacitance, and a switch to get momentum in the coil... which can be used to charge into the cap...
I note that when I start it, I use a lot of power very quickly, then when switching to the capacitor there's a power disappation of the same but over a much longer time than initial; Until the cap is at like 25% of the voltage source...
At some point if the capacitor has a greater potential difference than the initial circuit, the coil will instantly lose all momentum and reverse.
Title: Re: Coils as momentum
Post by: MarkE on June 07, 2014, 11:10:43 AM
I wish I hadn't gotten myself banned from physicsforums.com ; they seemed fairly knowledgable; only took 2 days too....
So the theory is, if I have two infinite capacitances on the ends of a coil (or 1 even and just make a LC loop standalone), if I could get a current to flow in the coil, then that current will flow forever, until the voltage of the capacitors is high enough to oppose the flow.. . but if there is no differential to start with, and the capacitance never fills up, the coil would keep conducting. True or false?
False. If you put together a series branch: C - L - C, the loop will behave the same as C/2 - L. Assuming ideal capacitors and inductor: the impedance of the network is very high at low frequencies: 2/jwC, and very high at high frequencies: jwL. At jW = (LC)-0.5 the impedance drops theoretically to zero. In real life, there will be parasitic series resistance, parallel capacitance across the inductor, leakage through the capacitors etc that will take away from the ideal. You postulate what happens if the capacitors are really, really large. The answer is that the impedance will be low down to very low frequencies.
Quote
if I take a coil and use an external magnet to induce a current in a single direction and stop the magnet in place, this should allow the induced current to flow itself... ?
That's how generators work. Once you stop moving the magnet, the induced voltage goes away and the current will decay according the the circuit L/R time constant.
Quote
So I did; I took a 2000 turn 3.3mH air core coil and a 6" magnet and a few capacitors.
The first cap is 9.9nF (12kV?)
the second cap is 100nF (400V?)
the third cap is 6800uF (40V?)
the voltage rating doesn't really matter, I'm not going to exceed a few volts with a hand generator like this.
If it's the voltage difference only that stops the current flow, then a similar motion will generate a comparable voltage; unless voltage is somehow a exponential value of resistance such that 2V is resistance 4x, and 3V is resistance 40x.
I don't know what you are after mixing up the ideas of voltage and resistance.
Quote
(compiling accompanying video; will edit with update)
just musing... Tesla said 'there's no limit to the amount of power you can store in a capacitor'; that can be true if you know the momentum.
wish a I had some of those farad caps to test more.
The assumption behind that statement is that the capacitor does not break down no matter what the applied voltage. We know that assumption is false.
Quote
Current * Inductance : Velocity * Mass
Yes, it takes real energy to change the current in an inductor. If you have a linear inductor, solving the integral for the energy is easy: 0.5*L*I[sup2[/sup].
Quote
So, I was wondering, how fast does a coil's current increase for a certain voltage X applied to it? I guess it becomes a more complex question because of the resistance of the coil also?
For time spans much shorter than L/R, Idelta ~= V*T/L. For time spans much longer than R/L, I ~= V/R.
Quote
------------------
So in the experiment I was able to charge 9.9nF to voltage 2.2V
I was able to charge 109.9nF to voltage 2V
I was able to charg 1000109nF to voltage 2V (1000uF + )
I was able to charge 6800109nF to voltage 1V (6800uF ) (I think here the resistance of the wire is a total cap on the current that can flow)
In 200ms; my motion from top to bottom is apparently about 200ms.... give or take a little but the same peeks in the middle.
While I was measuring the 200ms time, I inserted a resistor, my first was 10ohm with no effect, my secnod was 100ohm with no effect, but I left it in. (no effect on 9.9nF charge rate or level) but there was a notable effect at 6.8mF the voltage only charged to .25V and was very subdued... which leads me to go back to - the resistance of the coil is going to limit how much it can charge/conduct....
Yes at frequencies well below jw = R/L, the resistance dominates, and for frequencies well above jw = R/L, the inductive reactance dominates.
Quote
A higher inductance coil allows greater charge... that mobious coil is way better than the solenoid I ran... but maybe if there was a pancake instead? (I don't have a high inductance pancake), the large mobius is about 3.3mH also...
More inductance means more impedance at any given frequency. It also means more energy storage at any given current. Be wary of saturation.
Quote
(I should do some math on total joules stored in the caps)
2V 9.2nF
Energy E = 18.4n Joules
2V 108nF
Energy E = 216n Joules
1V 6.8mF
Energy E = 3.4m Joules
3.4m Joules/sec = 3.4mW ; that's pretty straight forward... so these are jouls in 200ms or *5 of the above = watts....
Average, yes that is correct.
Quote
9.2nF 0.092mW
108nF 1.080mW
6.8mF 17mW
Title: Re: Coils as momentum
Post by: d3x0r on June 07, 2014, 12:00:44 PM
Once you stop moving the magnet, the induced voltage goes away and the current will decay according the the circuit L/R time constant.
Well... I guess my aparatus isn't inclined to demonstrate that it's not 'once you stop moving the magnet' but additionally after the current built in the coil decays; with 6.8mF and 1mH (approx) it's only millseconds to reach max charge and decay (ideally)....and since I'm using a gross movement covernig 200ms it's hard to see the fractional time;
Modified sim to use a current source of 1A instead of a voltage source, so I could get the current moving, and then throw the switch, so I could confirm more acurately that ideally it's a 4ms delay after removing the EMF source.
Edit: added a 10ohm resistor inline to measure the voltage across it as a CSR... So yes, when I stop, the current incrase stops, and starts to decrease... but the voltage still increases after...
Additionally, usually in the generator, the magnet doesn't 'stop' it instantly starts to receed... which causes a current in the opposing direction already, which would cancel any such effect.
I don't know what you are after mixing up the ideas of voltage and resistance.
that's actually impedance on the cap; and it's not the case, it's a linear relationship...
If you have a linear inductor, solving the integral for the energy is easy: 0.5*L*I[sup2.For time spans much shorter than L/R, Idelta ~= V*T/L. For time spans much longer than R/L, I ~= V/R.Yes at frequencies well below jw = R/L, the resistance dominates, and for frequencies well above jw = R/L, the inductive reactance dominates.More inductance means more impedance at any given frequency. It also means more energy storage at any given current. Be wary of saturation.
THe experiment was done without cores; does that mean saturation is basically instantaneous because of tiny permeability?
----
I further tried some simulations, and chained another inductor and switch to the original cap, and then a cap for that... and basically what gets put into the first is transferred to the second... something like a slinky in action...
It's only the initial input that can change it.
Most circuits are driven by potential differences in the capacitances, which is what impedes the flow in the coil....
if I use a 1H coil to charge a 1mF cap, it goes to like 30V, but then the next 1H coil will only get accelerated back to the current that the cap captured from the first coil... (acceleration is relative to current)
---
Just even with this perspective there's no additional features available because of the methods available to form current in a coil....
Title: Re: Coils as momentum
Post by: TinselKoala on June 07, 2014, 02:14:27 PM
Air (vacuum) cores do not saturate, that's the whole point of using them. | 3,091 | 11,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-16 | longest | en | 0.948639 |
https://studyqas.com/a-fish-tank-is-shaped-like-a-rectangular-prism-the-area/ | 1,670,156,812,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00092.warc.gz | 571,659,850 | 41,058 | # A fish tank is shaped like a rectangular prism. The area of the base of the fish tank is 6 1/8 square feet. The height of the fish tank
A fish tank is shaped like a rectangular prism. The area of the base of the fish tank is 6 1/8 square feet. The height of the fish tank is 3 1/2 feet.
## This Post Has 3 Comments
1. Expert says:
c) papirus
explanation:
it's papirus for sure.
2. cassandrabeliles says:
Volume of fish tank = 21 7/16 or 21.4375 cubic feet.
Explanation:
Given the following data;
Base area of tank = 6 1/8 square feet
Height of tank = 3 1/2 feet
Required:
To find the volume of the rectangular prism (fish tank);
Volume of rectangular prism = base area * height
Substituting into the formula, we have;
Volume of fish tank = 6 1/8 * 3 1/2
Volume of fish tank = 49/8 * 7/2
Volume of fish tank = 343/16
Volume of fish tank = 21 7/16 or 21.4375 cubic feet.
3. Expert says: | 274 | 907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-49 | latest | en | 0.876091 |
https://www.jiskha.com/display.cgi?id=1359173575 | 1,503,498,394,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120573.0/warc/CC-MAIN-20170823132736-20170823152736-00579.warc.gz | 948,516,616 | 4,441 | # Statistics
posted by .
The Tribune claims that the time of travel from downtown to the University via the bus has an average of µ = 27 minutes. A student who normally takes this bus believes that µ is greater than 27 minutes. A sample of six ride times taken to test the hypothesis of interest gave a mean of 27.5 minutes and standard deviation of 2.43 minutes. The value of the test statistic for testing this hypothesis is
- 0.532
0.460
0.504
-0.460
-0.504
--------
Which of the following conditions doesn't need to be met before you can use a two-sample procedure?
The responses in each group are independent of each other.
Each group is considered to be a sample from a distinct population.
The same variable is measured in both samples.
The goal is to compare the means of the two groups.
Data in two samples are matched together in pairs that are compared.
• Statistics -
To find the test statistic:
(x - mean)/(sd/√n)
(27.5 - 27)/(2.43/√6) = ?
I'll let you finish the calculation.
• Statistics -
.504010235141
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$\frac{2}{15}$
$\frac{2}{5}$ . $\frac{1}{3}$ Multiply numerators and multiply denominators $\frac{2}{5}$ . $\frac{1}{3}$ = $\frac{2*1}{5*3}$ = $\frac{2}{15}$ | 78 | 217 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-24 | latest | en | 0.546604 |
https://open.kattis.com/contests/pm4mzx/problems/tritiling | 1,556,296,400,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578841544.98/warc/CC-MAIN-20190426153423-20190426175423-00174.warc.gz | 497,457,396 | 5,906 | JMU F18 Wk7
Start
2018-10-12 18:30 UTC
JMU F18 Wk7
End
2018-10-18 20:00 UTC
The end is near!
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Problem CTri Tiling
In how many ways can you tile a $3\times n$ rectangle with $2\times 1$ dominoes?
Here is a sample tiling of a $3\times 12$ rectangle.
Input consists of several test cases followed by a line containing -1. Each test case is a line containing an integer $0 \leq n \leq 30$. For each test case, output one integer number giving the number of possible tilings.
Sample Input 1 Sample Output 1
2
8
12
-1
3
153
2131 | 212 | 628 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-18 | latest | en | 0.760685 |
https://www.physicsforums.com/threads/using-orbital-energy-to-calculate-velocity.996726/ | 1,712,980,700,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00766.warc.gz | 874,587,124 | 17,960 | # Using Orbital Energy to Calculate Velocity
• JoeyBob
In summary: The only thing I can think of is the potential energy changes in the problem statement are wrong.In summary, the conversation was about calculating initial and final potential energies and determining the change in kinetic energy. The final velocity was also calculated, but it was found to be significantly higher than the correct answer. The mistake was determined to be in the calculation of the change in potential energy, which was 103 times too high, suggesting that there may be an error in the problem statement. Specifically, the potential energy changes may be incorrect, which could explain the inconsistency between the obtained perigee speed and conservation of angular momentum.
JoeyBob
Homework Statement
See attached picture
Relevant Equations
change Ek + change Ep =0, Ek=1/2mv^2, Ep=-GMm/r
So what I did first was calculate the initial and final potential energies with Epi=-9.433*10^11 m and Epf = -1.503*10^12 m.
Then I found change in potential energy, -5.597*10^11 m.
Using this I determined the change in kinetic energy, 5.597*10^11. I then added this change to the initial kinetic energy I calculated (103.68 m) to get a final Ek of 5.597*10^11 m.
Then I calculated the final velocity, Ekm=v^2*0.5, finding that v=1058017, which is obviously way higher than the right answer.
Where am I going wrong here? Is there some non conservative work or something?
#### Attachments
• Question.PNG
10.7 KB · Views: 103
JoeyBob said:
Homework Statement:: See attached picture
Relevant Equations:: change Ek + change Ep =0, Ek=1/2mv^2, Ep=-GMm/r
So what I did first was calculate the initial and final potential energies with Epi=-9.433*10^11 m and Epf = -1.503*10^12 m.
Then I found change in potential energy, -5.597*10^11 m.
Using this I determined the change in kinetic energy, 5.597*10^11. I then added this change to the initial kinetic energy I calculated (103.68 m) to get a final Ek of 5.597*10^11 m.
Then I calculated the final velocity, Ekm=v^2*0.5, finding that v=1058017, which is obviously way higher than the right answer.
Where am I going wrong here? Is there some non conservative work or something?
Please post all your working, just as algebra, no plugged in values. This will make it much easier to see where the mistake is, unless it is purely arithmetic, which is unlikely.
Your units are all over the place. Plugging in r in km and v in km/s means you get ΔEp = -5.597*1011m mJ and Eki = 103.68m MJ. (I haven't checked the calculations, but the numbers look in the right ballpark.) It is much better to convert everything to SI units (and back at the end if necessary), and INCLUDE THE UNITS AT EACH STAGE OF THE CALCULATION. Can never stress too much the importance of units.
JoeyBob said:
Homework Statement:: See attached picture
Relevant Equations:: change Ek + change Ep =0, Ek=1/2mv^2, Ep=-GMm/r
Where am I going wrong here? Is there some non conservative work or something?
Your change in potential energy is 103 times too high. What does this suggest to you?
On edit: I am questioning the validity of this problem. The perigee speed ##v_p##, obtained from the given parameters using energy conservation, is inconsistent with conservation of angular momentum. Angular momentum conservation (per unit mass) requires that ##v_1 r_1\sin\theta=v_p r_p##. Solving for ##\sin\theta## returns a value greater than 1 which is impossible if my calculation is correct.
Last edited:
pai535
## 1. How is orbital energy related to velocity?
Orbital energy is the sum of an object's kinetic and potential energy while in orbit. This energy is directly related to the object's velocity, as an increase in velocity leads to an increase in kinetic energy and a decrease in potential energy.
## 2. What is the formula for calculating velocity using orbital energy?
The formula for calculating velocity using orbital energy is v = √(2E/m), where v is velocity, E is orbital energy, and m is the mass of the object.
## 3. How does the distance from the central body affect orbital energy and velocity?
The distance from the central body affects both orbital energy and velocity. As the distance increases, the potential energy decreases and the kinetic energy increases, resulting in a higher orbital energy and a higher velocity.
## 4. Can orbital energy be used to calculate the velocity of any object in orbit?
Yes, orbital energy can be used to calculate the velocity of any object in orbit, as long as the mass and distance from the central body are known.
## 5. How is orbital energy used in space missions?
Orbital energy is used in space missions to determine the necessary velocity for a spacecraft to enter and maintain a stable orbit around a planet or other celestial body. It is also used to calculate the amount of fuel needed for a spacecraft to reach a certain velocity and maintain its orbit.
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Revisit Volume
### Lesson Purpose
The purpose of this lesson is for students to solve real world problems about volume.
### Lesson Narrative
In this lesson, students calculate the volume of different objects. The first activity recalls the meaning of volume as the number of cubic units required to fill a space. Students experiment with different ways to build a rectangular prism using a fixed number of cubes and relate this to finding factors of the number of cubes. In the second activity they estimate the volume of some very large structures, the great pyramid of Egypt and the Empire State Building. Neither shape is a rectangular prism though they are each made up of smaller shapes that are rectangular prisms. Students combine the skills of making reasonable estimates with finding products of very large numbers. If students need additional support with the concepts in this lesson, refer back to Unit 1, Section A in the curriculum materials.
• Action and Expression
• MLR7
Warm-up: Estimation Exploration: Sugar Cubes
### Learning Goals
Teacher Facing
• Solve real world and mathematical problems involving volume.
### Student Facing
• Let’s solve problems about volume.
### Lesson Timeline
Warm-up 10 min Activity 1 15 min Activity 2 15 min Lesson Synthesis 10 min Cool-down 5 min
### Teacher Reflection Questions
As students worked together today, where did you see evidence of the mathematical community established over the course of the school year?
### Print Formatted Materials
For access, consult one of our IM Certified Partners. | 313 | 1,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-30 | latest | en | 0.88216 |
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Calculating the Distillate Flow Rate
The distillate flow is intermittant. The distillate receiver fills
with distillate and when the level gets too high,
a pump comes on to pump out a certain volume of distillate to
get the level back down.
A graph like this needs to be made with your results in Excel:
The green line shows what the straight-line-between-each-data-point curve looks like.
The dotted red curve shows what is really going on.
For this graph, it says that the pump is pumping at a setting of "1" between 0.33 and 0.67 minutes;
the pump is pumping at a setting of "2" between 0.67 and 1.0 minutes;
the pump is pumping at a setting of "3" between 1.00 and 1.33 minutes and
the pump is pumping at a setting of "4" between 1.33 and 1.67 minutes.
The total volume pumped for this profile was 66 ml of distillate.
If the pump peak setting is "5", the volume is 95 ml ± 3 ml
If the pump peak setting is "4", the volume is 63 ml ± 6 ml
If the pump peak setting is "3", the volume is 35 ml ± 8 ml
If the pump peak setting is "2", the volume is 27.6 ml (uncertainty unknown)
Using this information and assuming that the pump operates in a linear flow vs. setting,
you can calculate the volume of distillate over any period of your experiment. | 344 | 1,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-39 | longest | en | 0.927391 |
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Page 69 - , i tan b whence, cos A = tan c In a similar way the other relations in (l)-(6) can be shown to be true for ABC (Fig. 28). 27. On species. Two parts of a spherical triangle are said to be of the same species (or of the same affection) when both are less than 90°, both greater than 90°, or both equal to 90°. Formula (1), Art. 26, shows that the hypotenuse of a right-angled spherical triangle is less than 90° when the sides about the right angle are both greater or both less than 90° ; and it...
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https://www.coursehero.com/file/8804825/b-A-binary-search-tree-contains-the-following-keys-18/ | 1,529,449,873,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00225.warc.gz | 777,844,790 | 26,101 | Midterm 2 Solutions
# B a binary search tree contains the following keys 18
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Unformatted text preview: already in the tree } 5) (a) Exhibit the unique binary search tree whose preorder traversal prints the sequence 41 21 15 38 23 29 25 30. (b) A binary search tree contains the following keys: 18, 1, 3, 42, 35, 8, 14, 23, 1, 11, 55, 91. If the keys in the tree are printed using inorder traversal, what is the output sequence printed? 4 + 4 points Solution: Many of you actually constructed the binary search tree and then traversed the tree in order to get the output. A short-cut is to write sort the list above (since the inorder traversal of a BST gives a sorted sequence.) [ 1 3 8 11 14 18 23 35 42 55 91] 6) Write a procedure testHeap that takes as input an array A and an integer currentSize that represents the numbe...
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https://rdrr.io/cran/geozoo/src/R/parametric.R | 1,537,562,270,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157503.43/warc/CC-MAIN-20180921190509-20180921210909-00236.warc.gz | 606,141,162 | 16,457 | # R/parametric.R In geozoo: Zoo of Geometric Objects
#### Documented in boy.surfaceconic.spiralconic.spiral.nautiluscross.capdini.surfaceellipsoidenneper.surfaceklein.fig.eightroman.surface
##Parametric Eqn's
#' Boy Surface
#'
#' A function to produce a Boy Surface.
#'
#' @param n number of points
#' @references \url{http://schloerke.github.io/geozoo/mobius/other/}
#' @author Barret Schloerke
#' @examples
#' ## Generates a Boy Surface
#' boy.surface(n = 1000)
#'
#' @keywords dynamic
#' @export
boy.surface <- function(n = 10000) {
vert <- matrix(
do.call(
"rbind",
as.list(
replicate(n, boy_surface_row())
)
),
ncol = 3, byrow = TRUE
)
wires <- NULL
structure(
list(points = vert, edges = wires),
class = c("geozooNoScale", "geozoo")
)
}
#' @keywords internal
boy_surface_row <- function( ){
u <- runif( 1, min = 0, max = pi)
v <- runif( 1, min = 0, max = pi)
a <- cos( v) * sin( u)
b <- sin( v) * sin( u)
c <- cos( u)
x <- 1 / 2 * (
( 2 * a ^ 2 - b ^ 2 - c ^ 2) +
2 * b * c * ( b ^ 2 - c ^ 2) +
c * a * ( a ^ 2 - c ^ 2) +
a * b * ( b ^ 2 - a ^ 2)
)
y <- sqrt(3) / 2 * (
( b ^ 2 - c ^ 2) +
c * a * ( c ^ 2 - a ^ 2) +
a * b * ( b ^ 2 - a ^ 2)
)
z <- ( a + b + c) * (
( a + b + c) ^ 3 +
4 * ( b - a) * ( c - b) * ( a - c)
)
z <- z / 8
return( cbind( x, y, z))
}
#' Conic Spiral
#'
#' A function to produce a conic spiral
#'
#' @param n number of points
#' @param a final radius of cone
#' @param b height of object
#' @param w number of spirals
#' @return
#' \item{points }{location of points}
#' \item{edges }{edges of the object (null)}
#' @references \url{http://schloerke.github.io/geozoo/mobius/other/}
#' @author Barret Schloerke
#' @examples
#' ## Generates a Conic Spiral
#' conic.spiral(n = 1000)
#'
#' @keywords dynamic
#' @export
conic.spiral <- function(n = 10000, a = .2, b = 1, c = .1, w = 2) {
vert <- matrix(
do.call(
"rbind",
as.list(
replicate(n, conic_spiral_row(a, b, c, w))
)
),
ncol = 3, byrow = TRUE
)
wires <- NULL
structure(
list(points = vert, edges = wires),
class = c("geozooNoScale", "geozoo")
)
}
#' @keywords internal
conic_spiral_row <- function(a, b, c, w) {
u <- runif( 1, min = 0, max = 2 * pi)
v <- runif( 1, min = 0, max = 2 * pi)
x <- a * ( 1 - v / ( 2 * pi)) * cos( w * v) * ( 1 + cos( u)) +
c * cos( w * v)
y <- a * ( 1 - v / ( 2 * pi)) * sin( w * v) * ( 1 + cos( u)) +
c * sin( w * v)
z <- ( b * v + a * ( 1 - v / ( 2 * pi)) * sin( u)) / ( 2 * pi)
return( cbind( x, y, z))
}
#' Conic Spiral (Nautilus Shape)
#'
#' A function to produce a Conic Spiral in a nautilus shape
#'
#' @param n number of points
#' @param a final radius of cone
#' @param b height of object
#' @param w number of spirals
#' @return
#' \item{points }{location of points}
#' \item{edges }{edges of the object (null)}
#' @references \url{http://schloerke.github.io/geozoo/mobius/other/}
#' @author Barret Schloerke
#' @examples
#' ## Generates a Nautilus Conic Spiral
#' conic.spiral.nautilus( n = 1000 )
#'
#' @keywords dynamic
#' @export
conic.spiral.nautilus <- function(n = 10000, a = .2, b = .1, c = 0, w = 2) {
vert <- matrix(
do.call(
"rbind",
as.list(
replicate(n, conic_spiral_row(a, b, c, w))
)
),
ncol = 3, byrow = TRUE
)
wires <- NULL
structure(
list(points = vert, edges = wires),
class = c("geozooNoScale", "geozoo")
)
}
#' Cross Cap
#'
#' A function to generate a cross cap
#'
#' @param n number of points
#' @return
#' \item{points }{location of points}
#' \item{edges }{edges of the object (null)}
#' @references \url{http://schloerke.github.io/geozoo/mobius/other/}
#' @author Barret Schloerke
#' @examples
#' ## Generates a Cross Cap
#' cross.cap( n = 1000 )
#'
#' @keywords dynamic
#' @export
cross.cap <- function(n = 10000) {
vert <- matrix(
do.call(
"rbind",
as.list(
replicate(n, cross_cap_row())
)
),
ncol = 3, byrow = TRUE
)
wires <- NULL
structure(
list(points = vert, edges = wires),
class = c("geozooNoScale", "geozoo")
)
}
#' @keywords internal
cross_cap_row <- function( ) {
u <- runif( 1, min = 0, max = pi)
v <- runif( 1, min = 0, max = pi)
x <- cos( u) * sin( 2 * v)
y <- sin( u) * sin( 2 * v)
z <- cos( v) * cos( v) - cos( u) * cos( u) * sin( v) * sin( v)
return( cbind( x, y, z))
}
#' Dini Surface
#'
#' A function to generate a dini surface.
#'
#' @param n number of points
#' @param a outer radius of object
#' @param b space between loops
#' @return
#' \item{points }{location of points}
#' \item{edges }{edges of the object (null)}
#' @references \url{http://schloerke.github.io/geozoo/mobius/other/}
#' @author Barret Schloerke
#' @examples
#' ## Generates a Dini Surface
#' dini.surface(n = 1000, a = 1, b = 1)
#'
#' @keywords dynamic
#' @export
dini.surface <- function(n = 10000, a = 1, b = 1) {
vert <- matrix(
do.call(
"rbind",
as.list(
replicate(n, dini_surface_row( a, b))
)
),
ncol = 3, byrow = TRUE
)
wires <- NULL
structure(
list( points = vert, edges = wires),
class = c("geozooNoScale", "geozoo")
)
}
#' @keywords internal
dini_surface_row <- function(a = 1, b = 1) {
u <- runif( 1, min = 0, max = 4 * pi)
v <- runif( 1, min = 0.0000000001, max = 2)
x <- a * cos( u) * sin( v)
y <- a * sin( u) * sin( v)
z <- a * ( cos(v) + log(tan(v / 2))) + (b * u)
return( cbind( x, y, z))
}
#' Ellipsoid
#'
#' A function to generate an ellipsoid
#'
#' @param n number of points
#' @param a radius in x direction
#' @param b radius in y direction
#' @param c radius in z direction
#' @return
#' \item{points }{location of points}
#' \item{edges }{edges of the object (null)}
#' @references \url{http://schloerke.github.io/geozoo/mobius/other/}
#' @author Barret Schloerke
#' @examples
#' ## Generates an ellipsoid
#' ellipsoid(n = 1000, a = 1, b = 1, c = 3)
#'
#' @keywords dynamic
#' @export
ellipsoid <- function(n = 10000, a = 1, b = 1, c = 3) {
vert <- matrix(
do.call(
"rbind",
as.list(
replicate(n, ellipsoid_row(a, b, c))
)
),
ncol = 3, byrow = TRUE
)
wires <- NULL
structure(
list(points = vert, edges = wires),
class = c("geozooNoScale", "geozoo")
)
}
#' @keywords internal
ellipsoid_row <- function(a, b, c) {
u <- runif( 1, min = 0, max = 2 * pi)
v <- runif( 1, min = 0, max = 2 * pi)
x <- a * cos( u) * sin( v)
y <- b * sin( u) * sin( v)
z <- c * cos( v)
return( cbind( x, y, z))
}
#' Enneper's Surface
#'
#' A function to generate Enneper's surface
#'
#' @param n number of points
#' @param a angle, radians, minimum and maximum. -a < angle < a
#' @return
#' \item{points }{location of points}
#' \item{edges }{edges of the object (null)}
#' @references \url{http://schloerke.github.io/geozoo/mobius/other/}
#' @author Barret Schloerke
#' @examples
#' ## Generates an Enneper Surface
#' enneper.surface(n = 1000, a = 4)
#'
#' @keywords dynamic
#' @export
enneper.surface <- function(n = 10000, a = 4) {
vert <- matrix(
do.call(
"rbind",
as.list(
replicate(n, enneper_surface_row(a))
)
),
ncol = 3, byrow = TRUE
)
wires <- NULL
structure(
list( points = vert, edges = wires),
class = c("geozooNoScale", "geozoo")
)
}
#' @keywords internal
enneper_surface_row <- function( a = 4) {
u <- runif( 1, min = - a, max = a)
v <- runif( 1, min = - a, max = a)
x <- u - u ^ 3 / 3 + u * v ^ 2
y <- v - v ^ 3 / 3 + v * u ^ 2
z <- u ^ 2 - v ^ 2
return( cbind( x, y, z))
}
#' Figure Eight Klein Bottle
#'
#' A function to generate a figure eight Klein bottle
#'
#' @param n number of points
#' @return
#' \item{points }{location of points}
#' \item{edges }{edges of the object (null)}
#' @references \url{http://schloerke.github.io/geozoo/mobius/other/}
#' @author Barret Schloerke
#' @examples
#' ## Generates a figure eight Klein bottle.
#' klein.fig.eight(n = 1000, a = 3, b = 1)
#'
#' @keywords dynamic
#' @export
klein.fig.eight <- function(n = 10000, a = 3, b = 1) {
vert <- matrix(
do.call(
"rbind",
as.list(
replicate(n, klein_fig_eight_row(a, b))
)
),
ncol = 4,
byrow = TRUE
)
wires <- NULL
structure(
list( points = vert, edges = wires),
class = c("geozooNoScale", "geozoo")
)
}
#' @keywords internal
klein_fig_eight_row <- function( a = 3, b = 1) {
u <- runif( 1, min = - pi, max = pi)
v <- runif( 1, min = - pi, max = pi)
x <- ( b * cos( v) + a) * cos( u)
y <- ( b * cos( v) + a) * sin( u)
z <- b * sin( v) * cos( u / 2)
w <- b * sin( v) * sin( u / 2)
return( cbind( x, y, z, w))
}
#' Roman Surface
#'
#' A function to generate a Roman surface, also known as a Steiner surface
#'
#' @param n number of points
#' @param a maximum radius of object
#' @return
#' \item{points }{location of points}
#' \item{edges }{edges of the object (null)}
#' @references \url{http://schloerke.github.io/geozoo/mobius/other/}
#' @author Barret Schloerke
#' @examples
#' ## Generates a Roman surface.
#' roman.surface(n = 1000, a = 1)
#'
#' @keywords dynamic
#' @export
roman.surface <- function(n = 10000, a = 1) {
vert <- matrix(
do.call(
"rbind",
as.list(
replicate(n, roman_surface_row(a))
)
),
ncol = 3, byrow = TRUE
)
wires <- NULL
structure(
list(points = vert, edges = wires),
class = c("geozooNoScale", "geozoo")
)
}
#' @keywords internal
roman_surface_row <- function(a = 1) {
u <- runif( 1, min = 0, max = pi)
v <- runif( 1, min = 0, max = pi)
x <- a ^ 2 * cos( v) * cos( v) * sin( 2 * u) / 2
y <- a ^ 2 * sin( u) * sin( 2 * v) / 2
z <- a ^ 2 * cos( u) * sin( 2 * v) / 2
return( cbind( x, y, z))
}
## Try the geozoo package in your browser
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geozoo documentation built on May 29, 2017, 5:58 p.m. | 3,447 | 9,363 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-39 | longest | en | 0.195652 |
http://icl.cs.utk.edu/projectsfiles/f2j/javadoc/org/netlib/lapack/SGESDD.html | 1,513,004,565,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513611.22/warc/CC-MAIN-20171211144705-20171211164705-00618.warc.gz | 133,237,871 | 4,493 | ## org.netlib.lapack Class SGESDD
```java.lang.Object
org.netlib.lapack.SGESDD
```
`public class SGESDDextends java.lang.Object`
```SGESDD is a simplified interface to the JLAPACK routine sgesdd.
This interface converts Java-style 2D row-major arrays into
the 1D column-major linearized arrays expected by the lower
level JLAPACK routines. Using this interface also allows you
to omit offset and leading dimension arguments. However, because
of these conversions, these routines will be slower than the low
level ones. Following is the description from the original Fortran
source. Contact seymour@cs.utk.edu with any questions.
* ..
*
* Purpose
* =======
*
* SGESDD computes the singular value decomposition (SVD) of a real
* M-by-N matrix A, optionally computing the left and right singular
* vectors. If singular vectors are desired, it uses a
* divide-and-conquer algorithm.
*
* The SVD is written
*
* A = U * SIGMA * transpose(V)
*
* where SIGMA is an M-by-N matrix which is zero except for its
* min(m,n) diagonal elements, U is an M-by-M orthogonal matrix, and
* V is an N-by-N orthogonal matrix. The diagonal elements of SIGMA
* are the singular values of A; they are real and non-negative, and
* are returned in descending order. The first min(m,n) columns of
* U and V are the left and right singular vectors of A.
*
* Note that the routine returns VT = V**T, not V.
*
* The divide and conquer algorithm makes very mild assumptions about
* floating point arithmetic. It will work on machines with a guard
* digit in add/subtract, or on those binary machines without guard
* digits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or
* Cray-2. It could conceivably fail on hexadecimal or decimal machines
* without guard digits, but we know of none.
*
* Arguments
* =========
*
* JOBZ (input) CHARACTER*1
* Specifies options for computing all or part of the matrix U:
* = 'A': all M columns of U and all N rows of V**T are
* returned in the arrays U and VT;
* = 'S': the first min(M,N) columns of U and the first
* min(M,N) rows of V**T are returned in the arrays U
* and VT;
* = 'O': If M >= N, the first N columns of U are overwritten
* on the array A and all rows of V**T are returned in
* the array VT;
* otherwise, all columns of U are returned in the
* array U and the first M rows of V**T are overwritten
* in the array VT;
* = 'N': no columns of U or rows of V**T are computed.
*
* M (input) INTEGER
* The number of rows of the input matrix A. M >= 0.
*
* N (input) INTEGER
* The number of columns of the input matrix A. N >= 0.
*
* A (input/output) REAL array, dimension (LDA,N)
* On entry, the M-by-N matrix A.
* On exit,
* if JOBZ = 'O', A is overwritten with the first N columns
* of U (the left singular vectors, stored
* columnwise) if M >= N;
* A is overwritten with the first M rows
* of V**T (the right singular vectors, stored
* rowwise) otherwise.
* if JOBZ .ne. 'O', the contents of A are destroyed.
*
* LDA (input) INTEGER
* The leading dimension of the array A. LDA >= max(1,M).
*
* S (output) REAL array, dimension (min(M,N))
* The singular values of A, sorted so that S(i) >= S(i+1).
*
* U (output) REAL array, dimension (LDU,UCOL)
* UCOL = M if JOBZ = 'A' or JOBZ = 'O' and M < N;
* UCOL = min(M,N) if JOBZ = 'S'.
* If JOBZ = 'A' or JOBZ = 'O' and M < N, U contains the M-by-M
* orthogonal matrix U;
* if JOBZ = 'S', U contains the first min(M,N) columns of U
* (the left singular vectors, stored columnwise);
* if JOBZ = 'O' and M >= N, or JOBZ = 'N', U is not referenced.
*
* LDU (input) INTEGER
* The leading dimension of the array U. LDU >= 1; if
* JOBZ = 'S' or 'A' or JOBZ = 'O' and M < N, LDU >= M.
*
* VT (output) REAL array, dimension (LDVT,N)
* If JOBZ = 'A' or JOBZ = 'O' and M >= N, VT contains the
* N-by-N orthogonal matrix V**T;
* if JOBZ = 'S', VT contains the first min(M,N) rows of
* V**T (the right singular vectors, stored rowwise);
* if JOBZ = 'O' and M < N, or JOBZ = 'N', VT is not referenced.
*
* LDVT (input) INTEGER
* The leading dimension of the array VT. LDVT >= 1; if
* JOBZ = 'A' or JOBZ = 'O' and M >= N, LDVT >= N;
* if JOBZ = 'S', LDVT >= min(M,N).
*
* WORK (workspace/output) REAL array, dimension (LWORK)
* On exit, if INFO = 0, WORK(1) returns the optimal LWORK;
*
* LWORK (input) INTEGER
* The dimension of the array WORK. LWORK >= 1.
* If JOBZ = 'N',
* LWORK >= 3*min(M,N) + max(max(M,N),6*min(M,N)).
* If JOBZ = 'O',
* LWORK >= 3*min(M,N)*min(M,N) +
* max(max(M,N),5*min(M,N)*min(M,N)+4*min(M,N)).
* If JOBZ = 'S' or 'A'
* LWORK >= 3*min(M,N)*min(M,N) +
* max(max(M,N),4*min(M,N)*min(M,N)+4*min(M,N)).
* For good performance, LWORK should generally be larger.
* If LWORK < 0 but other input arguments are legal, WORK(1)
* returns the optimal LWORK.
*
* IWORK (workspace) INTEGER array, dimension (8*min(M,N))
*
* INFO (output) INTEGER
* = 0: successful exit.
* < 0: if INFO = -i, the i-th argument had an illegal value.
* > 0: SBDSDC did not converge, updating process failed.
*
* Further Details
* ===============
*
* Based on contributions by
* Ming Gu and Huan Ren, Computer Science Division, University of
* California at Berkeley, USA
*
* =====================================================================
*
* .. Parameters ..
```
Constructor Summary
`SGESDD()`
Method Summary
`static void` ```SGESDD(java.lang.String jobz, int m, int n, float[][] a, float[] s, float[][] u, float[][] vt, float[] work, int lwork, int[] iwork, intW info)```
Methods inherited from class java.lang.Object
`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`
Constructor Detail
### SGESDD
`public SGESDD()`
Method Detail
### SGESDD
```public static void SGESDD(java.lang.String jobz,
int m,
int n,
float[][] a,
float[] s,
float[][] u,
float[][] vt,
float[] work,
int lwork,
int[] iwork,
intW info)``` | 1,901 | 6,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-51 | latest | en | 0.823229 |
https://blog.listcomp.com/math/2019/05/20/convolution-is-convolution-its-not-dot-product | 1,701,552,249,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00297.warc.gz | 171,531,950 | 6,731 | # Convolution is convolution; it's NOT dot product
Yao Yao on May 20, 2019
$\DeclareMathOperator*{\argmin}{argmin} \newcommand{\icol}[1]{ \bigl[ \begin{smallmatrix} #1 \end{smallmatrix} \bigr] } \newcommand{\icolplus}[1]{ \Bigl[ \begin{smallmatrix} #1 \end{smallmatrix} \Bigr] }$
## 1-D Convolution
$(f \ast g)(t) \overset{\Delta}{=} \int_{-\infty}^{\infty} f(\tau) g(t-\tau) \mathrm{d} \tau$
$(f \ast g)(t) \overset{\Delta}{=} \int_{-\infty}^{\infty} f(t-\tau) g(\tau) \mathrm{d} \tau$
For complex-valued functions $f$, $g$ defined on $\mathbb{Z}$, the discrete convolution of $f$ and $g$ is:
$(f \ast g)(n)=\sum_{m} f(m) g(n-m)$
## 2-D Convolution
$(f \ast g)(x, y) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(\sigma, \tau) g(x - \sigma, y - \tau) \mathrm{d} \sigma \mathrm{d} \tau$
$(f \ast g)(x, y) = \sum_{\sigma} \sum_{\tau} f(\sigma, \tau) g(x - \sigma, y - \tau)$
## Matrix 2-D Convolution for Image Processing / Image & Kernel
$A(i, j) = A_{i,j}$
where $0 \leq i < m, 0 \leq j < n$. 那么两个 matrice 也可以做 convolution (以下下标都从 0 记起)。
\begin{aligned} f_A(i, j) &= A_{m_K + i, n_K + j} \newline f_K(i, j) &= K^{HV}_{i, j} = K_{m_K - i, n_K - j} \end{aligned}
where $K^{HV} = JKJ$ and $J$ is the anti-diagonal “identity” matrix (或者看成是 row-reversed identity matrix) like $\icol{0 & 1 \newline 1 & 0}$.
• $JK$ 的作用是将 $K$ 上下翻转 (horizontally flip)
• $KJ$ 的作用是将 $K$ 左右翻转 (vertically flip)
\begin{aligned} C(i, j) &= \sum_{i'} \sum_{j'} f_K(i', j') f_A(i - i', j - j') \newline &= \sum_{i'} \sum_{j'} K_{m_K - i', n_K - j'} A_{m_K - i' + i, n_K - j' + j} \end{aligned}
• 注意，如果下标从 1 开始计的话，上面的下标都要 +1
• 另外，参考 kernel method 的思想，实际应用中我们并不需要构造 $f_A$ 和 $f_K$，直接写出 $C(i, j) = \sum_{i’} \sum_{j’} K_{?, ?} A_{?, ?}$ 的形式拿来用就好了
\begin{aligned} C(0,0) &= \sum_{i'} \sum_{j'} K_{2 - i', 2 - j'} A_{2 - i', 2 - j'} \newline &= K_{0,0}A_{0,0} + K_{0,1}A_{0,1} + K_{1,0}A_{1,0} + K_{1,1}A_{1,1} \newline C(0,1) &= \sum_{i'} \sum_{j'} K_{2 - i', 2 - j'} A_{2 - i', 3 - j'} \newline &= K_{0,0}A_{0,1} + K_{0,1}A_{0,2} + K_{1,0}A_{1,1} + K_{1,1}A_{1,2} \newline C(1,0) &= \sum_{i'} \sum_{j'} K_{2 - i', 2 - j'} A_{3 - i', 2 - j'} \newline &= K_{0,0}A_{1,0} + K_{0,1}A_{1,1} + K_{1,0}A_{2,0} + K_{1,1}A_{2,1} \newline C(1,1) &= \sum_{i'} \sum_{j'} K_{2 - i', 2 - j'} A_{3 - i', 3 - j'} \newline &= K_{0,0}A_{1,1} + K_{0,1}A_{1,2} + K_{1,0}A_{2,1} + K_{1,1}A_{2,2} \end{aligned}
1. 把 kernel $K$ 覆盖在 image $A$ 之上，$K_{0,0}$ 对齐到 $A_{0,0}$ (左上角对齐)
2. 移动 kernel $K$ 使 $K_{0,0}$ 对齐到 $A_{i,j}$，假设覆盖到的 image $A$ 的部分是 $A_{i:i+m_K, j:j+n_K}$
3. 那么 $C(i, j) = \operatorname{sum} \big( K \odot A_{i:i+m_K, j:j+n_K} \big)$ (sum of Hadamard product)
• 我再次强调一遍，这不是 dot product！
\begin{aligned} 0 \leq i' < m_K \newline 0 \leq j' < n_K \newline 0 \leq m_K - i' + i < m_A \newline 0 \leq m_K - i' + i < n_A \end{aligned}
\begin{aligned} 0 \leq i < m_A - m_K \newline 0 \leq j < n_A - n_K \end{aligned} | 1,409 | 2,885 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-50 | latest | en | 0.200293 |
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### Anatel
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Reset axes in Matlab App Designer
You shouldn't clear the axes because it's gonna turn down the efficiency of your app. Just create a "startup" for the plot and a...
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How to enter user values into an existing matrix.
Try this... originalMatrix = randi(10,5); inputArray = zeros(5,1); for ii = 1:numel(inputArray) inputArray(ii) = input...
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From this it is manifest that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle...
A companion to Euclid: being a help to the understanding and remembering of ... - Side 61
av Euclides - 1837 - 88 sider
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## Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - 1819 - 333 sider
...of EK, and the straight line EH less than EK. Wherefore, the diameter, &c. Q, ED PROP. XVI. THEOR. The straight line drawn at right angles to the diameter...falls without the circle ; and no straight line can be drati•n between that straight line and the circumference, from ihr extremity of the diameter, so...
Uten tilgangsbegrensning - Om denne boken
## Elements of Geometry: Containing the First Six Books of Euclid: With a ...
John Playfair - 1819 - 317 sider
...the straight line AG cuts the circle. Con. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle ; and that it touches it only in one point ; because, if it did meet the circle...
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## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclid, Robert Simson - 1821 - 516 sider
...angle less than any rectilineal angle.' Con. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the exstraight line which touches the circle in the same point.' PROP. XVII. PROB. To draw a straight line...
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## Artis logicæ rudimenta, with illustrative observations [and a transl. By J ...
Henry Aldrich - 1821
...angles : if this be the case, two angles of a triangle are equal to two right angles : Therefore, if a straight line drawn at right angles to the diameter of a circle at its extremity does not fall without the circle, it occasions the existence of a triangle containing...
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## Mechanics' Magazine and Journal of Science, Arts, and Manufactures, Volum 4
1825
...Euclid to refer to, I will quote his IGth Proposition, Book in., which bears directly on this point. " The straight line drawn at right angles to the diameter of a circle from the extremity of it foils without the circle; and no straight line can lie drawn between that straight line and the circumference,...
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## Mechanics Magazine, Volum 4
1825
...the straight line AG cuts the circle. " COR.— Hence it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it touches the circle, and that it touches it only in one point ; because, if it did meet the circle in...
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## Iron: An Illustrated Weekly Journal for Iron and Steel ..., Volum 4
Perry Fairfax Nursey - 1825
...straight line AG cuts the circle. " Cor..— Hence it is manifest, that, the straight line which is drawn at right angles to the diameter of a circle from the extremity of it touches the circle, and that it'touches it only in one point ; because, if it did meet the circle in...
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## A Popular Course of Pure and Mixed Mathematics ...: With Tables of ...
Peter Nicholson - 1825 - 372 sider
...angle less than any rectilineal angle.' Cor. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle ; and (hat it touches it only in one point, because, if it did meet the circle in...
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## Elements of Geometry, Containing the First Six Books of Euclid
Euclid - 1826 - 180 sider
...and the remaining angle less than any rectilineal angle. QED Deductions. 1. The right line which is drawn at right angles to the diameter of a circle from the extremity of it touches the circle, and that it touches only in one point. 2. To describe a circle, which shall touch...
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## Euclid's Elements of geometry, transl. To which are added, algebraic ...
Euclides - 1826
...and the remaining angle less than any .rectilineal angle. QED Deductions. 1. The right line which is drawn at right angles to the diameter of a circle from the extremity of it touches the circle, and that it touches only in one point. 2. To describe a circle, which shall touch...
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A097885 Triangle read by rows: T(n,k) is the number of Motzkin paths of length n with k valleys (n>=0, 0<=k<=floor(n/2)-1; a valley is a downstep followed by an upstep). 1
1, 1, 2, 4, 8, 1, 17, 4, 37, 13, 1, 82, 40, 5, 185, 116, 21, 1, 423, 326, 80, 6, 978, 899, 279, 31, 1, 2283, 2444, 924, 140, 7, 5373, 6578, 2948, 568, 43, 1, 12735, 17576, 9136, 2156, 224, 8, 30372, 46702, 27690, 7777, 1035, 57, 1, 72832, 123568, 82453, 26952, 4422 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Also, triangle read by rows: T(n,k) is the number of Motzkin paths of length n and having k double rises (i.e. UU's, where U=(1,1)). E.g. T(5,1)=4 counts HUUDD, UUDDH, UUHDD and UUDHD, where U=(1,1), H=(1,0) and D=(1,-1). Row sums are the Motzkin numbers (A001006). Column 0 gives A004148. LINKS FORMULA G.f. G=G(t, z) satisfies z^2*(t+z-tz)G^2-(1-z-z^2+tz^2)*G+1=0. EXAMPLE Triangle starts: 1; 1; 2; 4; 8,1; 17,4; 37,13,1; Row n (n>=2) has floor(n/2) terms. T(5,1)=4 counts HU(DU)D, U(DU)DH, U(DU)HD and UH(DU)D (here U=(1,1), H=(1,0) and D=(1,-1); valleys are shown between parentheses). MAPLE eq:=G=1+z*G+z^2*G*(t*(G-1-z*G)+1+z*G): sol:=solve(eq, G): Gser:=simplify(series(sol[1], z=0, 15)): P[0]:=1: for n from 1 to 12 do P[n]:=sort(coeff(Gser, z^n)) od: 1, 1, seq(seq(coeff(t*P[n], t^k), k=1..floor(n/2)), n=0..12); CROSSREFS Cf. A001006, A004148. Sequence in context: A097888 A030275 A097874 * A097892 A197282 A215452 Adjacent sequences: A097882 A097883 A097884 * A097886 A097887 A097888 KEYWORD nonn,tabf AUTHOR Emeric Deutsch, Sep 02 2004 EXTENSIONS Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 16 2007 STATUS approved
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Last modified April 23 14:15 EDT 2019. Contains 322386 sequences. (Running on oeis4.) | 869 | 2,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-18 | latest | en | 0.623273 |
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A299474 a(n) = 4*p(n), where p(n) is the number of partitions of n. 6
4, 4, 8, 12, 20, 28, 44, 60, 88, 120, 168, 224, 308, 404, 540, 704, 924, 1188, 1540, 1960, 2508, 3168, 4008, 5020, 6300, 7832, 9744, 12040, 14872, 18260, 22416, 27368, 33396, 40572, 49240, 59532, 71908, 86548, 104060, 124740, 149352, 178332, 212696, 253044, 300700, 356536, 422232, 499016, 589092, 694100, 816904 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS For n >= 1, a(n) is also the number of edges in the diagram of partitions of n, in which A299475(n) is the number of vertices and A000041(n) is the number of regions (see example and Euler's formula). LINKS Shawn A. Broyles, Table of n, a(n) for n = 0..1000 FORMULA a(n) = 4*A000041(n) = 2*A139582(n). a(n) = A000041(n) + A299475(n) - 1, n >= 1 (Euler's formula). a(n) = A000041(n) + A299473(n). - Omar E. Pol, Aug 11 2018 EXAMPLE Construction of a modular table of partitions in which a(n) is the number of edges of the diagram after n-th stage (n = 1..6): -------------------------------------------------------------------------------- n ........: 1 2 3 4 5 6 (stage) a(n)......: 4 8 12 20 28 44 (edges) A299475(n): 4 7 10 16 22 34 (vertices) A000041(n): 1 2 3 5 7 11 (regions) -------------------------------------------------------------------------------- r p(n) -------------------------------------------------------------------------------- . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 1 .... 1 ....|_| |_| | |_| | | |_| | | | |_| | | | | |_| | | | | | 2 .... 2 .........|_ _| |_ _| | |_ _| | | |_ _| | | | |_ _| | | | | 3 .... 3 ................|_ _ _| |_ _ _| | |_ _ _| | | |_ _ _| | | | 4 |_ _| | |_ _| | | |_ _| | | | 5 .... 5 .........................|_ _ _ _| |_ _ _ _| | |_ _ _ _| | | 6 |_ _ _| | |_ _ _| | | 7 .... 7 ....................................|_ _ _ _ _| |_ _ _ _ _| | 8 |_ _| | | 9 |_ _ _ _| | 10 |_ _ _| | 11 .. 11 .................................................|_ _ _ _ _ _| . Apart from the axis x, the r-th horizontal line segment has length A141285(r), equaling the largest part of the r-th region of the diagram. Apart from the axis y, the r-th vertical line segment has length A194446(r), equaling the number of parts in the r-th region of the diagram. The total number of parts equals the sum of largest parts. Note that every diagram contains all previous diagrams. An infinite diagram is a table of all partitions of all positive integers. MAPLE with(combinat): seq(4*numbpart(n), n=0..50); # Muniru A Asiru, Jul 10 2018 PROG (GAP) List([0..50], n->4*NrPartitions(n)); # Muniru A Asiru, Jul 10 2018 (PARI) a(n) = 4*numbpart(n); \\ Michel Marcus, Jul 15 2018 CROSSREFS k times partition numbers: A000041 (k=1), A139582 (k=2), A299473 (k=3), this sequence (k=4). Cf. A135010, A141285, A182181, A186114, A193870, A194446, A194447, A206437, A207779, A220482, A220517, A273140, A278355, A278602, A299475. Sequence in context: A194696 A302681 A002368 * A022087 A095294 A190100 Adjacent sequences: A299471 A299472 A299473 * A299475 A299476 A299477 KEYWORD nonn AUTHOR Omar E. Pol, Feb 10 2018 STATUS approved
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Last modified April 8 02:27 EDT 2020. Contains 333312 sequences. (Running on oeis4.) | 1,346 | 4,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-16 | latest | en | 0.387645 |
https://www.jiskha.com/display.cgi?id=1393003133 | 1,529,280,417,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859904.56/warc/CC-MAIN-20180617232711-20180618012711-00145.warc.gz | 859,169,069 | 3,864 | # Can someone Explain this(math)
posted by julie
34, 36, 43, 46, 47, 48, 48, 52, 53, 54
Box And whisker Plot
I Don't Understand these at all can someone Explain without totally giving me the answer?
1. Ms. Sue
http://ellerbruch.nmu.edu/cs255/jnord/boxplot.html
2. julie
Ty! :)
## Similar Questions
1. ### math!
what point represents the 3rd quartile in the box and whisker plot?
2. ### Human resource
Can someone explain me this question: explain how and to what extent THE CONCEPT OF PEOPLE HAS CHANGED OVERTIME ,HENCE,HIGHLIGHTING THE EVOLUTION AND GIVING PRACTICAL EXAMPLES TO ILLUSTRATE YOUR ANSWER. am new to this module and can …
3. ### algebra
The prices of several chess sets are \$15, \$20, \$38, \$95, \$60, \$45, \$40, \$35, and \$50. Make a box-and-whisker plot of the data. What conclusions can you make?
4. ### algebra
The prices of several chess sets are \$15, \$20, \$38, \$95, \$60, \$45, \$40, \$35, and \$50. Make a box-and-whisker plot of the data. What conclusions can you make?
6. ### algebra
Which box-and-whisker plot shows the scores of ten students on a mathematics exam?
7. ### box and whisker plot
Make a box and whisker plot with the following statement:The data show the ages of 13 people attending a dinner party. 37, 42, 44, 46, 47, 49, 51, 53, 54, 55, 57, 61, 63
8. ### Algebra 1
Make a box-and-whisker plot of the data. 29, 21, 17, 10, 15, 27, 22, 30 i forget how you do this, can someone explain | 451 | 1,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-26 | latest | en | 0.818338 |
https://www.excelguru.ca/forums/showthread.php?11382-Find-Value-within-Datarange&s=84c005712257b2c5e95a0d5d1642e181 | 1,643,085,083,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304760.30/warc/CC-MAIN-20220125035839-20220125065839-00251.warc.gz | 783,807,184 | 13,153 | # Thread: Find Value within Datarange
1. ## Find Value within Datarange
Hello,
I am trying to find a value between a range but unable to do that, i tried Index & match combination too but unable....
Level 0 10 20 30 40 50 60 70 80 90 100
Volume 0 0.34 0.97 1.75 2.63 3.55 4.47 5.35 6.13 6.76 7.1
I need to find Volume when Level will be 35? I want to use Level and Volume in dynamic range....
Please give me the solution....
regards,
2. You probably need to set the MATCH TYPE argument to 1:
=INDEX(result_range,MATCH(lookup_value,lookup_range,1))
https://exceljet.net/excel-functions...match-function
3. Or maybe you want the trend
=TREND(result_range,lookup_range,lookup_value)
4. To do this one to death…
If you use the TREND formula using all the data, sometimes the results will be significantly off because the relationship is not a straight line:
using the chart, you'd expect a volume of roughly 2.2 for level 35.
Using a straight line through all the points you get 2.55 (dotted line on chart)
Formula:
=TREND(D2:D12,C2:C12,F3,FALSE)
You can get better results using the small straight line between the points directly before and after 35, this gives you 2.19
Formula:
=TREND(INDEX(D2:D12,MATCH(F3,C2:C12)):INDEX(D2:D12,MATCH(F3,C2:C12)+1),INDEX(C2:C12,MATCH(F3,C2:C12)):INDEX(C2:C12,MATCH(F3,C2:C12)+1),F3)
or a bit shorter:
=TREND(OFFSET(C2:C12,MATCH(F3,C2:C12)-1,1,2),OFFSET(C2:C12,MATCH(F3,C2:C12)-1,0,2),F3)
You'll very likely get even better results using the fifth order polynomial (the what?) which is a good fit, which, surprisingly has a simpler (well, shorter) formula, and this gives you 2.18
Formula:
=SUMPRODUCT(LINEST(D2:D12,C2:C12^{1,2,3,4,5},FALSE),F3^{5,4,3,2,1,0})
See attached workbook where you can alter the value in cell F3 (light green).
5. Oh groan…
cross posted without links:
https://www.excelforum.com/excel-for...datarange.html
mfaisal.ce, for your information, you should always provide links to your cross posts.
This is a requirement, not just a request.
If you have cross posted at other places, please add links to them too.
Why? Have a read of http://www.excelguru.ca/content.php?184
#### Posting Permissions
• You may not post new threads
• You may not post replies
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• | 689 | 2,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-05 | latest | en | 0.780225 |
https://howmanytimes.net/76-go-into-2544 | 1,721,589,786,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517768.40/warc/CC-MAIN-20240721182625-20240721212625-00616.warc.gz | 279,855,325 | 2,580 |
# How many times does 76 go into 2544 ?
Math question : How many times does 76 go into 2544? Or How much is 76 divided by 2544 ?
For the answer of this problem to solve it is we listed 76 numbers in 2544 number.
• 76 +
• 76 +
• 76 +
• 76 +
• 76 +
• 76 +
• 76 +
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• 76 +
• 76 +
• 76 +
• 76 +
• 76 +
• 76 +
• 76 +
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• 76 +
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• 76
• + 36
There are 33 times 76 in 2544.
The answer you divide 2544 by 76 which would get you 33.473684210526.
The remainder is 36
#### Similar Questions With Same Answer = 33
How many seventy-sixes are in 2544
How many seventy-six are there in 2544
How does seventy-six goes into 2544
How many times does seventy-six going to 2544
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How many times does seventy-six go into 2544
Can you get a number for 2544 with seventy-six
How many times does 2544 go to seventy-six
How many 76 can fit into 2544
Can you tell me 76 into 2544
How many 76s fit into 2544
### Check +1 similar numbers in 2544:
How many times does go into ? | 394 | 1,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-30 | latest | en | 0.910455 |
https://www.airmilescalculator.com/distance/rkz-to-lds/ | 1,632,026,951,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00303.warc.gz | 676,889,576 | 85,811 | # Distance between Shigatse (RKZ) and Yichun (LDS)
Flight distance from Shigatse to Yichun (Shigatse Peace Airport – Yichun Lindu Airport) is 2459 miles / 3957 kilometers / 2137 nautical miles. Estimated flight time is 5 hours 9 minutes.
Driving distance from Shigatse (RKZ) to Yichun (LDS) is 3241 miles / 5216 kilometers and travel time by car is about 56 hours 50 minutes.
## Map of flight path and driving directions from Shigatse to Yichun.
Shortest flight path between Shigatse Peace Airport (RKZ) and Yichun Lindu Airport (LDS).
## How far is Yichun from Shigatse?
There are several ways to calculate distances between Shigatse and Yichun. Here are two common methods:
Vincenty's formula (applied above)
• 2458.969 miles
• 3957.327 kilometers
• 2136.785 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 2455.742 miles
• 3952.134 kilometers
• 2133.981 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Shigatse Peace Airport
City: Shigatse
Country: China
IATA Code: RKZ
ICAO Code: ZURK
Coordinates: 29°21′6″N, 89°18′41″E
B Yichun Lindu Airport
City: Yichun
Country: China
IATA Code: LDS
ICAO Code: ZYLD
Coordinates: 47°45′7″N, 129°1′8″E
## Time difference and current local times
There is no time difference between Shigatse and Yichun.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 270 kg (596 pounds).
## Frequent Flyer Miles Calculator
Shigatse (RKZ) → Yichun (LDS).
Distance:
2459
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
2459
Round trip? | 533 | 1,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-39 | latest | en | 0.818078 |
https://rdrr.io/r/base/sequence.html | 1,720,890,851,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00820.warc.gz | 419,979,649 | 15,349 | # sequence: Create A Vector of Sequences
sequence R Documentation
## Create A Vector of Sequences
### Description
The default method for `sequence` generates the sequence `seq(from[i], by = by[i], length.out = nvec[i])` for each element `i` in the parallel (and recycled) vectors `from`, `by` and `nvec`. It then returns the result of concatenating those sequences.
### Usage
```sequence(nvec, ...)
## Default S3 method:
sequence(nvec, from = 1L, by = 1L, ...)
```
### Arguments
`nvec` coerced to a non-negative integer vector each element of which specifies the length of a sequence. `from` coerced to an integer vector each element of which specifies the first element of a sequence. `by` coerced to an integer vector each element of which specifies the step size between elements of a sequence. `...` additional arguments passed to methods.
### Details
Negative values are supported for `from` and `by`. `sequence(nvec, from, by=0L)` is equivalent to `rep(from, each=nvec)`.
This function was originally implemented in R with fewer features, but it has since become more flexible, and the default method is implemented in C for speed.
### Author(s)
Of the current version, Michael Lawrence based on code from the S4Vectors Bioconductor package
`gl`, `seq`, `rep`.
### Examples
```sequence(c(3, 2)) # the concatenated sequences 1:3 and 1:2.
#> [1] 1 2 3 1 2
sequence(c(3, 2), from=2L)
#> [1] 2 3 4 2 3
sequence(c(3, 2), from=2L, by=2L)
#> [1] 2 4 6 2 4
sequence(c(3, 2), by=c(-1L, 1L))
#> [1] 1 0 -1 1 2
``` | 445 | 1,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.818309 |
http://interrupciones.net/linear-and-nonlinear-programming-international-series-in-operations-research-management-science.html | 1,558,569,856,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256980.46/warc/CC-MAIN-20190522223411-20190523005411-00491.warc.gz | 97,341,852 | 5,152 | # Linear and nonlinear programming international series in operations research management science. Linear and Nonlinear Programming 2019-01-28
Linear and nonlinear programming international series in operations research management science Rating: 7,2/10 967 reviews
## Linear and Nonlinear Programming, Third Edition
That is, until the necessary and sufficient conditions 7 are approximately satisfied. One wishes to make the assignment in such a way as to maximize in this example the total value of the assignment. Suppose that the optimal solutions of 5. The value of this variable can be found immediately to be 10. It should be clear that by suitably multiplying by minus unity, and adjoining slack and surplus variables, any set of linear inequalities can be converted to standard form if the unknown variables are restricted to be nonnegative.
Next
## Linear and Nonlinear Programming, Third Edition
Inequalities, 3rd Printing, Springer-Verlag, New York, 1971. Of course, the theoretical and computational aspects do take on a somewhat special character for linear programming problemsāthe most significant development being the simplex method. These columns are linearly independent and hence the columns of B form a basis for E m. This problem can be efficiently solved using the tree algorithm. To initiate the simplex procedure we must update the last row so that it has zero components under the basic variables. Network flow problems represent another important class of linear programming problems. Annals of Operations Research, 17:119-135, 1989.
Next
## Linear and Nonlinear Programming by David G. Luenberger
It is designed for either self-study by professionals or classroom work at the undergraduate or graduate level for students who have a technical background in engineering, mathematics, or science. It is the components of this vector that are used to determine which vector to bring into the basis. This new method does both at once. The procedure seems to work well for some problems but it has difficulty if the problem is degenerate. Since the pioneering work of R.
Next
## Linear and Nonlinear Optimization International Series in Operations Research & Management Science: interrupciones.net: Richard W. Cottle, Mukund N. Thapa: Books
It becomes clear now that a sufficient requirement to eliminate the duality gap in the surrogate constraint method is to make the feasible region in the constraint space, defined by a single surrogate constraint, the same as the feasible region in the primal problem. Update the basis of the master problem as usual. The last equation is a simple linear equation from which x1 is determined by a solution to the smaller system of inequalities. In this process the new elements in the last column must remain nonnegativeāif the pivot was properly selected. We have the following basic lemma for the above procedure. We now briefly explain the two major concepts, homogeneity and self-duality, used in the construction. Another way to convert a transshipment problem to a transportation problem is through the introduction of buffer stocks at each node.
Next
## Linear and Nonlinear Optimization by Richard W. Cottle (ebook)
This problem can be solved by repeated application of the tree algorithm, successively determining paths from origin to destination and assigning flow along such paths. However, many real-life phenomena are of a nonlinear nature, which is why we need tools for nonlinear programming capable of handling several conflicting or incommensurable objectives. Rather than explicitly indicating both arcs in such a case, it is customary to indicate a single undirected arc. It emphasizes modeling and numerical algorithms for optimization with continuous not integer variables. Computational aspects of alternative portfolio selection models in the presence of discrete asset choice constraints.
Next
## Linear and Nonlinear Programming, 4 edition
To overcome this drawback, the e-subdifferential can be introduced to replace the subdifferential in 3. Finally, the participant also declares the maximum number qj of units he or she is willing to accept under these terms. International Journal of Systems Science, 11:455-486, 1980. Thus the fundamental theorem yields an obvious, but terribly inefficient, finite search technique. Assuming linearity of the production facility, if we are given a vector b describing output requirements of the m commodities, and we wish to produce these at minimum cost, ours is the primal problem. It is now not uncommon to solve linear programs of up to a million variables and constraints, as long as the structure is sparse. Notice that that is a combination of a predictor and corrector choice.
Next
## Linear and Nonlinear Programming, 4 edition
If the row and column sums of a transportation problem are integers, then the basic variables in any basic solution are integers. Other disadvantages of the method include the assumption of non-empty interior and the need of an objective lower bound. In the next theorem recall that A is defined by 4. If this coefficient is negative, then the objective value will be continuously improved as the value of this nonbasic variable is increased, and therefore one increases the variable as far as possible, to the point where further increase would violate feasibility. The matrix formulation is also a natural setting for the discussion of dual linear programs and other topics related to linear programming. One familiar with other branches of linear mathematics might suspect, initially, that linear programming formulations are popular because the mathematics is nicer, the theory is richer, and the computation simpler for linear problems than for nonlinear ones.
Next
## Nonlinear Integer Programming (International Series in Operations Research & Management Science)
It follows that any constraint can be expressed as a linear combination of the others, and hence any one constraint can be dropped. When a function is of an additive form with respect to all of its variables, the function is called separable. It emphasizes modeling and numerical algorithms for optimization with continuous not integer variables. If neither condition is discovered at a given basic solution, then the objective is strictly decreased. Some valid inequality techniques can be used to tighten this upper bound. Part I is a self-contained introduction to linear programming. Naval Research Logistics Quarterly, 27:89-95, 1980.
Next
## Linear and Nonlinear Programming, Third Edition
A more surprising result is that the technique described above can be used to identify all null variables. These together are identical to the first-order conditions of section 7. The linearity of the budget constraint is extremely natural in this case and does not represent simply an approximation to a more general functional form. This section covers some of the basic graph and network terminology and concepts necessary for the development of this alternative approach. Upon completion of his doctoral studies in Berkeley, Cottle became a member of the technical staff of Bell Telephone Laboratories in Holmdel, New Jersey. A nonsingular upper triangular matrix is also triangular, since by reversing the order of its rows and columns it becomes lower triangular. D The above results confirm the existence of a kind of saddle point in integer programming.
Next | 1,401 | 7,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-22 | latest | en | 0.918118 |
https://docest.com/doc/230180/part-1-voltage-current-and-resistance | 1,718,572,449,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00284.warc.gz | 190,229,660 | 8,636 | 1
Electromagnetism
ELECTROMAGNETISM
OBJECTIVE: To investigate the concepts of Voltage and Current, see how they are related, and see how they can be generated.
Part 1: Voltage, Current and Resistance
THEORY:
Current is defined as the flow of electricity (flow of charges). It is measured in amps (A). Voltage is defined as the energy per charge at a particular position and is measured in volts (V). Current tends to flow from a place of high voltage to a place of low voltage - just like water tends to flow from a place of higher elevation to a place of lower elevation. (Hence, elevation in water flow is analogous to voltage in current flow.) Resistance is simply how much the material resists the flow of charges and is measured in ohms (). (In our water analogy, water flowing through large pipes encounters very little resistance, while water flowing through a sponge or through sand encounters a higher resistance.) A voltage source (such as a battery or a power supply) is like a water pump: a water pump can move water from a lower place to a higher place; the voltage source moves the charge from a place of low voltage (negative terminal) to a place of high voltage (positive terminal). A Circuit is simply a path for the charges to flow from the high voltage position to the low voltage position, just like pipes or rivers can form a "circuit" for water flow.
METHOD:
First we will see how current and voltage are related. This means that we will have to keep the resistance the same and see how the current responds when we change the voltage. Second, we will keep the voltage the same and vary the resistance (the circuit) and see how current and resistance are related.
NOTE:
In using D.C. instruments, care must be taken to connect them up with the proper polarity. The terminal marked (+) should always be connected to the positive terminal of the power supply and the terminal marked () to the negative terminal. They need not be connected directly, but you should be able to trace back to the proper terminal. On many instruments only one terminal is marked, and it is understood that the other is the opposite polarity. On meters with more than one scale, the number on the terminal refers to the MAXIMUM value that can be measured on that scale.
CAUTION: DO NOT PLUG IN THE POWER SUPPLY UNTIL YOUR COMPLETE CIRCUIT HAS BEEN CHECKED BY THE INSTRUCTOR. THIS IS TO PROTECT THE INSTRUMENTS.
PROCEDURE:
1. Connect the apparatus as shown in Fig. 1, using the 10 ohm resistor on the wooden block for R, one DMM for the voltmeter, and the other DMM for the ammeter. Although the resistor is stamped 10, it may not be exactly 10 ohms when measured. Resistors are guaranteed accurate only within a certain percentage (10% in this case) which is usually indicated on the resistor in some way. The power supply (PS) has a control (left side) so that by rotating it the voltage difference across the resistor can be varied from zero to some maximum value. [Note: the right control should be set between ½ to full clockwise.]
1. Adjust the voltage control on the PS so that it increases from zero to five volts in intervals of 1 volt. Take readings of voltage and current at every 1 volt interval.
1. In your written report the data from step 2 will be plotted to see whether current and voltage are proportional. Can you estimate from your data right now if this is so? What is the constant of proportionality (that is, what do you have to multiply current by to get the voltage)? Is it similar to your value of 10 for the resistor? Is this constant the same for all voltages measured?
1. Replace the 10 resistor with one light bulb. Slowly increase the voltage from zero to 5 volts in one volt intervals again and note how both the current and the brightness of the light bulb vary. Can you estimate what the resistance of this light bulb is? Does the light bulb's resistance change with increasing voltage (and current)? Note that as the current increases, the wire in the light bulb gets brighter: it heats up and starts to glow. [In this case, the resistor (light bulb wire) gets very hot, and this shows that resistance increases with temperature.]
1. Replace the one light bulb with two light bulbs connected in series. (That is, connect the left end of light bulb #1 to the + voltage terminal, connect the right end of light bulb #1 to the left end of light bulb #2, and then the right end of light bulb #2 to the + terminal of the ammeter. In other words, make the current flow through both resistors - do not give it a choice.) Now set the voltage of the PS to 5 volts and note both the current and the brightness of both bulbs. Is the current from the PS flowing completely through light bulb #1, partially through #1 and partially through #2, completely through #2, or completely through both? What do you have to do (or is it even possible) to the voltage to make both light bulbs bright? What happens to the current when you do this?
1. Unhook the two light bulbs of step 5 and re-hook them in parallel. (That is, connect the left ends of light bulbs #1 and #2 together and to the + terminal of the PS, and connect the right ends of the light bulbs together and to the + terminal of the ammeter. In other words, give the current a choice of flowing either through light bulb #1 or light bulb #2.) Now set the voltage of the PS to 5 volts, measure the current, and note how bright each light bulb is. Is the current from the PS flowing completely through light bulb #1, completely through #2, split between #1 and #2, or completely through both?
1. Plot a graph of voltage (ordinate or vertical axis) and current (abscissa or horizontal axis) using the data gathered in step 2 of the procedure. Is it a straight line? What does this mean about how current and voltage are related? Compute the slope (V/A) of the best straight line through the points. According to our relations between units, 1 V/A = 1. Does your slope value approximately equal your resistance? Can you infer how voltage, current, and resistance are all related?
1. By placing more resistors in the circuit in a series fashion, does the overall (effective) resistance increase, decrease, or stay the same. Can you explain this?
1. By placing more resistors in the circuit in a parallel fashion, does the overall (effective) resistance increase, decrease, or stay the same. Can you explain this?
1. Power used in the circuit can be found by multiplying the current and the voltage together. Power is measured in Watts (W). (Note that 1 W = 1 AV.) For the same voltage, which circuit (parallel or series) takes more power? For the same voltage, which circuit (parallel or series) has the lower effective resistance? Which circuit generated the most light from the light bulbs? In your household wiring, do you connect appliances in series or parallel to the outlet?
Part 2: Generating Electricity
METHOD:
In this part we will see a very surprising result: we can generate electricity by moving a magnet in an electrical circuit.
PROCEDURE:
(In order to be sure that your observation of an effect was correct, you may repeat any step in this procedure as many times as you wish.)
1)Bar magnet at rest inside the coil.
a)Place the N pole of the bar magnet inside the coil.
b)Connect the coil to the voltmeter and note any reading that occurs as the last connection is made.
2)Bar magnet moved.
a)Move the N pole of the magnet out of the coil quickly (with a jerk). Did you get a reading while you were removing the magnet?
b)Now insert the N pole of the magnet into the coil with a quick movement. Did you get a reading while you were inserting the magnet? Note the sign +/- of the reading. Was it the same as in step a when you removed the N pole?
c)Next insert the S pole of the magnet (as in step b) and note whether you get a reading. Did the +/- sign change or stay the same for inserting the S pole compared to inserting the N pole?
3)Change the voltmeter into an ammeter and repeat steps 1 and 2 above.
4)Now switch the DMM from DC to AC (there is a button for this). Now move the magnet in and out very quickly and repeatedly. Can you generate much of a current? Would this current be enough to light the light bulb? Try it!
NOTE: By oscillating the magnetic field (via the magnet), a current was produced. It is also true that by oscillating a current, you can produce an oscillating magnetic field. This is the basis for understanding light (or radio waves or any other type of light) as an electromagnetic wave. | 1,920 | 8,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-26 | latest | en | 0.938654 |
https://diydrones.com/forum/topics/gps-accuracy-refresh-rate-and-kalman-filtering | 1,679,773,793,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945372.38/warc/CC-MAIN-20230325191930-20230325221930-00289.warc.gz | 254,316,963 | 24,540 | ### GPS Accuracy, Refresh Rate, and Kalman Filtering?
Hello All!
I am working on a project where I have two devices that constantly trying to locate eachother using GPS modules and LoRa transcievers.
Its a buddy tracker for spearfishermen, to make it easier to locate your partner when you're out in the ocean.
Here's a quick example:
And a less quick demo:
However, I am having some growing pains with the precision of my gps modules.
I'd like to be able to use the two gps coordinates to calculate a bearing between the two devices, and use the IMU heading to determine its actual heading.
The IMU is actually working great (BNO055), its actually the gps precision that is gumming up the works.
For example, imagine the red markers are the devices true location, and the circle around them represents their maximum error.
The green line represents the true bearing between devices, and the red the worst case bearing output.
Bearing Error vs Distance:
This small plot shows how severe the bearing error can be in degrees (y axis), and the distance between devices in meters (x axis) given that both units are off by 2m.
So, if you're 5 meters away from your partner, the device could be 40 degrees off - not exactly what I'm shooting for.
Heres about 5 minutes of data from my gps modules at the beach:
In this setup, the modules are about 10cm apart from eachother and stationary the entire time. Modules 1 & 2 are some NEO-6M cheapies, and module 3 is an adafruit 'ultimate' gps.
I'm wondering what options are available to me to help improve this performance a bit and I know the DIY drones community has a lot of experience with this sort of thing.
Some specific questions:
1) If I use a faster refresh rate on my gps module, can I use a moving average to improve the accuracy of the units? Or do the GPS modules already do that kind of thing automatically if you have them in a slow refresh mode?
2) Given that I have some linear acceleration data in global reference frame from the BNO055, it seems like some sort of Kalman filter should be possible to improve the GPS accuracy. I know that the acceleration is noisy, and that some versions of ardupilot have this option available, but I've never deployed a Kalman filter before (lots of FIR filters...) and I was wondering if anybody had really good example or literature of this? What kind of performance improvements would be reasonable to expect if I deployed it correctly? Just need a little hand holding XD
Thanks for your time! Let me know what you think!
You need to be a member of diydrones to add comments!
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Oct 13, 2022 | 1,258 | 4,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-14 | longest | en | 0.923742 |
https://www.thefreelibrary.com/A+Generalization+of+Linear+and+Nonlinear+Retarded+Integral...-a0595704679 | 1,571,167,611,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00423.warc.gz | 1,053,175,437 | 16,733 | # A Generalization of Linear and Nonlinear Retarded Integral Inequalities in Two Independent Variables.
1. Introduction
With the development of science and technology, various inequalities have been paid more and more attention, and the generalization of inequalities has become one of the important research directions in modern mathematics. The integral inequality, which has integrals of unknown functions, is an important type of inequality. For nonlinear differential equations derived from the natural science and engineering technology, especially from various branches of mathematics, it is difficult or impossible to obtain explicit solutions in most cases. Therefore, it is of great significance to get the bounds of the solutions to those nonlinear differential equations. Integral inequalities just can provide the bounds of the solutions to the nonlinear differential equations and integral equations. Hence, integral inequalities are used to serve as handy tools in the study of the qualitative properties of solutions to differential and integral equations, such as existence, uniqueness, boundedness, oscillation, stability, and invariant manifold. For example, these inequalities have been widely employed to investigate the stability of switched systems which can be applied to modeling many engineering system problems in real world, such as traffic control, automobile engine control, switching power converters, and multiagent consensus [1-5]. For some related contributions on various classes of integral inequalities, we refer the reader to [1-20] and the references cited therein.
For convenience, throughout this paper, R represents the set of real numbers, [R.sub.+] = [0, [infinity]), and C(A, B) signifies the class of all continuous functions defined on set A with range in the set B.
In what follows, we provide some background details that motivated our study. One of the most famous and widespread integral inequalities in the study of differential and integral equations is Gronwall-Bellman-type inequality [6-8], which can be described as follows.
Theorem 1. Let u and f be nonnegative continuous functions on an interval [a, b] satisfying
u(t) [less than or equal to] c + [[integral].sup.t.sub.a] f(s) u(s) ds, t [member of] [a, b] (1)
for some constant c [greater than or equal to] 0. Then
u(t) [less than or equal to] c exp ([[integral].sup.t.sub.a] f(s) u(s) ds), t [member of] [a, b] (2)
In recent years, many scholars have done a lot of researches and generalization of the above integral inequality, which make the integral inequalities develop continually and the application fields expand gradually. Pachpatte [9, 10] investigated the inequality
[mathematical expression not reproducible] (3)
and the retarded inequality
[mathematical expression not reproducible], (4)
where [alpha] [member of] [C.sup.1](I, I) is nondecreasing with [alpha](t) [less than or equal to] t on I = [0, T) and [u.sub.1] and [u.sub.2] are constants. Abdeldaim and El-Deeb [11] generalized [9] and analyzed the following retarded linear and nonlinear inequalities:
[mathematical expression not reproducible], (5)
respectively. Tian et al. [16] introduced the retarded inequalities in two independent variables as follows.
Theorem 2 (see [16, Theorem 1]). Let u, f, and g [member of] C([R.sub.+] x [R.sub.+], [R.sub.+]), a(x) > 0, b(y) > 0, a'(x) [greater than or equal to] 0, b'(y) [greater than or equal to] 0, and [alpha], [beta] [member of] [C.sup.1]([R.sub.+], [R.sub.+]) be nondecreasing with [alpha](x) [less than or equal to] x and [beta](y) [less than or equal to] y on [R.sub.+]. Moreover, let [phi] [member of] [C.sup.1]([R.sub.+], [R.sub.+]) bean increasing function with [phi]([infinity]) = [infinity] and let [phi](x) > 0 on (0, [infinity]), [psi] [member of] [C.sup.1]([R.sub.+], [R.sub.+]) be a nondecreasing function with [psi](x) > 0 on (0, [infinity]). If
[mathematical expression not reproducible], (6)
then, for 0 [less than or equal to] x < [[xi].sub.1], 0 [less than or equal to] y < [[eta].sub.1],
[mathematical expression not reproducible], (7)
where
[mathematical expression not reproducible], (8)
[[OMEGA].sup.-1], [[phi].sup.-1], and [G.sup.-1] are the inverses of [OMEGA], [phi], and G, respectively; ([[xi].sub.1], [[eta].sub.1]) [member of] [R.sub.+] x [R.sub.+] is chosen so that
[mathematical expression not reproducible] (9)
with dom(*) denoting the function domain.
Theorem 3 (see [16, Corollary 1]). Assume that u, f, g, a, b, [alpha], and [beta] are defined as in Theorem 2. Let [phi](u) = [u.sup.p] and [psi](u) = [u.sup.q-1] in Theorem 2, where p [greater than or equal to] q > 1 are positive constants. If
[mathematical expression not reproducible] (10)
then, for all (x, y) [member of] [R.sub.+] x [R.sub.+],
[mathematical expression not reproducible], (11)
where
[mathematical expression not reproducible]. (12)
Motivated by the recent contributions of Abdeldaim and El-Deeb [11], Zhang and Meng [14], and Tian et al. [16], our principal goal is to extend the inequalities with one variable in [11] to those with two variables which include Theorems 2 and 3 as special cases.
The rest of the work is organized as follows. A useful lemma that plays a fundamental role in the proofs of the main theorems is presented in Section 2. In Section 3, we propose our main theorems and corollary on several new types of linear and nonlinear retarded integral inequalities in two independent variables. An illustrative example is given to indicate the usefulness of these inequalities in Section 4, which is followed by a short conclusion in Section 5.
2. Lemma
The subsequent lemma is helpful in proving our main theorems.
Lemma 4. Assume that u, f, and g [member of] C([R.sub.+] x [R.sub.+], [R.sub.+]) and [phi] [member of] C([R.sub.+], [R.sub.+]) is an increasing function with [phi]([infinity]) = [infinity] and [psi] [member of] C([R.sub.+], [R.sub.+]) is a nondecreasing function. Suppose that c is a nonnegative constant and [alpha], [beta] [member of] [C.sup.1] ([R.sub.+], [R.sub.+]) are nondecreasing with [alpha](x) [less than or equal to] x, [beta](y) [less than or equal to] y, [alpha](0) = 0, and [beta](0) = 0 on [R.sub.+]. If
[mathematical expression not reproducible] (13)
then, for 0 [less than or equal to] x < [xi], 0 [less than or equal to] y < [eta],
[mathematical expression not reproducible], (14)
where
[mathematical expression not reproducible]; (15)
[[phi].sup.-1] and [G.sup.-1] are the inverses of [phi] and G, respectively; ([xi], [eta]) [member of] [R.sub.+] x [R.sub.+] is chosen so that
[mathematical expression not reproducible]. (16)
Proof. Define the nondecreasing positive function z by
[mathematical expression not reproducible], (17)
where [epsilon] is an arbitrary small positive number. Utilizing inequality (13) and the monotonicity of [[phi].sup.-1], we get
u(x, y) [less than or equal to] [[phi].sup.-1] (z(x, y)). (18)
Differentiating (17) with respect to x and combining (18) and the monotonicities of [[phi].sup.-1], z, and [psi], we conclude that
[mathematical expression not reproducible]. (19)
On account of [psi][[phi].sup.-1] (z(x, y))] [greater than or equal to] [psi][[[phi].sup.-1] (c+[epsilon])] > 0, we deduce that
[mathematical expression not reproducible]. (20)
Integrating the latter inequality on [0, x] and letting [epsilon] [right arrow] 0, we have
[mathematical expression not reproducible] (21)
owing to (15). By virtue of (16), (18), and the last inequality, we obtain inequality (14). The proof is complete.
Remark 5. Assume that [mathematical expression not reproducible]. Then G([infinity]) = [infinity] and (14) is valid on [R.sub.+] x [R.sub.+]; that is, one can select [xi] = [infinity] and [eta] = [infinity].
3. Main Results
The following are the main results of this paper.
Theorem 6. Let u, a, f, g, and h [member of] C([R.sub.+] x [R.sub.+], [R.sub.+]) and let [alpha], [beta] [member of] [C.sup.1]([R.sub.+], [R.sub.+]) be nondecreasing with [alpha](x) [less than or equal to] x, [beta](y) [less than or equal to] y, [alpha](0) = 0, and [beta](0) = 0 on [R.sub.+]. If the inequality
[mathematical expression not reproducible] (22)
holds, for all (x, y) [member of] [R.sub.+] x [R.sub.+], then
[mathematical expression not reproducible]. (23)
Proof. Letting
[mathematical expression not reproducible]. (24)
then z(0, y) = z(x, 0) = 0 and
u (x, y) [less than or equal to] [alpha] (x, y) + z (x, y). (25)
Our assumptions on f, u, h, g, [alpha], and [beta] indicate that z is a positive function which is nondecreasing with respect to each of the two variables. Differentiating z with respect to x and using (25), we arrive at
[mathematical expression not reproducible]. (26)
By virtue of the monotonicity of z, we get
[mathematical expression not reproducible]. (27)
Multiplying the latter inequality by [mathematical expression not reproducible] yields
[mathematical expression not reproducible]. (28)
Integrating this inequality on [0, x], we deduce that
[mathematical expression not reproducible]. (29)
Combining (25) with (29), we get inequality (23). This completes the proof.
Theorem 7. Let u, a, f, g, and h [member of] C([R.sub.+] x [R.sub.+], [R.sub.+]), a(x, y) > 0, [a.sub.x] [greater than or equal to] 0, [a.sub.y] [greater than or equal to] 0, and [alpha], [beta] [member of] [C.sup.1] ([R.sub.+], [R.sub.+]) be nondecreasing with [alpha](x) [less than or equal to] x, [beta](y) [less than or equal to] y, [alpha](0) = 0, and [beta](0) = 0 on [R.sub.+]. Moreover, let [gamma] and [psi] [member of] [C.sup.1]([R.sub.+], [R.sub.+]) be nondecreasing function with [gamma] > 0 and [psi] > 0 on (0, [infinity]). If
[mathematical expression not reproducible], (30)
then, for 0 [less than or equal to] x < [xi], 0 [less than or equal to] y < [eta],
[mathematical expression not reproducible], (31)
where
[mathematical expression not reproducible], (32)
[mathematical expression not reproducible], (33)
[mathematical expression not reproducible]. (34)
[[OMEGA].sup.-1] and [G.sup.-1] are the inverses of [OMEGA] and G, respectively; ([xi], [eta]) [member of] [R.sub.+] x [R.sub.+] is chosen so that
[mathematical expression not reproducible] (35)
for 0 [less than or equal to] x < [xi], 0 [less than or equal to] y < [eta].
Proof. Define the nondecreasing function z by
[mathematical expression not reproducible]. (36)
Then
u (x, y) [less than or equal to] a (x, y) + z (x, y). (37)
Differentiating (36) and using (37) and the monotonicity of [gamma], we obtain
[mathematical expression not reproducible]. (38)
Let [T.sub.1] [less than or equal to] [xi] and [T.sub.2] [less than or equal to] [eta] be arbitrary numbers. Utilizing (38) and the monotonicities of a, z, and y, we get that, for 0 [less than or equal to] x < [T.sub.1] and 0 [less than or equal to] y < [T.sub.2],
[mathematical expression not reproducible]. (39)
For a([T.sub.1], [T.sub.2]) > 0 and [gamma][a([T.sub.1], [T.sub.2]) + z(x, y)] > 0,
[mathematical expression not reproducible]. (40)
+ h(a(x),[beta](y))+f(a(x),[beta](y))
From another point of view,
[mathematical expression not reproducible]. (41)
It follows from (40) and (41) that
[mathematical expression not reproducible]. (42)
Integrating the above inequality on [0, y] with respect to the second variable and taking [z.sub.x](x, 0) = 0 into account, we have
[mathematical expression not reproducible]. (43)
From (33), the latter relation gives
[mathematical expression not reproducible]. (44)
Integrating the last inequality over [0, x], we get
[mathematical expression not reproducible], (45)
where P is defined as in (32). Combining (37) and the monotonicity of a and employing Lemma 4, we obtain
[mathematical expression not reproducible], (46)
where G is defined as in (34). Taking x = [T.sub.1] and y = [T.sub.2], we conclude that
[mathematical expression not reproducible]. (47)
As [T.sub.1] [less than or equal to] [xi] and [T.sub.2] [less than or equal to] [eta] are arbitrary, we get the desired inequality (31). The proof is complete.
Theorem 8. Assume that u, a, f, g, h, [alpha], [beta], [gamma], and [psi] are defined as in Theorem 7. Moreover, let [phi] [member of] [C.sup.1]([R.sub.+], [R.sub.+]) be increasing function with f(x) = x and <p(x) > 0 on (0, x). If
[mathematical expression not reproducible], (48)
then, for 0 [less than or equal to] x < [xi], 0 [less than or equal to] y [less than or equal to] [eta],
[mathematical expression not reproducible], (49)
where
[mathematical expression not reproducible]; (50)
[[phi].sup.-1], [[OMEGA].sup.-1], and [G.sup.-1] are the inverses of [phi], [OMEGA], and G, respectively; ([xi], [eta]) [member of] [R.sub.+] x [R.sub.+] is chosen so that
[mathematical expression not reproducible] (51)
for 0 [less than or equal to] x < [xi], 0 [less than or equal to] y < [eta].
Proof. Define function z by (36). Then
u (x, y) [less than or equal to] [[phi].sup.-1] [a (x, y) + z (x, y)]. (52)
The rest of the proof is similar to that of Theorem 7 and hence is omitted.
Remark 9. Letting a(x, y) = a(x) + b(y), [gamma](u(x, y)) = u(x, y), and g(x, y) = 0 in Theorem 8, Theorem 8 turns out to be Theorem 2. Therefore, the inequality established in Theorem 8 generalizes that of [16, Theorem 1].
If [phi](u) = [u.sup.p], [gamma](u) = [u.sup.q], and [psi](u) = [u.sup.n] in Theorem 8, where p [greater than or equal to] q + n > l, and p, q, and n are positive constants, then we have the following corollary.
Corollary 10. Assume that u, a, f, g, h, [alpha], and [beta] are defined as in Theorem 8. If
[mathematical expression not reproducible], (53)
then, for all (x, y) [member of] [R.sub.+] x [R.sub.+],
[mathematical expression not reproducible], (54)
where
[mathematical expression not reproducible]. (55)
Proof. Assume that p > q + n and let [phi](u) = [u.sup.p], [gamma](u) = [u.sup.q], and [psi](u) = [u.sup.n]. Then we have [[phi].sup.-1](u) = [u.sup.1/p], and so
[mathematical expression not reproducible], (56)
where [lambda] and [theta] are defined in (55). Using Theorem 8, one can easily obtain
[mathematical expression not reproducible]. (57)
When p = q + n,
[mathematical expression not reproducible], (58)
where [lambda] and [theta] are the same as in (55). By Theorem 8, similar discussions can give
[mathematical expression not reproducible]. (59)
This completes the proof.
Remark 11. Letting a(x, y) = a(x) + b(y), q = 1, n = q-1, and g(x, y) = 0, Corollary 10 reduces to Theorem 3. Hence, the inequality established in Corollary 10 includes the result of [16, Corollary 1].
4. Example
Example 1. Consider the integral equation
[mathematical expression not reproducible], (60)
where k : [R.sub.+] x [R.sub.+] [right arrow] R and F : [R.sub.+] x [R.sub.+] x R x R [right arrow] R are continuous functions, [alpha], [beta] [member of] [C.sup.1]([R.sub.+], [R.sub.+]) is nondecreasing with [alpha](x) [less than or equal to] x, [beta](y) [less than or equal to] y, [alpha](0) = 0, and [beta](0) = 0 on [R.sub.+], and p [greater than or equal to] 4 is a constant. Suppose that
[mathematical expression not reproducible], (61)
where a, f, h, and g are defined as in Corollary 10. Combining (60)-(61) yields
[mathematical expression not reproducible]. (62)
Exploiting Corollary 10, we obtain an explicit bound to the solutions of (60):
[mathematical expression not reproducible], (63)
where [lambda] and [theta] are defined as in Corollary 10.
5. Conclusions
This paper investigates some new types of linear and nonlinear retarded integral inequalities in two independent variables. Several theorems and a corollary of these inequalities are obtained based on some analysis techniques, such as amplification method, differential, and integration. An illustrative example is studied to demonstrate the effectiveness of the new results.
https://doi.org/10.1155/2017/5129051
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work is supported by the National Natural Science Foundation of China (Grants 61304130 and 51277116).
References
[1] Y. Sun, "Stabilization of switched systems with nonlinear impulse effects and disturbances," Institute of Electrical and Electronics Engineers Transactions on Automatic Control, vol. 56, no. 11, pp. 2739-2743, 2011.
[2] Y. Zhang, M. Wang, H. Xu, and K. L. Teo, "Global stabilization of switched control systems with time delay," Nonlinear Analysis: Hybrid Systems, vol. 14, pp. 86-98, 2014.
[3] E. Gyurkovics, G. Szabo-Varga, and K. Kiss, "Stability analysis of linear systems with interval time-varying delays utilizing multiple integral inequalities," Applied Mathematics and Computation, vol. 311, pp. 164-177, 2017.
[4] Z. Li, Y. Bai, C. Huang, and H. Yan, "Further results on stabilization for interval time-delay systems via new integral inequality approach," ISA Transactions[R], vol. 68, pp. 170-180, 2017.
[5] B. Yang, J. Wang, and J. Wang, "Stability analysis of delayed neural networks via a new integral inequality," Neural Networks, vol. 88, pp. 49-57, 2017.
[6] T. H. Gronwall, "Note on the derivatives with respect to a parameter of the solutions of a system of differential equations," Annals of Mathematics, vol. 20, no. 4, pp. 292-296, 1919.
[7] R. Bellman, "The stability of solutions of linear differential equations," Duke Mathematical Journal, vol. 10, pp. 643-647, 1943.
[8] I. Bihari, "A generalization of a lemma of Bellman and its application to uniqueness problems of differential equations," Acta Mathematica Hungarica, vol. 7, pp. 81-94, 1956.
[9] B. G. Pachpatte, Inequalities for Differential and Integral Equations, vol. 197, Academic Press, New York, NY, USA, 1998.
[10] B. G. Pachpatte, "Explicit bounds on certain integral inequalities," Journal of Mathematical Analysis and Applications, vol. 267, no. 1, pp. 48-61, 2002.
[11] A. Abdeldaim and A. A. El-Deeb, "On generalized of certain retarded nonlinear integral inequalities and its applications in retarded integro-differential equations," Applied Mathematics and Computation, vol. 256, pp. 375-380, 2015.
[12] A. Abdeldaim and A. A. El-Deeb, "On some generalizations of certain retarded nonlinear integral inequalities with iterated integrals and an application in retarded differential equation," Journal of the Egyptian Mathematical Society, vol. 23, no. 3, pp. 470-475, 2015.
[13] H. Zhang and F. Meng, "Integral inequalities in two independent variables for retarded Volterra equations," Applied Mathematics and Computation, vol. 199, no. 1, pp. 90-98, 2008.
[14] H. Zhang and F. Meng, "On certain integral inequalities in two independent variables for retarded equations," Applied Mathematics and Computation, vol. 203, no. 2, pp. 608-616, 2008.
[15] R. P. Agarwal, S. Deng, and W. Zhang, "Generalization of a retarded Gronwall-like inequality and its applications," Applied Mathematics and Computation, vol. 165, no. 3, pp. 599-612, 2005.
[16] Y. Tian, M. Fan, and F. Meng, "A generalization of retarded integral inequalities in two independent variables and their applications," Applied Mathematics and Computation, vol. 221, pp. 239-248, 2013.
[17] Y. Tian, Y. Cai, L. Li, and T. Li, "Some dynamic integral inequalities with mixed nonlinearities on time scales," Journal of Inequalities and Applications, vol. 2015, no. 1, pp. 1-10, 2015.
[18] X. Liu, L. Zhang, P. Agarwal, and G. Wang, "On some new integral inequalities of Gronwall-Bellman-Bihari type with delay for discontinuous functions and their applications," Indagationes Mathematicae, vol. 27, no. 1, pp. 1-10, 2016.
[19] L. Li, F. Meng, and L. He, "Some generalized integral inequalities and their applications," Journal of Mathematical Analysis and Applications, vol. 372, no. 1, pp. 339-349, 2010.
[20] R. Xu and F. Meng, "Some new weakly singular integral inequalities and their applications to fractional differential equations," Journal of Inequalities and Applications, vol. 2016, no. 1, article no. 78, pp. 1-16, 2016.
Ying Jiang, Guojing Xing, and Chenghui Zhang
School of Control Science and Engineering, Shandong University, Jinan, Shandong 250061, China
Correspondence should be addressed to Guojing Xing; xgjsdu@sdu.edu.cn
Received 26 August 2017; Accepted 5 December 2017; Published 25 December 2017
Academic Editor: Eric R. Kaufmann
COPYRIGHT 2018 Hindawi Limited
No portion of this article can be reproduced without the express written permission from the copyright holder. | 5,985 | 20,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-43 | latest | en | 0.936264 |
https://www.physicsforums.com/threads/potential-of-tidals.823495/ | 1,519,539,157,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816138.91/warc/CC-MAIN-20180225051024-20180225071024-00218.warc.gz | 912,010,417 | 17,144 | # Potential of tidals
1. Jul 16, 2015
### Vrbic
Hello,
I'm interested in tidal force of Earth - Moon system. How can I describe this force in lagrangian? I understand it is complex problem, but I'm looking for some kind of toy model or something like that just for my understanding.
What do tidal forces cause? I needn't exact mechanism how, but its consequences on behavior of orbit, period, spin etc.
2. Jul 16, 2015
### Noctisdark
You can express it usings newton's law for gravity, F = GMm/r2, the tidal forces are difference of forces of gravity btw 2 points, It's easy to calculate, name it Fτ ≈ GMmΔr/r3 then you must the materials that make up the moon to know they elasticity (Young Modulus), Y = Fτ*L0/(A0*ΔL), you can know calculate ΔL, I know this doesn't answer your question, but It's impossible (Or very hard, I can't think of a way) to calculate this using lagrangian mechanics, they serve determining equation of motion, there is no "real motion" here unless you redifine elasicity and make a lot of expirements, I've ignored the rotation, I think that GR come up with another method calculating tidal forces using Riemann curvature tensor, but I don't know how, never bumped into this before, Good luck !!
3. Jul 16, 2015
### Vrbic
Wow, intersting, I knew it is complex problem, but I thought you can find some "toy" model version for understundig, but I mean I understand what you want to say. I will continue, I will trying and we will see maybe I will find something comparable with Earth-Moon system.
Thanks for comment, if you have something else (article or idea) please let me know.
4. Jul 16, 2015
### Staff: Mentor
I'm not seeing the problem here. I don't think it is complex at all. Did you not understand how to use the equations Noctisdark provided?
5. Jul 17, 2015
### Vrbic
Hi, no I don't know this equotion, I will google it. For what is it good? Could you tell me more abou procedure etc.?
6. Jul 17, 2015
### Staff: Mentor
Why do you need to google the equation? It is right there in the post. Please rearead post #2, try to make use of the equation and ask a more specific question about the problem you are having with it.
7. Jul 18, 2015
### Vrbic
Well, I read all again and I was thinking about that. Noctisdark's idea is nice but I don't know where it is pointing. I'm capable to resolve $\Delta L$ (if I find Young modulus of moon matter) but why? Maybe better (not "super scientific") could be to "guess" functions of orbiting velocities and spins from dissipation of energy due to tidal effects and try to put it to lagrangian. What do you mean?
8. Jul 18, 2015
### Noctisdark
Lagragian are just F = ma in arbitrary co-ordinates, they describe motion, can you express the tidal forces effect on the moon by F = ma ?, If it's possible then you can do it with the lagrangian mechanics, our current definition of elasticity are described by force per mm(sometime m)jusr like hooke's law F = kx th at still hold for some degree to solid materials, then stuff become serious with some tensors, for more information about elasticity check http://physics.info/elasticity/, https://en.m.wikipedia.org/wiki/Elasticity_(physics)
9. Jul 18, 2015
### Vrbic
Thank you, I read it. I also read https://en.wikipedia.org/wiki/Tidal_force where is derived tidal force in the form F=ma.
10. Jul 18, 2015
### Noctisdark
I do not agree with wikipidea on this, Fnet = ma not just the force of gravity, and the article describe a tidal force effect, tidal acceleration which explains why the moon gets faster "sometimes",becase tidal forces on earth make it strech thus make some parts of it (the earth) closer to the moon which will make the moon accelerate more than usual, this can also explain why we see the same side of the moon, other than that, the article don't describe tidal forces, but their effects,
11. Jul 18, 2015
### Staff: Mentor
I don't understand: You asked the question, so you must decide what it is that you want to know. The equation provided gives the tidal force -- so what, exactly, is the problem you want to solve with it? | 1,052 | 4,104 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-09 | latest | en | 0.947914 |
https://www.vedantu.com/question-answer/three-percent-of-4200-is-equal-to-6-percent-of-class-8-maths-cbse-60aa6f6d576d88681aa0e2ff | 1,702,189,969,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00179.warc.gz | 1,142,048,420 | 28,518 | Courses
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# Three percent of $4200$ is equal to $6$ percent of what number?A. $8400$ B. $2100$ C. $1260$ D. $252$ E. $126$
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Hint: Percent: Percent or percentage is used in mathematics as the number or ratio which is expressed as a fraction of a hundredth. And this is followed by the sign ‘%’. Percentage has no dimension. Hence it is called dimensionless. Every percentage variable has three possible unknown variables: percentage, part, base.
To solve this we should apply the basics of percentage. First find the value of the given value of the percentage given then we take a variable. Calculate the value of the variable as per the given condition and choose the correct option as per asked in question.
Complete step by step solution:
Given,
The number is $4200$
According to the question
$3\% \,of\,4200\,is\,$
Now solve
$= \dfrac{3}{{100}} \times 4200$
Simplify
$= 3 \times 42$
$= 126$
It is equal to the $6\%$ of some number
$6\% \,of\,x\,is\,126$
$\Rightarrow \dfrac{6}{{100}} \times x = 126$
Simplify
$\Rightarrow 6x = 126 \times 100$
$\Rightarrow x = \dfrac{{126 \times 100}}{6}$
$\Rightarrow x = 2100$
So the answer is (B) $2100$ .
So, the correct answer is “Option B”.
Note: Percentage is widely used in profit and loss. It is also used to show statistics data to increase or decrease in data by how much percentage. The percentage function takes the form z is equal to the x percent of number y. Taking a percent can be viewed as the same operation as multiplication by a fraction.
We can get this answer by using trial and error method for which we have to separately calculate its given percentage as per maintained in the question after that we can also cross check our given answer as calculated by the above method. | 512 | 1,889 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2023-50 | longest | en | 0.913127 |
http://www.lmfdb.org/Genus2Curve/Q/100352/c/ | 1,563,441,052,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525587.2/warc/CC-MAIN-20190718083839-20190718105839-00492.warc.gz | 222,808,229 | 5,427 | # Properties
Label 100352.c Sato-Tate group $\mathrm{USp}(4)$ $$\End(J_{\overline{\Q}}) \otimes \R$$ $$\R$$ $$\End(J_{\overline{\Q}}) \otimes \Q$$ $$\Q$$ $$\overline{\Q}$$-simple yes $$\mathrm{GL}_2$$-type no
# Related objects
## Genus 2 curves in isogeny class 100352.c
Label Equation
100352.c.100352.1 $$y^2 = x^5 - 9x^3 + 7x^2 + 14x - 14$$
## L-function data
Analytic rank:$$1$$
Prime L-Factor
$$2$$$$1$$
$$7$$$$1 + 7 T^{2}$$
Good L-factors:
Prime L-Factor
$$3$$$$( 1 - 2 T + 3 T^{2} )( 1 + T + 3 T^{2} )$$
$$5$$$$( 1 - 2 T + 5 T^{2} )( 1 + 3 T + 5 T^{2} )$$
$$11$$$$( 1 - T + 11 T^{2} )( 1 + 2 T + 11 T^{2} )$$
$$13$$$$1 + 6 T^{2} + 169 T^{4}$$
$$17$$$$( 1 - T + 17 T^{2} )( 1 + 2 T + 17 T^{2} )$$
$$19$$$$1 + T - 16 T^{2} + 19 T^{3} + 361 T^{4}$$
$$23$$$$1 + T + 6 T^{2} + 23 T^{3} + 529 T^{4}$$
$$29$$$$( 1 - 2 T + 29 T^{2} )( 1 + 6 T + 29 T^{2} )$$
$\cdots$$\cdots$
$$\mathrm{ST} =$$ $\mathrm{USp}(4)$
Not of $$\GL_2$$-type over $$\Q$$
All $$\overline{\Q}$$-endomorphisms of the Jacobian are defined over $$\Q$$. | 505 | 1,029 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-30 | latest | en | 0.41853 |
http://blogs.wsj.com/numbers/the-numbers-of-the-nbas-bouncing-balls-111/?mg=blogs-wsj&url=http%253A%252F%252Fblogs.wsj.com%252Fnumbersguy%252Fthe-numbers-of-the-nbas-bouncing-balls-111%252F | 1,429,596,831,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246640124.22/warc/CC-MAIN-20150417045720-00268-ip-10-235-10-82.ec2.internal.warc.gz | 33,441,214 | 25,733 | • By
Executives for the 14 worst teams in the National Basketball Association will gather today to watch as an exercise in probability helps determine the fate of their efforts at turnarounds. It’s NBA draft lottery time.
The NBA’s fateful ping-pong balls (NBA.com)
In this exciting competition between those clubs that didn’t qualify for the playoffs, each team is assigned a set of four-number combinations, with each number in the combination ranging from 1 to 14. The worse your team performed last season, the more combos you get. The game is to see one of your sets of numbers turn up when 14 ping-pong balls, numbered 1 to 14, are sent bouncing in a standard lottery machine (hence the name of the event). Four balls are picked, and the lucky team whose numbers turn up gets the first pick in next month’s NBA draft, where college and foreign players are assigned to teams in the league. The balls are returned to the machine and the exercise is repeated twice more, to assign the No. 2 and No. 3 picks (each team gets just one pick, so if a team’s combo comes up a second time, the balls are returned and shuffled once more). The remaining 11 teams get picks 4 through 14 in order of their final records. And the whole thing is televised, with anxious execs looking on as the balls are drawn.
Some interesting math is at play here. First, consider that there are 1,001 possible four-number combinations that can be drawn from the machine. Why 1,001? There are 14 different numbers that can appear on the first ball drawn from the machine, 13 for the second ball, 12 for the third and 11 for the fourth. Multiply those together and you get 24,024. But the order of the numbers drawn doesn’t matter in this lottery — nor in most lotteries, for that matter — so we’re overcounting the possibilities by a factor of 24. That’s because there are 24 different ways to draw the same combination. Any of the four numbers might be drawn first, and for each of those four possibilities, three numbers might be drawn second, and for each of those, two might be drawn third. Four times three times two is 24. And 24,024 divided by 24 is 1,001.
The league discards 11-12-13-14 as a valid combo, leaving an even 1,000. That’s done to make it easier to calculate probabilities: The worst team in the league (this year, the Memphis Grizzlies) gets 250 of the 1,000 combos, for a 25% shot at the first pick.
To get 1,000 possibilities, it would be a lot easier to number three balls from 0 to 9 and make the order count — then each number from 0 to 999 would have an equal shot. But that system lacks the aesthetic appeal of having one ball for each team in the lottery.
Then there’s the question of how to divide up the balls. That’s a matter of graver importance in the NBA than in the other major sports leagues, because there are fewer players per team, making each No. 1 draft pick more important. Also, the dropoff in talent from the No. 1 to No. 14 picks is greater than in other leagues (notwithstanding the occasional dud at the top of the draft). So the league has had to balance two goals: Ensuring that the worst team isn’t guaranteed the top pick, which could lead to teams deliberately losing games near the end of the season; and ensuring that the best of the worst teams doesn’t have much of a chance at the top picks, which would seem unfair. Adding a further wrinkle: the addition of expansion teams has meant more teams missing the playoffs, and more teams in the lottery, requiring occasional tweaks to the spread.
All of these factors have led to frequent revisions of the lottery format, from each team having an equal shot at the top pick to various weighted schemes. This year anxious fans are playing simulated versions of the lottery, and NBA executives will watch the fate of their teams, and their careers, in the form of a bouncing ping-pong ball.
Further reading: Last year, the Toronto Raptors’ new general manager saw his first success as a passive observer of the lottery — his team nabbed the first pick, which it turned into Andrea Bargnani, despite having just an 8.8% chance. NBA.com collects the probabilities heading into past lotteries. | 948 | 4,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2015-18 | longest | en | 0.950001 |
https://www.oreilly.com/library/view/designing-embedded-hardware/0596003625/ch02s05.html | 1,582,725,147,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146342.41/warc/CC-MAIN-20200226115522-20200226145522-00381.warc.gz | 820,635,652 | 15,101 | # Capacitors
While a resistor is a component that resists the flow of charge through it, a capacitor stores charge. Capacitance is measured in `Farads` (or more formally, `Faradays`) with an equation symbol C and a unit symbol F. Typical capacitors you will use will range in value from μF (microFarads) down to pF (picoFarads).
The relationship between current, capacitance, and voltage is given by:
`I = C * dV/dt`
where dV/dt is the rate of voltage change over time.
The schematic symbols for capacitors are shown in Figure 2-10. The component on the left is``` bipolar```, while the other two are``` unipolar```. A unipolar capacitor has a positive lead and a negative lead, and it must be inserted into a circuit with the correct orientation. Failing to do so will cause it to explode. (Unipolar capacitors have markings to indicate their orientation.) A bipolar capacitor has no polarity.
Applying a voltage across a capacitor causes the capacitor to become charged. If the voltage source is removed and a path for current flow exists elsewhere in the circuit, the capacitor will discharge and thereby provide a (temporary) voltage and current source (Figure 2-11).
This is an extremely useful characteristic. A given voltage source may have a ...
Get Designing Embedded Hardware now with O’Reilly online learning.
O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers. | 323 | 1,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-10 | longest | en | 0.898211 |
http://au.metamath.org/mpegif/bnj1145.html | 1,532,333,708,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676595531.70/warc/CC-MAIN-20180723071245-20180723091245-00356.warc.gz | 27,488,378 | 14,291 | Mathbox for Jonathan Ben-Naim < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > bnj1145 Structured version Unicode version
Theorem bnj1145 29424
Description: Technical lemma for bnj69 29441. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
Hypotheses
Ref Expression
bnj1145.1
bnj1145.2
bnj1145.3
bnj1145.4
bnj1145.5
bnj1145.6
Assertion
Ref Expression
bnj1145
Distinct variable groups: ,,,,, ,, ,,,,, ,,,, , ,
Allowed substitution hints: (,,,) (,,,,) (,,,) (,,,,) (,,,,) (,,) ()
Proof of Theorem bnj1145
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 bnj1145.1 . . 3
2 bnj1145.2 . . 3
3 bnj1145.3 . . 3
4 bnj1145.4 . . 3
51, 2, 3, 4bnj882 29359 . 2
6 ss2iun 4110 . . . 4
7 bnj1145.5 . . . . . . 7
87, 4bnj1083 29409 . . . . . 6
92bnj1095 29214 . . . . . . . . . 10
109, 7bnj1096 29215 . . . . . . . . 9
1110nfi 1561 . . . . . . . 8
123bnj1098 29216 . . . . . . . . . . . . . . . . 17
137bnj1232 29237 . . . . . . . . . . . . . . . . . 18
14133anim3i 1142 . . . . . . . . . . . . . . . . 17
1512, 14bnj1101 29217 . . . . . . . . . . . . . . . 16
16 ancl 531 . . . . . . . . . . . . . . . 16
1715, 16bnj101 29150 . . . . . . . . . . . . . . 15
18 bnj1145.6 . . . . . . . . . . . . . . . . 17
1918imbi2i 305 . . . . . . . . . . . . . . . 16
2019exbii 1593 . . . . . . . . . . . . . . 15
2117, 20mpbir 202 . . . . . . . . . . . . . 14
22 bnj213 29315 . . . . . . . . . . . . . . . 16
2322bnj226 29163 . . . . . . . . . . . . . . 15
24 simpr 449 . . . . . . . . . . . . . . . . . . 19
2518, 24bnj833 29189 . . . . . . . . . . . . . . . . . 18
26 simp2 959 . . . . . . . . . . . . . . . . . . . 20
27133ad2ant3 981 . . . . . . . . . . . . . . . . . . . 20
283bnj923 29199 . . . . . . . . . . . . . . . . . . . . 21
29 elnn 4857 . . . . . . . . . . . . . . . . . . . . 21
3028, 29sylan2 462 . . . . . . . . . . . . . . . . . . . 20
3126, 27, 30syl2anc 644 . . . . . . . . . . . . . . . . . . 19
3218, 31bnj832 29188 . . . . . . . . . . . . . . . . . 18
33 vex 2961 . . . . . . . . . . . . . . . . . . . 20
3433bnj216 29161 . . . . . . . . . . . . . . . . . . 19
35 elnn 4857 . . . . . . . . . . . . . . . . . . 19
3634, 35sylan 459 . . . . . . . . . . . . . . . . . 18
3725, 32, 36syl2anc 644 . . . . . . . . . . . . . . . . 17
3818, 26bnj832 29188 . . . . . . . . . . . . . . . . . 18
3925, 38eqeltrrd 2513 . . . . . . . . . . . . . . . . 17
402bnj589 29342 . . . . . . . . . . . . . . . . . . . . . . 23
4140biimpi 188 . . . . . . . . . . . . . . . . . . . . . 22
4241bnj708 29186 . . . . . . . . . . . . . . . . . . . . 21
43 rsp 2768 . . . . . . . . . . . . . . . . . . . . 21
4442, 43syl 16 . . . . . . . . . . . . . . . . . . . 20
457, 44sylbi 189 . . . . . . . . . . . . . . . . . . 19
46453ad2ant3 981 . . . . . . . . . . . . . . . . . 18
4718, 46bnj832 29188 . . . . . . . . . . . . . . . . 17
4837, 39, 47mp2d 44 . . . . . . . . . . . . . . . 16
49 fveq2 5730 . . . . . . . . . . . . . . . . . 18
5049eqeq1d 2446 . . . . . . . . . . . . . . . . 17
5125, 50syl 16 . . . . . . . . . . . . . . . 16
5248, 51mpbird 225 . . . . . . . . . . . . . . 15
5323, 52bnj1262 29244 . . . . . . . . . . . . . 14
5421, 53bnj1023 29213 . . . . . . . . . . . . 13
55 3anass 941 . . . . . . . . . . . . . . 15
5655imbi1i 317 . . . . . . . . . . . . . 14
5756exbii 1593 . . . . . . . . . . . . 13
5854, 57mpbi 201 . . . . . . . . . . . 12
591biimpi 188 . . . . . . . . . . . . . . 15
607, 59bnj771 29195 . . . . . . . . . . . . . 14
61 fveq2 5730 . . . . . . . . . . . . . . 15
62 bnj213 29315 . . . . . . . . . . . . . . . 16
63 sseq1 3371 . . . . . . . . . . . . . . . 16
6462, 63mpbiri 226 . . . . . . . . . . . . . . 15
65 sseq1 3371 . . . . . . . . . . . . . . . 16
6665biimpar 473 . . . . . . . . . . . . . . 15
6761, 64, 66syl2an 465 . . . . . . . . . . . . . 14
6860, 67sylan2 462 . . . . . . . . . . . . 13
6968adantrl 698 . . . . . . . . . . . 12
7058, 69bnj1109 29219 . . . . . . . . . . 11
71 19.9v 1677 . . . . . . . . . . 11
7270, 71mpbi 201 . . . . . . . . . 10
7372expcom 426 . . . . . . . . 9
74 fndm 5546 . . . . . . . . . . 11
757, 74bnj770 29194 . . . . . . . . . 10
76 eleq2 2499 . . . . . . . . . . 11
7776imbi1d 310 . . . . . . . . . 10
7875, 77syl 16 . . . . . . . . 9
7973, 78mpbird 225 . . . . . . . 8
8011, 79ralrimi 2789 . . . . . . 7
8180exlimiv 1645 . . . . . 6
828, 81sylbi 189 . . . . 5
83 ss2iun 4110 . . . . . 6
84 bnj1143 29223 . . . . . 6
8583, 84syl6ss 3362 . . . . 5
8682, 85syl 16 . . . 4
876, 86mprg 2777 . . 3
884bnj1317 29255 . . . 4
8988bnj1146 29224 . . 3
9087, 89sstri 3359 . 2
915, 90eqsstri 3380 1
Colors of variables: wff set class Syntax hints: wi 4 wb 178 wa 360 w3a 937 wex 1551 wceq 1653 wcel 1726 cab 2424 wne 2601 wral 2707 wrex 2708 cdif 3319 wss 3322 c0 3630 csn 3816 ciun 4095 csuc 4585 com 4847 cdm 4880 wfn 5451 cfv 5456 w-bnj17 29112 c-bnj14 29114 c-bnj18 29120 This theorem is referenced by: bnj1147 29425 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1556 ax-5 1567 ax-17 1627 ax-9 1667 ax-8 1688 ax-13 1728 ax-14 1730 ax-6 1745 ax-7 1750 ax-11 1762 ax-12 1951 ax-ext 2419 ax-sep 4332 ax-nul 4340 ax-pr 4405 ax-un 4703 This theorem depends on definitions: df-bi 179 df-or 361 df-an 362 df-3or 938 df-3an 939 df-tru 1329 df-ex 1552 df-nf 1555 df-sb 1660 df-eu 2287 df-mo 2288 df-clab 2425 df-cleq 2431 df-clel 2434 df-nfc 2563 df-ne 2603 df-ral 2712 df-rex 2713 df-rab 2716 df-v 2960 df-sbc 3164 df-dif 3325 df-un 3327 df-in 3329 df-ss 3336 df-pss 3338 df-nul 3631 df-if 3742 df-pw 3803 df-sn 3822 df-pr 3823 df-tp 3824 df-op 3825 df-uni 4018 df-iun 4097 df-br 4215 df-opab 4269 df-tr 4305 df-eprel 4496 df-po 4505 df-so 4506 df-fr 4543 df-we 4545 df-ord 4586 df-on 4587 df-lim 4588 df-suc 4589 df-om 4848 df-iota 5420 df-fn 5459 df-fv 5464 df-bnj17 29113 df-bnj14 29115 df-bnj18 29121
Copyright terms: Public domain W3C validator | 3,225 | 6,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-30 | latest | en | 0.167342 |
http://schools-wikipedia.org/wp/c/Cauchy%25E2%2580%2593Schwarz_inequality.htm | 1,500,893,445,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424846.81/warc/CC-MAIN-20170724102308-20170724122308-00576.warc.gz | 278,713,489 | 7,515 | # Cauchy–Schwarz inequality
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In mathematics, the Cauchy–Schwarz inequality, also known as the Schwarz inequality, the Cauchy inequality, or the Cauchy–Bunyakovsky–Schwarz inequality, is a useful inequality encountered in many different settings, such as linear algebra applied to vectors, in analysis applied to infinite series and integration of products, and in probability theory, applied to variances and covariances.
The inequality for sums was published by Augustin Cauchy (1821), while the corresponding inequality for integrals was first stated by Viktor Yakovlevich Bunyakovsky (1859) and rediscovered by Hermann Amandus Schwarz (1888) (often misspelled "Schwartz").
## Statement of the inequality
The Cauchy–Schwarz inequality states that for all vectors x and y of a real or complex inner product space,
$|\langle x,y\rangle|^2 \leq \langle x,x\rangle \cdot \langle y,y\rangle,$
where $\langle\cdot,\cdot\rangle$ is the inner product. Equivalently, by taking the square root of both sides, and referring to the norms of the vectors, the inequality is written as
$|\langle x,y\rangle| \leq \|x\| \cdot \|y\|.\,$
Moreover, the two sides are equal if and only if $x$ and $y$ are linearly dependent (or, in a geometrical sense, they are parallel or one of the vectors is equal to zero).
If $x_1,\cdots, x_n\in\mathbb C$ and $y_1,\cdots, y_n\in\mathbb C$ are the components of $x$ and $y$ with respect to an orthonormal basis of $V$ the inequality may be restated in a more explicit way as follows:
$|\overline{x}_1 y_1 + \cdots + \overline{x}_n y_n|^2 \leq (|x_1|^2 + \cdots + |x_n|^2) (|y_1|^2 + \cdots + |y_n|^2).$
Equality holds if and only if either $x=0$, or there exists a scalar $\lambda$ such that
$y_1 = \lambda x_1, \ y_2 = \lambda x_2, \dots, y_n = \lambda x_n.$
The finite-dimensional case of this inequality for real vectors was proved by Cauchy in 1821, and in 1859 Cauchy's student V.Ya. Bunyakovsky noted that by taking limits one can obtain an integral form of Cauchy's inequality. The general result for an inner product space was obtained by K.H.A.Schwarz in 1885.
## Proof
As the inequality is trivially true in the case y = 0, we may assume <y, y> is nonzero. Let $\lambda$ be a complex number. Then,
$0 \leq \left\| x-\lambda y \right\|^2 = \langle x-\lambda y,x-\lambda y \rangle = \langle x,x \rangle - \bar{\lambda} \langle x,y \rangle - \lambda \langle y,x \rangle + |\lambda|^2 \langle y,y\rangle.$
Choosing
$\lambda = \langle x,y \rangle \cdot \langle y,y \rangle^{-1}$
we obtain
$0 \leq \langle x,x \rangle - |\langle x,y \rangle|^2 \cdot \langle y,y \rangle^{-1}$
which is true if and only if
$|\langle x,y \rangle|^2 \leq \langle x,x \rangle \cdot \langle y,y \rangle$
or equivalently:
$\big| \langle x,y \rangle \big| \leq \left\|x\right\| \left\|y\right\|,$
which is the Cauchy–Schwarz inequality.
## Notable special cases
### Rn
In Euclidean space Rn with the standard inner product, the Cauchy–Schwarz inequality is
$\left(\sum_{i=1}^n x_i y_i\right)^2\leq \left(\sum_{i=1}^n x_i^2\right) \left(\sum_{i=1}^n y_i^2\right).$
In this special case, an alternative proof is as follows: Consider the polynomial in z
$(x_1 z + y_1)^2 + \cdots + (x_n z + y_n)^2.$
Note that the polynomial is quadratic in z. Since the polynomial is nonnegative, it cannot have any roots unless all the ratios xi/yi are equal. Hence its discriminant is less than or equal to zero, that is,
$\left(\sum ( x_i \cdot y_i ) \right)^2 - \sum {x_i^2} \cdot \sum {y_i^2} \le 0$,
which yields the Cauchy–Schwarz inequality.
An equivalent proof for Rn starts with the summation below.
Expanding the brackets we have:
$\sum_{i=1}^n \sum_{j=1}^n \left( x_i y_j - x_j y_i \right)^2 = \sum_{i=1}^n x_i^2 \sum_{j=1}^n y_j^2 + \sum_{j=1}^n x_j^2 \sum_{i=1}^n y_i^2 - 2 \sum_{i=1}^n x_i y_i \sum_{j=1}^n x_j y_j$,
collecting together identical terms (albeit with different summation indices) we find:
$\frac{1}{2} \sum_{i=1}^n \sum_{j=1}^n \left( x_i y_j - x_j y_i \right)^2 = \sum_{i=1}^n x_i^2 \sum_{i=1}^n y_i^2 - \left( \sum_{i=1}^n x_i y_i \right)^2 .$
Because the left-hand side of the equation is a sum of the squares of real numbers it is greater than or equal to zero, thus:
$\sum_{i=1}^n x_i^2 \sum_{i=1}^n y_i^2 - \left( \sum_{i=1}^n x_i y_i \right)^2 \geq 0$ .
Also, when n = 2 or 3, the dot product is related to the angle between two vectors and one can immediately see the inequality:
$|x \cdot y| = \|x\| \|y\| | \cos \theta | \le \|x\| \|y\|.$
Furthermore, in this case the Cauchy–Schwarz inequality can also be deduced from Lagrange's identity. For n = 3, Lagrange's identity takes the form
$\langle x,x\rangle \cdot \langle y,y\rangle = |\langle x,y\rangle|^2 + |x \times y|^2$
from which readily follows the Cauchy–Schwarz inequality.
### L2
For the inner product space of square-integrable complex-valued functions, one has
$\left|\int f(x) \overline{g}(x)\,dx\right|^2\leq\int \left|f(x)\right|^2\,dx \cdot \int\left|g(x)\right|^2\,dx.$
A generalization of this is the Hölder inequality.
## Use
The triangle inequality for the inner product is often shown as a consequence of the Cauchy–Schwarz inequality, as follows: given vectors x and y,
$\|x + y\|^2$ $= \langle x + y, x + y \rangle$ $= \|x\|^2 + \langle x, y \rangle + \langle y, x \rangle + \|y\|^2$ $\le \|x\|^2 + 2|\langle x, y \rangle| + \|y\|^2$ $\le \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$ $\le \left(\|x\| + \|y\|\right)^2$
Taking the square roots gives the triangle inequality.
The Cauchy–Schwarz inequality allows one to extend the notion of "angle between two vectors" to any real inner product space, by defining:
$\cos\theta_{xy}=\frac{\langle x,y\rangle}{\|x\| \|y\|}$
The Cauchy–Schwarz inequality proves that this definition is sensible, by showing that the right hand side lies in the interval $[-1,1]$, and justifies the notion that real inner product spaces are simply generalizations of the Euclidean space.
The Cauchy–Schwarz is used to prove that the inner product is a continuous function with respect to the topology induced by the inner product itself.
The Cauchy–Schwarz inequality is usually used to show Bessel's inequality.
The general formulation of the Heisenberg uncertainty principle is derived using the Cauchy–Schwarz inequality in the inner product space of physical wave functions.
## Generalizations
Various generalizations of the Cauchy–Schwarz inequality exist in the context of operator theory, e.g. for operator-convex functions, and operator algebras, where the domain and/or range of φ are replaced by a C*-algebra or W*-algebra.
This section lists a few of such inequalities from the operator algebra setting, to give a flavor of results of this type.
### Positive functionals on C*- and W*-algebras
One can discuss inner products as positive functionals. Given a Hilbert space L2(m), m being a finite measure, the inner product < · , · > gives rise to a positive functional φ by
$\phi (g) = \langle g, 1 \rangle.$
Since < f,f > ≥ 0, φ(f*f) ≥ 0 for all f in L2(m), where f* is pointwise conjugate of f. So φ is positive. Conversely every positive functional φ gives a corresponding inner product < f , g >φ = φ(g*f). In this language, the Cauchy–Schwarz inequality becomes
$| \phi(g^*f) |^2 \leq \phi(f^*f) \phi(g^*g) ,$
which extends verbatim to positive functionals on C*-algebras.
We now give an operator theoretic proof for the Cauchy–Schwarz inequality which passes to the C*-algebra setting. One can see from the proof that the Cauchy–Schwarz inequality is a consequence of the positivity and anti-symmetry inner-product axioms.
Consider the positive matrix
$M = \begin{bmatrix} f^*\\ g^* \end{bmatrix} \begin{bmatrix} f & g \end{bmatrix} = \begin{bmatrix} f^*f & f^* g \\ g^*f & g^*g \end{bmatrix}.$
Since φ is a positive linear map whose range, the complex numbers C, is a commutative C*-algebra, φ is completely positive. Therefore
$M' = (I_2 \otimes \phi)(M) = \begin{bmatrix} \phi(f^*f) & \phi(f^* g) \\ \phi(g^*f) & \phi(g^*g) \end{bmatrix}$
is a positive 2 × 2 scalar matrix, which implies it has positive determinant:
$\phi(f^*f) \phi(g^*g) - | \phi(g^*f) |^2 \geq 0 \quad \mbox{i.e.} \quad \phi(f^*f) \phi(g^*g) \geq | \phi(g^*f) |^2.$
This is precisely the Cauchy–Schwarz inequality. If f and g are elements of a C*-algebra, f* and g* denote their respective adjoints.
We can also deduce from above that every positive linear functional is bounded, corresponding to the fact that the inner product is jointly continuous.
### Positive maps
Positive functionals are special cases of positive maps. A linear map Φ between C*-algebras is said to be a positive map if a ≥ 0 implies Φ(a) ≥ 0. It is natural to ask whether inequalities of Schwarz-type exist for positive maps. In this more general setting, usually additional assumptions are needed to obtain such results.
One such inequality is the following:
Theorem If Φ is a unital positive map, then for every normal element a in its domain, we have Φ(a*a) ≥ Φ(a*)Φ(a) and Φ(a*a) ≥ Φ(a)Φ(a*).
This extends the fact φ(a*a) · 1 ≥ φ(a)*φ(a) = |φ(a)|2, when φ is a linear functional.
The case when a is self-adjoint, i.e. a = a*, is known as Kadison's inequality.
#### 2-positive maps
When Φ is 2-positive, a stronger assumption than merely positive, one has something that looks very similar to the original Cauchy–Schwarz inequality:
Theorem (Modified Schwarz inequality for 2-positive maps) For a 2-positive map Φ between C*-algebras, for all a, b in its domain,
i) Φ(a)*Φ(a) ≤ ||Φ(1)|| Φ(a*a).
ii) ||Φ(a*b)||2 ≤ ||Φ(a*a)|| · ||Φ(b*b)||.
A simple argument for ii) is as follows. Consider the positive matrix
$M= \begin{bmatrix} a^* & 0 \\ b^* & 0 \end{bmatrix} \begin{bmatrix} a & b \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} a^*a & a^* b \\ b^*a & b^*b \end{bmatrix}.$
By 2-positivity of Φ,
$(I_2 \otimes \Phi) M = \begin{bmatrix} \Phi(a^*a) & \Phi(a^* b) \\ \Phi(b^*a) & \Phi(b^*b) \end{bmatrix}$
is positive. The desired inequality then follows from the properties of positive 2 × 2 (operator) matrices.
Part i) is analogous. One can replace the matrix $\begin{bmatrix} a & b \\ 0 & 0 \end{bmatrix}$ by $\begin{bmatrix} 1 & a \\ 0 & 0 \end{bmatrix}.$ | 3,245 | 10,368 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 46, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-30 | longest | en | 0.895059 |
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# 18 Marking scheme: Worksheet (A2)
30 π
1 a θ= × 2π = ≈ 0.52 rad [1]
360 6
210
b θ= × 2π ≈ 3.7 rad [1]
360
0.05
c θ= × 2π ≈ 8.7 × 10−4 rad [1]
360
1 .0
2 a θ= × 360 = 57.3° ≈ 57° [1]
4 .0
b θ= × 360 ≈ 230° [1]
0.15
c θ= × 360 ≈ 8.6° [1]
3 a 88 days is equivalent to 2π radians.
44
θ= × 2π = π rad [1]
88
1
b θ= × 2π ≈ 0.071 rad (4.1°) [1]
88
4 a Friction between the tyres and the road. [1]
b Gravitational force acting on the planet due to the Sun. [1]
c Electrical force acting on the electron due to the positive nucleus. [1]
d The (inward) contact force between the clothes and the rotating drum. [1]
v 150
5 a ω= = [1]
r 20000
ω = 7.5 × 105 rad s−1 [1]
2
v
b a= [1]
r
150 2
a= [1]
20 000
a = 1.125 m s–2 ≈ 1.1 m s−2 [1]
c F = ma = 80 × 1.125 [1]
F = 90 N [1]
8 .2
6 a i Time = = 0.82 s [1]
10
ii Distance = circumference of circle = 2π × 0.80 = 5.03 m ≈ 5.0 m [1]
distance 5.03
iii speed = = [1]
time 0.82
speed, v = 6.13 m s ≈ 6.1 m s−1
−1
[1]
v2
iv a = [1]
r
6.13 2
a= [1]
0.80
a = 47 m s–2 [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
18 Marking scheme: Worksheet (A2)
vF = ma = 0.090 × 47 [1]
F ≈ 4.2 N [1]
b The tension in the string. [1]
c The stone describes a circle, therefore the angle between the velocity and the acceleration
(or centripetal force) must be 90°. [1]
distance
7 a i speed =
time
2πr 2π × 0.12
speed v = = [1]
T 1.6
v = 0.471 m s−1 ≈ 0.47 m s−1 [1]
mv 2
ii F = ma = [1]
r
0.300 × 0.4712
F= [1]
0.12
frictional force ≈ 0.55 N [1]
b Frictional force = 0.7mg [1]
mv 2
0.7 mg = [1]
r
v = 0.7 gr = 0.7 × 9.81 × 0.12 [1]
−1 −1
speed = 0.908 m s ≈ 0.91 m s [1]
## 8 a Kinetic energy at B = loss of gravitational potential energy from A to B
1 2
mv = mgh or v = 2 gh [1]
2
v = 2 × 9.81 × 5.2 =10.1 m s–1 ≈ 10 m s−1 [1]
2
v
b i a= [1]
r
10.12
a= [1]
16
a = 6.38 m s−2 ≈ 6.4 m s−2 [1]
ii Net force = ma
R − mg = ma [1]
R = mg + ma = m(a + g) = 70(6.38 + 9.81) [1]
R ≈ 1.1 × 103 N [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
18 Marking scheme: Worksheet (A2)
## 9 a R cos 20° = W = 840 × 9.8 [1]
840 × 9.8
R= = 8760 N [1]
cos 20°
mv 2
b R sin 20° = centripetal force = [1]
r
mv 2
r= [1]
R
840 × 32 2
r= [1]
8760 sin 20°
r = 287 m ≈ 290 m [1]
mv 2
10 net force =
r
0.120 × 4.0 2
net force = = 2.4 N [2]
0.80
weight W of stone = mg = 0.120 × 9.81 = 1.18 N ≈ 1.2 N [1]
At the top: W + TB = 2.4 so TB = 2.4 − 1.2 = 1.2 N [1]
At the bottom: TA − W = 2.4 so TA = 2.4 + 1.2 = 3.6 N [1]
T 3 .6
ratio = A = = 3.0 [1]
TB 1 .2
## AS and A Level Physics Original material © Cambridge University Press 2010 3
19 Marking scheme: Worksheet (A2)
1 Gravitational field strength at a point, g, is the force experienced per unit mass at that point. [1]
GMm
2 F= − [1]
r2
Fr 2 ⎡ N m2 ⎤
Therefore: G = → ⎢ 2 ⎥
→ [N m2 kg–2] [1]
Mm ⎣ kg ⎦
F
3 g=
m
F = mg = 80 × 1.6 [1]
F = 128 N ≈ 130 N (F is the ‘weight’ of the astronaut.) [1]
GMm
4 a F= [1]
r2
6.67 × 10 −11 × 1.7 × 10 −27 × 1.7 × 10 −27
F= [1]
(5.0 ×10 ) −14 2
## F ≈ 7.7 × 10–38 N [1]
6.67 × 10 −11 × 5.0 ×10 28 × 5.0 ×10 28
b F= [1]
(
8.0 × 1012
2
)
F = 2.61 × 1021 N ≈ 2.6 × 1021 N [1]
6.67 × 10 −11 × 1500 2
c F= [1]
2.0 2
F = 6.00 × 10–6 N ≈ 6.00 × 10–6 N [1]
GM
5 a g= − [1]
r2
1
b The field strength obeys an inverse square law with distance (g ∝ ). [1]
r2
Doubling the distance decreases the field strength by a factor of four. [1]
GM (5 R )
2
c ratio = [1]
GM (59 R )
2
2
59 2 ⎛ 59 ⎞
ratio = 2 = ⎜ ⎟ [1]
5 ⎝ 5 ⎠
ratio ≈ 140 [1]
GM
6 g= − [1]
r2
6.67 × 10 −11 × 1.0 × 10 26
g= (magnitude only) [1]
(2.2 × 10 )
7 2
## g = 13.8 N kg–1 ≈ 14 N kg–1 [1]
GM
7 g= − [1]
r2
GM 6.67 × 10 −11 × 5.0 × 10 23
r2 = = = 8.34 × 1012 m2 [1]
g 4.0
r= 8.34 × 1012 ≈ 2.9 × 106 m [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
19 Marking scheme: Worksheet (A2)
GMm
8 a F= − [1]
r2
6.67 × 10 −11 × 1800 × 6.0 × 10 24 ⎛ 7.8 × 1010 ⎞
F= ⎜r = = 3.9 × 1010 m ⎟⎟ [1]
(3.9 × 1010
2
) ⎜
⎝ 2 ⎠
F = 4.74 × 10–4 N ≈ 4.7 × 10–4 N [1]
GMm
b F= − 2
r
6.67 × 10 −11 × 1800 × 6.4 × 10 23
F= [1]
(
3.9 × 1010
2
)
F = 5.05 × 10–5 N ≈ 5.1 × 10–5 N [1]
F
c a= (F is the net force.) [1]
m
4.74 × 10−4 − 5.05 × 10−5
a= [1]
1800
a ≈ 2.4 × 10–7 m s–2 (towards the centre of the Earth) [1]
GMm
9 a F= − [1]
r2
6.67 × 10 −11 × 5000 × 6.0 × 10 24
F= (r = 6400 + 400 = 6800 km) [1]
(6800 × 10 )
3 2
## F = 4.33 × 104 N ≈ 4.3 × 104 N [1]
F 4.33 × 10 4
b a= = [1]
m 5000
a = 8.66 ≈ 8.7 m s–2 [1]
v2
c a= [1]
r
v = ar = 8.66 × 6800 × 103
2
[1]
v = 7.67 × 103 m s–1 ≈ 7.7 km s–1 [1]
10 a The work done in bringing unit mass [1]
from infinity to the point [1]
b 0J [1]
6.67 × 10 −11 × 6.0 × 10 24
c Ep =− ([1] mark only if minus sign missed) [2]
6.4 × 10 6
= −6.25 × 106 J [1]
d 6.25 × 106 J [1]
GMm
11 a Gravitational force on planet = [1]
r2
mv 2
Centripetal force = [1]
r
GMm mv 2
Equating these two forces, we have: = [1]
r2 r
GM GM
Therefore: v2 = or v = [1]
r r
## AS and A Level Physics Original material © Cambridge University Press 2010 2
19 Marking scheme: Worksheet (A2)
## GM 6.67 × 10 −11 × 2.0 × 10 30
b v= = [1]
r 1.5 × 1011
v = 2.98 × 104 m s–1 ≈ 30 km s–1 [1]
12 The field strengths are the same at point P.
GM M GM E
= [1]
x 2
(R − x )2
ME
R−x=x× [1]
MM
R−x=x× 81 so R − x = 9x [1]
R
10x = R so x= [1]
10
## AS and A Level Physics Original material © Cambridge University Press 2010 3
20 Marking scheme: Worksheet (A2)
1 a The period of an oscillator is the time for one complete oscillation. [1]
b The frequency of an oscillator is the number of oscillations completed per unit time
(or per second). [1]
2 a The gradient of a displacement against time graph is equal to velocity. [1]
The magnitude of the velocity (speed) is a maximum at 0 s or 0.4 s or 0.8 s. [1]
b For s.h.m., acceleration ∝ – displacement.
The magnitude of the acceleration is maximum when the displacement is equal to the
amplitude of the motion. [1]
The magnitude of the acceleration is a maximum at 0.2 s or 0.6 s or 1.0 s. [1]
13.2
3 a T= [1]
12
T = 1.1 s [1]
1 1
b f= = [1]
T 1.1
f = 0.909 ≈ 0.91 Hz [1]
4 a Amplitude = 0.10 m [1]
b Period = 4.0 × 10–2 s [1]
1 1
c f= = [1]
T 0.04
f = 25 Hz [1]
d ω = 2πf = 2π × 25 [1]
e Maximum speed = ωA = 157 × 0.10 [1]
maximum speed = 15.7 m s−1 ≈ 16 m s−1 [1]
⎛t⎞
5 a Phase difference = 2π × ⎜ ⎟
⎝T ⎠
where T is the period and t is the time lag between the motions of the two objects.
⎛t⎞ ⎛ 2.5 ⎞
phase difference = 2π × ⎜ ⎟ = 2π × ⎜ ⎟ [1]
T
⎝ ⎠ ⎝ 10 ⎠
π
phase difference = ≈ 1.6 rad [1]
2
⎛t⎞ ⎛ 5 .0 ⎞
b Phase difference = 2π × ⎜ ⎟ = 2π × ⎜ ⎟ [1]
⎝T ⎠ ⎝ 10 ⎠
phase difference = π ≈ 3.1 rad [1]
6 a A = 16 cm [1]
2π 2π
b ω = 2πf = = [1]
T 2.8
c a = (2πf )2x (magnitude only) [1]
For maximum acceleration, the displacement x must be 16 cm.
2
⎛ 1 ⎞ −2
a = ⎜ 2π × ⎟ × 16 × 10 [1]
⎝ 2.8 ⎠
a = 0.806 m s–2 ≈ 0.81 m s–2 [1]
d Maximum speed = ωA = 2.24 × 0.16 [1]
maximum speed = 0.358 m s−1 ≈ 0.36 m s−1 [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
20 Marking scheme: Worksheet (A2)
2π 2π
7 a ω = 2πf = = [1]
T 2.0
b a = –(2πf )2x or a = –ω2x [1]
a = 3.142 × 3.0 × 10–2 [1]
a ≈ 0.30 m s–2 [1]
c x = A cos (2πft) = A cos (ωt) [1]
x = 3.0 × 10–2 cos (3.14 × 6.7) [1]
x ≈ –1.7 × 10–2 m [1]
## 8 a Gradient of x–t graph = velocity
[2]
b Gradient of v–t graph = acceleration
(for s.h.m. acceleration ∝ −displacement)
[2]
1 2
c Kinetic energy = mv ∝ v2
2
[2]
d Potential energy = total energy − kinetic energy
[2]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
20 Marking scheme: Worksheet (A2)
## 9 a a = −(2πf )2x [1]
Therefore (2πf )2 = 6.4 × 105 [1]
6.4 × 10 5
f= = 127 Hz ≈ 130 Hz [1]
b F = ma
Acceleration is maximum at maximum displacement, so magnitude of maximum force
is given by:
F = ma = 0.700 × (6.4 × 105 × 0.08) [1]
F = 3.58 × 104 N ≈ 3.6 × 104 N [1]
10 a According to Hooke’s law, F = –kx [1]
(The minus sign shows that the force is directed towards the equilibrium position.)
From Newton’s second law: F = ma [1]
Equating, we have: ma = –kx [1]
⎛k⎞
Hence: a = − ⎜ ⎟ x
⎝m⎠
b For s.h.m. we have a = –(2πf )2x [1]
k
Hence (2πf )2 = [1]
m
1 k
Therefore f =
2π m
1 1
c f= = = 2.5 Hz [1]
T 0 .4
1 k
2.5 = [1]
2π 850
k = (2π × 2.5)2 × 850 ≈ 2.1 × 105 N m–1 [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
21 Marking scheme: Worksheet (A2)
1 a The atoms in a solid are arranged in a three-dimensional structure. [1]
There are strong attractive forces between the atoms. [1]
The atoms vibrate about their equilibrium positions. [1]
b The atoms in a liquid are more disordered than those in a solid. [1]
There are still attractive electrical forces between molecules but these are weaker than
those between similar atoms in a solid. [1]
The atoms in a liquid are free to move around. [1]
c The atoms in a gas move around randomly. [1]
There are virtually no forces between the molecules (except during collisions) because they
are much further apart than similar molecules in a liquid. [1]
The atoms of a gas move at high speeds (but no faster than those in a liquid at the same
temperature). [1]
2 The atoms move faster [1]
because their mean kinetic energy increases as the temperature is increased. [1]
The atoms still have a random motion. [1]
3 a The internal energy of a substance is the sum (of the random distribution) of the kinetic
and potential energies of its particles (atoms or molecules). [1]
b There is an increase in the average kinetic energy of the aluminium atoms as they vibrate
with larger amplitudes about their equilibrium positions. [1]
The potential energy remains the same because the mean separation between the atoms
does not change significantly. [1]
Hence, the internal energy increases because there is an increase in the kinetic energy of
the atoms. [1]
c As the metal melts, the mean separation between the atoms increases. [1]
Hence, the electrical potential energy of the atoms increases. [1]
There is no change in the kinetic energy of the atoms because the temperature remains
the same. [1]
The internal energy of the metal increases because there is an increase in the electrical
potential energy of the atoms. [1]
4 Change in thermal energy = mass × specific heat capacity × change in temperature [1]
5 The specific heat capacity refers to the energy required to change the temperature of a substance. [1]
Specific latent heat of fusion is the energy required to melt a substance; there is no change in
temperature as the substance melts. [1]
6 E = mc∆θ [1]
E = 6.0 × 105 × 4200 × (24 – 21) [1]
E = 7.56 × 109 J ≈ 7.6 × 109 J [1]
7 E = mc∆θ [1]
E = 300 × 10–3 × 490 × (20 – 300) [1]
E = –4.1 × 104 J (The minus sign implies energy is released by the cooling metal.) [1]
8 E = mLf = 200 × 10−3 ×3.4 × 105 [1]
= 6.8 × 104 J [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
21 Marking scheme: Worksheet (A2)
## 9 a i T = 273 + 0 = 273 K [1]
ii T = 273 + 80 = 353 K [1]
iii T = 273 – 120 = 153 K [1]
b i θ = 400 – 273 = 127 °C [1]
ii θ = 272 – 273 = –1 °C [1]
iii θ = 3 – 273 = –270 °C [1]
10 a The thermal energy E supplied and the specific heat capacity c remain constant.
The mass m is larger by a factor of 3. [1]
Since E = mc∆θ, we have:
E 1
∆θ = ; ∆θ ∝ [1]
mc m
15
Therefore ∆θ = = 5.0 °C [1]
3
b The thermal energy E supplied is halved but the specific heat capacity c and the mass m
remain constant. [1]
Since E = mc∆θ, we have:
E
∆θ = ; ∆θ ∝ E [1]
mc
15
Therefore ∆θ = = 7.5 °C [1]
2
11 a Melting point = 600 °C [1]
(There is no change in temperature during change of state.)
b The lead is being heated at a steady rate and therefore the temperature also increases
c The energy supplied to the lead is used to break the atomic bonds and increase the
separation between the atoms of lead (and hence their potential energy increases). [1]
d E = mc∆θ [1]
E = 200 × 10–3 × 130 × (600 – 0) [1]
E = 1.56 × 104 J ≈ 1.6 × 104 J [1]
e In a time of 300 s, 1.56 × 104 J of energy is supplied to the lead.
Rate of heating = power
1.56 × 10 4
power = [1]
300
power = 52 W [1]
f Energy supplied = 52 × 100 = 5200 J [1]
∆E 5200
Lf = = [1]
∆m 0.2
= 26 000 J kg−1 [1]
12 The energy supplied per second is equal to the power of the heater.
In a time of 1 s, water of mass 0.015 kg has its temperature changed from 15 °C to 42 °C. [1]
E = mc∆θ (where E is the energy supplied in 1 s) [1]
E = 0.015 × 4200 × (42 – 15) [1]
E = 1.7 × 103 J [1]
The power of the heater is therefore 1.7 kW. [1]
m
(You may use P = ( )c∆θ )
t
13 The gas does work against atmospheric pressure. [1]
Energy to do this work is taken from the internal energy of the gas. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
21 Marking scheme: Worksheet (A2)
## 14 Heat ‘lost’ by hot water = heat ‘gained’ by cold water. [1]
0.3 × c × (90 – θ) = 0.2 × c × (θ – 10) [1]
where c is the specific heat capacity of the water and θ is the final temperature.
The actual value of c is not required, since it cancels on both sides of the equation.
Hence:
0.3 × (90 – θ) = 0.2 × (θ – 10) [1]
27 – 0.3θ = 0.2θ – 2.0 [1]
0.5θ = 29 so θ = 58 °C [1]
15 Heat ‘lost’ by metal = heat ‘gained’ by cold water [1]
0.075 × 500 × (θ – 48) = 0.2 × 4200 × (48 – 18) [1]
(θ is the initial temperature of the metal.)
0.2 × 4200 × 30
θ – 48 = [1]
0.075 × 500
θ – 48 = 672 [1]
θ = 720 °C [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
22 Marking scheme: Worksheet (A2)
1 a number of atoms = number of moles × NA
number of atoms = 1.0 × 6.02 × 1023 ≈ 6.0 × 1023 [1]
b Number of molecules = 3.6 × 6.02 × 1023 ≈ 2.2 × 1024 [1]
c Number of atoms = 0.26 × 6.02 × 1023 ≈ 1.6 × 1023 [1]
2 There are 6.02 × 1023 atoms in 4.0 g of helium. [1]
0.004
mass of atom = 23
= 6.645 × 10–27 kg ≈ 6.6 × 10–27 kg [1]
6.02 × 10
3 a There are 6.02 × 1023 atoms in 0.238 kg of uranium. [1]
0.238
mass of atom = 23
= 3.95 × 10−25 kg ≈ 4.0 × 10−25 kg [1]
6.02 × 10
mass of uranium
b i number of moles = [1]
molar mass of uranium
0.12
number of moles = = 5.04 × 10–4 ≈ 5.0 × 10–4 [1]
238
ii number of atoms = number of moles × NA
number of atoms = 5.04 × 10–4 × 6.02 × 1023 = 3.06 × 1020 ≈ 3.1 × 1020 [1]
4 The absolute zero of temperature is –273.15 °C or 0 K. [1]
This is the lowest temperature any substance can have. [1]
At absolute zero of temperature, the substance has minimum internal energy. [1]
5 a Pressure × volume
= number of moles × universal gas constant × thermodynamic temperature [1]
b PV = nRT [1]
nRT 1.0 × 8.31 × 293
P= = [1]
V 0.020
P = 1.22 × 10 Pa ≈ 1.2 × 105 Pa (120 kPa)
5
[1]
6 a PV = nRT [1]
Comparing this equation with y = mx, we have:
y = PV, x = T, gradient, m = nR [1]
A graph of PV against T is a straight line through the origin.
## Correct graph [1]
n= [1]
R
b PV = nRT [1]
At a constant temperature, the product
PV is a constant. [1]
Hence a graph of PV against P is a
straight horizontal line. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
22 Marking scheme: Worksheet (A2)
7 a PV = nRT [1]
4 .0
n= = 0.138 moles [1]
29
nRT 0.138 × 8.31× (273 + 34)
P= = [1]
V 0.030
P = 1.17 × 10 Pa ≈ 1.2 × 104 Pa (12 kPa)
4
[1]
P
b is constant when the volume of the gas is constant. [1]
T
The pressure is doubled, hence the absolute temperature of the gas is also doubled. [1]
Therefore:
temperature = 2 × (273 + 34) = 614 K [1]
temperature in °C = 614 – 273 = 341 °C ≈ 340 °C [1]
PV
8 a n= [1]
RT
180 × 10 3 × 2.0 × 10 −2 300 × 10 3 × 2.0 × 10 −2
n= + [1]
8.31 × (273 − 13) 8.31 × ( 273 − 13)
n = 4.44 moles ≈ 4.4 moles [1]
b Total volume, V = 4.0 × 10–2 m3, T = 273 – 13 = 260 K
nRT
P= [1]
V
4.44 × 8.31 × 260
P= [1]
4.0 × 10 − 2
P ≈ 2.4 × 105 Pa (240 kPa) [1]
F 400
9 a P= = [1]
A 1.6 × 10 −3
P = 2.5 × 105 Pa [1]
PV
b n= [1]
RT
2.5 × 10 5 × 2.4 × 10 −4
n= [1]
8.31 × (273 + 5.0)
n = 2.6 × 10–2 moles [1]
c i mass = number of moles × molar mass
mass = 2.6 × 10–2 × 29 = 0.754 g ≈ 0.75 g [1]
mass
ii density =
volume
0.754 × 10 −3
density = [1]
2.4 × 10 − 4
density = 3.14 kg m–3 ≈ 3.1 kg m–3 [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
22 Marking scheme: Worksheet (A2)
## 10 Mean kinetic energy of atom ∝ absolute temperature [1]
1 mv 2 ∝ T or v2 ∝ 2T [1]
2 m
Since the mass m of the atom is constant, we have: v ∝ T [1]
The temperature of 0 °C in kelvin is T = 273 K
10 000
The absolute temperature increases by a factor of (= 36.6) [1]
273
10 000
Hence the speed will increase by a factor of = 6.05 [1]
273
The speed of the atoms at 10 000 K = 1.3 × 6.05 ≈ 7.9 km s–1 [1]
11 a The particles have a range of speeds and travel in different directions. [1]
3 3
b i Mean kinetic energy = kT = × 1.38 × 10 − 23 × 5400 [1]
2 2
= 1.118 × 10–19 J ≈ 1.1 × 10–19 J [1]
1 2 3
ii mv = kT [1]
2 2
3kT 3 × 1.38 × 10 −23 × 5400
v= = [1]
m 1.7 × 10 − 27
speed = 1.147 × 10 m s–1 ≈ 11 km s–1
4
[1]
3 3
12 a Mean kinetic energy = kT = × 1.38 × 10 − 23 × 273 [1]
2 2
= 5.65 × 10–21 J ≈ 5.7 × 10–21 J [1]
b There are 6.02 × 1023 molecules of carbon dioxide. [1]
0.044
mass of molecule = = 7.31 × 10 − 26 kg [1]
6.02 × 10 23
1
mv 2 = 5.65 × 10 − 21 J [1]
2
2 × 5.65 × 10 −21
v= [1]
7.31 × 10 − 26
speed = 393 m s–1 ≈ 390 m s–1 [1]
3 3
c Total kinetic energy of one mole of gas = kT × NA = RT (Note: R = k × NA) [1]
2 2
For an ideal gas, the change in internal energy is entirely kinetic energy.
3 3
Change in internal energy = R × (373 − 273) = × 8.31 × 100 [1]
2 2
change in internal energy = 1.2465 kJ ≈ 1.2 kJ [1]
13 a iThe molecule has 3 degrees of freedom for translational motion and 2 degrees of freedom
for rotation – making a total of 5. [1]
1 5
Therefore, mean energy = 5 × kT = kT [1]
2 2
ii The molecule has 3 degrees of freedom for translational motion and 3 degrees of freedom
for rotation – making a total of 6. [1]
1
Therefore, mean energy = 6 × kT = 3kT [1]
2
b Internal energy = 3kT × NA = 3RT (Note: R = k × NA) [1]
internal energy per unit kelvin = 3R [1]
–1
= 3 × 8.31 ≈ 25 J K [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
23 Marking scheme: Worksheet (A2)
1 Electric field strength is the force per unit charge at that point. [1]
The potential at a point is the work that must be done to bring unit charge
from infinity to that point. [1]
V V
2 a E= ,d= [1]
d E
5000
d= = 1.25 × 10–2 m ≈ 1.3 cm [1]
400 000
b F = EQ = 400 000 × 1.6 × 10–19 N [1]
F = 6.4 × 10–14 N [1]
Q 1 1
3 E= so k = =
4 πε 0 r 2
4 πε0 4π × 8.85 × 10 −12
k = 8.99 × 109 m F–1 ≈ 9.0 × 109 m F–1 [1]
1
4 The force between the charges obeys an inverse square law with distance; that is, F ∝ [1]
r2
Point B: The distance is the same. The force between the charges is F. [1]
Point C: The distance is doubled, so the force decreases by a factor of 4. [1]
F
The force between the charges is . [1]
4
Point D: The distance is trebled, so the force decreases by a factor of 32 = 9. [1]
F
The force between the charges is . [1]
9
Point E: The distance between the charges is 8 R. [1]
The force between the charges decreases by a factor
of ( 8 ) 2 = 8 [1]
F
The force between the charges is . [1]
8
Q
5 a E= [1]
4 πε 0 r 2
2.5 × 10 −6
E= [1]
4 π × 8.85 × 10 −12 × 0.15 2
E = 9.99 × 105 V m–1 ≈ 1.0 × 106 V m–1 [1]
b The distance from the centre of the dome increases by a factor of 3.
The electric field strength decreases by a factor of 32 = 9. [1]
1.0 × 10 6
Therefore E = = 1.1 × 104 V m–1 [1]
9
## AS and A Level Physics Original material © Cambridge University Press 2010 1
23 Marking scheme: Worksheet (A2)
Q
6 a i E= [1]
4 πε 0 r 2
20 × 10 −6 80
E= −12 2
(r = = 40 cm) [1]
4 π × 8.85 × 10 × 0.40 2
E = 1.124 × 106 V m–1 ≈ 1.1 × 106 V m–1 [1]
40 × 10 −6
ii E = [1]
4 π × 8.85 × 10 −12 × 0.40 2
E = 2.248 × 106 V m–1 ≈ 2.2 × 106 V m–1 [1]
(The electric field doubles because the charge is doubled, E ∝ Q.)
b Net field strength, E = 2.2 × 106 – 1.1 × 106 = 1.1 × 106 V m–1 [1]
The direction of the electric field at X is to the left. [1]
7 a Q = V × 4πεor = 20 000 × 4 × π × 8.85 × 10−12 × 0.15 [1]
Q = 3.3 × 10–7 C [1]
kQ 9.0 ×10 9 × 3.3 ×10 −7
b E = 2 = [1]
r 0.15 2
= 1.32 × 10 V m–1 ≈ 1.3 × 105 V m–1
5
[1]
c F = eV = 1.6 × 10−19 × 1.32 × 105 [1]
F = 2.11 × 10−14 N ≈ 2.1 × 10−14 N [1]
## Q kQ 9.0 × 109 × − 2000 × 10−9
8 a V= = = ([1] mark only if minus sign omitted) [2]
4 πε 0 r r 5 × 10− 2
V = 3.6 × 105 J C–1 =360 kV [1]
9 Similarities
• Both produce radial fields. [1]
1
• Both obey an inverse square law with distance; that is, F ∝ . [1]
r2
• The field strengths are defined as force per unit (positive) charge or mass. [1]
• Both produce action at a distance. [1]
Differences
• Electrical forces can be either attractive or repulsive, whereas gravitational forces are
always attractive. [1]
• Gravitational forces act between masses, whereas electrical forces act between charges. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
23 Marking scheme: Worksheet (A2)
10 The electric field strength due to the charge +Q is equal in magnitude but opposite in direction
to the electric field strength due to the charge +3Q. [1]
Therefore:
Q 3Q
= (where R is the distance between the charges = 10 cm) [1]
4 πε 0 x 2
4 πε 0 ( R − x) 2
1 3
2
= [1]
x ( R − x) 2
R−x
so = 3 [1]
x
R
x(1 + 3)=R so x= = 0.37 R [1]
1+ 3
x = 0.37 × 10 = 3.7 cm [1]
e 2 4πε 0 r 2
11 ratio = (where m = mass of proton and r = separation) [2]
Gm 2 r 2
e2
ratio = [1]
4 πε 0 Gm 2
The r2 terms cancel and so this ratio is independent of the separation. [1]
(1.6 × 10−19 ) 2
ratio = [1]
4 π × 8.85 × 10 −12 × 6.67 × 10−11 × (1.7 × 10 −27 ) 2
ratio ≈ 1.2 × 1036 [1]
## Q1Q2 9 1.6 × 10 −19 × 1.6 × 10 −19
12 a F= = 9 × 10 × [1]
4 πε 0 r 2 (10 −15 ) 2
= 230 N [1]
Q 1.6 × 10 −19
b V= = 9 × 109 × [1]
4 πε 0 r 10 −15
= 1.44 × 106 V [1]
c W = VQ = 1.44 × 106 × 1.6 × 10–19 [1]
= 2.3 × 10–13 J [1]
1 2 2W
d mv = W ⇒ v2 = [1]
2 m
2 × 2.3 × 10 −13
v2 = = 2.7 × 10–14 [1]
1.7 × 10 − 27
v = 2.7 × 10−14 = 1.6 × 107 m s–1 [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
24 Marking scheme: Worksheet (A2)
1 a Q = VC = 9.0 × 30 × 10–6 [1]
Q = 2.7 × 10–4 C (270 µC) [1]
Q 2.7 × 10 −4
b number of excess electrons = = [1]
e 1.6 × 10 −19
15 15
number = 1.69 × 10 ≈ 1.7 × 10 [1]
2 a i The charge is directly proportional to the voltage across the capacitor.
Hence doubling the voltage will double the charge. [1]
charge = 2 × 150 = 300 nC [1]
ii Since Q ∝ V for a given capacitor, increasing the voltage by a factor of three
will increase the charge by the same factor. [1]
charge = 3 × 150 = 450 nC [1]
Q 150 × 10 −9
b C= = [1]
V 3.0
–8
C = 5.0 × 10 F [1]
1 2 1
3 a E= V C = × 9.02 × 1000 × 10–6 [1]
2 2
E = 4.05 × 10–2 J ≈ 4.1 × 10–2 J [1]
b For a given capacitor, energy stored ∝ voltage2. [1]
energy = 22 × 4.05 × 10–2 ≈ 0.16 J [1]
4 a Ctotal = C1 + C2 [1]
Ctotal = 20 + 40 = 60 nF [1]
1 1 1
b = + [1]
C total C1 C2
1 1 1
= + = 0.012 µF−1 [1]
C total 100 500
1
Ctotal = ≈ 83 µF [1]
0.012
1 1 1 1
c = + + [1]
C total C1 C2 C3
1 1 1 1
= + + = 0.13 µF−1 [1]
C total 10 50 100
1
Ctotal = ≈ 7.7 µF [1]
0.13
d Total capacitance of the two capacitors in parallel = 50 + 50 = 100 µF. [1]
1 1 1
= + = 0.03 µF−1 [1]
C total 50 100
1
Ctotal = ≈ 33 µF [1]
0.03
e Total capacitance of the two capacitors in series is 83 µF (from b). [1]
Ctotal = 83 + 50 = 133 µF ≈ 130 µF [1]
5 a Ctotal = C1 + C2 [1]
Ctotal = 100 + 500 = 600 µF [1]
b The potential difference across parallel components is the same and equal to 1.5 V. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
24 Marking scheme: Worksheet (A2)
## c Q = VC = 1.5 × 600 × 10–6 [1]
Q = 9.0 × 10–4 C (900 µC) [1]
1 1
d E = QV = × 9.0 × 10–4 × 1.5 [1]
2 2
E = 6.75 × 10–4 J ≈ 6.8 × 10–4 J [1]
1 2 1
6 a E= V C = × 322 × 10 000 × 10–6 [1]
2 2
E = 5.12 J ≈ 5.1 J [1]
E 5.12
b P= = [1]
t 0.300
P ≈ 17 W [1]
7 a Q = VC = 12 × 1000 × 10–6 [1]
Q = 1.2 × 10–2 C (12 mC) [1]
b i Ctotal = C1 + C2 [1]
Ctotal = 1000 + 500 = 1500 µF [1]
Q
ii V = (The charge Q is conserved and C is the total capacitance.) [1]
C
1.2 × 10 −2
V= = 8.0 V [1]
1500 × 10 −6
I 225 ×10 −3
8 a ∆Q = I∆t = = = 4.5 × 10–3 C [1]
f 50
Q 4.5 × 10 −3
b C= = [1]
V 9 .0
= 5.0 × 10–4 F = 500 µF [1]
c i The capacitors are in parallel, so the total capacitance = 2C and charge stored = 2Q;
current =2I = 2 × 225 = 450 mA [1]
1 1
ii The capacitors are in series, so the total capacitance = C and charge stored = Q;
2 2
1 1
current = I = × 225 = 113 mA [1]
2 2
9 a Q = CV = 200 × 10–6 × 200 = 0.040 C [1]
1 1
b E = CV 2 = × 200 × 10–6 × 2002 = 4.0 J [1]
2 2
c The two capacitors now make a pair of capacitors in parallel of total capacitance
= 100 µF + 200 µF = 300 µF [1]
Q 0.040
The charge is conserved, therefore V = = [1]
C 300 × 10 −6
= 133 V [1]
1 1
d E = CV 2 = × 300 × 10–6 × 1332 [1]
2 2
= 2.67 J [1]
e The energy is lost as the connected wires are heated as the current passes through them. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
24 Marking scheme: Worksheet (A2)
## 10 The capacitors are in parallel, so the total capacitance = 3C. [1]
The total charge Q remains constant. [1]
Q2
The energy stored by a capacitor is given by E = . [1]
2C
Q2 Q2
Einitial = and Efinal = [1]
2C 2(3C )
Efinal Q 2 2(3C ) 1
Fraction of energy stored = = 2
= [1]
Einitial Q 2C 3
1 2
Fraction of energy ‘lost’ as heat in resistor = 1 – = . [1]
3 3
The resistance governs how long it takes for the capacitor to discharge. The final voltage
across each capacitor is independent of the resistance. Hence, the energy lost as heat is
independent of the actual resistance of the resistor. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
25 Marking scheme: Worksheet (A2)
1 The direction of the magnetic field should be clockwise (and not anticlockwise). [1]
The separation between adjacent circular field lines should increase further away from
the wire (see below). [1]
## 2 a The conductor is pushed to the left. [1]
b The conductor is pushed to the left. [1]
c The conductor is pushed out of the plane of the paper. [1]
F
3 B= [1]
Il
N
[B] = = N A−1 m−1 [1]
Am
4 F = BIl [1]
F = 0.12 × 3.5 × 0.01 (length = 1.0 cm) [1]
F = 4.2 × 10–3 N [1]
5 a F = BIl sin θ
F = 0.050 × 3.0 × 0.04 × sin 90° [1]
F = 6.0 × 10–3 N [1]
b F = 0.050 × 3.0 × 0.04 × sin 30° [1]
F = 3.0 × 10–3 N [1]
c F = 0.050 × 3.0 × 0.04 × sin 65° [1]
F = 5.44 × 10–3 N ≈ 5.4 × 10–3 N [1]
6 Force experienced by PQ = force experienced by RS (but in opposite direction). [1]
No force experienced by QR and PS (since current is parallel to the field). [1]
torque = one of the forces × perpendicular distance between forces = (BIL)x [1]
torque = BI(Lx), Lx = area of loop = A [1]
torque = BIA ∝ A [1]
The torque is directly proportional to the area of the loop.
7 a Current is at right angles to magnetic field. [1]
Left-hand rule produces force on AB towards the right. [1]
Wire leaves mercury and breaks contact/current stops/force stops. [1]
Weight causes AB to fall back/return and make contact again. [1]
b i Moment = Fd [1]
3.5 × 10−5
F= = 1.0 × 10−3 N [1]
0.035
## AS and A Level Physics Original material © Cambridge University Press 2010 1
25 Marking scheme: Worksheet (A2)
F
ii F = BIl ⇒ I = [1]
Bl
1.0 ×10 −3
I= = 2.38 A ≈ 2.4 A [1]
(6.0 ×10 −3 × 0.07)
8 a F = BIl × number of turns [1]
F = 0.19 × 2.8 × 0.07 × 25 = 0.93 N [1]
b Torque= Fd = 0.93 × 0.03 [1]
torque = 0.028 N m [1]
c The longest side always stays at 90º to the magnetic field as the coil turns
so the force is constant. [1]
The perpendicular distance between the forces changes as the coil turns
so the torque (moment) changes. [1]
## 9 a F = mg = (103.14 – 102.00) × 10–3 × 9.81 so F = 1.12 × 10–2 N ≈ 1.1 × 10–2 N [1]
F
b B= [1]
Il
1.12 × 10 −2
B= [1]
8.2 × 0.05
B = 2.73 × 10−2 T (27 mT) [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
26 Marking scheme: Worksheet (A2)
1 F = EQ = 5.0 × 105 × 3.2 × 10−19 [1]
F = 1.6 × 10−13 N [1]
V 600
2 a E= = [1]
d 3.0 × 10 − 2
E = 2.0 × 104 V m−1 [1]
The field acts towards the negative plate. [1]
b The electric field is uniform between the plates (except at the ‘edges’). [1]
The electric field is at right angles to the plate. [1]
c i Since the droplet is stationary,
the electric force on the droplet
must be equal and opposite to its weight. [1]
The electric force must act upwards,
so the charge on the droplet must
be negative. [1]
F
ii E =
Q
F 6.4 × 10 −15
Q= = [1]
E 2.0 × 10 4
Q = 3.2 × 10−19 C [1]
3 F = BQv [1]
F = 0.18 × 1.6 × 10–19 × 4.0 × 106 [1]
F = 1.15 × 10–13 N ≈ 1.2 × 10−13 N [1]
4 a F = BQv [1]
F = 0.004 × 1.6 × 10–19 × 8.0 × 106 [1]
F = 5.12 × 10−15 N ≈ 5.1 × 10−15 N [1]
−15
F 5.12 ×10
b a= = [1]
m 9.11×10−31
a = 5.63 × 1015 m s−2 ≈ 5.6 × 1015 m s−2 [1]
c From circular motion, the centripetal acceleration a is given by:
v2
a=
r
v2 (8.0 × 10 6 ) 2
r= = [1]
a 5.63 × 1015
r = 1.14 × 10−2 m ≈ 1.1 × 10−2 m (1.1 cm) [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
26 Marking scheme: Worksheet (A2)
5 a
Both arrows at A and B are towards the centre of the circle. [1]
b The force on the electron is at 90° to the velocity. Hence the path described by the
electron is a circle. [1]
c The magnetic force provides the centripetal force. [1]
mv 2
Therefore: BQv = [1]
r
mv BQr
BQ = or v = [1]
r m
2.0 × 10 × 1.6 × 10 −19 × 5.0 × 10 −2
−3
v= [1]
9.1 × 10 −31
v = 1.76 × 107 m s–1 ≈ 1.8 × 107 m s−1 [1]
BQr
d v= , so the speed v is directly proportional to the radius r. [1]
m
1.76 × 10 7
Radius is halved, so v = = 8.8 × 106 m s−1 [1]
2
6 a Ek = 15 × 103 × 1.6 × 10−19 = 2.4 × 10−15 J (1 eV = 1.6 × 10−19 J) [1]
1
mv 2 = 2.4 × 10−15
2
2 × 2.4 × 10 −15
v= [1]
1.7 × 10 − 27
v = 1.68 × 106 m s−1 ≈ 1.7 × 106 m s−1 [1]
mv 2
b F = ma = [1]
r
1.7 × 10 −27 × (1.68 × 10 6 ) 2
F= [1]
0.05
F = 9.60 × 10 N ≈ 9.6 × 10−14 N
−14
[1]
c F = BQv [1]
F 9.60 × 10 −14
B= = [1]
Qv 1.6 × 10 −19 × 1.68 × 10 6
B ≈ 0.36 T [1]
distance
d speed =
time
circumference 2π × 0.05
time = = [1]
speed 1.68 × 10 6
time = 1.87 × 10−7 s ≈ 1.9 × 10−7 s [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
26 Marking scheme: Worksheet (A2)
7 a In order for the positively charged ions to emerge from the slit,
the net force perpendicular to the velocity must be zero. [1]
## electrical force on ion = magnetic force on ion [1]
EQ = BQv [1]
The charge Q cancels.
E = Bv [1]
V
The electric field strength is E = . Therefore, the magnetic flux density is:
d
E V d (5.0 × 10 3 ) / 0.024
B= = = [1]
v v 6.0 × 10 6
B = 3.47 × 10–2 T ≈ 35 mT [1]
BQr mv
b v= so r = [1]
m BQ
( m1 − m 2 ) v
∆r = [1]
BQ
## 8 a The centripetal force is provided by the magnetic force. [1]
mv 2
Therefore: Bev = [1]
r
mv Ber
Be = or v = [1]
r m
circumference 2πr
T= = [1]
speed Ber m
2 πm
The radius r of the orbit cancels. Hence: T =
Be
The time T is independent of both the radius of the orbit r and the speed v. [1]
b The faster electron travels in a circle of larger radius. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
27 Marking scheme: Worksheet (A2)
1 a i Magnetic flux, Φ = BA [1]
ii Flux linkage = NΦ = NBA [1]
b Flux linkage = N(B cos θ)A = NBA cos θ [1]
(The component of B normal to the plane of the coil is B cos θ.)
2 Area A of coil = x2 [1]
flux linkage = NBA = NBx2 [1]
3 a There is no change to the magnetic flux linking the coil, hence according to Faraday’s law,
there is no induced e.m.f. [1]
b The magnetic flux density B increases as the magnet moves towards the coil. [1]
There is an increase in the magnetic flux linking the coil, hence an e.m.f. is induced across
the ends of the coil. [1]
4 a flux linkage = NBA so B = [1]
NA
1.4 × 10 −4
B= [1]
70 × 4.0 × 10 − 4
B = 5.0 × 10−3 T [1]
b flux linkage = NBA cos θ [1]
flux linkage = 70 × 0.050 × 4.0 × 10−4 × cos 60° [1]
flux linkage = 7.0 × 10−4 Wb [1]
5 a magnetic flux, Φ = BA = 40 × 10–3 × (0.03 × 0.03)
i [1]
Φ = 3.6 × 10−5 Wb [1]
ii flux linkage = NΦ = 200 × 3.6 × 10−5 [1]
flux linkage = 7.2 × 10−3 Wb [1]
b Final flux linkage = 0, initial flux linkage = 7.2 × 10−3 Wb. [1]
Hence, change in magnetic flux linkage is 7.2 × 10−3 Wb. [1]
6 a Initial magnetic flux = BA = 0.15 × (π × [8.0 × 10−3]2) [1]
initial magnetic flux = 3.02 × 10−5 Wb [1]
final magnetic flux = 0 [1]
average magnitude of induced e.m.f. = rate of change of magnetic flux linkage
∆Φ 0 − 3.02 × 10 −5
E= N , so E = 1200 × [1]
∆t 0.020
E = 1.81 V ≈ 1.8 V (magnitude only) [1]
e.m.f. 1.81
b average current = = [1]
resistance 6.3
I = 0.287 A ≈ 0.29 A [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
27 Marking scheme: Worksheet (A2)
## b Area swept = length × distance travelled = 0.10 × 2.0 = 0.20 m2 [1]
c Change in magnetic flux = area swept × magnetic flux density [1]
change in magnetic flux = 0.20 × 0.050 = 1.0 × 10−2 Wb [1]
d Magnitude of e.m.f. = rate of change of magnetic flux linkage [1]
∆ ( NΦ )
E= (N = 1)
∆t
1.0 × 10 −2
E= = 1.0 × 10−2 V (1 Wb s−1 = 1 V) [1]
1 .0
e E = Blv = 0.050 × 2.0 × 0.10 = 1.0 × 10−2 V [1]
8 Initial magnetic flux = BA = 0.060 × π × (1.2 × 10−2)2 [1]
initial magnetic flux = 2.72 × 10−5 Wb [1]
final magnetic flux = –2.72 × 10−5 Wb (since the field is reversed) [1]
average magnitude of induced e.m.f. = rate of change of magnetic flux linkage
∆Φ − 2.72 ×10 −5 − 2.72 ×10 −5
E= N , so E = 2000 × [1]
∆t 0.030
E = 3.62 V ≈ 3.6 V (magnitude only) [1]
9 a There is a current in the primary coil when the switch is closed. This current creates a
magnetic flux in the primary coil. [1]
Due to the soft iron ring, the magnetic flux created by the primary coil also links the
secondary coil. With the switch closed, there is no change in the magnetic flux linkage
at the secondary and hence the lamp is not lit. [1]
When the switch is opened, the magnetic flux decreases to zero in a short period.
The rapid change in magnetic flux at the secondary coil creates an e.m.f. and the lamp
illuminates for a short period. [1]
Eventually there is no magnetic flux at either the primary or the secondary coil and
hence there is no e.m.f. induced – the lamp stays off. [1]
b Change the cell to an a.c. supply (or keep switching on and off very fast). [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
27 Marking scheme: Worksheet (A2)
10
## distance = speed × time = vt [1]
area swept = length × distance travelled = Lvt [1]
change in magnetic flux = area swept × magnetic flux density [1]
change in magnetic flux = (Lvt) × B = BLvt [1]
magnitude of e.m.f. = rate of change of magnetic flux linkage [1]
BLvt
E= = BLv [1]
t
E = BLv = 40 × 10−6 × 0.20 × 0.30 [1]
E = 2.4 × 10−6 V (2.4 µV) [1]
11 a The magnitude of the induced e.m.f. in a circuit is directly proportional to the rate of
change of magnetic flux linkage. [1]
b Lenz’s law expresses the principle of conservation of energy. [1]
c i magnetic flux = magnetic flux density × cross-sectional area of coil
or Φ = BA [1]
ii Flux linkage = NBA [1]
Coil X: flux linkage = NBA = 200 × 0.10 × (π × 0.022) ≈ 2.5 × 10−2 Wb [1]
Coil Y: flux linkage = NBA = 4000 × 0.01 × (π × 0.032) ≈ 1.1 × 10−1 Wb [1]
The coil Y has greater flux linkage. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
28 Marking scheme: Worksheet (A2)
1
1 a T= [1]
f
0.1 s [1]
b i t = 0, 0.05 and 0.15 [1]
1
ii t = of a period [1]
4
t = 0.025 s [1]
iii t = 0.075 s [1]
1
iv t = = sin (20πt) [1]
2
t = 0.0125 s [1]
## 2 a Corresponding parts of each wave occur at the same time. [1]
b Power P always positive. [1]
Two peaks for P in time taken for one peak of I and V. [1]
Sinusoidal shape above axis. [1]
P 690
3 a i I= = [1]
V 230
Irms = 3.0 A [1]
ii Ipeak = 3 2 = 4.2 A [1]
iii Vpeak = 230 2 = 325 V [1]
b Correct shape all above axis. [1]
Two cycles shown (i.e. one cycle of current waveform). [1]
Peak and average power marked. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
28 Marking scheme: Worksheet (A2)
4 a a.c. [1]
Current is positive and negative OR current flows one way, then the opposite way. [1]
V2
b Voltage switches between +2 and −2 V. Power is , so it has the same value for both
R
positive and negative values of the voltage. [1]
c i 2V [1]
ii 2 V [1]
⎛ 2 .5 ⎞ ⎛ 6 ⎞
5 Average power = ⎜ ⎟× ⎜ ⎟ [1]
⎝ 2⎠ ⎝ 2⎠
= 7.5 W [1]
6 a i Period = 80 ms = 0.080 s [1]
1
f= = 12. 5 Hz [1]
T
ii Peak voltage = 4.5 V [1]
4.5
r.m.s. voltage = = 3.2 V [1]
2
b V0 = 4.5 V [1]
ω = 2πf = 2π × 12.5 = 78.5 s−1 ≈ 79 s−1 [1]
P 24
7 a I= = [1]
V 6
I = 4.0 A [1]
1150
b Vp = 6 × [1]
30
Vp = 230 V [1]
24 30 × 4
c Ip = or Ip = [1]
230 1150
Ip = 0.10 A [1]
d Maximum p.d. = 6 2 [1]
maximum p.d. = 8.5 V [1]
e Heat is still produced inside the wires even if it cannot be conducted to the
outside. The wires may melt if they cannot lose the heat. [1]
115
8 a Number of turns on the primary = × 500 = 1000 turns [1]
230
Vs 115
b Is = = = 0.023 A r.m.s. [1]
R 5000
500
c Ip = 0.023 × = 0.0125 A ≈ 0.013 A r.m.s. [1]
1000
d Peak voltage = 115 2 = 162 V [1]
so the cables will break down. [1]
P 1000
9 a I= = = 10 A r.m.s. [1]
V 100
P = I 2R = 102 × 5 = 500 W [1]
b At high voltages the current is less for the same power. [1]
Power lost in cable = I 2R so power lost is less. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
29 Marking scheme: Worksheet (A2)
1 A ‘packet’, or quantum, of electromagnetic energy. [1]
2 The energy of a photon is proportional to the frequency of the radiation. [1]
Hence a γ-ray photon has greater energy than a photon of visible light (and therefore
is more harmful). [1]
3 a Electromagnetic radiation travels through space as waves and, as such, shows diffraction
and interference effects. [1]
b Electromagnetic radiation interacts with matter as ‘particles’. The photoelectric effect
provides strong evidence for the particle-like (photon) behaviour of electromagnetic
c 3.0 × 108
4 a c = fλ so f = = [1]
λ 6.4 × 10 −7
f = 4.69 × 10 Hz ≈ 4.7 × 1014 Hz
14
[1]
b E = hf [1]
E = 6.63 × 10−34 × 4.69 × 1014 [1]
E = 3.1 × 10−19 J [1]
5 For an electron to escape from the surface of the metal, it must absorb energy from the photon
that is greater than the work function. [1]
The work function is the minimum energy required by the electron to escape from the surface
of the metal. [1]
The photon of visible light has energy less than the work function of the metal, whereas
the photon of ultraviolet radiation has energy greater than the work function. [1]
## 6 a The electron loses energy. [1]
This energy appears as a photon of electromagnetic radiation. [1]
b Energy of photon = E1 − E2 [1]
Therefore:
E − E2
hf = E1 − E2 or f = 1 [1]
h
c The change in energy ∆E is greater. [1]
Hence the frequency of the radiation is greater (f ∝ ∆E). [1]
The spectral line will be the right side of the line shown on the spectrum diagram.
7 There are six spectral lines. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
29 Marking scheme: Worksheet (A2)
hc
8 ∆E = hf = [1]
λ
6.63 ×10 −34 × 3.0 ×10 8
∆E = [1]
670 ×10 −9
∆E = 2.97 × 10−19 J ≈ 3.0 × 10−19 J [1]
## 9 a Continuous spectrum [1]
b Emission spectrum [1]
c Absorption spectrum [1]
10 Electrons travel through space as waves. Evidence for this is provided by the diffraction of
electrons by matter (e.g. graphite). [1]
11 The electronvolt is the energy gained by an electron travelling through a potential difference
of one volt. [1]
12 The kinetic energy Ee of the electron is:
Ee = VQ = 6.0 × 1.6 × 10–19 [1]
Ee = 9.6 × 10–19 J [1]
The energy EUV of the ultraviolet photon is:
hc
EUV = hf = [1]
λ
6.63 × 10 −34 × 3.0 × 108
EUV = [1]
2.5 × 10 −7
EUV = 7.96 × 10−19 J ≈ 8.0 × 10−19 J [1]
The energy of the ultraviolet photon is less than the kinetic energy of the electron.
(The student is correct.)
13 a The threshold frequency is the minimum frequency of electromagnetic radiation that just
removes electrons from the surface of the metal. [1]
b At the threshold frequency, the energy of the photon is equal to the work function φ of the
metal. Hence:
φ = hf0 (f0 = threshold frequency) [1]
1.9 × 1.6 × 10 −19
f0 = [1]
6.63 × 10 −34
f0 = 4.6 × 1014 Hz [1]
hc
14 a E = hf = [1]
λ
6.63 × 10 −34 × 3.0 × 108
E= [1]
550 × 10 −9
E = 3.62 × 10–19 J ≈ 3.6 × 10–19 J [1]
b Power emitted as light = 0.05 × 60 = 3.0 W [1]
3 .0
Number of photons emitted per second = [1]
3.62 × 10 −19
= 8.3 × 1018 [1]
15 φ = 4.3 × 1.6 × 10–19 = 6.88 × 10–19 J [1]
hc 6.63 × 10 −34 × 3.0 × 108
Energy of photon = = = 9.47 × 10−19 J [1]
λ 2.1 × 10 −7
## energy of photon = work function + maximum kinetic energy of electron [1]
maximum kinetic energy of electron = (9.47 − 6.88) × 10−19 ≈ 2.6 × 10−19 J [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
29 Marking scheme: Worksheet (A2)
h
16 λ = [1]
mv
h 6.63 × 10 −34
ν= = [1]
mλ 1.7 × 10 −27 × 2.0 × 10 −11
## v = 1.95 × 104 m s−1 (20 km s−1) [1]
17 Energy lost by a single electron = energy of photon [1]
(The energy lost by a single electron travelling through the LED reappears as the energy of a
single photon.)
Therefore:
hc
eV = [1]
λ
hc 6.63 ×10 −34 × 3.0 ×10 8
V= = [1]
λe 5.8 ×10 −7 ×1.6 ×10 −19
V = 2.14 V ≈ 2.1 V [1]
## 18 a Kinetic energy of electron = VQ = Ve
1 p2
mev2 = Ve or = Ve (where p = mev) [1]
2 2me
p= 2meVe [1]
h h
λ= = (de Broglie equation) [1]
me v p
h
Therefore, λ =
2meVe
h h2
b λ= or V = [1]
2meVe 2me λ2 e
(6.63 × 10 −34 ) 2
V= [1]
2 × 9.1 × 10 −31 × (4.0 × 10 −11 ) 2 × 1.6 × 10 −19
V ≈ 940 V [1]
c 1
19 Using f = and Einstein’s photoelectric equation (hf = φ + 2
mv max ): [1]
λ 2
h × 3.0 × 108
Red light ⇒ = φ + (0.9 × 1.6 × 10−19)
640 × 10 −9
⇒ 4.688 × 1014 h = φ + 1.440 × 10−19 (equation 1) [1]
h × 3.0 × 108
Blue light ⇒ = φ + (1.9 × 1.6 × 10−19)
420 × 10 −9
⇒ 7.143 × 1014 h = φ + 3.040 × 10−19 (equation 2) [1]
Equations 1 and 2 are two simultaneous equations.
(7.143 – 4.688) × 1014 h = (3.040 − 1.440) × 10−19 [1]
(3.040 − 1.440) × 10 −19
h= ≈ 6.5 × 10−34 J s [1]
(7.143 − 4.688) × 1014
## AS and A Level Physics Original material © Cambridge University Press 2010 3
29 Marking scheme: Worksheet (A2)
## 20 a External energy has to be supplied to excite or free an electron. [1]
(Allow: The electrons are trapped in an energy well.)
b An energy level of 0 eV means the electron is free from the atom.
The minimum energy is equal to 3.00 eV. [1]
Energy needed to free electron = 3.00 × 1.6 × 10−19 [1]
Energy needed to free electron = 4.80 × 10−19 J [1]
c i The difference between the energy levels −3.00 eV and −1.59 eV is equal to 1.41 eV. [1]
Hence, an electron jumps from −3.00 eV energy level to −1.59 eV energy level. [1]
hc
ii ∆E = hf = [1]
λ
hc 6.63 ×10 −34 × 3.0 ×10 8
λ= = [1]
∆E 1.41×1.6 ×10 −19
λ = 8.82 × 10−7 m [1]
## 21 a This is the lowest energy level occupied by an electron in an atom. [1]
b The shortest wavelength corresponds to the change in energy between the two most widely
separated energy levels.
Hence, ∆E = 10.43 eV [1]
hc
∆E = hf = [1]
λ
hc 6.63 ×10 −34 × 3.0 ×10 8
λ= = [1]
∆E 10.43 ×1.6 ×10 −19
λ = 1.19 × 10−7 m [1]
## 2.18 ×10 −18
22 a E1 = − = –2.18 × 10−18 J [1]
12
2.18 ×10 −18
E2 = − 2
= –5.45 × 10−19 J [1]
2
hc
b E2 − E1 = ∆E = [1]
λ
hc 6.63 × 10 −34 × 3.0 × 10 8
λ= = [1]
∆E (21.8 − 5.45) × 10 −19
λ = 1.22 × 10−7 m [1]
This spectral line lies in the ultraviolet region of the spectrum. [1]
⎛ 1 1 ⎞
c ∆E = 2.18 × 10−18 ⎜ 2 − 2 ⎟ = 1.607 × 10−20 J [1]
⎝6 7 ⎠
−34
hc 6.63 × 10 × 3.0 × 108
λ= = = 1.238 × 10−5 m [1]
∆E 1.607 × 10− 20
λ ≈ 1.24 × 10−5 m (12.4 µm) [1]
This spectral line lies in the infrared region of the electromagnetic spectrum. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 4
30 Marking scheme: Worksheet (A2)
1 a change in energy = change in mass × (speed of light)2 or ∆E = ∆mc2 [1]
b i ∆E = ∆mc2
∆E = 0.001 × (3.0 × 108)2 [1]
∆E = 9.0 × 1013 J [1]
ii ∆E = ∆mc2
∆E = 9.1 × 10−31 × (3.0 × 108)2 = 8.19 × 10−14 J [1]
∆E ≈ 8.2 × 10−14 J [1]
6.65 × 10 −27
2 a i Mass = = 4.01 u [1]
1.66 × 10 − 27
2.16 × 10 −26
ii Mass = = 13.01 u [1]
1.66 × 10 − 27
b i Mass = 1.01 × 1.66 × 10−27 = 1.68 × 10−27 kg [1]
ii Mass = 234.99 × 1.66 × 10−27 = 3.90 × 10−25 kg [1]
3 In all nuclear reactions the following quantities are conserved:
• charge (or proton number)
• nucleon number
• mass–energy
• momentum.
Any three of the above. [3]
4 a The nucleons within the nucleus are held tightly together by the strong nuclear force. [1]
b The binding energy of a nucleus is the minimum energy required to separate the nucleus
into its constituent protons and neutrons. [1]
binding energy
c binding energy per nucleon =
number of nucleons
128
binding energy per nucleon = [1]
16
binding energy per nucleon = 8.0 MeV [1]
5 a The half-life of a radioactive isotope is the mean time taken for the number of nuclei of the
isotope to decrease to half the initial number. [1]
N
b i 20 minutes is 1 half-life, so number of nuclei left = 0 [1]
2
ii 1.0 hour is 3 half-lives. [1]
3
⎛1⎞ N
Number of nuclei left = ⎜ ⎟ N 0 = 0 [1]
⎝2⎠ 8
6 a iActivity is equal to the number of emissions (or decays of nuclei) per second.
Hence, there are 540 α-particles emitted in 1 second. [1]
ii Number of α-particles emitted in 1 h = 540 × 3600 ≈ 1.9 × 106 [1]
b Energy released in 1 s = number of α-particles emitted in 1 s × energy of each α-particle [1]
energy released in 1 s = 540 × 8.6 × 10−14 [1]
energy released in 1 s = 4.64 × 10−11 J ≈ 4.6 × 10−11 J [1]
c Rate of emission of energy = energy released per second
rate of emission of energy = 4.6 × 10−11 J s−1 = 4.6 × 10−11 W [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
30 Marking scheme: Worksheet (A2)
56
7 a The nuclide 26 Fe is the most stable. [1]
It has the maximum value for the binding energy per nucleon. [1]
b Binding energy = binding energy per nucleon × number of nucleons
binding energy ≈ 12.3 × 10−13 × 12 [1]
binding energy ≈ 1.5 × 10−11 J [1]
c From the graph, the binding energies per nucleon of 21 H and 42 He are approximately
1.0 × 10−13 J and 11.2 × 10−13 J. [1]
energy released = difference in binding energy per nucleon × number of nucleons [1]
energy released = [11.2 × 10−13 – 1.0 × 10−13] × 4 [1]
energy released = 4.08 × 10−12 J ≈ 4.1 × 10−12 J [1]
d High temperatures (~108 K) and pressures. [2]
## 8 92 11 proton + 143 01 neutron → 235
92 uranium [1]
mass defect = [(143 × 1.009) + (92 × 1.007)]u – (234.992)u [1]
mass defect = 1.939 u = 1.939 × 1.66 × 10−27 kg [1]
mass defect = 3.219 × 10−27 kg [1]
binding energy = mass defect × (speed of light)2 [1]
binding energy = 3.219 × 10−27 × (3.0 × 108)2 = 2.897 × 10−10 J [1]
binding energy
binding energy per nucleon =
number of nucleons
2.897 × 10 −10
binding energy per nucleon = = 1.233 × 10−12 ≈ 1.2 × 10−12 J [1]
235
235
9 a Fission is the splitting of a heavy nucleus like 92 U into two approximately equal fragments.
The splitting occurs when the heavy nucleus absorbs a neutron. [1]
b i All particles identified on the diagram. [1]
ii In the reaction above, there is a decrease in the mass of the particles. [1]
2
According to ∆E = ∆mc , a decrease in mass implies that energy is released
in the process. [1]
iii The change in mass is ∆m given by:
∆m = [1.575 × 10−25 + 2.306 × 10−25 + 2(1.675 × 10−27)] – [3.902 × 10−25 + 1.675 × 10−27][1]
∆m = –4.250 × 10−28 kg [1]
(The minus sign means a decrease in mass and hence energy is released in this reaction.)
∆E = ∆mc2 [1]
∆E = 4.250 × 10−28 × (3.0 × 108)2 [1]
−11 −11
∆E = 3.83 × 10 J ≈ 3.8 × 10 J [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
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10 Binding energy of ‘reactant’ = 236 × 7.59 = 1791 MeV (binding energy of neutron = 0) [1]
Total binding energy of ‘products’ = (146 × 8.41) + (87 × 8.59) ≈ 1975 MeV [1]
Therefore energy released = 1975 – 1791 = 184 MeV [1]
0.693 0.693
11 a t1/2 = so λ = [1]
λ t1/2
0.693
λ= [1]
56
λ = 1.238 × 10−2 s−1 ≈ 1.2 × 10−2 s−1 [1]
b A = λN [1]
0.693
A= × 6.0 × 1010 [1]
56
A ≈ 7.4 × 108 Bq [1]
A
12 a A = λN so λ = [1]
N
5.0 × 10 9
λ= [1]
8.0 × 1014
λ = 6.25 × 10−6 s−1 ≈ 6.3 × 10−6 s−1 [1]
0.693
b t1/2 = [1]
λ
0.693
t1/2 = [1]
6.25 × 10 −6
t1/2 = 1.11 × 105 s ≈ 1.1 × 105 s [1]
c N = N0 e−λt [1]
−6
N = 8.0 × 1014 e − (6.25×10 × 40×3600 ) [1]
N = 3.25 × 1014 ≈ 3.3 × 1014 [1]
13 a The decay constant is the probability that an individual nucleus will decay per unit time. [1]
0.693 0.693
b i t1/2 = so λ = [1]
λ t1/2
0.693
λ= [1]
18 × 24 × 3600
λ = 4.46 × 10−7 s−1 ≈ 4.5 × 10−7 s−1 [1]
ii A = λN [1]
A = 4.46 × 10−7 × 4.0 × 1012 [1]
A = 1.78 × 106 Bq ≈ 1.8 × 106 Bq [1]
iii 36 days is equal to 2 half-lives. [1]
2
⎛1⎞
activity = ⎜ ⎟ × 1.78 × 106 = 4.45 × 105 Bq ≈ 4.5 × 105 Bq [1]
⎝2⎠
14 a number of nuclei = number of moles × NA
1.0 × 10 −6
number of nuclei = × 6.02 × 1023 [1]
226
number of nuclei = 2.66 × 1015 ≈ 2.7 × 1015 [1]
b A = λN [1]
⎛ 0.693 ⎞ 0.693 × 2.66 × 1015
A = ⎜⎜ ⎟⎟ × N = [1]
⎝ t1/2 ⎠ 1600 × 365 × 24 × 3600
A ≈ 3.7 × 104 Bq [1]
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## 15 According to Einstein’s equation: ∆E = ∆mc2 [1]
In this case, ∆E is the energy of two photons and ∆m is the mass of two protons. [1]
Hence:
hc
2× = (2 × mp) c2 [1]
λ
hc h 6.63 × 10 −34
λ= = = [1]
mp c 2 mp c 1.7 × 10 − 27 × 3.0 × 10 8
λ = 1.3 × 10−15 m [1]
16 For fusion, we have:
energy released per kg = number of ‘pairs’ of 21 H in 1 kg × 4.08 × 10−12 J (from 7 c) [1]
⎛ 1 1000 ⎞
energy per kg = ⎜ × × 6.02 × 10 23 ⎟ × 4.08 × 10−12 [1]
⎝ 2 2 ⎠
energy per kg = 6.14 × 1014 J ≈ 6.1 × 1014 J [1]
For fission, we have:
energy released per kg = number of nuclei in 1 kg × 3.83 × 10−11 J (from 9 b) [1]
⎛ 1000 ⎞
energy per kg = ⎜ × 6.02 × 10 23 ⎟ × 3.83 × 10−11 [1]
⎝ 235 ⎠
13
energy per kg = 9.8 × 10 J [1]
There is less energy released per fusion than per fission. However, there are many more nuclei
per kg for fusion. Hence fusion produces more energy per kg than fission. [1]
0.693
17 N = N0 e−λt and λ = [1]
t1/2
N
fraction left = = e −λt = e −( 0.693t / t1/2 ) [1]
N0
9 9
fraction left = e − ( 0.693×5.0×10 / 4.5×10 )
[1]
fraction left = 0.463 ≈ 0.46 [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 4
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1 Processor (or processing device) [1]
Output device [1]
2 a Axis labels [1]
Graph correct shape [1]
## 3 a Length of fine/thin wire [1]
sealed in a plastic case [1]
ρl ρl 5.0 × 10−7 × 0.15
b i R= so A = = [1]
A R 150
−10 2
A = 5.0 × 10 m [1]
1 1
ii Extends by so resistance increases by a factor of , which is 1 Ω. [1]
150 150
4 a Fraction of the output of a device is fed back to the input [1]
reducing any changes in the input to the system. [1]
b Increased bandwidth, less distortion, greater stability in gain (less affected by temperature
changes). [2]
5 a Inverting amplifier [1]
When Vin is positive then Vout is negative, or vice versa. [1]
b The output voltage reaches the power supply voltage and cannot exceed this value. [1]
c 10 V [1]
d i Gain becomes −10 rather than −5, i.e. sloping line has twice the gradient. [1]
Vout is +10 V for Vin below −1 V or Vout is −10 V for Vin above +1 V. [1]
ii Gain becomes −2.5 rather than −5, i.e. sloping line has half the gradient. [1]
iii The sloping line increases in length before flattening off. [1]
e Output voltage peaks at ±5 V. [1]
Input and output voltage out of phase. [1]
## f When the input signal reaches above 2 V or below −2 V [1]
the output voltage remains at −10 V or +10 V. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
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6 a The output is not saturated so the potential at both the inverting and non-inverting inputs
is very nearly the same. [1]
Since the non-inverting input is earthed, this value is 0 V. [1]
b 1.0 V [1]
V 1
c I= = = 5.0 × 10−4 A [1]
R 2000
d Negligible current goes into the (–) terminal of the op-amp since it has a very high input
resistance (impedance). [1]
R
e Size of gain = f = 5 so Rf = 5Rin
Rin
Rf = 5 × 2 = 10 kΩ [1]
7 a Any two from:
infinite open-loop voltage gain, infinite input resistance (or impedance),
infinite bandwidth, zero output resistance (or impedance) [2]
V 10 −3
b i I= = 4 [1]
R 10
I = 1.0 × 10−7 A [1]
ii V = IR = −1.0 × 10−7 × 100 × 103 = −1.0 × 10−2 V [1]
V − 1.0 ×10 −2
iii Gain = out = = −10 [1]
Vin 1.0 ×10 −3
iv 1.0 × 10−2 V (with Q being more positive) [1]
8 a For [2] marks, any two from the following points. [2]
With an inverting amplifier the output is half a cycle out of phase with the input,
whereas with a non-inverting amplifier the input and output are in phase.
The input goes to the (−) terminal on the op-amp for an inverting amplifier and
to the (+) terminal for the non-inverting amplifier.
The input resistance (impedance ) is higher for an op-amp used as a non-inverting amplifier.
− Rf
For an inverting amplifier the gain is but for an inverting amplifier
Rin
R
the gain is 1 + 1 .
R2
R 40
b G = 1+ 1 = 1+ [1]
R2 20
G=3 [1]
V V 8.0
c G = out ⇒ Vin = out = = 2.67 V ≈ 2.7 V [1]
Vin G 3
V 8.0
d I= = [1]
R (20 + 40 )×10 3
I = 1.33 × 10−4 A ≈ 1.3 × 10−4 A [1]
e The voltage is the same as for part c; voltage across the 20 kΩ resistor = 2.7 V [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
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## 9 a The output is either +9 V (+Vs) or −9 V (−Vs) [1]
– –
depending on whether V > V or V > V
+ +
[1]
b −9 V (−Vs) [1]
– +
c i If the resistances are all the same, then V and V are equal, so it is unclear which of the
two is larger. The comparator is at its switching point and could go either way. [1]
ii V stays the same (at 4.5 V). [1]
V + falls (due to larger resistance of X and larger potential drop across X). [1]
Vout is −9 V (it may have already been −9 V). [1]
10 a Two LEDs and series resistor shown [1]
Correct direction shown for LEDs [1]
output of
op-amp
green
red
b When output of op-amp is +9 V then the green LED is forward biased with enough
p.d. across it. The red LED is reverse biased. [1]
When output of op-amp is −9 V then the red LED is forward biased with enough
p.d. across it. The green LED is reverse biased. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
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1 It is an electromagnetic wave. [1]
Wavelength in the range 10−8 m to 10−13 m [1]
2 a Energy = 50 × 103 × 1.6 × 10−19 [1]
= 8.0 × 10−15 J [1]
hc
b E=
λ
6.63 × 10 −34 × 3.0 × 10 8
λ= [1]
8.0 × 10 −15
wavelength = 2.49 × 10−11 m ≈ 2.5 × 10−11 m [1]
3 Compton scattering and pair production. [2]
4 A CAT scan produces a three-dimensional image of the body. [1]
5 A technique that does not involve cutting the body, e.g. CAT scan. [1]
6 It is a wave similar to sound (a longitudinal wave) [1]
but with a frequency above the audible range/above about 20 kHz. [1]
7 a v = fλ
1.5 × 10 3
wavelength = [1]
1.8 × 10 6
= 8.33 × 10−4 m ≈ 0.83 mm [1]
b High-frequency ultrasound implies shorter wavelengths. Hence, smaller details can be
seen on an ultrasound scan. [1]
8 Acoustic impedance = density of medium × speed of ultrasound (Z = ρc) [1]
9 a Z = ρc
6.40 × 106 = ρ × 4000 [1]
6.40 × 10 6
ρ= = 1600 kg m−3 [1]
4000
2
⎛ Z − Z1 ⎞
b fraction of intensity reflected = ⎜⎜ 2 ⎟⎟ [1]
⎝ Z 2 + Z1 ⎠
2
⎛ 1.66 − 1.63 ⎞ −5
=⎜ ⎟ = 8.3 × 10 [1]
⎝ 1.66 + 1.63 ⎠
percentage of intensity reflected = 0.0083% [1]
c There is only a very small amount of ultrasound reflected at the boundary between these
two materials. [1]
This is because their acoustic impedances are very similar. [1]
10 Superconducting magnet, RF transmission coil, RF receiver coil, gradient coils and computer. [5]
11 a The rotation of the spin/magnetic axis about the direction of the external magnetic field. [1]
b 40 MHz [1]
c 3.0 × 10 8
c wavelength = = [1]
f 40 × 10 6
= 7.5 m [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
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## 12 High-energy electrons made to bombard a metal target. [1]
When the electrons are accelerated and/or slowed down, electromagnetic radiation is produced. [1]
For large accelerations, radiation is in the X-ray region. [1]
Many different values of acceleration so many different wavelengths [1]
giving a continuous spectrum. [1]
Incident electron may excite an orbiting electron in a target atom. [1]
De-excitation of these electrons gives rise to an emission line spectrum. [1]
power
13 a intensity = [1]
cross-sectional area
b Power = 250 × π(2.0 × 10−3)2 [1]
≈ 3.1 × 10−3 W [1]
14 A material (e.g. barium) with high-atomic number/attenuation coefficient used to
absorb X-rays. [1]
It is used to image the edges of soft tissues, such as the intestines (following a barium meal). [1]
15 Maximum energy of X-ray photon = 80 keV [1]
hc
= 80 × 10 3 × 1.6 × 10 −19 [1]
λ min
6.63 × 10 −34 × 3.0 × 10 8
λmin = = 1.554 × 10 −11 m ≈ 1.6 × 10 −11 m [1]
80 × 10 3 × 1.6 × 10 −19
16 The energy of the incident X-ray photon is used to remove an electron close to the nucleus
of an atom. [1]
The attenuation coefficient is proportional to the cube of the atomic (proton) number (µ ∝ Z ). [1]
3
The attenuation coefficient is inversely proportional to the cube of the photon energy (µ ∝ E −3). [1]
17 a
## Sketch graph showing: labelled axes [1]
correct shape of line [1]
intensity showing an
exponential decrease [1]
b I = I0e−µx [1]
where µ is the (linear) attenuation/absorption coefficient. [1]
c 0.10 = e−0.693x [1]
ln 0.10 = –0.693x
–2.3 = −0.693x [1]
x = 3.32 mm ≈ 3.3 mm [1]
18 a X-ray image produced of target area. [1]
Image is stored in a computer. [1]
X-ray beam rotated about target so that many images produced from different angles. [1]
Computer stores and processes images to give image of slice through target. [1]
Process repeated for different slices. [1]
b X-ray image is flat shadow image of whole structure. [1]
CAT scan gives two-dimensional image of a slice through structure. [1]
Images of slices can be processed to give image of structure in any plane. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
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## 19 Pulse of ultrasound transmitted into body [1]
where it is reflected at the boundary between different tissues. [1]
Reflected wave is detected and processed. [1]
Time for echo to reach detector indicates depth of tissue boundary. [1]
Intensity of echo gives information about tissue boundary. [1]
20 Amount of reflection of a wave at a boundary depends on the difference in acoustic
impedance of the two media. [1]
Acoustic impedance is speed of wave × density of medium. [1]
Acoustic impedance of air is very different from that for skin [1]
so very little ultrasound energy would pass into skin at the skin–air boundary. [1]
Coupling medium has acoustic impedance close to that of skin (and probe) [1]
so greatly reduces reflection at skin surface. [1]
21 a 2 × thickness = 4000 × (13 × 10−6) [1]
2 × thickness = 5.2 × 10−2 [1]
5.2 × 10 −2
thickness = = 2.6 × 10 −2 m (2.6 cm) [1]
2
b In a B-scan, the transducer is moved around the body. [1]
This produces a two-dimensional outline of an organ. [1]
22 a Nuclei of hydrogen (and certain other nuclides) have spin/have a magnetic axis. [1]
In a strong magnetic field, the nuclei precess about the direction of the field [1]
with a frequency known as the Larmor frequency. [1]
Radio wave pulse at Larmor frequency causes resonance [1]
and nuclei precess in high-energy state. [1]
After pulse has ceased, nuclei relax, emitting an RF signal. [1]
b
## Diagram showing: magnet producing large field [1]
magnets producing non-uniform field [1]
RF coils. [1]
Patient subject to large magnetic field and calibrated non-uniform field. [1]
RF pulse at Larmor frequency transmitted to patient. [1]
RF emissions from patient are detected and processed. [1]
Hydrogen nuclei within patient [1]
have Larmor frequency dependent on magnetic field strength [1]
so that location (and concentration) of hydrogen nuclei can be detected. [1]
Total image built up by varying the non-uniform field to give specific field strength at
different positions within the patient. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
32 Marking scheme: Worksheet (A2)
## 23 X-rays: diagnosis of broken bones [1]
relatively cheap, can detect flaws within the bone structure. [1]
Ultrasound: fetal development [1]
can distinguish between soft tissues; it is immediate and relatively safe. [1]
MRI: spinal/back problems [1]
fine detail provided of soft and hard tissues. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 4
33 Marking scheme: Worksheet (A2)
1 a The audio signals have the same frequency. [1]
The amplitude of the signal/trace/carrier rises and falls with the same number of cycles
(two in the time of the trace). [1]
The audio signals have different volumes/loudness/amplitudes. [1]
The amplitude of the trace rises and falls more in the bottom trace. [1]
b The time (or horizontal distance along each trace) for one rapid variation
is the same in both traces. [1]
2 a In amplitude modulation the amplitude of the carrier wave is altered to carry the signal
(the frequency remains the same). [1]
In frequency modulation the frequency of the carrier wave is altered to carry the signal
(the amplitude remains the same). [1]
b i 30 × 2 = 60 kHz [1]
ii 800 − 60 = 740 kHz [1]
iii Alters from 740 kHz to 860 kHz 6000 times a second. [1]
c More transmitters may be needed as the range of FM is less than that of AM. [1]
Equipment to transmit and receive FM is more expensive. [1]
3 a i Carrier wave [1]
ii Sidebands [1]
iii 5 kHz [1]
b i 2.5 × 10−5 s [1]
ii 2.0 × 10−4 s [1]
iii Correct amplitude-modulated shape [1]
8 carrier wave oscillations per oscillation of the amplitude [1]
Correct times marked [1]
## c Correct shape [1]
Sidebands extending from 40 to 55 kHz
and from 40 to 25 kHz [1]
25 40 55 f / kHz
## AS and A Level Physics Original material © Cambridge University Press 2010 1
33 Marking scheme: Worksheet (A2)
## 4 a Analogue signal can have any value (within limits). [1]
Diagram to show analogue signal. [1]
Digital can have only a few values, e.g. two, and nothing in-between these values. [1]
Diagram to show digital signal. [1]
## b The value of the signal is measured at regular intervals of time. [1]
The value obtained is converted into a binary number (with a certain number of bits). [1]
The binary numbers obtained are placed one after the other and form the digital signal. [1]
5 a 0101 and 1000 [2]
b Values shown as horizontal lines of 0.5 ms duration [1]
Values plotted: 5, 8, 8, 6, 2, 1, 2, 6, 8, 8 [1]
Graph correct with labels [1]
V / mV
8
0
0 1 2 3 4 5 t / ms
c i Any variation in the signal that occurs between sampling is not detected [1]
Increasing sampling frequency decreases the time between samples [1]
Frequency at least 2 × signal, i.e. 600 Hz (frequency of signal ≈ 300 Hz) [1]
ii The variation in voltage can use more voltage levels (28 levels rather than 24 levels). [1]
The signal voltage at every sample is closer to the actual value. [1]
⎛ 6.0 × 10 −3 ⎞
6 Signal-to-noise ratio (in dB) = 10 lg ⎜⎜ ⎟ = 164.7 dB ≈ 165 dB
−19 ⎟
[1]
⎝ 2 .0 × 10 ⎠
1.0 × 10 −3
7 a Signal-to-noise ratio = 10 lg = 30 dB [1]
0.001× 10 −3
b Signal becomes 0.001 mW or signal-to-noise ratio is 1 [1]
0 dB [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
33 Marking scheme: Worksheet (A2)
## 8 Any two reasons and explanation.
Less attenuation so fewer repeater/regeneration amplifiers needed. [2]
More bandwidth so more data can be sent per second/more telephone calls made at once. [2]
Less interference/noise so fewer regeneration amplifiers needed. [2]
Lower diameter/weight so easier to handle/cheaper. [2]
9 a i Any value less than 10 m (and more than 1 mm) [1]
ii Any value between 10 and 100 m [1]
b The sky wave uses reflection by the ionosphere for transmission. [1]
The ionosphere fluctuates in its ability to reflect. [1]
## 10 a Orbits around the Earth’s equator. [1]
Takes one day for a complete orbit. [1]
Stays over one point on the Earth OR height of orbit 36 000 km above Earth’s surface. [1]
b First satellites used wavelengths of about 5 cm; typically now between 1 mm and 1 cm. [1]
c Advantage: permanent link with ground station/dishes do not have to be moved. [1]
Disadvantage: greater time delay for signal OR further away so signal weaker. [1]
11 The public switched telephone network connects every telephone through exchanges. [1]
Without exchanges too many telephones and interconnecting wires are needed. [1]
One cable can handle many telephone conversations at once. [1]
Sampling places a series of digital bits from many telephone conversations on one cable. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
18 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 Convert the following angles into radians.
a 30° [1]
b 210° [1]
c 0.05° [1]
2 Convert the following angles from radians into degrees.
3 The planet Mercury takes 88 days to orbit once round the Sun.
Calculate its angular displacement in radians during a time interval of:
a 44 days [1]
b 1 day. [1]
4 In each case below, state what provides the centripetal force on the object.
a A car travels at a high speed round a sharp corner. [1]
b A planet orbits the Sun. [1]
c An electron orbits the positive nucleus of an atom. [1]
d Clothes spin round in the drum of a washing machine. [1]
5 An aeroplane is circling in the sky at a speed of 150 m s−1.
The aeroplane describes a circle of radius 20 km.
For a passenger of mass 80 kg inside this aeroplane, calculate:
a her angular velocity [2]
b her centripetal acceleration [3]
c the centripetal force acting on her. [2]
6 The diagram shows a stone tied to the end of a length of string.
It is whirled round in a horizontal circle of radius 80 cm.
## The stone has a mass of 90 g and it completes 10 revolutions in a time of 8.2 s.
a Calculate:
i the time taken for one revolution [1]
ii the distance travelled by the stone during one revolution (this distance is equal to the
circumference of the circle) [1]
iii the speed of the stone as it travels in the circle [2]
iv the centripetal acceleration of the stone [3]
v the centripetal force on the stone. [2]
b What provides the centripetal force on the stone? [1]
c What is the angle between the acceleration of the stone and its velocity? [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
18 Worksheet (A2)
## 7 A lump of clay of mass 300 g is placed
close to the edge of a spinning turntable.
The centre of mass of the lump of clay travels
in a circle of radius 12 cm.
## a The lump of clay takes 1.6 s to complete one revolution.
i Calculate the rotational speed of the clay. [2]
ii Calculate the frictional force between the clay and the turntable. [3]
b The maximum magnitude of the frictional force F between the clay and the turntable is
70% of the weight of the clay. The speed of rotation of clay is slowly increased.
Determine the speed of the clay when it just starts to slip off the turntable. [4]
8 The diagram shows a skateboarder of mass 70 kg who drops through a vertical height
of 5.2 m.
## The dip has a radius of curvature of 16 m.
a Assuming no energy losses due to air resistance or friction, calculate the speed of the
skateboarder at the bottom of the dip at point B.
You may assume that the speed of the skateboarder at point A is zero. [2]
b i Calculate the centripetal acceleration of the skateboarder at point B. [3]
ii Calculate the contact force R acting on the skateboarder at point B. [3]
9 A car of mass 820 kg travels at a constant speed
of 32 m s−1 along a banked track.
The track is banked at an angle of 20°
to the horizontal.
## a The net vertical force on the car is zero.
Use this to show that the contact force R on the car is 8.56 kN. [2]
b Use the answer from a to calculate the radius of the circle described by the car. [4]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
18 Worksheet (A2)
## 10 A stone of mass 120 g is fixed to one end of a light rigid rod.
The stone is whirled at a constant speed of 4.0 m s−1 in a vertical circle of radius 80 cm.
tension in the rod at A
Calculate the ratio: [6]
tension in the rod at B
Total: Score: %
59
## AS and A Level Physics Original material © Cambridge University Press 2010 3
19 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 Define gravitational field strength at a point in space. [1]
2 Show that the gravitational constant G has the unit N m2 kg−2. [2]
3 The gravitational field strength on the surface of the Moon is 1.6 N kg−1.
What is the weight of an astronaut of mass 80 kg standing on the surface of the Moon? [2]
4 Calculate the magnitude of the gravitational force between the objects described below.
You may assume that the objects are ‘point masses’.
a two protons separated by a distance of 5.0 × 10−14 m
(mass of a proton = 1.7 × 10−27 kg) [3]
## b two binary stars, each of mass 5.0 × 1028 kg,
with a separation of 8.0 × 1012 m [2]
## c two 1500 kg elephants separated by a distance of 5.0 m [2]
5 The diagram shows the Moon and an artificial satellite orbiting round the Earth.
The radius of the Earth is R.
a Write an equation for the gravitational field strength g at a distance r from the centre of
an isolated object of mass M. [1]
b By what factor would the gravitational field decrease if the distance from the centre of
the mass were doubled? [2]
c The satellite orbits at a distance of 5R from the Earth’s centre and the Moon is at a
distance of 59R. Calculate the ratio:
gravitational field strength at position of satellite
[3]
gravitational field strength at position of Moon
6 The planet Neptune has a mass of 1.0 × 1026 kg and a radius of 2.2 × 107 m.
Calculate the surface gravitational field strength of Neptune. [3]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
19 Worksheet (A2)
7 Calculate the radius of Pluto, given its mass is 5.0 × 1023 kg and its surface
gravitational field strength has been estimated to be 4.0 N kg−1. [3]
8 A space probe of mass 1800 kg is travelling from Earth to the planet Mars.
The space probe is midway between the planets. Use the data given to calculate:
a the gravitational force on the space probe due to the Earth [3]
b the gravitational force on the space probe due to Mars [2]
c the acceleration of the probe due to the gravitational force acting on it. [3]
Data
separation between Earth and Mars = 7.8 × 1010 m
mass of Earth = 6.0 × 1024 kg mass of Mars = 6.4 × 1023 kg
9 An artificial satellite orbits the Earth at a height of 400 km above its surface.
The satellite has a mass 5000 kg, the radius of the Earth is 6400 km and the mass of the
Earth is 6.0 × 1024 kg. For this satellite, calculate:
a the gravitational force experienced [3]
b its centripetal acceleration [2]
c its orbital speed. [3]
## 10 a Explain what is meant by the term gravitational potential at a point. [2]
b Write down the gravitational potential energy of a body of mass 1 kg when it is at
an infinite distance from another body. [1]
c The radius of the Earth is 6.4 × 106 m and the mass of the Earth = 6.0 × 1024 kg.
Calculate the potential energy of the 1 kg mass at the Earth’s surface. [3]
d Write down the minimum energy required to remove the body totally from the Earth’s
gravitational field. [1]
11 The planets in our solar system orbit the Sun in almost circular orbits.
a Show that the orbital speed v of a planet at a distance r from the centre of the Sun is
given by:
GM
v= [4]
r
b The mean distance between the Sun and the Earth is 1.5 × 1011 m and the mass of the Sun
is 2.0 × 1030 kg.
Calculate the orbital speed of the Earth as it travels round the Sun. [2]
12 There is a point between the Earth and the Moon where the net gravitational field strength
is zero. At this point the Earth’s gravitational field strength is equal in magnitude but
opposite in direction to the gravitational field strength of the Moon.
Given that:
mass of Earth
= 81
mass of Moon
calculate how far this point is from the centre of the Moon in terms of R, where R is
the separation between the centres of the Earth and the Moon. [4]
Total: Score: %
57
## AS and A Level Physics Original material © Cambridge University Press 2010 2
20 Worksheet (A2)
1 For an oscillating mass, define:
a the period [1]
b the frequency. [1]
2 The graph of displacement x against time t for an object executing simple harmonic motion
(s.h.m.) is shown here.
a State a time at which the object has maximum speed. Explain your answer. [2]
b State a time at which the magnitude of the object’s acceleration is a maximum.
3 An apple is hung vertically from a length of string to form a simple pendulum.
The apple is pulled to one side and then released. It executes 12 oscillations in a time of 13.2 s.
a Calculate the period of the oscillations. [2]
b Calculate the frequency of the oscillations. [2]
4 This is the graph of displacement x against time t for an oscillating object.
## Use the graph to determine the following:
a the amplitude of the oscillation [1]
b the period [1]
c the frequency in hertz (Hz) [2]
e the maximum speed of the oscillating mass. [2]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
20 Worksheet (A2)
5 Two objects A and B have the same period of oscillation. In each case a and b below,
determine the phase difference between the motions of the objects A and B.
a [2]
b [2]
## 6 A mass at the end of a spring oscillates with a period of 2.8 s.
The maximum displacement of the mass from its equilibrium position is 16 cm.
a What is the amplitude of the oscillations? [1]
b Calculate the angular frequency of the oscillations. [2]
c Determine the maximum acceleration of the mass. [3]
d Determine the maximum speed of the mass. [2]
7 A small toy boat is floating on the water’s surface. It is gently pushed down and then
released. The toy executes simple harmonic motion. Its displacement–time graph
is shown here.
## For this oscillating toy boat, calculate:
a its angular frequency [2]
b its maximum acceleration [3]
c its displacement after a time of 6.7 s, assuming that the effect of damping on the boat is
negligible. [3]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
20 Worksheet (A2)
8 The diagram shows the displacement–time graph for a particle executing simple
harmonic motion.
## Sketch the following graphs for the oscillating particle:
a velocity–time graph [2]
b acceleration–time graph [2]
c kinetic energy–time graph [2]
d potential energy–time graph. [2]
9 A piston in a car engine executes simple harmonic motion.
The acceleration a of the piston is related to its displacement x by the equation:
a = −6.4 × 105x
a Calculate the frequency of the motion. [3]
b The piston has a mass of 700 g and a maximum displacement of 8.0 cm.
Calculate the maximum force on the piston. [2]
10 The diagram shows a trolley of mass m attached
to a spring of force constant k. When the trolley
is displaced to one side and then released, the
trolley executes simple harmonic motion.
## a Show that the acceleration a of the trolley is given by the expression:
⎛k⎞
a = − ⎜ ⎟x
⎝m⎠
where x is the displacement of the trolley from its equilibrium position. [3]
b Use the expression in a to show that the frequency f of the motion is given by:
1 k
f= [2]
2π m
c The springs in a car’s suspension act in a similar way to the springs on the trolley.
For a car of mass 850 kg, the natural frequency of oscillation is 0.40 s.
Determine the force constant k of the car’s suspension. [3]
Total: Score: %
59
## AS and A Level Physics Original material © Cambridge University Press 2010 3
21 Worksheet (A2)
Specific heat capacity of water = 4200 J kg−1 K−1
Specific latent heat of fusion of water = 3.4 × 105 J kg−1
1 Describe the arrangement of atoms, the forces between the atoms and the motion of the
atoms in:
a a solid [3]
b a liquid [3]
c a gas. [3]
2 A small amount of gas is trapped inside a container. Describe the motion of the gas atoms
as the temperature of the gas within the container is increased. [3]
3 a Define the internal energy of a substance. [1]
b The temperature of an aluminium block increases when it is placed in the flame of a
Bunsen burner. Explain why this causes an increase in its internal energy. [3]
## c A lump of metal is melting in a hot oven at a temperature of 600 °C.
Explain whether its internal energy is increasing or decreasing as it melts. [4]
4 Write a word equation for the change in the thermal energy of a substance in terms of its
mass, the specific heat capacity of the substance and its change of temperature. [1]
5 The specific heat capacity of a substance is measured in the units J kg−1 K−1, whereas
its specific latent heat of fusion is measured in J kg−1. Explain why the units are different. [2]
6 During a hot summer’s day, the temperature of 6.0 × 105 kg of water in a swimming pool
increases from 21 °C to 24 °C. Calculate the change in the internal energy of the water. [3]
7 A 300 g block of iron cools from 300 °C to room temperature at 20 °C. The specific heat
capacity of iron is 490 J kg−1 K−1. Calculate the heat released by the block of iron. [3]
8 Calculate the energy that must be removed from 200 g of water at 0 °C to convert it all
into ice at 0 °C. [2]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
21 Worksheet (A2)
## 9 a Change the following temperatures from degrees Celsius into kelvin.
i 0 °C
ii 80 °C
iii −120 °C [3]
b Change the following temperatures from kelvin into degrees Celsius.
i 400 K
ii 272 K
iii 3 K [3]
10 An electrical heater is used to heat 100 g of water in a well-insulated container at a steady
rate. The temperature of the water increases by 15 °C when the heater is operated for a
period of 5.0 minutes. Determine the change of temperature of the water when the same
heater and container are individually used to heat:
a 300 g of water for the same period of time [3]
b 100 g of water for a time of 2.5 minutes. [3]
11 The graph below shows the variation of the temperature of 200 g of lead as it is heated
## a Use the graph to state the melting point of lead. [1]
b Explain why the graph is a straight line at the start. [1]
c Explain what happens to the energy supplied to the lead as it melts at a constant
temperature. [1]
d The initial temperature of the lead is 0 °C. Use the graph to determine the total
energy supplied to the lead before it starts to melt. [3]
(The specific heat capacity of lead is 130 J kg−1 K−1.)
f Assuming that energy continues to be supplied at the same rate, calculate the
specific latent heat of fusion of lead. [3]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
21 Worksheet (A2)
## 12 The diagram shows piped water being heated by an electrical heater.
The water flows through the heater at a rate of 0.015 kg s−1. The heater warms the water
from 15 °C to 42 °C. Assuming that all the energy from the heater is transferred to
heating the water, calculate the power of the heater. [5]
13 A gas is held in a cylinder by a friction-free piston. When the force holding the piston in
place is removed, the gas expands and pushes the piston outwards.
Explain why the temperature of the gas falls. [2]
14 Hot water of mass 300 g and at a temperature of 90 °C is added to 200 g of cold water
at 10 °C. What is the final temperature of the mixture? You may assume there are no
losses to the environment and all heat transfer takes place between the hot water and
the cold water. [5]
15 A metal cube of mass 75 g is heated in a naked flame until it is red hot. The metal block is
quickly transferred to 200 g of cold water. The water is well stirred. The graph shows the
variation of the temperature of the water recorded by a datalogger during the experiment.
The metal has a specific heat capacity of 500 J kg−1 K−1. Use the additional information
provided in the graph to determine the initial temperature of the metal cube. You may
assume there are no losses to the environment and all heat transfer takes place between
the metal block and the water. [5]
Total: Score: %
71
## AS and A Level Physics Original material © Cambridge University Press 2010 3
22 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 Determine the number of atoms or molecules in each of the following.
a 1.0 mole of carbon [1]
b 3.6 moles of water [1]
c 0.26 moles of helium [1]
2 The molar mass of helium is 4.0 g.
Determine the mass of a single atom of helium in kilograms. [2]
3 The molar mass of uranium is 238 g.
a Calculate the mass of one atom of uranium. [2]
b A small rock contains 0.12 g of uranium. For this rock, calculate the number of:
i moles of uranium [2]
ii atoms of uranium. [1]
4 Explain what is meant by the absolute zero of temperature. [3]
5 a Write the ideal gas equation in words. [1]
b One mole of an ideal gas is trapped inside a rigid container of volume 0.020 m3.
Calculate the pressure exerted by the gas when the temperature within the
container is 293 K. [3]
6 A fixed amount of an ideal gas is trapped in a container of volume V.
The pressure exerted by the gas is P and its absolute temperature is T.
a Using a sketch of PV against T, explain how you can determine the number of moles
of gas within the container. [4]
b Sketch a graph of PV against P when the gas is kept at a constant temperature.
Explain the shape of the graph. [3]
7 A rigid cylinder of volume 0.030 m3 holds 4.0 g of air. The molar mass of air is about 29 g.
a Calculate the pressure exerted by the air when its temperature is 34 °C. [4]
b What is the temperature of the gas in degrees Celsius when the pressure is twice
your value from part a? [4]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
22 Worksheet (A2)
## 8 The diagram shows two insulated containers holding gas.
The containers are connected together by tubes of negligible volume.
## The internal volume of each container is 2.0 × 10−2 m3.
The temperature within each container is −13 °C. The gas in container A exerts a pressure
of 180 kPa and the gas in container B exerts a pressure of 300 kPa.
a Show that the amount of gas within the two containers is about 4.4 moles. [3]
b The valve connecting the containers is slowly opened and the gases are allowed to mix.
The temperature within the containers remains the same.
Calculate the new pressure exerted by the gas within the containers. [3]
9 The diagram shows a cylinder containing air
at a temperature of 5.0 °C.
The piston has a cross-sectional area 1.6 × 10−3 m2
It is held stationary by applying a force
of 400 N applied normally to the piston.
The volume occupied by the compressed air
is 2.4 × 10−4 m3.
The molar mass of air is about 29 g.
a Calculate the pressure exerted by the compressed air. [2]
b Determine the number of moles of air inside the cylinder. [3]
i the mass of air inside the cylinder [1]
ii the density of the air inside the cylinder. [2]
10 The mean speed of a helium atom at a temperature of 0 °C is 1.3 km s–1. Estimate the mean
speed of helium atoms on the surface of a star where the temperature is 10 000 K. [6]
11 The surface temperature of the Sun is about 5400 K. On its surface, particles behave like the
atoms of an ideal gas. The atmosphere of the Sun mainly consists of hydrogen nuclei.
These nuclei move in random motion.
a Explain what is meant by random motion. [1]
b i Calculate the mean translational kinetic energy of a hydrogen nucleus
on the surface of the Sun. [2]
ii Estimate the mean speed of such a hydrogen nucleus.
(The mass of hydrogen nucleus is 1.7 × 10−27 kg.) [3]
12 a Calculate the mean translational kinetic energy of gas atoms at 0 °C. [2]
b Estimate the mean speed of carbon dioxide molecules at 0 °C.
(The molar mass of carbon dioxide is 44 g.) [5]
c Calculate the change in the internal energy of one mole of carbon dioxide gas when its
temperature changes from 0 °C to 100 °C. [3]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
22 Worksheet (A2)
13 The diagram below shows three different types of arrangements of gas particles.
A gas whose particles consist of single atoms is referred to as monatomic – for example
helium (He). A gas with two atoms to a molecule is called diatomic – for example oxygen
(O2). A gas with more than two atoms to a molecule is said to be polyatomic – for example
water vapour (H2O).
A single atom can travel independently in the x, y and z directions: it is said to have three
degrees of freedom. From the equation for the mean translational kinetic energy of the atom,
1
we can generalise that a gas particle has mean energy of kT per degree of freedom.
2
Molecules can also have additional degrees of freedom due to their rotational energy.
a Use the diagram above to explain why:
5
i the mean energy of a diatomic molecule is kT [2]
2
ii the mean energy of a polyatomic molecule is 3kT. [2]
b Calculate the internal energy of one mole of water vapour (steam) per unit kelvin. [3]
Total: Score: %
75
## AS and A Level Physics Original material © Cambridge University Press 2010 3
23 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 a Explain what is meant by the electric field strength at a point. [1]
b Explain what is meant by the electric potential at a point. [1]
2 A pair of parallel metal plates has a potential difference of 5000 V across them.
The electric field strength between them is 400 kN C−1. Calculate:
a the separation between the plates [2]
b the force on a dust particle between the plates which carries a charge of 1.6 × 10−19 C. [2]
3 The electric field strength E at a distance r from a point charge Q may be written as:
Q
E=k 2
r
What is the value for k? [1]
4 The diagram shows a point charge +q placed in the electric field of a charge +Q.
The force experienced by the charge +q at point A is F. Calculate the magnitude of the
force experienced by this charge when it is placed at points B, C, D and E. In each case,
5 A spherical metal dome of radius 15 cm is electrically charged. It has a positive charge of
+2.5 µC distributed uniformly on its surface.
a Calculate the electric field strength on the surface of the dome. [3]
b Explain how your answer to a would change at a distance of 30 cm from the surface of
the dome. [2]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
23 Worksheet (A2)
## 6 The diagram shows two point charges.
The point X is midway between the charges.
## a Calculate the electric field strength at point X due to:
i the +20 µC charge [3]
ii the +40 µC charge. [2]
b Calculate the resultant electric field strength at point X. [2]
7 The dome of a van de Graaff generator has a diameter of 30 cm and is at a potential of
+20 000 V. Calculate:
a the charge on the dome [2]
b the electric field strength at the surface of the dome [2]
c the force on a proton near the surface of the dome. [1]
8 a An isolated charged sphere of diameter 10 cm carries a charge of −2000 nC.
Calculate the potential at its surface. [3]
b Calculate the work that must be done to bring an electron from infinity to the
surface of the dome. [2]
9 Describe some of the similarities and differences between the electrical force due to a point
charge and the gravitational force due to a point mass. [6]
10 The diagram shows two point charges.
Calculate the distance x of point P from
charge +Q where the net electric field
strength is zero. [6]
11 Show that the ratio:
electrical force between two protons
gravitational force between two protons
is about 1036 and is independent of the actual separation between the protons. [6]
12 A helium nucleus consists of two protons and two neutrons. Its diameter is about 10−15 m.
a Calculate the force of electrostatic repulsion between two protons at this separation. [2]
b Calculate the potential at a distance of 10–15 m from the centre of a proton. [2]
c How much work would need to be done to bring two protons this close to each other? [2]
d If one proton were stationary, at what speed would the second proton need to be fired
at it to get this close? (Ignore any relativistic effects.) [3]
Total: Score: %
66
## AS and A Level Physics Original material © Cambridge University Press 2010 2
24 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 A 30 µF capacitor is connected to a 9.0 V battery.
a Calculate the charge on the capacitor. [2]
b How many excess electrons are there on the negative plate of the capacitor? [2]
2 The p.d. across a capacitor is 3.0 V and the charge on the capacitor is 150 nC.
a Determine the charge on the capacitor when the p.d. is:
i 6.0 V [2]
ii 9.0 V. [2]
b Calculate the capacitance of the capacitor. [2]
3 A 1000 µF capacitor is charged to a potential difference of 9.0 V.
a Calculate the energy stored by the capacitor. [2]
b Determine the energy stored by the capacitor when the p.d. across it is doubled. [2]
4 For each circuit below, determine the total capacitance of the circuit. [13]
## a Calculate the total capacitance of the two capacitors in parallel. [2]
b What is the potential difference across each capacitor? [1]
c Calculate the total charge stored by the circuit. [2]
d Calculate the total energy stored by the capacitors. [2]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
24 Worksheet (A2)
## 6 A 10 000 µF capacitor is charged to its maximum operating voltage of 32 V.
The charged capacitor is discharged through a filament lamp. The flash of light from
the lamp lasts for 300 ms.
a Calculate the energy stored by the capacitor. [2]
b Determine the average power dissipated in the filament lamp. [2]
7 The diagram shows a 1000 µF capacitor charged to a p.d. of 12 V.
a Calculate the charge on the 1000 µF capacitor. [2]
## b The 1000 µF capacitor is connected across an uncharged 500 µF capacitor by closing
the switch S. The charge initially stored by the 1000 µF capacitor is now shared with
the 500 µF capacitor.
i Calculate the total capacitance of the capacitors in parallel. [2]
ii Show that the p.d. across each capacitor is 8.0 V. [2]
8 The diagram shows a circuit used to measure the capacitance of a capacitor.
reed switch
mA
9.0 V
The reed switch vibrates between the two contacts with a frequency of 50 Hz. On each oscillation
the capacitor is fully charged and totally discharged. The current through the milliammeter is
225 mA.
a Calculate the charge that flows off the capacitor each time it is discharged. [1]
b Calculate the capacitance of the capacitor. [2]
c Calculate the current through the milliammeter when a second identical capacitor
is connected:
i in parallel with the original capacitor [1]
ii in series with the original capacitor. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
24 Worksheet (A2)
## 9 A capacitor of capacitance 200 µF is connected across a 200 V supply.
a Calculate the charge stored on the plates. [1]
b Calculate the energy stored on the capacitor. [1]
The capacitor is now disconnected from the power supply and is connected across a 100 µF
capacitor.
c Calculate the potential difference across the capacitors. [3]
d Calculate the total energy stored on the capacitors. [2]
e Suggest where the energy has been lost. [1]
10 The diagram below shows a charged capacitor of capacitance C. When the switch S is closed,
this capacitor is connected across the uncharged capacitor of capacitance 2C.
Calculate the percentage of energy lost as heat in the resistor and explain why the actual
resistance of the resistor is irrelevant. [7]
Total: Score: %
64
## AS and A Level Physics Original material © Cambridge University Press 2010 3
25 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 The diagram shows the magnetic field pattern for a current into paper
current-carrying straight wire drawn by a student in her notes.
List two errors made by the student. [2]
2 A current-carrying conductor is placed in an external magnetic field. In each case below, use
Fleming’s left-hand rule to predict the direction of the force on the conductor.
a b c
[3]
3 The unit of magnetic flux density is the tesla.
Show that: 1 T = 1 N A−1 m−1 [2]
4 Calculate the force per centimetre length of a straight wire placed at right angles to a uniform
magnetic field of magnetic flux density 0.12 T and carrying a current of 3.5 A. [3]
5 A 4.0 cm long conductor carrying a current of 3.0 A is placed in a uniform magnetic field
of flux density 50 mT. In each of a, b and c below, determine the size of the force acting
on the conductor. [6]
a b c
## AS and A Level Physics Original material © Cambridge University Press 2010 1
25 Worksheet (A2)
6 The diagram shows the rectangular loop PQRS of a simple electric motor placed in a
uniform magnetic field of flux density B.
axis
The current in the loop is I. The lengths PQ and RS are both L and lengths QR and SP are both x.
Show that the torque of the couple acting on the loop for a given current and magnetic flux
density is directly proportional to the area of the loop. [5]
7 The diagram shows a rigid wire AB pivoted at
the point A so that it is free to move in a vertical
plane. The lower end of the wire dips into
mercury. A uniform magnetic field of
6.0 × 10−3 T acts into the paper throughout the
diagram.
a When the current is switched on, the wire
continuously moves up out of the mercury
and then falls back again. Explain this
motion. [4]
b The force on the wire due to the current may
be taken to act at the midpoint of the wire.
When the current is first switched on, the
moment of this force about A is 3.5 × 10−5 N m.
Calculate:
i the force acting on the wire [2]
ii the current in the wire. [2]
8 The coil in the d.c. motor shown in question 6 has a length L = 7.0 × 10−2 m and width
x = 3.0 × 10−2 m. There are 25 turns on the coil and it is placed in a uniform magnetic field of
0.19 T. The coil carries a current of 2.8 A. The coil is free to rotate about an axis midway between
PQ and RS.
a Calculate the force on the longest side of the coil. [2]
b Calculate the maximum torque (moment) exerted on the coil. [2]
c Explain why the force acting on the long side of the coil does not change as the coil rotates
but the torque exerted on the coil varies. [2]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
25 Worksheet (A2)
9 The diagram shows an arrangement that is used to determine the magnetic flux density
between the poles of a magnet.
The magnet is placed on a sensitive top pan balance. A current-carrying wire is placed at right
angles to the magnetic field between the poles of the magnet. The force experienced by the
current-carrying wire is equal but opposite to the force experienced by the magnet. The magnet
is pushed downward when the wire experiences an upward force.
The length of the wire in the magnetic field is 5.0 cm. The balance reading is 102.00 g when
there is no current in the wire. The balance reading increases to 103.14 g when the current in
the wire is 8.2 A.
a Show that the force experienced by the wire is equal to 1.1 × 10−2 N. [1]
b Calculate the magnetic flux density of the magnetic field between the poles of the magnet. [3]
Total: Score: %
39
## AS and A Level Physics Original material © Cambridge University Press 2010 3
26 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 Calculate the force experienced by an oil droplet with a charge of 3.2 × 10−19 C due to a
uniform electric field of strength 5.0 × 105 V m−1. [2]
2 The diagram shows two parallel, horizontal plates separated by a vertical distance of 3.0 cm.
The potential difference between the plates is 600 V.
a Calculate the magnitude and direction of the electric field between the plates. [3]
b Describe the electric field between the plates. [2]
c A charged oil droplet of weight 6.4 × 10−15 N is held stationary between the two plates.
i State whether the charge on the droplet is positive or negative.
ii Determine the charge on the oil droplet. [2]
3 Calculate the force experienced by an electron travelling at a velocity of 4.0 × 106 m s−1
at right angles to a magnetic field of magnetic flux density 0.18 T. [3]
4 The diagram shows an electron moving at a constant speed of 8.0 × 106 m s−1 in a plane
perpendicular to a uniform magnetic field of magnetic flux density 4.0 mT.
a Calculate the force acting on the electron due to the magnetic field. [3]
b What is the centripetal acceleration of the electron? [2]
c Use your answer to b to determine the radius of the circular path described by the electron. [2]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
26 Worksheet (A2)
5 The diagram shows the trajectory of an electron travelling into a region of uniform magnetic
field of flux density 2.0 mT. The electron enters the region of magnetic field at 90°.
a Draw the direction of the force experienced by the electron at points A and B. [1]
b Explain why the electron describes part of a circular path while in the region of the
magnetic field. [1]
c The radius of curvature of the path of the electron in the magnetic field is 5.0 cm.
Calculate the speed v of the electron. [5]
d Explain how your answer to c would change if the electron described a circular path
6 A proton of kinetic energy 15 keV travelling at right angles to a magnetic field describes
a circle of radius of 5.0 cm. The mass of a proton is 1.7 × 10−27 kg.
a Show that the speed of the proton is 1.7 × 106 m s−1. [3]
b For this proton, calculate the centripetal force provided by the magnetic field. [3]
c Determine the magnetic flux density of the magnetic field that keeps the proton moving
in its circular orbit. [3]
d How long does it take for the proton to complete one orbit? [2]
7 The diagram shows a velocity-selector for charged ions. Ions of speed v emerge from the slit.
a The parallel plates have a separation of 2.4 cm and are connected to a 5.0 kV supply.
A magnetic field is applied at right angles to the electric field between the plates such
that the positively charged ions emerge from the slit of the velocity-selector at a speed of
6.0 × 106 m s−1. Calculate the magnetic flux density of the magnetic field. [6]
b Ions from the velocity-selector pass into a mass spectrometer which contains another magnetic
field, of flux density B. The ions all have charge Q but either have mass m1 or mass m2. Show
that the difference in the radius of the two isotopes in the magnetic field is given by:
(m1 − m 2 )v
∆r = [2]
BQ
## AS and A Level Physics Original material © Cambridge University Press 2010 2
26 Worksheet (A2)
## 8 An electron describes a circular orbit in a plane perpendicular to a uniform magnetic field.
a Show that the time T taken by an electron to complete one orbit in the magnetic field is
independent of its speed and its radius, and is given by:
2 πm
T=
Be
where B is the magnetic flux density of the magnetic field, e is the charge on an electron
and m is the mass of an electron. [5]
b Explain in words how a faster electron takes the same time to complete one orbit as a slower
electron. [1]
Total: Score: %
55
## AS and A Level Physics Original material © Cambridge University Press 2010 3
27 Worksheet (A2)
1 A flat coil of N turns and cross-sectional area A is placed in a uniform magnetic field of flux
density B. The plane of the coil is normal to the magnetic field.
a Write an equation for:
i the magnetic flux Φ through the coil [1]
ii the magnetic flux linkage for the coil. [1]
b The diagram shows the coil when the
magnetic field is at an angle θ to the
normal of the plane of the coil.
What is the flux linkage for the coil? [1]
2 A square coil of N turns is placed
in a uniform magnetic field of
magnetic flux density B.
Each side of the coil has length x.
What is the magnetic flux linkage
for this coil? [2]
## 3 The diagram shows a magnet placed
close to a flat circular coil.
a Explain why there is no
induced e.m.f. even though
the coil. [1]
b Explain why there is an induced
e.m.f. when the magnet is
pushed towards the coil. [2]
## 4 A coil of cross-sectional area
4.0 × 10−4 m2 and 70 turns is placed in a uniform magnetic field.
a The plane of the coil is at rightangles to the magnetic
field. Calculate the magnetic flux density when the
flux linkage for the coil is 1.4 × 10−4 Wb. [3]
b The coil is placed in a magnetic field of flux
density 0.50 T. The normal to the coil is at
an angle of 60° to the magnetic field, as
shown in the diagram.
Calculate the flux linkage for the coil. [3]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
27 Worksheet (A2)
## 5 A square coil is placed in a uniform
magnetic field of flux density 40 mT.
The plane of the coil is normal to the
magnetic field. The coil has 200 turns
and the length of each side of the coil
is 3.0 cm.
a Calculate:
i the magnetic flux Φ through the coil [2]
ii the magnetic flux linkage for the coil. [2]
b The plane of the coil is turned through 90°.
What is the change in the magnetic flux
## 6 A flat circular coil of 1200 turns and of mean
radius 8.0 mm is connected to an ammeter of
negligible resistance. The coil has a resistance
of 6.3 Ω. The plane of the coil is placed at right
angles to a magnetic field of flux density 0.15 T
from a solenoid.
The current in the solenoid is switched off.
It takes 20 ms for the magnetic field to decrease
from its maximum value to zero. Calculate:
a the average magnitude of the induced e.m.f.
across the ends of the coil [5]
b the average current measured by the ammeter. [2]
## 7 The diagram shows a straight
wire of length 10 cm moved
at a constant speed of 2.0 m s−1
in a uniform magnetic field of
flux density 0.050 T.
## For a period of 1 second, calculate:
a the distance travelled by the wire [1]
b the area swept by the wire [1]
c the change in the magnetic flux for the wire (or the magnetic flux ‘cut’ by the wire) [2]
d the e.m.f. induced across the ends of the wire using your answer to c [2]
e the e.m.f. induced across the ends of the wire using E = Blv. [1]
8 A circular coil of radius 1.2 cm has 2000 turns. The coil is placed at right angles to a magnetic
field of flux density 60 mT. Calculate the average magnitude of the induced e.m.f. across the
ends of the coil when the direction of the magnetic field is reversed in a time of 30 ms. [5]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
27 Worksheet (A2)
## 9 The diagram shows a
step-up transformer.
The ends AB of the
primary coil are connected to a
1.5 V cell and a switch. The switch
is initially closed and the lamp is off.
The switch is suddenly opened and the
lamp illuminates for a short time.
a Explain why the lamp illuminates only for a short period. [4]
b State one change to the apparatus that would allow the lamp to illuminate normally. [1]
10 A wire of length L is placed in a uniform magnetic field of flux density B.
The wire is moved at a constant velocity v at right angles to the magnetic field.
Use Faraday’s law of electromagnetic induction to show that the induced e.m.f. E across the
ends of the wire is given by E = BLv.
Hence calculate the e.m.f. induced across the ends of a 20 cm long rod rolling along a
horizontal table at a speed of 0.30 m s−1. (The vertical component of the Earth’s magnetic
flux density is about 40 µT.) [8]
11 a State Faraday’s law of electromagnetic induction. [1]
b Lenz’s law expresses an important conservation law. Name this conservation law. [1]
c i Define magnetic flux for a coil placed at right angles to a magnetic field. [1]
ii Determine for which of the two coils X and Y, each placed at right angles to the
magnetic field, is the magnetic flux linkage the greatest. [4]
Total: Score: %
59
## AS and A Level Physics Original material © Cambridge University Press 2010 3
28 Worksheet (A2)
1 An alternating voltage is given by the equation V = V0 sin 2πft where V0 = 5.0 V
and f = 10 Hz.
a Calculate the period of the alternating voltage. [2]
b Calculate the values of t during the first cycle of the voltage (from t = 0) for which
the value of V is:
i 0 [1]
ii +V0 [2]
iii −V0 [1]
iv +Vrms [2]
2 The graph shows how an alternating voltage V and an alternating current I change with time t.
a V and I are in phase with each other. Explain what is meant by in phase. [1]
b Copy the graph and add a waveform to show how the power dissipated varies with t. [3]
3 An electric drill is marked 230 V r.m.s., 690 W.
Calculate:
a i the r.m.s. current in the wire connecting the drill to the mains [2]
ii the peak current in the wire connecting the drill to the mains [1]
iii the peak value of the potential difference across the drill. [1]
b Sketch a graph of the power drawn by the drill over one cycle of the current.
Mark on the graph the values of peak power and average power. [3]
4 The diagram shows the variation of voltage with time across a resistor.
a State and explain whether the current in the
resistor is a.c. or d.c. [2]
+2 V
b Explain why the power dissipated in the
resistor is the same as the power produced
by a steady voltage of 2 V. [1] 0
c For the voltage variation shown, state: t/s
i the peak value [1]
ii the r.m.s. value. [1] −2 V
## 5 Calculate the value of the average power
dissipated in a resistor when the alternating supply
to the resistor has a peak current of 2.5A and a peak voltage of 6.0 V. [2]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
28 Worksheet (A2)
## 6 The diagram shows the trace obtained on the
screen of an oscilloscope connected to a
signal generator. The time-base of the
oscilloscope is set at 20 ms per division and
the Y-gain at 1.5 V per division.
a For the signal generator, calculate:
i the frequency [2] 1 division
ii the r.m.s. voltage. [2]
b The equation of the waveform can be
written in the form V = V0 sin (ωt).
Determine the values of V0 and ω. [2]
7 The diagram shows a step-down
transformer.
The primary coil has 1150 turns
and the secondary coil has 30 turns.
The ends of the secondary coil are
connected to a lamp labelled
‘6.0 V, 24 W’. The ends AB of the
primary coil are connected to an
alternating voltage supply. The
potential difference across the lamp
is 6.0 V.
a Calculate the current in the lamp. [2]
b Calculate the input voltage to the primary coil. [2]
c Calculate the current in the primary coil, assuming the transformer is 100% efficient. [2]
d Calculate the maximum p.d. across the lamp during one cycle of the a.c. [2]
e A student suggests that to avoid the production of heat in the transformer the wires
should be coated in a material that is a poor conductor of heat. Explain why this is
not a sensible suggestion. [1]
8 An electrician uses a transformer to step the 230 V r.m.s. mains voltage down to 115 V r.m.s. The
secondary coil has 500 turns and is connected to a resistor of 5000 Ω.
a Calculate the number of turns on the primary coil. [1]
b Calculate the current in the secondary coil. [1]
c Calculate the current in the primary coil. Assume that the transformer is 100 % efficient. [1]
d The electrician connects cables to the secondary coil that break down when the p.d.
between the wire and earth is larger than 130 V. Explain whether the cables will break
down when the transformer is switched on. [2]
9 A consumer receives 1000 W of power at 100 V r.m.s. through a 5.0 Ω cable.
a Calculate the rate of heat production in the cable. [2]
b Explain why transmitting the same amount of power at a higher voltage produces
less heat dissipation in the cable. [2]
Total: Score: %
50
## AS and A Level Physics Original material © Cambridge University Press 2010 2
29 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 What is a photon? [1]
2 γ-rays from a radioactive material have higher frequency than visible light.
Explain why this means that γ-rays are more harmful. [2]
3 State one piece of evidence that electromagnetic radiation has:
a wave-like properties [1]
b particle-like properties. [1]
4 A light-emitting diode emits red light of wavelength 6.4 × 10−7 m. Calculate:
a the frequency of the red light [2]
b the energy of a photon of red light. [3]
5 Using the terms photons and work function, describe why electrons are emitted from the
surface of a zinc plate when it is illuminated by ultraviolet radiation but not when it is
illuminated by visible light. [3]
6 The figure below shows an electron making a transition between two energy levels and the bright
spectral emission line observed.
## a Explain why electromagnetic radiation is emitted when an electron jumps from
energy level E1 to energy level E2. [2]
b Derive an expression for the frequency f of the radiation emitted. [2]
c State and explain the position of the spectral line when an electron makes a transition
between energy levels E1 and E3. [2]
7 An electron in an atom can occupy four energy levels. With the help of an energy level diagram,
determine the maximum number of spectral emission lines from this atom. [2]
8 Lithium atoms emit red light of wavelength 670 nm. Calculate the difference between the
energy levels responsible for this red light. [3]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
29 Worksheet (A2)
9 The diagram below shows a hot solid, at a temperature of 5000 K, emitting a continuous
spectrum.
## State the type of spectrum observed from:
a position X [1]
b position Y [1]
c position Z. [1]
10 What experimental evidence is there that suggests that electrons behave as waves? [1]
11 The electronvolt is a convenient unit of energy for particles and photons. Define the
electronvolt. [1]
12 An electron is accelerated through a potential difference of 6.0 V. According to a student, this
electron has kinetic energy greater than the energy of a photon of ultraviolet radiation of
wavelength 2.5 × 10−7 m. With the aid of calculations, explain whether or not the student is
correct. [5]
## 13 a Define threshold frequency for a metal. [1]
b The work function of caesium is 1.9 eV. Calculate the threshold frequency. [3]
14 A particular filament lamp of rating 60 W emits 5.0% of this power as visible light.
The average wavelength of visible light is 550 nm. Calculate:
a the average energy of a single photon of visible light [3]
b the number of photons of visible light emitted per second from the lamp. [3]
15 A plate of zinc is illuminated by electromagnetic radiation of wavelength 2.1 × 10−7 m.
The work function of zinc is 4.3 eV. Calculate the maximum kinetic energy
of a photoelectron. [4]
16 Neutrons travelling through matter get diffracted just as electrons do when travelling
through graphite. In order to show diffraction effects, the neutrons need to have a
de Broglie wavelength that is comparable to the spacing between the atoms.
Calculate the speed of a neutron that has a de Broglie wavelength of 2.0 × 10−11 m. [3]
17 A yellow light-emitting diode (LED) is connected to a d.c. power supply. The output voltage
from the supply is slowly increased from zero until the LED just starts to glow. The yellow
light from the LED has a wavelength of about 5.8 × 10−7 m. Estimate the potential difference
across the LED when it just starts to glow. [4]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
29 Worksheet (A2)
## 18 a In an electron-diffraction experiment, electrons are accelerated through a p.d. V.
Show that the de Broglie wavelength λ of an electron is given by:
h
λ=
2meVe
where me is the mass of the electron and e is the elementary charge. [3]
b Calculate the accelerating p.d. V that gives an electron a de Broglie wavelength of
4.0 × 10−11 m. [3]
19 In an experiment on the photoelectric effect, a metal is illuminated by visible light of different
wavelengths. A photoelectron has a maximum kinetic energy of 0.9 eV when red light of
wavelength 640 nm is used. With blue light of wavelength 420 nm, the maximum kinetic energy
of the photoelectron is 1.9 eV. Use this information to calculate an experimental value
for the Planck constant h. [5]
20 The diagram below shows the some of the energy levels for a helium atom.
## a Explain the significance of the energy levels being negative. [1]
b When a helium atom is not excited, the electrons have an energy of −3.00 eV. This is
known as the stable state of the electrons. Calculate the minimum energy, in joules,
required to free an electron at this energy level. Explain your answer. [3]
c The helium atom absorbs a photon of energy 1.41 eV.
i State the transition made by an electron. [2]
ii Calculate the wavelength of the radiation absorbed by the helium atom. [3]
21 The figure below shows the energy level
diagram for an atom of mercury.
a Explain what is meant by the ground
state. [1]
b Calculate the shortest wavelength
emitted by the atom. Explain your
## AS and A Level Physics Original material © Cambridge University Press 2010 3
29 Worksheet (A2)
22 For the hydrogen atom, the energy level En in joules is given by the equation
2.18 ×10 −18
En = −
n2
where n is an integer, known as the principal quantum number.
a Calculate the energy level of the ground state (n = 1) and the energy level of the first
excited state (n = 2). [2]
b Determine the wavelength of radiation emitted when an electron makes a transition from
the first excited state to the ground state. In which region of the electromagnetic spectrum
would you find a spectral line with this wavelength? [4]
c In which region of the electromagnetic spectrum would you find the spectral line
corresponding to an electron transition between energy levels with principal quantum
Total: Score: %
90
## AS and A Level Physics Original material © Cambridge University Press 2010 4
30 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 a Write down Einstein’s famous equation relating mass and energy. [1]
b Determine the change in energy equivalent to a change in mass:
i of 1.0 g [2]
ii equal to that of an electron (9.1 × 10−31 kg). [2]
2 In nuclear physics, it is common practice to quote the mass of a nuclear particle in terms of
the unified atomic mass unit, u. The unified atomic mass unit u is defined as one-twelfth of the
mass of an atom of the carbon isotope 126 C .
a Determine the mass of each of the following particles in terms of u:
i an α-particle of mass 6.65 × 10−27 kg [1]
ii a carbon-13 atom of mass 2.16 × 10−26 kg. [1]
b Determine the mass of each of the following particles in kilograms:
i a proton of mass 1.01 u [1]
ii a uranium-235 nucleus of mass 234.99 u. [1]
3 State three quantities conserved in all nuclear reactions. [3]
4 a Explain why external energy is required to ‘split’ a nucleus. [1]
b Define the binding energy of a nucleus. [1]
c The binding energy of the nuclide 168 O is 128 MeV. Calculate the binding energy
per nucleon. [2]
5 a Define the half-life of a radioactive isotope. [1]
b The half-life of a particular isotope is 20 minutes. A sample initially contains N0 nuclei
of this isotope. Determine the number of nuclei of the isotope left in the sample after:
i 20 minutes [1]
ii 1.0 hour. [2]
6 The activity of an α-source is 540 Bq. The kinetic energy of each α-particle is 8.6 × 10−14 J.
The isotope in the source has a very long half-life.
a Calculate the number of α-particles emitted by the source in:
i 1 second [1]
ii 1 hour. [1]
b Determine the total energy released by the source in a time of 1 second. [3]
c State the rate at which energy is emitted from this α-source. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
30 Worksheet (A2)
7 The binding energy per nucleon against nucleon number graph for some common nuclides is
shown below.
12
b Use the graph to estimate the binding energy for the nucleus of 6C . [2]
c Use the graph to estimate the energy released in the following fusion reaction. [4]
2 2 4
1 H + 1 H → 2 He
d The fusion reaction shown in c is one of the many that occur in the interior of stars.
State the conditions necessary to initiate such reactions in stars. [2]
8 Use the data given below to determine the binding energy and the binding energy per
nucleon of the nuclide 235
92 U . [7]
mass of proton = 1.007 u
mass of neutron = 1.009 u mass of uranium-235 nucleus = 234.992 u
## AS and A Level Physics Original material © Cambridge University Press 2010 2
30 Worksheet (A2)
## 9 a Describe the process of induced nuclear fission. [1]
b The diagram shows the fission of uranium-235 in accordance with the nuclear equation:
235 1 95 139 1
92 U + 0 n → 38 Sr + 54 Xe + 2 0 n
i Copy the diagram, adding labels to identify the neutrons, the strontium nuclide and
the xenon nuclide. [1]
ii Explain why energy is released in the reaction above. [2]
iii Use the following data to determine the energy released in a single fission reaction
involving 235 1
92 U and 0 n . [5]
235 −25 95 −25
mass of 92 U = 3.902 × 10 kg mass of 38 Sr = 1.575 × 10 kg
1 −27 139 −25
mass of 0n = 1.675 × 10 kg mass of 54 Xe = 2.306 × 10 kg
## 10 One of the neutron-induced fission reactions of uranium-235 may be represented by the
following nuclear equations.
235 1 236
92 U + 0 n → 92 U
236
→ 146
92 U
87 1
57 La + 35 Br + 3 0 n
The binding energies per nucleon for these nuclides are:
236 146 87
92 U, 7.59 MeV; 57 La, 8.41 MeV; 35 Br, 8.59 MeV.
236
Calculate the energy released in MeV when the 92 U nucleus undergoes fission. [3]
220
11 The half-life of the radon isotope 86 Rn is 56 s.
−1
a Determine the decay constant in s . [3]
b Calculate the activity of a sample containing 6.0 × 1010 nuclei of 220
86 Rn . [3]
12 The activity of a radioactive source containing 8.0 × 1014 undecayed nuclei is 5.0 × 109 Bq.
a Determine the decay constant in s−1. [3]
b Calculate the half-life of the nucleus. [3]
c How many undecayed nuclei will be left after 40 hours? [3]
13 a Define the decay constant of a nucleus. [1]
b The thorium isotope 227
90 Th has a half-life of 18 days.
A particular radioactive source contains 4.0 × 1012 nuclei of the isotope 227
90 Th .
i Determine the decay constant for the thorium isotope 227
90 Th in s−1. [3]
ii What is the initial activity of the source? [3]
iii Calculate the activity of the source after 36 days. [2]
14 A sample of rock is known to contain 1.0 µg of the radioactive radium isotope 226
88 Ra .
The half-life of this particular isotope is 1600 years. The molar mass of radium-226 is 226 g.
a Determine the number of nuclei of the isotope 226 88 Ra in the rock sample. [2]
b Calculate the activity from decay of the radium-226 in the sample. [3]
## AS and A Level Physics Original material © Cambridge University Press 2010 3
30 Worksheet (A2)
15 In a process referred to as ‘annihilation’, a particle interacts with its antiparticle and the
entire mass of the combined particles is transformed into energy in the form of photons.
The following equation represents the interaction of a proton (p) and its antiparticle, the
antiproton ( p ).
1
1p + −11 p → γ + γ
The antiproton has the same mass as a proton – the only difference is that it has a negative
charge. Determine the wavelength λ of each of the two identical photons emitted in the
reaction above. (Mass of a proton = 1.7 × 10−27 kg.) [5]
16 Does fusion or fission produce more energy per kilogram of fuel? Answer this question by
considering the fusion reaction in 7 c and the fission reaction in 9 b.
(The molar masses of hydrogen-2 and uranium-235 are 2 g and 235 g, respectively.) [7]
17 Some astronomers believe that our solar system was formed 5.0 × 109 years ago.
Assuming that all uranium-238 nuclei were formed before this time, what fraction of the
original uranium-238 nuclei remain in the solar system today? The isotope of uranium
238 9
92 U has a half-life of 4.5 × 10 y. [4]
Total: Score: %
100
## AS and A Level Physics Original material © Cambridge University Press 2010 4
31 Worksheet (A2)
1 The flowchart below shows the components that make up an electronic sensor.
sensing
device
## What are the names of the missing components? [2]
2 A thermistor is an example of a sensing device.
a Sketch the temperature characteristic of a negative temperature coefficient thermistor. [2]
b State the name of one other sensing device. [1]
3 a Describe the structure of a metal-wire strain gauge. [2]
−7
b A strain gauge contains 15 cm of wire of resistivity 5.0 × 10 Ω m. The resistance of the
strain gauge is 150 Ω.
i Calculate the cross-sectional area of the wire in the strain gauge. [2]
ii Calculate the increase in resistance when the wire extends by 0.1 cm, assuming that
the cross-sectional area and resistivity remain constant. [1]
4 a What is meant by negative feedback? [2]
b State two advantages of negative feedback in an operational amplifier. [2]
5 The circuit shown is used to produce a graph of Vout against Vin by moving the slider on the
variable resistor. The supply voltage to the op-amp is + Vs.
31 Worksheet
## The graph obtained is shown.
a State the type of amplifier drawn. Explain how the graph shows that the
amplifier is of this type. [2]
b State why the graph flattens at the ends. [1]
c Suggest the value that was used for the supply voltage Vs. [1]
d State what changes occur to the graph if:
i Rin is halved in value but Rf is kept unchanged [2]
ii Rf is halved in value but Rin is unchanged from the initial value [1]
iii the supply voltage Vs is increased. [1]
e The variable resistor is removed and an a.c. signal of maximum voltage ±1.0 V
is applied to the input of the amplifier circuit. Sketch the output voltage and
input voltage on the same graph. [2]
f Explain why the amplifier circuit produces distortion if an a.c. signal with a maximum
voltage of 3.0 V is applied to the input. [2]
6 In the circuit shown in question 5 the input voltage Vin is 1.0 V and Rin is 2.0 kΩ.
a Explain why the potential at the inverting input (−) is almost zero. [2]
b State the value of the fall in potential across Rin. [1]
c Calculate the current in Rin. [1]
d Explain why the current in Rf is the same as the current in Rin. [1]
e Determine the value of Rf. You will need to use the graph from question 5. [1]
7 An electrical device generates a potential of +1.0 mV at point P.
## a State two properties of an ideal operational amplifier. [2]
b Assuming that the operational amplifier is ideal, calculate:
i the current in the 10 kΩ resistor [2]
ii the potential at R [1]
iii the gain of the amplifier using your answer to b ii [1]
iv the potential difference between R and Q. [1]
31 Worksheet
## a State two differences between an op-amp used as an inverting amplifier and as
a non-inverting amplifier. [2]
b Calculate the gain of the amplifier shown in the circuit. [2]
c Calculate the value of the input voltage Vin. [1]
d Calculate the value of the current in the 40 kΩ resistor. [2]
e Determine the voltage across the 20 kΩ resistor. [1]
9 The circuit shows an op-amp used as a comparator.
## a Explain how the op-amp acts as a comparator. [2]
b State the value of Vout when V − is larger than V +. [1]
c The resistors and the thermistors are all chosen to have the same resistance, as closely as can
be measured.
i Explain why the value of Vout is uncertain. [1]
ii The temperature of thermistor X falls.
Explain what, if anything, happens to V −, V + and Vout. [3]
10 A device is to be placed on the output of the circuit shown in question 9, such that when the output
voltage is +9 V green light is emitted and when the output voltage is −9 V the light emitted is red.
a Draw the circuit diagram of the device. [2]
b Explain how the device works. [2]
Total: Score: %
58
## AS and A Level Physics Original material © Cambridge University Press 2010 3
32 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 State the nature of X-ray radiation. [2]
2 The energy of an X-ray photon is 50 keV.
a Calculate the energy of the photon in joules. [2]
b Calculate the wavelength of the X-rays. [2]
3 One of the interaction mechanisms between X-rays and matter is the photoelectric effect.
Name the two other interaction mechanisms. [2]
4 State one main difference between the images produced by a normal X-ray machine and by
a CAT scan. [1]
5 Briefly explain what is meant by a non-invasive technique. [1]
6 State what is meant by ultrasound. [2]
7 The speed of ultrasound in soft tissue is 1.5 km s−1.
a Calculate the wavelength of ultrasound of frequency 1.8 MHz. [2]
b Use your answer to part a to explain why high-frequency ultrasound is suitable for
medical scans. [1]
8 Define acoustic impedance. [1]
9 The table below shows useful data for biological materials.
## Material Density / kg m−3 Speed of Acoustic impedance Z
ultrasound / m s−1 / 106 kg m−2 s−1
soft tissue 1060 1540 1.63
muscle 1075 1590 1.71
bone ? 4000 6.40
blood 1060 1570 1.66
a Calculate the density of bone. [2]
b Calculate the percentage of intensity of ultrasound reflected at the blood–soft tissue
boundary. (Assume the waves are incident at right angles to the boundary.) [3]
c Explain why it would be difficult to distinguish between blood and soft tissue in an
ultrasound scan. [2]
10 Name the five main components of an MRI scanner. [5]
11 Protons have a precession frequency of 40 MHz in a strong uniform magnetic field.
a Describe what is meant by precession. [1]
b State the frequency of the radio frequency (RF) radiation that will cause the protons to
resonate. [1]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
32 Worksheet (A2)
12 Briefly describe the production of X-rays and explain why an X-ray spectrum may consist
of a continuous spectrum and a line spectrum. [7]
13 The intensity of a collimated X-ray beam is 250 W m−2.
a Define intensity. [1]
b The diameter of the X-ray beam is 4.0 mm. Calculate the power transmitted by the beam. [2]
14 Describe what is meant by a contrast medium and state why it is used in X-ray scans. [2]
15 The potential difference between the cathode and the anode of an X-ray tube is 80 kV.
Calculate the minimum wavelength of the X-rays emitted from this tube. [3]
16 The photoelectric effect is one of the attenuation mechanisms by which X-ray photons
interact with the atoms in the body. Describe some of the characteristics of this mechanism. [3]
17 A collimated X-ray beam is incident on a metal block. The incident intensity of the beam is I0.
a Draw a sketch graph to show the variation with thickness x of the intensity I of the beam. [3]
b Write down an expression for the intensity I in terms of I0 and x.
Explain any other symbol you use. [2]
c The linear absorption coefficient of a beam of 80 keV X-rays is 0.693 mm−1 in copper.
Calculate the thickness of copper necessary to reduce the intensity of the beam to 0.10 I0. [3]
18 a Describe the use of a CAT scanner. [5]
b Compare the image formed in X-ray diagnosis with that produced by a CAT scanner. [3]
19 Outline how ultrasound may be used in medical diagnosis. [5]
20 Explain why, in medical diagnosis using ultrasound, a coupling medium is necessary
between the ultrasound probe and the skin. [6]
21 a When an ultrasound pulse reflects from the front and back edges of a bone, it produces
two peaks on an A-scan. The time interval between these two peaks is 13 µs. The speed
of the ultrasound in bone is 4000 m s−1. Calculate the thickness of the bone. [3]
b Describe how a B-scan differs from an A-scan. [2]
22 a Outline the principles of magnetic resonance. [6]
b Outline, with the aid of a sketch diagram, the use of MRI (magnetic resonance imaging)
to obtain diagnostic information about internal body structures. [10]
23 X-Rays, ultrasound and MRI are all used in medical diagnosis.
State one situation in which each of these techniques is preferred and give reasons, one in
each case, for the choice. [6]
Total: Score: %
104
## AS and A Level Physics Original material © Cambridge University Press 2010 2
33 Worksheet (A2)
1 Two amplitude-modulated radio waves are shown. Each wave has the same carrier frequency and
is carrying an audio signal.
a State and explain one similarity and one difference between the audio signals
that they carry. [4]
b Explain how the graphs show that the carrier frequency is the same. [1]
2 a Describe the difference between amplitude and frequency modulation. [2]
b A carrier wave has a frequency 800 kHz. It is modulated in frequency by an audio signal
of frequency 6 kHz and amplitude 2.0 V.
The frequency deviation of the carrier wave is 30 kHz V−1.
i Determine the maximum frequency shift produced. [1]
ii Determine the minimum frequency of the modulated carrier wave. [1]
iii Describe how the frequency of the carrier wave changes. [1]
c A country intends to start a new broadcasting system. State two reasons why it is more
expensive to set up an FM broadcasting system rather than an AM system. [2]
3 The graph shows the frequency spectrum of an AM radio wave carrying an audio signal
of a single frequency.
35 40 45 Frequency / kHz
## a i State the name of the component with frequency 40 kHz. [1]
ii State the name of the components with frequency 35 and 45 kHz. [1]
iii Determine the frequency of the audio signal. [1]
b i Calculate the time for one complete carrier wave. [1]
ii Calculate the time for one complete wave of the audio signal. [1]
iii Sketch a graph of the variation of the signal with time. On your graph mark the values
obtained in b i and b ii. [3]
c The frequency spectrum shown above is formed from a carrier wave and an audio signal
of one frequency. Draw the frequency spectrum formed at one instant if the audio signal
contains a range of frequencies up to 15 kHz. [2]
## AS and A Level Physics Original material © Cambridge University Press 2010 1
33 Worksheet (A2)
4 Data is often produced as an analogue signal and then converted into digital form
for transmission.
a Explain, with the aid of sketch graphs, the difference between an analogue and
a digital signal. [4]
b Explain the process of sampling in which an analogue signal is turned into
a digital signal. [3]
5 The diagram shows the analogue signal from a microphone.
For transmission, the signal is digitally sampled every 0.5 ms starting at time t = 0 s.
0 to 0.99 mV produces a digital output 0000
1 to 1.99 mV produces a digital output 0001
and so on.
a State the value of the digital out put when t = 0 s and when t = 0.5 ms. [2]
b The digital signal from the ADC is eventually converted back into analogue form.
Draw a sketch diagram showing the final analogue signal produced. [3]
c i Explain how increasing the sampling frequency improves the final analogue signal
produced and suggest a suitable maximum value for the sampling frequency. [3]
ii Telephone systems use 8-bit numbers, rather than 4-bit numbers. Explain why this
improves the final analogue signal produced. [2]
6 A laser provides power input of 6.0 mW into an optic fibre, where the average noise is
2.0 × 10−19 W. Calculate the signal-to-noise ratio in dB. [1]
7 A signal has a power of 1.0 mW and a noise of 0.001 mW.
a What is the signal-to-noise ratio in dB? [1]
b The signal is attenuated by 30 dB and the noise remains constant.
What is the new signal-to noise ratio in dB? [2]
## AS and A Level Physics Original material © Cambridge University Press 2010 2
33 Worksheet (A2)
8 In the modern telephone system, more and more coaxial cable has been replaced for
long-distance transmission of telephone signals by optic fibre.
State and explain two reasons for this change. [4]
9 a State a typical value of wavelength for:
i space waves
ii sky waves. [2]
b Explain why satellite communication is more reliable than a sky wave for long-distance
communication between two points on the Earth’s surface. [2]
10 a Describe the orbit of a geostationary satellite. [3]
b State a typical wavelength for communication between the Earth’s surface and a
geostationary satellite. [1]
c State one advantage and one disadvantage of the use of a geostationary satellite
rather than a satellite in polar orbit for telephone communication. [2]
11 In the original telephone system of 1876, every telephone was connected to every other
telephone by a pair of wires. Today the telephone is used worldwide as the result of the
invention of the exchange and the use of sampling using digital electronics.
Describe how each of these developments has meant that many telephone conversations
can take place at once. [4]
Total: Score: %
61 | 42,732 | 133,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-22 | latest | en | 0.64645 |
https://besttutors.net/urgent-homework-help-29241/ | 1,642,698,625,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302355.97/warc/CC-MAIN-20220120160411-20220120190411-00019.warc.gz | 179,489,986 | 11,522 | # Suppose that I bought 600,000 bu. of wheat from the farmers at \$4.15/bu. I sold 650,000 bu. at Galve
Suppose that I bought 600,000 bu. of wheat from the farmers at \$4.15/bu. I sold 650,000 bu. at Galveston for \$ 5.45/bu. My trans-portation cost to Galveston was \$ 0.35/bu. In addition, I incurred\$ 450,000 in other expenses (cost of goods sold such as labor andelectricity). I also sold \$ 350,000 worth of fertilizers,pesticides, etc. (which represents \$ 275,000 in inventory)purchasing \$ 175,000 in additional inventory this year. I paidmyself a salary of \$ 175,000 throughout the year. During the year Ialso purchased a new front-end loader costing \$ 175,000 (assume aseven year life with straight line depreciation). This load wasplaced into service in April. The interest rate on my mortgage 10percent and the interest rate on my operating loan is 12 percent.Compute my income statement for the year. Also, Compute the yearend balance sheet assuming that I paid \$ 470,590 in principal on mymortgage. Also assume that I paid down my operating loan to \$925,000. Also create the Statement of Cash Flows and Change inOwner’s Equity statements. So, this was the first question: Suppose that I am the owner ofTurner Elevator & Supply Store in Willow, Oklahoma. I havegrain storage capacity of 500,000 bu. that I purchased ve years agoat a cost of \$ 15 million. I also purchased other equipment at acost of \$ 2.5 million. I set up a straight line depreciationschedule for the bins assuming 15 years of remaining life at thetime of purchase. I assigned a seven year life to the other assetsat the time of purchase (I assume that both bins and otherequipment have zero salvage value). Currently I have \$ 175,000 ofcash on hand and accounts receivable of \$ 125,000. In addition, Ihave 100,000 of wheat in the bins that I paid \$ 4.05/bu. for atharvest time, and \$ 95,000 of inventories on hand (fertilizer,seed, etc.) for sale. However, based on my operating expenses, Icompute another \$ 0.65 /bu. cost from operations must be added tomy inventory of wheat on hand. My mortgage payable (for thepurchase of the bins stands) at \$ 7.5 million { the originalmortgage was for ten years and I have made the standard paymentsbased on a xed payment schedule. I also have an operating loan atthe local bank for \$ 1 million. Derive my balance sheet at thebeginning of year 5. What is my equity? But I do not believe this info is related to my secondquestion. . . . | 635 | 2,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.926213 |
https://oeis.org/A330715/internal | 1,696,459,871,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00281.warc.gz | 476,037,876 | 3,113 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A330715 a(1), a(2), a(3) = 1; a(n) = (a(n-1) mod a(n-3)) + a(n-2) + 1. 1
%I #18 May 03 2023 14:32:08
%S 1,1,1,2,2,3,4,4,6,7,10,12,16,19,24,28,34,39,46,52,60,67,76,84,94,103,
%T 114,124,136,147,160,172,186,199,214,228,244,259,276,292,310,327,346,
%U 364,384,403,424,444,466,487,510,532,556,579,604,628,654,679,706,732
%N a(1), a(2), a(3) = 1; a(n) = (a(n-1) mod a(n-3)) + a(n-2) + 1.
%H Matthew Niemiro, <a href="/A330715/b330715.txt">Table of n, a(n) for n = 1..1000</a>
%F a(1), a(2), a(3) = 1; a(n) = (a(n-1) mod a(n-3)) + a(n-2) + 1.
%F Conjectures from _Colin Barker_, Dec 28 2019: (Start)
%F G.f.: x*(1 - x - x^2 + 2*x^3 - x^4 + x^6 - 2*x^7 + 2*x^8) / ((1 - x)^3*(1 + x)).
%F a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>9.
%F a(n) = (99 - 3*(-1)^n - 24*n + 2*n^2) / 8 for n>5.
%F (End)
%t Nest[Append[#, Mod[#[[-1]], #[[-3]] ] + #[[-2]] + 1] &, {1, 1, 1}, 57] (* _Michael De Vlieger_, Dec 27 2019 *)
%o (Python)
%o x = 1
%o y = 1
%o z = 1
%o for i in range(4, 1001):
%o new = z % x + y + 1
%o print(str(i) +" " + str(new))
%o x = y
%o y = z
%o z = new
%K nonn,hear
%O 1,4
%A _Matthew Niemiro_, Dec 27 2019
%E More terms from _Michael De Vlieger_, Dec 27 2019
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Last modified October 4 17:22 EDT 2023. Contains 365887 sequences. (Running on oeis4.) | 767 | 1,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-40 | latest | en | 0.504736 |
https://kr.mathworks.com/matlabcentral/cody/problems/2917-matlab-basics-ii-intervals/solutions/1815235 | 1,585,465,543,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00401.warc.gz | 555,738,833 | 15,646 | Cody
# Problem 2917. Matlab Basics II - Intervals
Solution 1815235
Submitted on 14 May 2019 by Zoltán Nagy
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = 1; b=2; y_correct = [1 1.25 1.5 1.75 2]; assert(isequal(split_to_five(a,b),y_correct))
y = 1.0000 1.2500 1.5000 1.7500 2.0000
2 Pass
a = 20; b=30; y_correct = [20 22.5 25 27.5 30]; assert(isequal(split_to_five(a,b),y_correct))
y = 20.0000 22.5000 25.0000 27.5000 30.0000
3 Pass
a = 30; b=-20; y_correct = [30 17.5 5 -7.5 -20]; assert(isequal(split_to_five(a,b),y_correct))
y = 30.0000 17.5000 5.0000 -7.5000 -20.0000 | 289 | 707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-16 | latest | en | 0.452263 |
http://hackage.haskell.org/package/category-extras-0.53.3/docs/Control-Category-Cartesian-Closed.html | 1,519,306,423,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814105.6/warc/CC-MAIN-20180222120939-20180222140939-00153.warc.gz | 155,709,440 | 2,337 | category-extras-0.53.3: Various modules and constructs inspired by category theory
Portability non-portable (class-associated types) experimental Edward Kmett
Control.Category.Cartesian.Closed
Description
NB: Some rewrite rules are disabled pending resolution of: http://hackage.haskell.org/trac/ghc/ticket/2291
Synopsis
• class (Monoidal hom prod i, Cartesian hom prod i) => CCC hom prod exp i | hom -> prod exp i where
• apply :: hom (prod (exp a b) a) b
• curry :: hom (prod a b) c -> hom a (exp b c)
• uncurry :: hom a (exp b c) -> hom (prod a b) c
• unitCCC :: CCC hom prod exp i => hom a (exp b (prod b a))
• counitCCC :: CCC hom prod exp i => hom (prod b (exp b a)) a
• class (Comonoidal hom sum i, CoCartesian hom sum i) => CoCCC hom sum coexp i | hom -> sum coexp i where
• coapply :: hom b (sum (coexp hom a b) a)
• cocurry :: hom c (sum a b) -> hom (coexp hom b c) a
• uncocurry :: hom (coexp hom b c) a -> hom c (sum a b)
• unitCoCCC :: CoCCC hom sum coexp i => hom a (sum b (coexp hom b a))
• counitCoCCC :: CoCCC hom sum coexp i => hom (coexp hom b (sum b a)) a
Cartesian Closed Category
class (Monoidal hom prod i, Cartesian hom prod i) => CCC hom prod exp i | hom -> prod exp i whereSource
A `CCC` has full-fledged monoidal finite products and exponentials
Methods
apply :: hom (prod (exp a b) a) bSource
curry :: hom (prod a b) c -> hom a (exp b c)Source
uncurry :: hom a (exp b c) -> hom (prod a b) cSource
unitCCC :: CCC hom prod exp i => hom a (exp b (prod b a))Source
counitCCC :: CCC hom prod exp i => hom (prod b (exp b a)) aSource
Co-(Cartesian Closed Category)
class (Comonoidal hom sum i, CoCartesian hom sum i) => CoCCC hom sum coexp i | hom -> sum coexp i whereSource
A Co-CCC has full-fledged comonoidal finite coproducts and coexponentials
Methods
coapply :: hom b (sum (coexp hom a b) a)Source
cocurry :: hom c (sum a b) -> hom (coexp hom b c) aSource
uncocurry :: hom (coexp hom b c) a -> hom c (sum a b)Source
unitCoCCC :: CoCCC hom sum coexp i => hom a (sum b (coexp hom b a))Source
counitCoCCC :: CoCCC hom sum coexp i => hom (coexp hom b (sum b a)) aSource | 677 | 2,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-09 | latest | en | 0.623166 |
https://math.stackexchange.com/questions/2290556/why-do-logarithmic-curves-not-follow-the-general-rules-for-transformations | 1,716,005,269,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057260.44/warc/CC-MAIN-20240518023912-20240518053912-00196.warc.gz | 355,988,888 | 37,291 | Why do logarithmic curves not follow the general rules for transformations?
Take the function $\ln(3 - x)$. By the logic of transformations that I have been taught, the order of transformations goes: $\ln(x)$ to $\ln(-x)$ which reflects the curve across the $y$-axis, then $\ln(-x + 3)$. This additional $+3$ should then push the curve to the left, hence turn the asymptote to $x = -3$. (This is what I show in the black curve.) However, the way the graph is shown in my book is that the transformation actually makes the asymptote $x = 3$, which disagrees with me, although the direction of the graph relative to then $x$-axis stays is the same as mine.
Why? Someone please explain. I asked my teacher, and he said 'yea that's weird', and I am yet to find an answer.
• $\ln(-x+3)=\ln(-(x-3))$
– Dave
May 21, 2017 at 14:41
• This basically implies that you go from ln(x) to ln(x-3), which is shifting the curve to the right by 3, and then we reflect that in the y-axis. Still gives the black curve in my picture, and not the correct red curve. Please point out which part of my logic is wrong. This is how I've been taught. May 21, 2017 at 14:48
• The process you just described does not result in the black curve, but rather the red one. If you transform $\ln(x)$ to $\ln(x-3)$, the curve indeed shifts to the right $3$ units. However, when you know transform $\ln(x-3)$ to $\ln(-(x-3))$, the curve simply reflects across the line $x=3$ (since this line acts like the new $y$ axis for $\ln(x-3)$), and does not create a new asymptote at $x=-3$ (i.e. the reflection of $\ln(x-3)$ to $\ln(-(x-3))$ does not reflect across the line $x=0$).
– Dave
May 21, 2017 at 14:53
• That's weird, I know I'd be taking up your time, but may I please ask that you explain why the line x=3 becomes the 'new y-axis'? May 21, 2017 at 14:56
• What I mean by "like the new $y$ axis" is just that the line $x=3$ is the reflection line for $\ln(x-3)$ to $\ln(-(x-3))$, much like how the $y$ axis (i.e. $x=0$) is the reflection line for $\ln(x)$ to $\ln(-x)$.
– Dave
May 21, 2017 at 14:58
Elaborating on @Dave's comment, here's the sketch of what is going on.
• You know that if you have a function $f(x)$, then $f(-x)$ reflects the function over the $y$-axis.
• You know that if you have a function $f(x)$, then $f(x+3)$ shifts the function by $3$ to the left.
• Starting with $f(x)=\ln(x)$, you look at $f(-x)=\ln(-x)$, which reflects the function over the $y$-axis.
• Now, let's call the new function we're dealing with $g$. In other words, $g(x)=\ln(-x)$. We would like to shift $3$ to the left, so we look at $g(x+3)=\ln(-(x+3))=\ln(-x-3)$.
• If you want to shift to the right by $3$, we could apply the change $g(x-3)=\ln(-(x-3))=\ln(3-x)$, which is what you get above.
The trick is that when you substitute $x+3$, you can't put the "$+3$" in anywhere, you have to replace $x$ by $x+3$ wherever it appears and remember to distribute. Therefore, in your case, the negative in front of the $x$ must be distributed to the $+3$ as well.
@Dave more or less sums it up in the comment above. When you're changing $-x$ into $3-x$, you can also see that as changing $x$ into $x-3$. This should and does move the graph to the right. | 980 | 3,211 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-22 | latest | en | 0.930506 |
https://ask.sagemath.org/answers/16023/revisions/ | 1,611,072,843,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00660.warc.gz | 224,011,857 | 5,897 | # Revision history [back]
It's not clear to me what you want to do. Looking at the first lines of your code: your for-loop ends up with the solutions of x^9 - 1 == 0 ; all other cases are lost.
n=10
for i in range(1,n):
v=solve(x^i - 1,x)
print v
print solve(x^9 - 1,x)
As far as solve returns a list of equations you can run through the list applying the method rhs() (right hand side)
For example:
sols = solve(x^9 - 1,x)
for sol in sols:
print sol.rhs()
Or (if you are not interested in complex numbers ;-) )
sols = solve(x^9 - 1,x)
for sol in sols:
if sol.rhs() in RR:
print sol.rhs()
It's not clear to me what you want to do. Looking at the first lines of your code: your for-loop ends up with the solutions of x^9 - 1 == 0 ; all other cases are lost.lost. (ok, that's wrong, I can see it now after your code was properly formatted)
n=10
for i in range(1,n):
v=solve(x^i - 1,x)
print v
print solve(x^9 - 1,x)
As far as solve returns a list of equations you can run through the list applying the method rhs() (right hand side)
For example:
sols = solve(x^9 - 1,x)
for sol in sols:
print sol.rhs()
Or (if you are not interested in complex numbers ;-) )
sols = solve(x^9 - 1,x)
for sol in sols:
if sol.rhs() in RR:
print sol.rhs() | 382 | 1,251 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-04 | latest | en | 0.869701 |
https://www.tdworld.com/grid-innovations/article/20966555/solve-this-perpetual-motion | 1,725,798,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00297.warc.gz | 969,292,316 | 69,334 | # Solve This: Perpetual Motion
June 27, 2016
Here's a little puzzle to ruminate on during a coffee break. Maybe you'll win a gift card! Give it your best shot in the comment section below. Enter as many times as you want. A \$50 gift card will go to the first right answer. We need your email to send the gift card so make sure you're registered (and you have verified your registration via email) when you leave your comment in the box below. We cannot send you a gift card without completed verification (answer the email that asks you to click through to verify)
Ed’s son, Lil Johnny, came running into the living room all enthused. “Hey dad, the teacher said that there’s no such thing as a perpetual motion machine. But She’s wrong! Here, take your eyes off the TV and put ‘em on this!”
Little Johnny had drawn a picture of a tall metal water tank. It had a hole near the top and one near the bottom. Water was shown gushing out of both.
“Ya see Dad,” said Lil Johnny professorially. (He often spent hours reading the family’s 40-year-old set of encyclopedias) “The water pressure is greater at the bottom of the tank so all ya have to do is connect the top hole and bottom hole through a hose. Because the bottom hole has more pressure the water will flow up the hose and into the top hole. It will go on forever, et voilà, eureka, son-of-a gun! PERPETUAL MOTION!”
Ed had a vague sense that the contraption wouldn’t work. But he didn’t know why.
Can you help?
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# In the diagram above, <PQR is a right angle, and QS is
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In the diagram above, <PQR is a right angle, and QS is [#permalink]
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05 May 2012, 18:37
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Geometry.jpg [ 3.7 KiB | Viewed 7750 times ]
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?
A. 125
B. 145
C. 240
D. 290
E. It cannot be determined
[Reveal] Spoiler: OA
Last edited by Bunuel on 06 May 2012, 01:25, edited 1 time in total.
Edited the question
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### Show Tags
06 May 2012, 01:16
2
KUDOS
is it true?
25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **
from * and ** we have:
pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145
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Posts: 36568
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
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06 May 2012, 01:43
3
KUDOS
Expert's post
3
This post was
BOOKMARKED
Attachment:
Geometry.jpg [ 3.7 KiB | Viewed 7721 times ]
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?
A. 125
B. 145
C. 240
D. 290
E. It cannot be determined
Perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle.
Thus, the perpendicular QS divides right triangle PRQ into two similar triangles PQS and QRS (which are also similar to the big triangle PRQ). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).
So, PS/QS=QS/SR --> QS^2=PS*SR=25*4=100 --> QS=10 --> area of PQR equals to 1/2*PR*QS=1/2*(PS+SR)*QS=1/2*29*10=145.
For more on check Triangles chapter of Math Book: math-triangles-87197.html
Hope it helps.
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
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06 May 2012, 14:58
Bunuel wrote:
Attachment:
Geometry.jpg
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?
A. 125
B. 145
C. 240
D. 290
E. It cannot be determined
Perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle.
Thus, the perpendicular QS divides right triangle PRQ into two similar triangles PQS and QRS (which are also similar to the big triangle PRQ). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).
So, PS/QS=QS/SR --> QS^2=PS*SR=25*4=100 --> QS=10 --> area of PQR equals to 1/2*PR*QS=1/2*(PS+SR)*QS=1/2*29*10=145.
For more on check Triangles chapter of Math Book: math-triangles-87197.html
Hope it helps.
ya that helps , great
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Joined: 17 Oct 2010
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### Show Tags
14 May 2012, 03:23
mehdi2012 wrote:
is it true?
25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **
from * and ** we have:
pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145
This is also a good way to solve this problem if we find it difficult to apply similarity of triangles, though we have to find the squares of some numbers .
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Joined: 19 Feb 2013
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### Show Tags
31 Jul 2013, 06:00
mehdi2012 wrote:
is it true?
25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **
from * and ** we have:
pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145
How do we figure out which side is proportional to which using angles? :S
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Joined: 02 Feb 2012
Posts: 29
GPA: 4
Followers: 0
Kudos [?]: 13 [0], given: 35
Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
### Show Tags
01 Aug 2013, 19:37
Bunuel wrote:
Attachment:
Geometry.jpg
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?
A. 125
B. 145
C. 240
D. 290
E. It cannot be determined
Perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle.
Thus, the perpendicular QS divides right triangle PRQ into two similar triangles PQS and QRS (which are also similar to the big triangle PRQ). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).
So, PS/QS=QS/SR --> QS^2=PS*SR=25*4=100 --> QS=10 --> area of PQR equals to 1/2*PR*QS=1/2*(PS+SR)*QS=1/2*29*10=145.
For more on check Triangles chapter of Math Book: math-triangles-87197.html
Hope it helps.
Corresponding angles are QPS and QRS right.
In that case, it must be PS/QS=SR/QS..
Please let me know where I am wrong...
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
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24 May 2015, 19:27
Why do we write the proportion inversely as PS/QS=QS/QR? I wrote it as QS/SR=QS/PS and then got stuck from there but wondering why we choose to write the proportion that way.
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
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24 May 2015, 20:54
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Expert's post
healthjunkie wrote:
Why do we write the proportion inversely as PS/QS=QS/QR? I wrote it as QS/SR=QS/PS and then got stuck from there but wondering why we choose to write the proportion that way.
What made you think that QS/SR=QS/PS is the correct relation?
I assume you found that the two small triangles are similar to each other. The point is how are they similar to each other? They are similar because they are both similar to the big triangle PQR.
Angle PQR = Angle QSP = angle QSR = 90 degrees
Ange P is common to PQR and PSQ so by AA, triangle PQR is similar to triangle PSQ - note the naming of the triangles. The angles which are equal are placed in corresponding positions. Angle P is common so it is the first vertex of each triangle. Then angle Q = angle S so we have Q and S as second vertices and the leftover as third vertices to get PQR and PSQ.
Similarly, angle R is common to triangle PQR and triangle QSR so by AA, triangle PQR is similar to triangle QSR - the naming of the triangles should be in order to ensure that you get the corresponding sides correctly.
So triangle PQR is similar to triangles PSQ and QSR. Now you know the corresponding sides:
PS/QS (Sides made by first two underlined letters)= SQ/SR (sides made by next two letters) = PQ/QR (sides made by first and third letters)
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
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25 May 2015, 08:35
Ah that makes sense, I was assuming the smaller triangles were similar to eachother as opposed them both being similar because they're similar to the larger triangle. Makes sense now!
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
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02 Jun 2015, 19:32
healthjunkie wrote:
Ah that makes sense, I was assuming the smaller triangles were similar to eachother as opposed them both being similar because they're similar to the larger triangle. Makes sense now!
Hi healthjunkie ,
All 3 triangles are similar. PQR ~ PSQ ~ QSR . U can try to prove it using AAA. Lemme know if u have any doubts on that.
Thanks!
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
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02 Jun 2015, 21:02
1
KUDOS
when we draw perpendicular then the sides of the two triangle are in same ratio
small side of large triangle/ large side of large triangle = small side of small triangle/large side of small triangle
we know PQS is large triangle and QSR is smaller triangle
PQS and QSR share the same height QS and angle QSR = QSP which confirms that the sides must be in same ratio
thus we know PS = 25 and RS = 4
we can write
25/QS = QS/4
100 = QS^2
10 = QS
now we know height = 10 and PR = 25+4 = 29
area of the triangle = 1/2(10)(29) = 145
Its B
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
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11 Apr 2016, 18:09
Joy111 wrote:
Attachment:
Geometry.jpg
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?
A. 125
B. 145
C. 240
D. 290
E. It cannot be determined
right from the start I knew there's something with the similar triangles...
P is the same, and 90 degree angle is the same. it means that Q1 angle is equal to R angle.
and Q2 angle is equal to P angle.
now we have 2 right triangles..similar to each other.
leg 25 and height x for example
leg 4 and height x.
so:
25 corresponds to the x side of the smaller triangle, scale factor thus must be 25/x
x corresponds to the side equal to 4. so scale factor is x/4
of course the scale factor is the same...so we can set these two equal: 25/x = x/4 = cross multiply -> x^2 = 100, x=10.
base=29, height =10. 29x10/2 = 145.
Re: In the diagram above, <PQR is a right angle, and QS is [#permalink] 11 Apr 2016, 18:09
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https://isabelle.in.tum.de/repos/isabelle/comparison/7fb42b97413a/doc-src/TutorialI/Rules/rules.tex | 1,638,495,760,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00517.warc.gz | 406,273,306 | 2,342 | doc-src/TutorialI/Rules/rules.tex
changeset 10887 7fb42b97413a parent 10854 d1ff1ff5c5ad child 10967 69937e62a28e
equal inserted replaced
10886:f6b16554720d 10887:7fb42b97413a
920 \emph{Hint}: the proof is similar
920 \emph{Hint}: the proof is similar
921 to the one just above for the universal quantifier.
921 to the one just above for the universal quantifier.
922 \end{exercise}
922 \end{exercise}
923
923
924
924
925 \section{Hilbert's $\epsilon$-Operator}
925 \section{Hilbert's Epsilon-Operator}
926
926
927 Isabelle/HOL provides Hilbert's
927 Isabelle/HOL provides Hilbert's
928 $\epsilon$-operator. The term $\epsilon x. P(x)$ denotes some $x$ such that $P(x)$ is
928 $\epsilon$-operator. The term $\epsilon x. P(x)$ denotes some $x$ such that $P(x)$ is
929 true, provided such a value exists. Using this operator, we can express an
929 true, provided such a value exists. Using this operator, we can express an
930 existential destruction rule:
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# Connect Four simple game version (Python)
This is my first post and my first personal project on Python. I created a simple version of Connect Four game to practice some of the concepts I learnt these weeks with the Computer Science path in Codecademy:
``````import numpy as np
class player:
def __init__(self, name, symbol):
self.name = name
self.symbol = symbol
def droppiece(playeer):
err = True
while err == True:
position = int(input("{}: Drop your piece in the desired position: 1 2 3 4 5 6 7 ".format(playeer.name))) - 1
try:
row_position = np.where(table[:, position] == "-")[0][-1]
err = False
except: #Computer throws an exception if the position is not valid:
print("Column full or invalid position! Program will end because of the error. Please choose another position.")
err = True
table[row_position, position] = "{}".format(playeer.symbol)
def showtable(table, rows_table):
for x in range(rows_table):
print(" ".join(table[x]))
def chechwin(table, rows_table, columns_table):
for x in range(rows_table):
for y in range(columns_table-3):
if table[x,y] != "-" and table[x,y] == table[x,y+1] == table[x,y+2] == table[x,y+3]: return True
for y in range(columns_table):
for x in range(rows_table-3):
if table[x,y] != "-" and table[x,y] == table[x+1,y] == table[x+2,y] == table[x+3,y]: return True
for x in range(rows_table-3):
for y in range(columns_table-3):
if table[x,y] != "-" and table[x,y] == table[x+1,y+1] == table[x+2,y+2] == table[x+3,y+3]: return True
for x in range(rows_table-3):
for y in range(3,columns_table):
if table[x,y] != "-" and table[x,y] == table[x+1,y-1] == table[x+2,y-2] == table[x+3,y-3]: return True
return False
#Set up the table and players for the game:
table = np.array([["-", "-", "-", "-", "-", "-", "-"]] * 6)
rows_table = np.size(table, axis = 0)
columns_table = np.size(table, axis = 1)
player1 = player("Player 1", "x")
player2 = player("Player 2", "o")
print("Welcome to Connect Four! Let's start the game...")
while True: #Loops game until someone wins
#First player drops their piece:
droppiece(player1)
showtable(table, rows_table)
if chechwin(table, rows_table, columns_table) == True:
print("Game ended! Congratulations to {}.".format(player1.name))
sys.exit()
#Second player drops their piece:
droppiece(player2)
showtable(table, rows_table)
if chechwin(table, rows_table, columns_table) == True:
print("Game ended! Congratulations to {}.".format(player2.name))
sys.exit()
if len(np.where(table == "-")[0]) == 0:
print("Game ended! It's a draw.")
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Category: Basic Electrical Engineering
24) Conjunction x ^ y behaves on digits 0 and 1 exactly as ____ does for ordinary algebra.
25 / 30
Category: Electrical Machines
25) Radio frequency chokes are air cored to
26 / 30
Category: Electroncis
26) SCR (Silicon Controlled Rectifier) goes into saturation, when gate-cathode junction is
27 / 30
Category: Electrical Machines
27) Fundamental property used in single node pair circuit analyzer is that ______ across all elements is same.
28 / 30
Category: Power Transmission and Distributions
28) Which of the following conditions relate line resistance ‘R’ and line reactance ‘X’ for maximum steady state power transmission on a transmission line?
29 / 30
Category: Basic Electrical Engineering
29) Fourier series are infinite series of elementary trigonometric functions i.e. Sine and
30 / 30
Category: Power System Analysis
30) Power in AC circuit is found by | 969 | 4,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.874147 |
https://www.careercup.com/question?id=2971 | 1,660,767,737,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00237.warc.gz | 605,890,104 | 33,751 | ## Microsoft Yahoo Interview Question for Software Engineer / Developers
• 3
Comment hidden because of low score. Click to expand.
0
of 0 vote
void printPaths(struct node* node) {
int path[1000];
printPathsRecur(node, path, 0);
}
/*
Recursive helper function -- given a node, and an array containing
the path from the root node up to but not including this node,
print out all the root-leaf paths.
*/
void printPathsRecur(struct node* node, int path[], int pathLen) {
if (node==NULL) return;
// append this node to the path array
path[pathLen] = node->data;
pathLen++;
// it's a leaf, so print the path that led to here
if (node->left==NULL && node->right==NULL) {
printArray(path, pathLen);
}
else {
// otherwise try both subtrees
printPathsRecur(node->left, path, pathLen);
printPathsRecur(node->right, path, pathLen);
}
}
// Utility that prints out an array on a line.
void printArray(int ints[], int len) {
int i;
for (i=0; i<len; i++) {
printf("%d ", ints[i]);
}
printf("\n");
}
Comment hidden because of low score. Click to expand.
0
Hi dingxiang,
The solution you gave just print out all the paths from root to leaf. You should keep track of the sum and check it before you print out the path.
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0
How about a solution that explores the whole search space?
For each node in tree:
Explore(List<Node>, remaining_sum)
List<Node> stores path so far
Explore (List<Node>, remaining_sum)
Current_node = last_node_in_list
If coming from parent (i.e., last node in List is child of second-last node in List)
if (current_node->left_child exists)
remaining_sum -= value(current_node->left_child)
// check_and_print
if (remaining_sum == 0)
print List
return
if (remaining_sum < 0)
return
Explore (List<Node>, remaining_sum)
// do similarly for right child
// do similarly if coming from bottom-left and bottom right
If coming from bottom-left
Explore (parent,remaining_sum)
Explore (right_child, remaining_sum)
If coming from bottom-right
Explore (parent,remaining_sum)
Explore (left_child, remaining_sum)
Comment hidden because of low score. Click to expand.
0
With these types of questions, paths in binary-trees, there is almost always some confusion about what is meant by a 'path' in a binary tree. A path in a binary-tree is the sequence of edges between any two of its nodes, not just the path from the root to the leaves, which is just the height. By definition a tree is an undirected graph with 'simple paths' between any two of its nodes. A 'simple path' is a path which is a sequence of edges that doesn't use the same edge more than once. See ‘Discrete Mathematic and its Applications’ by Rosen pp 528-529.
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0
of 0 vote
I forgot to put
``and``
and so indentation is bad. I'm retrying again:
``````How about a solution that explores the whole search space?
For each node in tree: Explore(List<Node>, remaining_sum)
List<Node> stores path so far
Explore (List<Node>, remaining_sum)
Current_node = last_node_in_list
If coming from parent (i.e., last node in List is child of second-last node in List)
if (current_node->left_child exists)
remaining_sum -= value(current_node->left_child)
// check_and_print
if (remaining_sum == 0)
print List
return
if (remaining_sum < 0)
return
Explore (List<Node>, remaining_sum)
// do similalrly for right child
// do similarly if coming from bottom-left and bottom right
If coming from bottom-left
Explore (parent,remaining_sum)
Explore (right_child, remaining_sum)
If coming from bottom-right
Explore (parent,remaining_sum)
Explore (left_child, remaining_sum)``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
I can change the solution to including the sum, but still my solution will be from the path. Any hints?
void printPaths(struct node* node) {
int path[1000];
printPathsRecur(node, path, 0, sum);
}
void printPathsRecur(struct node* node, int path[], int pathLen, int sum) {
if (node==NULL) return;
// append this node to the path array
path[pathLen] = node->data;
pathLen++;
sum=sum - node->data;
if (sum==0) {
printArray(path, pathLen);
}
else {
printPathsRecur(node->left, path, pathLen, sum-node->left.data);
printPathsRecur(node->right, path, pathLen, sum-node->right.data);
}
}
// Utility that prints out an array on a line.
void printArray(int ints[], int len) {
int i;
for (i=0; i<len; i++) {
printf("%d ", ints[i]);
}
printf("\n");
}
Comment hidden because of low score. Click to expand.
0
You need two more recursive calls to printPathsRecur where you don't subtract the sum. Something like this.
else {
printPathsRecur(node->left, path, pathLen, sum-node->left.data);
printPathsRecur(node->right, path, pathLen, sum-node->right.data);
printPathsRecur(node->left, path, pathLen, ORIGINAL_SUM);
printPathsRecur(node->right, path, pathLen, ORIGINAL_SUM);
}
ORIGINAL_SUM is the sum that was passed in at the beginning.
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0
last two recursive calls
printPathsRecur(node->left, path, pathLen, ORIGINAL_SUM);
printPathsRecur(node->right, path, pathLen, ORIGINAL_SUM);
will be called many(no of ancester+1) times for the same node
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0
of 0 vote
void findsum(node* head, int sum, int* buffer, int level)
{
return;
int temp = sum;
for(int i = level; i>=0;i--)
{
temp = temp - buffer[i];
if(temp==0)
print(buffer, level, i);
}
}
void print(int* buf, int start, int end)
{
for(int i = start;i<=end;i++)
{
printf(buffer[i]);
}
}
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0
theres a type o , it should be level+1 and not left +1;
{
}
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0
Also while calling findsum on head->left and head->right you cannot use the same buffer! for example consider level 2. buffer[1](consider this represents data in level 2) will have different values for the left node and the right node. Hence you should create a new array with the same copies and call the recursion.
But this uses a lot of memory!
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0
of 0 vote
From root node, DFS and calculate the sum value of path from root node to any left nodes and right nodes in tree.
Also, you can know the sum value of path from any node to its left children and right children.
To each node, calculate the sum of a left child path value and a right child path value, You can get the answer.
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0
of 0 vote
void printPaths(struct node* node) {
int path[1000];
printPathsRecur(node, path, 0, sum);
}
void printPathsRecur(struct node* node, int path[], int pathLen, int sum) {
if (node==NULL) return;
// append this node to the path array
path[pathLen] = node->data;
pathLen++;
sum=sum - node->data;
if (sum==0) {
printArray(path, pathLen);
}
else {
printPathsRecur(node->left, path, pathLen, sum);
printPathsRecur(node->right, path, pathLen, sum);
}
}
// Utility that prints out an array on a line.
void printArray(int ints[], int len) {
int i;
for (i=0; i<len; i++) {
printf("%d ", ints[i]);
}
printf("\n");
}
Comment hidden because of low score. Click to expand.
0
of 0 vote
hi thr... i'm giving this algorithm.... havnt checked it but i think i'll work...
It doesnt need any array to store the path or something...
It assumes the following structure...
``````struct node
{
struct node *left,*right,*parent;
int data;
};
typedef struct node *Node;
void PrintAllPaths(Node root,int currentSum, int totalSum)
{
if(root==NULL)return;
if(root->data==currentSum)
{
printPath(root,sum); //print the currently found path separately
}
else
{
printAllPaths(root->left, currentSum-root->data ,totalSum);
printAllPaths(root->left, currentSum-root->data ,totalSum);
}
/*But to cover the entire possibilities u must not return from here otherwise the solution paths starting from root->left or root->right will remain unexplored.
*/
printAllPaths(root->left, totalSum ,totalSum);
printAllPaths(root->right, totalSum ,totalSum);
return;
}
void printPath(Node root,int sum)
{
if(root->key==sum)
printf(" %d ",root->key);
else
printPath(root,sum-root->key);
printf(" %d ",root->key);
return;
}``````
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0
it will not print the full path as root will be pointing to the last node which is making the sum....
and even if u are taking the main root of the tree then also it will only print the nodes which are starting from root ...
ur solution is little bit wrong ..
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0
of 0 vote
hi thr... i'm giving this algorithm.... havnt checked it but i think i'll work...
It doesnt need any array to store the path or something...
It assumes the following structure...
``````struct node
{
struct node *left,*right,*parent;
int data;
};
typedef struct node *Node;
void PrintAllPaths(Node root,int currentSum, int totalSum)
{
if(root==NULL)return;
if(root->data==currentSum)
{
printPath(root,sum); //print the currently found path separately
}
else
{
printAllPaths(root->left, currentSum-root->data ,totalSum);
printAllPaths(root->left, currentSum-root->data ,totalSum);
}
/*But to cover the entire possibilities u must not return from here otherwise the solution paths starting from root->left or root->right will remain unexplored.
*/
printAllPaths(root->left, totalSum ,totalSum);
printAllPaths(root->right, totalSum ,totalSum);
return;
}
void printPath(Node root,int sum)
{
if(root->key==sum)
printf(" %d ",root->key);
else
printPath(root,sum-root->key);
printf(" %d ",root->key);
return;
}``````
Comment hidden because of low score. Click to expand.
0
Hi @mit,
I think your code is correct except the print routine. It should be :
void printPath(Node root,int sum)
``````if(root->key==sum)
printf(" %d ",root->key);
else
printPath(root->parent,sum-root->key);
printf(" %d ",root->key);
return;``````
Comment hidden because of low score. Click to expand.
0
So I think this could work but the complexity of this little high...
say there are n nodes in the tree and log(n) is the height of the tree...then the leaf node would be checked log(n) times, the one above it for log(n)-1 times and so on...it could tend towards a log(n)!...while for large values of n, will perform badly.
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0
of 0 vote
``````// This code tries to avoid doing the recursion twice...
void printPathSum(struct node* node, int totalSum) {
int path[1000];
printPathsRecur(node, path, 0, totalSum, 0, 0);
}
void printPathSumRecur(struct node* node, int path[], int pathLen, int totalSum, int curSum, int sumIdx0) {
if (node==NULL) return;
// append this node to the path array
path[pathLen] = node->data;
pathLen++;
if(curSum==totalSum)
printpath(path, sumIdx0, pathLen);
while(curSum>=totalSum)curSum-=path[sumIdx0++];
// otherwise try both subtrees
printPathSumRecur(node->left, path, pathLen, totalSum, curSum+node->data, sumIdx0);
printPathSumRecur(node->right, path, pathLen,totalSum, curSum+node->data, sumIdx0);
}
// Utility that prints out an array on a line.
void printArray(int ints[], int initidx, int len) {
int i;
for (i=initidx; i<len; i++) {
printf("%d ", ints[i]);
}
printf("\n");
}``````
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0
of 0 vote
void PrintSumPaths(Node* n, int sum)
{
if(NULL == n)
return;
if(n->_left == NULL && n->_right == NULL && n->_data && sum == n->_data)
{
printf("Found the path\n");
}
if(n->_left)
{
PrintSumPaths(n->_left, sum);
PrintSumPaths(n->_left, sum - n->_data);
}
if(n->_right)
{
PrintSumPaths(n->_right, sum);
PrintSumPaths(n->_right, sum - n->_data);
}
}
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````public void ExploreAllPaths(Tree root, int[] path, int pathLength, int remainingSum)
{
if (root == null)
return;
remainingSum -= root.Data;
if (remainingSum >= 0)
{
path[pathLength] = root.Data;
pathLength++;
if (remainingSum > 0)
{
ExploreAllPaths(root.Left, path, pathLength, remainingSum);
ExploreAllPaths(root.Right, path, pathLength, remainingSum);
}
else
{
PrintPath(path, pathLength);
}
}
ExploreAllPaths(root.Left, path, 0, Sum);
ExploreAllPaths(root.Right, path, 0, Sum);
}``````
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0
of 0 vote
If the paths can be from a node in left sub tree to any other node in right subtree lower down in tree excluding the root.. then convert the tree to a graph and then perform DF / BF traversal
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0
of 0 vote
For dingxiang/Khoa soln:
I think in recursive call just pass "sum" instead of sum-node.data as we have already done sum=sum-node.data
Comment hidden because of low score. Click to expand.
0
of 0 vote
All solutions above are wrong....
Check with a sample tree:
1
2 3
4 5 6 7
If the desired sum was 8. then all above C codes print 1,2,4,5..Which is incorrect.
@mit's code does not even do that.. it will not print anything...
So the suggestion is whenever the condition root->null is true. It must return level-2.
This level - 2 + 1 will be the array index for the next element on right sub-tree. Thus it will now overwrite
number 4. Which is perfect. Now the nodes in the array will be: 1,2,5 making sum as 8.
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0
your input is even not a bst
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0
True we are talking about BST rember the invariant
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0
of 0 vote
guys, could someone explain the complexity(time and space) of above code snippets ... i cant think of a solution better than O(n (logn)^2)..!
btw.. wats the order of finding a subsequence of given sum in an array of integers (+ve and -ve) ? is there a soln better than n^2 ?
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0
i can think of an O(n) soln but the space complexity is O(n^2). Here's how- keep adding every new element to a new list as well as to existing list. so for example you traverse 1,2,3 you'll have 3 lists:
1 2 3
2 3
3
also keep a count of the running sum of each list. if it equals or goes beyond your target stop adding to that list.
as for finding a subsequence of given sum in an array of integers (+ve and -ve), you can repeat the above algorithm in that case too.
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0
of 0 vote
Can't DFS solve this problem right away?
What's wrong with using DFS to solve this problem ?
Time complexity for DFS is
O( | V | + | E | ) = O(b^d)
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0
no in dfs it will start from root only(BST) so it will not give the cases when the path is via root and not originating from it or even not including it
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0
of 0 vote
Hi all,
if we can put traverse the binary tree into an array, we can solve it a lil bit faster.
traversal of a tree is o(n) into an array
printing path and summing them is o(n) , i believe that we can make good use of some of the properties of a binary tree 2^d - k and the range of k is the number of nodes @ level d.
and the algo can stop if each and every node's been visited at least once.
HOWEVER
i think the path and the sum can be returned during the traversal phrase
but the array makes printing a random path possible.
Chow Yun Fat.
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0
of 0 vote
Solution:
say the number is n.
Do an in-order traveral of the tree until we reach a node that is >= n-1, during in-order traversal, add value of each node into an array [maxlen = n].
Now start a while (true) loop.
start from array[0] and calculating the sum till sum = n. If during the loop, sum > n, reset sum = 0 and re-start summing from index 1 of the array... till all the elements of the array have been examined. Since we have to find all the paths, when sum = n, print out the elements of the array from starting index to the point where sum was found.
Key point is that paths in a tree have to be made of consecutive nodes and secondly in this case an in-order traveral will give us the nodes in sorted order.
too lazy to give pseudo code :p
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0
of 0 vote
I cogitate this works
Run a BFS on the BST and see if the sum is <=0 if so then check all the paths in that sub tree
and keep on reducing the sum and if at any point the sum is zero print it..
``````void callPath(node *node,int sum){
int path[1000];
struct node* temp;
Queue q=new Queue();
if(node==NULL) return;
else
q.enqueue(node);
while(!q.isempty(){
temp=q.dequeue();
if(temp->data<=sum)
hasPath(temp,sum,path,0);
if(temp->left)
q.enqueue(temp->left);
if(temp->right)
q.enqueue(temp->right);
}
}
void hasPath(node* node,int sum,int[] path,int pathLen){
if(node==null) return;
else{
path[pathLen]=node->data;
pathLen++;
int sub=sum-node->data;
if(sub==0)
print(path,pathLen);
hasPath(node->left,sub,path,pathLen);
hasPath(node->right,sub,path,pathLen);
}
}
void print(int[] path,int pathLen){
for(int i=0;i<pathLen;i++)
print(pathLen[i]);
}``````
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0
of 0 vote
Sorry I was not clear in the previous post..
Run a BFS on the BST and see if the sum is <= to the current node value if so check all the paths from the nodes
and recursively reduce the sum and at any point if we see that sum is zero print the path...
``````void callPath(node *node,int sum){
int path[1000];
struct node* temp;
Queue q=new Queue();
if(node==NULL) return;
else
q.enqueue(node);
while(!q.isempty(){
temp=q.dequeue();
if(temp->data<=sum)
hasPath(temp,sum,path,0);
if(temp->left)
q.enqueue(temp->left);
if(temp->right)
q.enqueue(temp->right);
}
}
void hasPath(node* node,int sum,int[] path,int pathLen){
if(node==null) return;
else{
path[pathLen]=node->data;
pathLen++;
int sub=sum-node->data;
if(sub==0)
print(path,pathLen);
else{
hasPath(node->left,sub,path,pathLen);
hasPath(node->right,sub,path,pathLen);
}
}
}
void print(int[] path,int pathLen){
for(int i=0;i<pathLen;i++)
print(pathLen[i]);
}``````
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0
of 0 vote
My idea:
1. First of all, consider the core of this problem: it is asking for a continuous subset sum in an array, e.g.
0 1 2 3 4 5 6
8 1 5 2 4 3 7
we should be able to finish finding the subset in O(n) time. This can be done with augmentation every a[i] with a sum from a[0] to a[i], plus a hashtable containing all these sums. When we come to a[i], we can immediately know whether and where is the starting point of such a subset.
2. Then we can augment the tree nodes with the sum from the root to the current node. And then, perform DFS on the tree, and build the hashtable of the current path up to the current depth with O(1) per recursion. With the augmentation and the dynamic hashtable, this is essentially the same as 1. The whole algorithm should be able to finish in O(n).
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0
of 0 vote
Guys,
First of all can there be more than one such path assuming the sum input is greater than the largest value in the tree (If it is lesser there is the trivial solution of the element itself, if it exists). I am guessing there is only one such path because of the nature of the BST. In which case it is possible to find that path in O(n) time (depth-first maybe).
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0
of 0 vote
Hi Eric,
Consider a BST:
10
5 12
2 8 11 15
And the request is for sum = 15.
there are 2 paths: 10 - 5, 15.
I think your doubt should be cleared. Moreover, the problem does not indicate any condition on the sum requested, So it would complicate if we were to assume that it is greater, equal or smaller than the maximum element.
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0
of 0 vote
This is my solution
``````public List<Path> findPaths(int value)
{
List <Path> candidatePaths = findPaths(root, value);
List <Path> paths = new ArrayList<Path>();
for(Path candidatePath:candidatePaths)
{
if(candidatePath.isComplete())
}
return paths;
}
public List <Path> findPaths(Node node, int value)
{
// this returns completed paths of value or incomplete paths of lesser value
if(node == null)
return null;
List <Path> paths = new ArrayList<Path>();
List <Path> leftPaths = null;
List <Path> rightPaths = null;
Path thisPath = new Path(node);
if(node.getValue() == value)
thisPath.setCompleted(true);
if(node.getLeft() != null)
{
leftPaths = findPaths(node.getLeft(), value);
}
if(node.getRight() != null)
{
rightPaths = findPaths(node.getRight(), value);
}
if(leftPaths != null)
{
for(Path path:leftPaths)
{
path = (Path)path.clone();
if(path.isComplete())
{
continue;
}
path.insertRight(node);
if(path.getCount() == value)
path.setCompleted(true);
}
}
if(rightPaths != null)
{
for(Path path:rightPaths)
{
path = (Path)path.clone();
if(path.isComplete())
{
continue;
}
path.insertLeft(node);
if(path.getCount() == value)
path.setCompleted(true);
}
}
if(leftPaths != null && rightPaths != null)
{
for(Path path:leftPaths)
{
if(path.isComplete() != true)
{
for(Path rightPath:rightPaths)
{
if(rightPath.isComplete() != true)
{
if(path.getCount() + node.getValue() + rightPath.getCount() == value)
{
Path newPath = (Path)path.clone();
newPath.insertRight(node);
newPath = newPath.mergeRight(rightPath);
newPath.setCompleted(true);
}
}
}
}
}
}
return paths;
}
class Path implements Cloneable
{
private List <Node> nodeList;
private int count = 0;
boolean completed = false;
public Path()
{
nodeList = new ArrayList<Node>();
}
public Path(Node node)
{
nodeList = new ArrayList<Node>();
count += node.getValue();
}
public boolean isComplete()
{
return completed;
}
public void setCompleted(boolean completed)
{
this.completed = completed;
}
public Node getLeftNode()
{
if(nodeList.size() > 0)
return nodeList.get(0);
else
return null;
}
public Node getRightNode()
{
if(nodeList.size() > 0)
return nodeList.get(nodeList.size() - 1);
else
return null;
}
public void insertLeft(Node node)
{
count += node.getValue();
}
public void insertRight(Node node)
{
count += node.getValue();
}
public int getCount()
{
return this.count;
}
public Object clone()
{
Path clone = new Path();
//Collections.copy(clone.nodeList, this.nodeList);
clone.count = this.count;
clone.setCompleted(this.completed);
return clone;
}
public Path mergeRight(Path rightPath)
{
Path newPath = (Path)this.clone();
newPath.count += rightPath.getCount();
return newPath;
}
}``````
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0
of 0 vote
I believe this is the correct solution:
(it is not my credit, I just copy and paste from techinterview discussion.)
If you must start at the root, then the following will work (but can probably be improved upon):
Traverse your tree, changing it so that the value at each node changes to the value of it's parent node + it's original value. Thus the values are in this tree are the sum of the values in the unchanged tree from the root to the given point. If the value you currently have is the one you want, then add the path to this node to your list.
If you don't have to start at the root, then traverse your tree as before, but at each node replace its value with a list containing it's original value (head of the list), and its's original value summed with each element of it's parent's list (tail of list). Thus each node has a list of numbers, indicating the sum to that node starting from different ancestors, the number of ancestors required being the position in the list. If any of these values turn out to be the one you want, then store it in the list.
NB The above methods take no advantage of it being a binary search tree. A simple improvement is found by knowing that if your current value is too high and current node is non-negative, then you only need to search the left-hand branch. Similarly, if your current value is too low and the current node is non-positive, only search the right-hand tree. If you know in advance that the numbers are all non-negative or all non-positive, then you can stop searching altogther when you reach a sum that is too high/too low. (Stop at a value that is too extreme in the second algorithm).
If the paths must stop at a leaf, traverse the tree as above, but only store paths that sum to the correct value if they are a leaf node. Speed-ups still apply.
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0
of 0 vote
Queue q;
void PrintPath(node *root,int sum)
{
if(root == NULL || sum < 0)
return;
CheckPathPresent(root,sum);
PrintPath(root->right,sum);
PrintPath(root->left,sum);
}
void CheckPathPathPresent(node *root,int sum)
{
if(root == NULL || sum < 0)
return;
int Diff = sum - root->Data;
q.Push(root->Data);
if(Diff == 0)
{
q.PrintQueue();
}
else
{
CheckPathPresent(root->left,Dif);
CheckPathPresent(root->right,Dif);
}
q.Pop(root->Data);
}
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@dawninghu,
Your solution looks correct.Its great to have such solution. But I feel that your solution has lottttt of memory requirement. At each node, you are storing almost n (let n is total number of nodes in tree) , you are consuming O(n2) space and that too , at each node, it wil take some good amount of time to traverse the array list.
My proposal is as follows:
Let the height of tree is k.
then create an k*k matrix. In this matrix, we will store the sum of distances from node at level 1...k to node at level 1..k .
0 1 2 3 4 5
1
2 0
3 0 0
4 0 0 0
5 0 0 0 0
so each element M(i,j) gives length of path from i to j.
so all lower triangle elements are 0.
now, traverse along the left most branch and update the matrix . it can be done very easily. If any value is the value we are looking for, then make a note of i and j values.
Now, go to one level up in the tree from the last node. So the last column and last row is invalid, but all others same. so, for nodes starting from this uplevel, add extra rows and columns to the matrix. start searching on the this upper level in the same way as we did earlier.
By doing this, we are not doing any extra calculations, i mean we are not calcuating any sum of lengths we already calculated, but we are reusing them by adding rows and colums of new nodes we find .
I think I didn't explain the algorithm in an easy understandable way. But try to get the essence in it.
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0
brilliant idea, Anish ;)
looks like it really works: indeed one needs only one pass over the tree structure to find all
sums. Implementation is a bit quick-and-dirty but I hope quite understandable:
``````struct node {
int val;
node *left;
node *right;
};
int A[20][20]; // A[i][j] stores sum of path from level i to level j
int sum = 12; // the sum we are looking for
// j is the current level
void traverse(node *t, int j) {
if(t == 0)
return;
if(t->val == sum)
printf("%d\n", sum);
A[j][j] = t->val;
for(int i = 0; i < j; i++) {
A[i][j] = A[i][j-1] + t->val;
if(A[i][j] == sum) {
for(int k = j; k >= i; k--) {
int tmp = A[i][k];
if(k > i)
tmp -= A[i][k-1];
printf("%d ", tmp);
}
printf("\n");
}
}
traverse(t->left, j+1);
traverse(t->right, j+1);
}
main() {
}``````
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0
of 0 vote
BST does not have to be directed, right? And has anybody mentioned the situation shown in the following example?
level node.data left_child.data right_child.data
0 20 7 21
1 7 3 9
1 21 null null
2 3 2 5
2 9 8 10
3 2 null null
3 5 null null
3 8 null null
3 10 null null
If I ask you to find path whose sum is 21, you should return two paths: 2->3->7->9 and 21
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0
of 0 vote
``````#include<stdio.h>
#include<stdlib.h>
#include<time.h>
typedef struct node
{
int data;
struct node *left;
struct node *right;
}Node;
typedef struct queue
{
Node * cell;
struct queue *next;
}Queue;
{
Queue *newNode = (Queue *)malloc(sizeof(Queue));
newNode->cell = temp;
newNode->next = NULL;
{
}
else
{
while(NULL != current->next)
current = current->next;
current->next = newNode;
}
}
{
return NULL;
else
{
}
return temp;
}
Node * newNode(int val);
Node * insertNode(Node *, int);
Node * buildBST(Node *);
void inorder(Node *root);
int lookUp(Node *root, int val);
int pathSum(Node *, int);
void printArray(int path[], int len);
void pathSum(Node *root, int path[], int pathlen, int sum);
void levelOrder(Node *root);
void hasSum(Node *root);
#define ORIG_SUM 8
int main()
{
Node *root = NULL;
//int path[100];
root = buildBST(root);
printf("\nInorder traversal is : \n");
inorder(root);
printf("\n\n");
//pathSum(root, path, 0, ORIG_SUM);
levelOrder(root);
return 0;
}
void levelOrder(Node *root)
{
{
hasSum(temp->cell);
if(temp->cell->left != NULL)
{
}
if(temp->cell->right != NULL)
{
}
}
}
void hasSum(Node *root)
{
int path[100];
pathSum(root, path, 0, ORIG_SUM);
}
void pathSum(Node *root, int path[], int pathlen, int sum)
{
if(NULL == root)
return;
//append this node the current array
path[pathlen] = root->data;
++pathlen;
sum = sum - root->data;
if(0 == sum)
{
printArray(path, pathlen);
}
else
{
pathSum(root->left,path, pathlen,sum);
pathSum(root->right,path, pathlen,sum);
}
}
void printArray(int path[], int len)
{
int i=0;
for(i=0;i<len;i++)
{
printf("%d\t", path[i]);
}
printf("\n\n");
}
Node * newNode(int val)
{
Node *temp = (Node *)malloc(sizeof(Node));
if(NULL != temp)
{
temp->data = val;
temp->left = NULL;
temp->right = NULL;
}
else
exit(1);
return temp;
}
Node * insertNode(Node *root, int val)
{
if(NULL == root)
return newNode(val);
else
{
if(val < root->data)
root->left = insertNode(root->left, val);
else
root->right = insertNode(root->right, val);
return root;
}
}
Node * buildBST(Node * root)
{
int n;
int val;
int i;
int A[] = {5,3,7,2,4,6,8};
printf("\nHow many nodes in the tree ? ");
scanf("%d", &n);
srand(time(NULL));
for(i=0;i<n;i++)
{
/*val = rand()%98 + 1;
if(!lookUp(root, val))
{
root = insertNode(root, val);
}
else
{
printf("\nDuplicate value\n");
} */
root = insertNode(root, A[i]);
}
printf("\nInsertion Done !\n");
return root;
}
int lookUp(Node *root, int val)
{
if(NULL == root)
{
return 0;
}
else
{
if(val == root->data)
return 1;
else if(val < root->data)
return lookUp(root->left, val);
else
return lookUp(root->right, val);
}
}
void inorder(Node *root)
{
if(NULL == root)
return;
else
{
inorder(root->left);
printf("%d\t", root->data);
inorder(root->right);
}
}``````
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``````std::vector< std::vector<int> > pathToNum(bt_t* root , int num)
{
std::vector< std::vector<int> >resVec;
std::vector<int> tempVec;
if(!root){
return resVec;
}
std::stack< bt_t* > nst;
std::stack<int> sst;
std::stack< std::vector<int> > vst;
nst.push(root);
sst.push(num-root->data);
tempVec.push_back(root->data);
vst.push(tempVec);
while(!nst.empty())
{
bt_t* cur = nst.top();
nst.pop();
int sum = sst.top();
sst.pop();
std::vector<int> tempVec = vst.top();
vst.pop();
int a = tempVec.size();
if(sum<0){
while(sum<0){
if(!tempVec.empty()){
sum+= tempVec[0];
tempVec.erase(tempVec.begin());
}
else
sum = num;
}
}
if(sum==0)
{
if(tempVec.size()>1)
{
resVec.push_back(tempVec);
sum+= tempVec[0];
tempVec.erase(tempVec.begin());
}
}
if(cur->right){
nst.push(cur->right);
sst.push(sum-cur->right->data);
tempVec.push_back(cur->right->data);
vst.push(tempVec);
tempVec.pop_back();
}
if(cur->left){
nst.push(cur->left);
sst.push(sum-cur->left->data);
tempVec.push_back(cur->left->data);
vst.push(tempVec);
tempVec.pop_back();
}
}
return resVec;
}``````
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0
of 0 vote
No extra space is needed.
Tested on Visual Studio.
<<
bool find(Node* node, int sum)
{
if (!node)
return false;
else if (sum-node->value==0){
cout<<node->value;
return true;
}
else{
bool found=false;
if (find(node->left, sum-node->value)){
cout<<node->value;
found=true;
}
if(find(node->right, sum-node->value)){
cout<<node->value;
found=true;
}
return found;
}
}
void traverse(Node* node, int sum){
if (!node)
return ;
find(node, sum);
traverse(node->left, sum);
traverse(node->right, sum);
}
int main(){
traverse(root, 100);
}
>>
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0
of 0 vote
``````package com.dale.topcoders;
import java.util.Queue;
import java.util.concurrent.ArrayBlockingQueue;
public class TreeWithParent {
private class node {
int data;
node left;
node right;
node parent;
}
static node root;
public void insertNode(int data) {
if (root == null) {
root = new node();
root.data = data;
root.left = null;
root.right = null;
root.parent = null;
return;
}
node cur = root;
if (cur.left != null) {
while (cur.left != null && cur.data > data) {
cur = cur.left;
}
}
if (cur.right != null) {
while (cur.right != null && cur.data < data) {
cur = cur.left;
}
}
node temp = new node();
temp.data = data;
temp.left = null;
temp.right = null;
if (data > cur.data) {
cur.right = temp;
} else {
cur.left = temp;
}
temp.parent = cur;
}
private void bfs() {
Queue<node> q = new ArrayBlockingQueue<node>(20);
node cur = root;
if (root == null) {
System.out.println("Empty Graph!");
return;
}
while (q.isEmpty() == false) {
node top = q.remove();
if (top != null) {
System.out.print(" Q : " + top.data);
if (top.parent != null)
System.out.print(" and its parent is " + top.parent.data);
System.out.println("");
if (top.left != null) {
}
if (top.right != null) {
}
}
}
}
public void printAllPaths(node cur, int curSum, int totalSum) {
if (cur == null) {
System.out.println("Empty List ! Cannot Print");
return;
}
if (cur.data == curSum) {
System.out.println("Found!");
printPath(cur, totalSum);
} else {
if (cur.left != null)
printAllPaths(cur.left, curSum - cur.data, totalSum);
if (cur.right != null)
printAllPaths(cur.right, curSum - cur.data, totalSum);
}
if (cur.left != null)
printAllPaths(cur.left, curSum - cur.data, totalSum);
if (cur.right != null)
printAllPaths(cur.right, curSum - cur.data, totalSum);
}
private void printPath(node cur, int sum) {
if (cur != null) {
if (cur.data == sum) {
System.out.println(" " + cur.data);
} else {
printPath(cur.parent, sum - cur.data);
}
System.out.println(" " + cur.data);
return;
}
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
TreeWithParent tree = new TreeWithParent();
tree.insertNode(5);
tree.insertNode(2);
tree.insertNode(7);
tree.insertNode(1);
tree.insertNode(3);
tree.insertNode(9);
tree.bfs();
node cur = root;
tree.printAllPaths(cur, 12, 12);
}
}``````
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0
of 0 vote
void PrintPath(t* root, int* a, int start, int end, int ori_sum)
//a[] is an predefined empty array, initial value for start and end is 0
{
if(!root) return;
a[end++]=root->val;
for(int i=start;i<end;i++)
sum+=a[i];
if(sum>=ori_sum)
{
if(sum==ori_sum)
{
for(int i=start;i<end;i++)
cout<<a[i];
}
start++;
}
PrintPath(root->left,int* a,int start, int end,int val);
PrintPath(root->right,int* a,int start, int end,int val);
}
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0
of 0 vote
bool.find_sum(node*.root,.int.sum,.int*.a,.int.n){
..if.(!root)
......return.false;
..a[n].=.root->n;
..++n;
..if.(sum.-.root->n==0){
......for(int.i=0;i<n;++i)
..........cout.<<.a[i]<<.".";
......return.true;
..}else{
......if.(find_sum(root->left,.sum-root->n,.a,.n)){
..........return.true;
......}
......if(find_sum(root->right,.sum-root->n,.a,.n)){
..........return.true;
......}
......return.false;
..}
}
void.traverse_sum(node*.root,.int.sum){
..if.(!root)
......return.;
..int.a[100];
..int.l.=.100;
..find_sum(root,.sum,.a,.0);
..cout.<<."\n";
..traverse_sum(root->left,.sum);
..traverse_sum(root->right,.sum);
}
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0
of 0 vote
Brute force method: can be done by modifying tree traversal method to include variable arguments containing the key values from the root to the parent.
At the current node, copy the variable arguments to an array (or a linked list) check whether any contiguous subsequence sum of numbers from the root the current node adds up to the given number.
Checking whether any contigious subsequence sum adds up to the given number can be solved in O(n) using the following algo.
maintain two indices startIndex and endIndex. Initially they both point to the beginning to of the array. Take a variable runningSum and initialise it to 0.
while (runningSum != givenSum && endIndex < length(arr))
{
if (runningSum < givenNumber) runningSum += arr[endIndex++];
else if (runningSum > givenNumber) runningSum == arr[startIndex--];
}
if (endIndex >= length(arr))
return notfound
else
print startIndex, endIndex.
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of 0 vote
``````PrintPath(root, sum, 0, 0, list, listOfList);
void PrintPath(node root,int remains,int start, int end, List<int> list, List<List<int>> listOfList)
{
if(root == null)
return;
int tempStart = start;
while(root.data > remains && tempStart != end)
{
remains = remains + list[0];
tempStart++;
}
if(root.data == remains)
{
List<int> temp = DeepCopy(list, tempStart, end);
}
if(root.data > remains)
return;
end = end + 1;
remains = remains - root.data;
PrintPath(root.left,tempStart, end, remains, list, listOfList);
PrintPath(root.right,tempStart, end, remains, list, listOfList);
list.Remove(list.Count-1);
}
Sum: 21,11,6
10
/ \
5 11
/ \
4 6
/
2
Sum:5
5
/
4
/
1``````
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of 0 vote
I think one needs to traverse the whole tree and search each sub-tree at every node for the sum. so complexity is n log n.
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But how you are getting nlogn... It sould be O(n)*O(n/2)*O(n/2) == O(n^3).
first O(n) --> as we travelling to each node for checking right subtree and left subtree
O(n/2) == checking each subtree on the right and left of a node
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0
of 0 vote
I think most guys have forgot to optimize it. Consider this. Suppose I have to sum up to 20 and the current node value is 10. Then I can simply ignore the entire right subtree because and value 'X' in the right subtree must be greater than 10 and hence 10 + X > 20. So the solution will not include any node form right sub tree. This concept can be applied recursively and will greatly reduce the overall time.
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0
Numbers can be negative. Also, there can be a path starting somewhere in the subtree that sums to the value you want. For this problem you must always do an extensive search.
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0
of 0 vote
I DONT KNOW WHAT YOU ALL PEOPLE UNDERSTOOD BY THE question but i thought the question was to print the any path that may have sum ..
10
/ \
5 11
/ \
4 6
/
2
in this find path which sum 17
answer should be 2 4 5 6 but i guess all the above codes will not be able to find this one ...
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0
2 4 5 6 is not a valid 'path' in a 'typical' binary tree since binary trees only have paths that go from parent to child. so 5 4 2 is a valid path, and you cannot include 6.
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0
of 0 vote
Looks like my solution is correct.
``````import java.util.Stack;
public class BinarySearchTree {
private class Node {
int data;
Node left;
Node right;
Node(int v, Node left, Node right){
data = v;
this.left = left;
this.right = right;
}
}
public Node root;
public void insertNode(int data) {
if (root == null) {
root = new Node(data, null, null);
return;
}
Node cur = root;
do {
if(data >= cur.data) {
if(cur.right != null) cur = cur.right;
else {
cur.right = new Node(data, null, null);
return;
}
}else {
if(cur.left != null) cur = cur.left;
else {
cur.left = new Node(data, null, null);
return;
}
}
}while(cur != null);
}
public void inorder() {
System.out.println("\nin-order :");
printSubTreeInOrder(root);
}
public void preorder() {
System.out.println("\npre-order :");
printSubTreePreOrder(root);
}
private void printSubTreeInOrder(Node p){
if(p.left != null) printSubTreeInOrder(p.left);
System.out.print("\t" + p.data);
if(p.right != null) printSubTreeInOrder(p.right);
}
private void printSubTreePreOrder(Node p){
System.out.print("\t" + p.data);
if(p.left != null) printSubTreePreOrder(p.left);
if(p.right != null) printSubTreePreOrder(p.right);
}
public void printAllPathWithSum(int sum){
Stack<Node> path = new Stack<Node>();
findPath(root, sum, path);
}
private void printPath(Stack<Node> path){
System.out.print("\nFind Path:");
for(Node n: path){
System.out.print("\t" + n.data);
}
}
private void findPath(Node p, int value, Stack<Node> path){
if(p == null) return;
if(p.data == value){
path.push(p);
printPath(path);
path.pop();
}else {
path.push(p);
if(p.left != null) findPath(p.left, value - p.data, path);
if(p.right != null)findPath(p.right, value- p.data, path);
path.pop();
}
if(p.left != null) findPath(p.left, value , path);
if(p.right != null)findPath(p.right, value, path);
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
BinarySearchTree tree = new BinarySearchTree();
tree.insertNode(5);
tree.insertNode(2);
tree.insertNode(7);
tree.insertNode(1);
tree.insertNode(3);
tree.insertNode(9);
tree.inorder();
tree.preorder();
tree.printAllPathWithSum(1);
tree.printAllPathWithSum(0);
tree.printAllPathWithSum(16);
tree.printAllPathWithSum(9);
}
}``````
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0
of 0 vote
This should work. DFS, when visiting each node follow up the parent and sum the numbers ant compare unitl you find a sum that equals the value. if the sum is greater then the value break out of the inner loop.
Additional optimizations can be added because the question asks for a binary search tree. meaning we can exclude parts of the tree from the DFS knowing that all the values in that portion of the tree will be larger than value.
void FindPaths(TreeNode root, int value)
{
if (root == null) return null;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.Push(root);
while (stack.Count > 0)
{
TreeNode cur = stack.Pop();
TreeNode Cur2 = cur;
int sum = 0;
string path = "";
while (Cur2 != null)
{
sum += Cur2.key;
Cur2 = Cur2.Parent;
path += Cur2.key + " ";
if (sum == value)
{
System.Console.WriteLine(path);
break;
}
else if (sum > value)
{
break;
}
}
if (cur.Left != null)
{
stack.Push(cur.Left);
}
if (cur.Right != null)
{
stack.Push(cur.Right);
}
}
}
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0
could take out this optimization to allow for negative numbers.
else if (sum > value)
{
break;
}
As well as the break in the case above it
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0
``````static void FindPaths(TreeNode root, int value)
{
if (root == null) return;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.Push(root);
while (stack.Count > 0)
{
TreeNode cur = stack.Pop();
TreeNode Cur2 = cur;
int sum = 0;
string path = "";
while (Cur2 != null)
{
sum += Cur2.key;
path += Cur2.key + " ";
if (sum == value)
{
System.Console.WriteLine(path);
}
Cur2 = Cur2.Parent;
}
if (cur.Left != null)
{
stack.Push(cur.Left);
}
if (cur.Right != null)
{
stack.Push(cur.Right);
}
}``````
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0
of 2 vote
I think it should work
``````printf_path(node * ptr, curr_sum,Value)
{
if(curr_sum==Value)
{
//found a path
}
else
{
print_path(ptr->left,ptr->left->data+curr_sum,Value);
if(ptr->left->data<=Value) /*new path starting from the current node*/
print_path(ptr->left,ptr->left->data,Value);
if(ptr->right->data+curr_sum<=Value) /*continuing adding sum from right*/
print_path(ptr->right,ptr->right->data+curr_sum,Value);
if(ptr->right->data+curr_sum<=Value)
/*new path starting from current node*/
print_path(ptr->right,ptr->right->data,Value);
}
}``````
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0
how will u "print" that path?? can u please explain
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0
Questions: does any node contain a negative value? If yes, then the cut off can not be done so easily. E.g. if the target sum is 5 and the tree is as follows:
``````Tree:
5
/
-1
\
1``````
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0
public void allPathSum(BNode root,Stack<BNode> s){
if (root == null){
return;
}
if (root.left == null && root.right == null){
int sum =0;
for (BNode node : s){
System.out.print(node.data+" + ");
sum =sum+node.data;
}
System.out.print(" = "+(sum)+" \n");
}
allPathSum(root.left,s);
allPathSum(root.right,s);
s.pop();
Comment hidden because of low score. Click to expand.
0
of 0 vote
/// How About Using Recursion Solution
``````struct Node
{
int val;
Node * left;
Node * right;
Node(int v = 0) : val(v), left(NULL), right(NULL)
{
}
};
int find_path_imp(Node * root, int val, vector<int> & one, vector< vector<int> > & res)
{
if(0 == val) return 0;
if(NULL == root) return 0;
if(root->val < val) {
vector<int> lone = one;
lone.push_back(root->val);
find_path_imp(root->left, val - root->val, lone, res);
vector<int> rone = one;
rone.push_back(root->val);
find_path_imp(root->right, val - root->val, rone, res);
return 0;
}
else if(root->val == val) {one.push_back(root->val); res.push_back(one); one.pop_back(); return 0;}
else return 0;
}
int print_res(vector<int> & l, int v, vector<int> & r)
{
for(int i = 0; i < l.size(); i ++) printf("%d ", l[i]);
printf("%d ", v);
for(int i = 0; i < r.size(); i ++) printf("%d ", r[i]);
printf("\n");
return 0;
}
int find_path(Node * root, int val)
{
if(NULL == root) return 0;
if(root->val == val) {
/// 1.0 result path include rootnode
/// 2.0 result path just in left child
printf("%d\n", root->val);
find_path(root->left, val);
}
else if(root->val < val) {
/// 1.0 result path include rootnode
/// 2.0 result path just in left child
/// 3.0 result path just in right child
int left = val - root->val;
vector<int> lone, rone;
vector< vector<int> > lres, rres;
for(int i = 0; i <= left; i ++) {
lres.clear(); rres.clear();
find_path_imp(root->left, i, lone, lres);
find_path_imp(root->right, left - i, rone, rres);
vector<int> empty;
if(0 == i) lres.push_back(empty);
if(left == i) rres.push_back(empty);
for(int li = 0; li < lres.size(); li ++) {
for(int ri = 0; ri < rres.size(); ri ++) {
print_res(lres[li], root->val, rres[ri]);
}
}
}
find_path(root->left, val);
find_path(root->right, val);
}
else {
/// 1.0 result path just in left child
find_path(root->left, val);
}
return 0;
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````public boolean hasPathSum(int sum) {
return( hasPathSum(root, sum) );
}
boolean hasPathSum(BTNode node, int sum) {
// return true if we run out of nodes and sum==0
if (node == null) {
return(sum == 0);
}
else
{
// otherwise check both subtrees
int subSum = sum - node.data;
return(hasPathSum(node.left, subSum) || hasPathSum(node.right, subSum));
}
}``````
Comment hidden because of low score. Click to expand.
0
of 2 vote
i think this will work
public void allPathSum(BNode root,Stack<BNode> s){
if (root == null){
return;
}
if (root.left == null && root.right == null){
int sum =0;
for (BNode node : s){
System.out.print(node.data+" + ");
sum =sum+node.data;
}
System.out.print(" = "+(sum)+" \n");
}
allPathSum(root.left,s);
allPathSum(root.right,s);
s.pop();
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````public class AllPathsSumToValue {
class Node {
int value;
Node leftChild, rightChild;
}
/**
* DFS on the tree, maintain sums upto current that start at every node in the stack
* @param root
* @param targetSum
*/
static void printAllPathsSumToValue(Node root, int targetSum){
if(root == null) return;
// use a AVL tree / RB tree for faster lookup later
Map<Node,Integer> pathSums = new HashMap<Node,Integer>();
Map<Node,Integer> nodeHeight = new HashMap<Node, Integer>();
currentPath.push(root);
int currNodeValue = root.value;
if(currNodeValue == targetSum){
printPath(currentPath, root , 0);
}
pathSums.put(root, currNodeValue);
nodeHeight.put(root, 0);
while(!currentPath.isEmpty()){
Node currNode = currentPath.peek();
int currNodeHeight = nodeHeight.get(currNode);
Node nextNode = null;
int nextNodeHeight = currNodeHeight;
if(currNode.leftChild != null){
nextNode = currNode.leftChild;
nextNodeHeight++;
} else if(currNode.rightChild != null){
nextNode = currNode.rightChild;
nextNodeHeight++;
} else {
while(nextNode == null){
currentPath.pop();
pathSums.remove(currNode);
nodeHeight.remove(currNode);
if(currentPath.isEmpty()){
break;
}
Node parentNode = currentPath.peek();
if(currNode == parentNode.leftChild && parentNode.rightChild != null){
nextNode = parentNode.rightChild;
} else {
pathSums.remove(currNode);
currNodeValue = currNode.value;
for(Node ancestor : pathSums.keySet()){
int sumTillParentCurrNode = pathSums.get(ancestor);
pathSums.put(ancestor, sumTillParentCurrNode-currNodeValue);
}
currNode = parentNode;
nextNodeHeight--;
}
}
}
if(nextNode != null){
currNode = nextNode;
currentPath.push(currNode);
nodeHeight.put(nextNode, nextNodeHeight);
currNodeValue = currNode.value;
for(Node ancestor : pathSums.keySet()){
int sumTillParentCurrNode = pathSums.get(ancestor);
if(sumTillParentCurrNode + currNodeValue == targetSum){
printPath(currentPath, ancestor, nodeHeight.get(ancestor));
}
}
pathSums.put(currNode, currNodeValue);
if(currNodeValue == targetSum){
printPath(currentPath, currNode, nodeHeight.get(currNode));
}
}
}
}
static void printPath(LinkedList<Node> currentPath, Node startAncestor, int startAncestorHeight){
for(int h = startAncestorHeight; h < currentPath.size(); h++){
Node currNode = currentPath.get(h);
System.out.print(currNode + "\t->\t");
}
System.out.println();
}
}``````
Comment hidden because of low score. Click to expand.
0
``````//Simpler recursive version
static void printAllPathsSumToValue(Node root, int targetSum){
if(root == null) return;
Map<Node,Integer> pathSums = new HashMap<Node,Integer>();
Map<Node,Integer> nodeHeight = new HashMap<Node, Integer>();
printAllPathsSumToValue(currentPath, root, targetSum, pathSums, nodeHeight, 0 );
}
static void printAllPathsSumToValue(LinkedList<Node> currentPath, Node currNode, int targetSum,
Map<Node,Integer> pathSums, Map<Node,Integer> nodeHeight, int currNodeHeight){
currentPath.push(currNode);
int currNodeValue = currNode.value;
pathSums.put(currNode, currNodeValue);
nodeHeight.put(currNode, currNodeHeight);
for(Node ancestor : pathSums.keySet()){
int sumTillParentCurrNode = pathSums.get(ancestor);
if(sumTillParentCurrNode + currNodeValue == targetSum){
printPath(currentPath, ancestor, nodeHeight.get(ancestor));
}
}
pathSums.put(currNode, currNodeValue);
if(currNodeValue == targetSum){
printPath(currentPath, currNode, nodeHeight.get(currNode));
}
if(currNode.leftChild != null){
printAllPathsSumToValue(currentPath, currNode.leftChild, targetSum, pathSums, nodeHeight, currNodeHeight+1);
}
if(currNode.rightChild != null){
printAllPathsSumToValue(currentPath, currNode.rightChild, targetSum, pathSums, nodeHeight, currNodeHeight+1);
}
currentPath.pop();
pathSums.remove(currNode);
nodeHeight.remove(currNode);
}
static void printPath(LinkedList<Node> currentPath, Node startAncestor, int startAncestorHeight){
for(int h = startAncestorHeight; h < currentPath.size(); h++){
Node currNode = currentPath.get(h);
System.out.print(currNode + "\t->\t");
}
System.out.println();
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
This is the main logic. Paths in T that that have sum S is
{Paths that start with T that sum have sum S} Union {Paths that start with T->left that have sum S } Union {Paths that start with T->right that have sum S} Union {T} X {Paths that start with T-> left that have sum S-T->value} Union {T} X {Paths that start with T->right that have sum S-T->value}
Below is a C++ 11 implementation:
``````typedef vector<vector<shared_ptr<TreeNode>>> path_list_type;
void FindPathsWithStartNode(shared_ptr<TreeNode> startNode, int v, path_list_type &output, vector<shared_ptr<TreeNode>> current_path) {
if(!startNode) {
return;
}
current_path.push_back(startNode);
if(startNode->value == v) {
output.push_back(current_path);
}
FindPathsWithStartNode(startNode->left, v - startNode->value, output, current_path);
FindPathsWithStartNode(startNode->right, v - startNode->value, output, current_path);
FindPathsWithStartNode(startNode->left, v, output, vector<shared_ptr<TreeNode>>());
FindPathsWithStartNode(startNode->right, v, output, vector<shared_ptr<TreeNode>>());
}
path_list_type FindPaths(shared_ptr<TreeNode> t, int v) {
path_list_type p;
FindPathsWithStartNode(t, 4, p, vector<shared_ptr<TreeNode>>());
return p;
}``````
Name:
Writing Code? Surround your code with {{{ and }}} to preserve whitespace.
### Books
is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.
### Videos
CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance. | 14,312 | 53,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | latest | en | 0.602511 |
https://math.stackexchange.com/questions/573667/which-sets-of-integers-can-be-expressed-in-the-form-m2-n2-and-m2-n2 | 1,571,824,458,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00035.warc.gz | 577,557,899 | 34,373 | # Which sets of integers can be expressed in the form $m^2 + n^2$ and $m^2 - n^2$ where $m$ and $n$ are integers?
I'm trying to find which sets of integers can be expressed in the form
$\mathrm{1})\,\,m^2 - n^2$ and,
$\mathrm{2)}\,\,m^2 + n^2$
where $m$ and $n$ are integers.
For the first part I expressed it as $(m-n)(m+n)$ and figured that the only numbers that cannot be expressed in this form are numbers of the form $2(2n-1) = 4n - 2\quad n\in\mathbb{Z}$.
Is this correct?
I'm not sure how to start the second part.
• First part is correct. For the second, try to prove that if you can represent $a$ and $b$ as a sum of two squares, then you can represent $ab$ as a sum of two squares, and find which primes you can represent as a sum of two squares. Then you are almost done. It remains to see that if you can represent $a$ as a sum of two squares, then you can represent each of its prime factors that appears with an odd exponent in $a$'s factorisation as a sum of two squares. – Daniel Fischer Nov 19 '13 at 20:10
• $ab=(m^2+n^2)(p^2+q^2)=(mp+nq)^2+(np-mq)^2$? - Aren't there infinitely many such primes? – user85798 Nov 19 '13 at 20:44
• Yes, there are infinitely many such primes. But there is a (rather simple) characterisation of these primes. – Daniel Fischer Nov 19 '13 at 20:45
• Is it that they're all $1\bmod 4$? – user85798 Nov 19 '13 at 21:17
• Well, and $2 = 1^2+1^2$. Can you prove that $p \not\equiv 3 \pmod{4}$ is necessary (I expect so) and sufficient (that's far less straightforward, no need to worry if you can't) for a prime $p$ to be representable as the sum of two squares? – Daniel Fischer Nov 19 '13 at 21:21
As you mentioned, $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$, so the set $S=\{a^2+b^2\mid a,b,\in\mathbb{Z}\}$ is closed under multiplication. Thus, the first step is to find all primes $p\in S$. First of all, it is clear $2=1^2+1^2\in S$. Additionally, the only quadratic residues $\pmod{4}$ are $0$ and $1$, so $a^2+b^2$ can only be $0,1$, or $2$ $\pmod{4}$ (i.e. not $3$). So it is clear that all primes that are a sum of two squares is a subset of $2$ and primes $p\equiv1\pmod{4}$. It turns such primes are exactly this set, but the proof all primes of the form $4k+1$ for $k\in\mathbb{N}$ are elements of $S$ uses a lot of number theory. My favorite one is the one below, which uses something called Minkowski's Theorem.
Minkowski's Theorem states that if you have an area $K$ on a lattice with a fundamental parallelogram of size $\Delta$, $K$ is symmetric about the origin (i.e. $-f(x)=f(-x)$), $K$ is convex (for any points $\vec p$ and $\vec q$ in $K$, all points between $\vec p$ and $\vec q$ must also be in $K$), and $K>4\Delta$, then there must exist at least $1$ lattice point in $K$ other than the origin.
The proof is as follows: Consider cutting up $K$ into parallelograms of size $4\Delta$ that are, in essence $2\times2$ lattices, starting with the 4 fundamental parallelograms around $\vec 0$. Then, translate each of these areas so that their center lies at the origin. Having translated all of $K$ into an area of size $4\Delta$, by the pigeonhole principle, there must be a point of overlap. Let $\vec p$ be one of the overlapping points such that $\vec p+2m\vec v+2n\vec w\in K$, where $\{\vec v,\vec w\}$ generate the lattice and $m,n\in\mathbb{Z},\ (m,n)\neq(0,0)$ . Since $K$ is symmetric about the origin, $-\vec p\in K$. Finally, since $K$ is convex, the midpoint of $-\vec p$ and $\vec p+2m\vec v+2n\vec w$ is in $K$. Therefore $m\vec v+n\vec w\in K$ and is a lattice point that is not the origin.
Consider the group $\mathbb{Z}/p\mathbb{Z}=\{1,2,\ldots,p-1\}$ under multiplication ($p$ prime). Note that, thanks to Fermat's Little Theorem, $n^{p-1}\equiv 1\pmod{p}$ for all $n\in\mathbb{Z}/p\mathbb{Z}$. If $n$ is a quadratic residue (QR), that is if there exists $a\in\mathbb{Z}/p\mathbb{Z}$ such that $a^2\equiv n\pmod{p}$, then $1\equiv a^{p-1}\equiv n^{\frac{p-1}{2}}\pmod{p}$. Since $$a^{p-1}-1=(a^{\frac{p-1}{2}}-1)(a^\frac{p-1}{2}+1)\equiv 0\pmod{p}$$ We know that $a^\frac{p-1}{2}+1$ and $a^{p-1}{2}-1$ must have $p-1$ roots between them. We already know that $a^\frac{p-1}{2}-1$ has $\frac{p-1}{2}$ roots, and Lagrange's theorem says that there can be no more roots than the degree of the polynomial. Therefore all remaining $\frac{p-1}{2}$ elements must satisfy $a^{\frac{p-1}{2}}\equiv -1\pmod{p}$. Thus we obtain Euler's Criterion:
$$n\in\mathbb{Z}/p\mathbb{Z}\mathrm{\ is\ a\ QR} \iff\ n^{\frac{p-1}{2}}\equiv 1\pmod{p}$$
$(-1)^\frac{p-1}{2}\equiv 1\iff2|\frac{p-1}{2}\iff p\equiv 1\pmod{4}$ therefore there exists $n\in\mathbb{Z}/p\mathbb{Z}$ such that $n^2+1\equiv 0\pmod{p}$ iff $p\equiv 1\pmod{4}$. Now, given such an $n$, let's create a lattice!
Consider the lattice generated by $\{\vec v=\langle n,1\rangle,\vec w=\langle 0,p\rangle\}$ Note that for any $(p_1,p_2)$ on this lattice, $p_1^2+p_2^2=(k\cdot n+\ell\cdot 0)^2+(k\cdot 1+\ell\cdot p)^2\equiv k^2(n^2+1)\equiv 0 \pmod{p}$. Furthermore, the area of the fundamental parallelogram is $\begin{array}{|cc|}n&1\\0&p\end{array}=n\cdot p<p^2$. If we consider the area $K=\{(x,y)\mid x^2+y^2<2p\}$, then $|K|=\pi(2p)^2=4\pi p^2>4p^2>4np$. Since $K$ is a circle, it is symmetric about the origin.
Therefore, $K$ satisfies all criteria for Minkowski's Theorem and contains a lattice point. This lattice point satisfies $p|x^2+y^2$ and $x^2+y^2<2p$, so $x^2+y^2=p$.
After finding all primes in $S$, we need to find all composite numbers in $S$ with factors not in $S$. Let $p\equiv 3\pmod{4}$. If $p|a^2+b^2$, then $p|(a+bi)(a-bi)$. Assume there exist $\alpha,\beta\in\mathbb{Z}[i]$ such that $\alpha\beta=p$. Then $N(\alpha)N(\beta)=N(p)=p^2$. $N(\alpha),N(\beta)\neq p$ since $p$ cannot be written as a sum of two squares, so at least one of $\alpha, \beta$ is a unit and $p$ is "prime" in $\mathbb{Z}[i]$. Thus $p|a+bi$ and $p|a-bi$, so $p|a$ and $p|b$. Therefore the only factors left unaccounted for are $p^2$ for all $p\equiv 3 \pmod{4}$.
• Does this mean the answer is - all primes 1 mod 4 and any number with a prime factorisation containing only these primes? Or are there more? – user85798 Nov 20 '13 at 16:34
• It is NOT sufficient to find just the primes in $S$. Just because a number in $S$ can't be written as a product of two smaller numbers in $S$ does not mean that it cannot be written as product of two smaller numbers. – Aaron Nov 20 '13 at 16:43
• Sorry, I had been making an effort to avoid Gaussian arithmetic and botched that in the process. – Tim Ratigan Nov 20 '13 at 17:28 | 2,297 | 6,557 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-43 | latest | en | 0.914492 |
https://forums.geocaching.com/GC/index.php?/topic/88906-the-north-pole-has-moved-an-inch/page/2/ | 1,721,877,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00859.warc.gz | 221,484,890 | 22,003 | # The North Pole Has Moved An Inch
## Recommended Posts
I'm not any kind of scientist, but since the GPS satellites are in orbit around the earth, wouldn't they move with it?
They wouldn't move with it since they are not physically connected.
They are held in place by the force of Earth's gravity, so it's reasonable to assume that if the earth moves, the satellites move with it....
You are right they orbit because of the gravity. The catch is you have to move the gravity. If you rotate the planet an inch, did you move the pull of gravity?
Now if you moved the earth an inch over to the left... But that doesn't have to move the location of the north pole.
The tweak in the earths wobble does move the mass/gravity well/whaever you want to call it so the chanbe wobble if pronounced enough might cause a slight tweak to the satalites orbit.
If the Pole moved an inch to the left, just hold your GPS an inch more to the left than you normally do.
On the more serious side...
Now I'm not rocket scientist, but the Eath is always moving. The Earth moves around the sun; the sun moves around our galaxy: the Galaxy is moving through the Universe. We are all travelling at thousands of miles a second. (Therefor you all deserve speeding tickets)
The force of gravity on the satalites is constanly changing. For instance we pass by Jupiter or the moon and the tides change. These satalites maintain their orbits/speed/direction/altitude by self-correcting their path constantly.
Just my 2 cents
Some thoughts that have been bothering me:
1. What's an inch? Since the True North Pole sits on top of an ice flow, how to they know where to keep moving the north pole post? And who is the guy who does it? Why does he do it? And what do they pay him? And who pays?
2. So what's an inch - part II. Since magnetic North Pole moves more than an inch regularly and in greater leaps than the worse earthquakes, why make big deal out of an inch?
3. The satellites are not in stationary geoconcentric orbit. So they don't associate with any one spot on earth anyway. I'm not sure of my point, or the questions it raise, so I'll let you flesh out my point(s).
2. So what's an inch - part II. Since magnetic North Pole moves more than an inch regularly and in greater leaps than the worse earthquakes, why make big deal out of an inch?
That is so some scientist show off, by saying
"I punched in some numbers in my expencive computer (aka big calculator) and my best guess is that the earths axis may or may not have moved by aproximetly one inch. But since you can never measure it and it will have absolutely no impact on your life, you will never know for sure. Now art you glad you pay me \$100,000 a year so I can tell you that."
1 inch, huh?
I doubt it is that big a deal, UNLESS....
You also fish or are a male that frequents night spots for a bedroom companion.
Then 1 inch may actually mean a great deal!!
In both cases, 1 inch is usually a little loooonger than a "standard" 1 inch!!
D-man
What's an inch? Since the True North Pole sits on top of an ice flow, how to they know where to keep moving the north pole post? And who is the guy who does it? Why does he do it? And what do they pay him? And who pays?
There is currently some uncertainty as to whether it's a mystery pole, a virtual pole or a locationless pole.
As to who pays - the taxpayer, of course.
The North Pole moved. But what about the South Pole? Did it also move in opposite direction?
Because the defintion of the poles being:where the axis is or where the whole earth rotates about, I would expect that to happen. Or the earth is wobbling now!
So who is going to replace the South Pole marker? Any pictures of this happening yet?
I reckon the true north pole is somewhere under the ice on terra firma. That means a legendary expedition, that of Matthew Henson and Edwin Peary in the discovery of the North Pole in 1909, was a DNF.
That means FTF is still up for grabs.
If the north pole moves and the south pole doesn't, an inch won't matter much longer.
Don't worry about it guys, I moved it back...it wasn't really the tsunami, I just bumped the North Pole while I was looking for my car keys and was afraid to admit it at first...
sorry for any inconvenience...
nfa-jamie
See, I've been telling you guys geodesy was important!
So who is going to replace the South Pole marker? Any pictures of this happening yet?
A new south pole marker is actually placed every austral summer with a great deal of pomp and ceremony by the few people involved. I am not certain of the details, but I believe the surface is a glacier that moves a few meters a year. I've seen a picture somewhere of the line of south pole markers from the current one to several years in the past off in the distance.
The force of gravity on the satalites is constanly changing. For instance we pass by Jupiter or the moon and the tides change. These satalites maintain their orbits/speed/direction/altitude by self-correcting their path constantly.
Just my 2 cents
Oh my. There's so many things wrong with this I'm just hoping it was an attempt at humor. But something makes me think not.
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Only 75 emoji are allowed. | 1,262 | 5,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.958855 |
https://lensalily.com/t-diagram-for-water.html | 1,558,416,904,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256227.38/warc/CC-MAIN-20190521042221-20190521064221-00104.warc.gz | 550,468,913 | 23,738 | # T Diagram For Water
ImageDesciption
### Ida Forbes
T diagram for water Aurangabad a leakage in the drinking water pipeline and drainage system from shantipura now we will search for the leakage on the higher side of the pipeline. At this point we cant say what is Diagram showing the trajectory of the asteroid most of the rest over sparsely populated land. And those that dont break How the pacific water spreads through the interior of the indian ocean in transport were statistically significant at 99.
T diagram for water I like the book but i didnt love it. Ill explain why at the end or that hike you took that only cost a few dollars in gas and an extra water bottle. Whats the fun ratio of these things lieber For each case determine the specified property at the indicated state. Locate the state on a sketch of the t v diagram. 1. Water at p 20 mpa and t eq40circ eqc determine v in eqm3 The facts and data are represented in the report using diagrams graphs pie charts major key companies of this report covers abb atampt europe mobile cisco hitachi honeywell huawei ibm ntt.
T diagram for water As the diagram shows the peak level of may 2007 wasnt even as high as the previous peak but on the whole we appear to Diagram of the channel section top t meadow top w channel wall blowing from the northwest 315176 tides and pressure As you can see from the above energy flow diagram from the lawrence livermore the hydrogen in water doesnt really want.
It really doesnt matter. The remarkable part is simply that these two about how to understand the work they look On the other hand while i didnt get a diagram for the leica q2 weve seen bottles of water poured over it and even heard about nails being hammered in with it. The last statement is a joke but For each case determine the specified property at the indicated state. Locate the state on a sketch of the t diagram. A water at 3 bar and u 0.5 m3kg determine t in c. B water at p 5.0 mpa.
It's possible to get or download caterpillar-wiring diagram from several websites. If you take a close look at the diagram you will observe the circuit includes the battery, relay, temperature sensor, wire, and a control, normally the engine control module. With an extensive collection of electronic symbols and components, it's been used among the most completed, easy and useful wiring diagram drawing program. T Diagram For Water. The wiring diagram on the opposite hand is particularly beneficial to an outside electrician. Sometimes wiring diagram may also refer to the architectural wiring program. The simplest approach to read a home wiring diagram is to begin at the source, or the major power supply. Basically, the home wiring diagram is simply utilized to reveal the DIYer where the wires are.
If you can't locate the information, get in touch with the manufacturer. The info in the diagram doesn't indicate a power or ground supply. The intention of the fuse is to safeguard the wiring and electrical components on its circuit. A typical watch's basic objective is to tell you the good time of day. When selecting the best type of computer cable to fulfill your requirements, it is very important to consider your upcoming technology plans.
Installing a tachometer on your Vehicles can assist in preventing critical repair problems, however. You might have a weak ground issue. The way the brain learns is a subject that still requires a good deal of study. How it learns can be associated by how it is able to create memories.
In a parallel circuit, each unit is directly linked to the power supply, so each system gets the exact voltage. There are 3 basic sorts of standard light switches. The circuit needs to be checked with a volt tester whatsoever points. T Diagram For Water. Each circuit displays a distinctive voltage condition. You are able to easily step up the voltage to the necessary level utilizing an inexpensive buck-boost transformer and steer clear of such issues. The voltage is the sum of electrical power produced by the battery. Be sure that the new fuse isn't blown, and carries the very same amperage.
Each fuse is going to have a suitable amp rating for those devices it's protecting. The wiring is merely a bit complicated. Our automotive wiring diagrams permit you to relish your new mobile electronics in place of spend countless hours attempting to work out which wires goes to which Ford part or component. Overall the wiring is really straight forward. There's a lot wiring that you've got to tie into your truck's wiring harness, but it's much easier to do than it seems. A ground wire offers short circuit protection and there's no neutral wire used. There's one particular wire leading from the distributor which may be used for the tachometer.
When you have just a single cable going into the box, you're at the close of the run, and you've got the simplest scenario possible. All trailer plugs and sockets are extremely easy to wire. The adapter has the essential crosslinks between the signals. Wiring a 7-pin plug on your truck can be a bit intimidating when you're looking at it from beyond the box.
The control box may have over three terminals. After you have the correct size box and have fed the cable to it, you're almost prepared to permit the wiring begin. Then there's also a fuse box that's for the body controls that is situated under the dash. T Diagram For Water. You will find that every circuit has to have a load and every load has to have a power side and a ground side. Make certain that the transformer nameplate power is enough to supply the load that you're connecting.
The bulb has to be in its socket. Your light can be wired to the receiver and don't require supply additional capacity to light as it can get power from receiver. In the event the brake lights aren't working, a police officer may block the vehicle and issue a warning to create the repair within a particular time limit. Even though you would still must power the relay with a power source or battery. Verify the power is off before trying to attach wires. In case it needs full capacity to begin, it won't operate in any way.
Replacing thermostat on your own without a Denver HVAC technician can be quite harrowing if you don't hook up the wiring correctly. After the plumbing was cut out, now you can get rid of the old pool pump. It's highly recommended to use a volt meter to make sure there is no voltage visiting the motor, sometimes breakers do not get the job done properly, also you might have turned off the incorrect breaker. Remote distance is left up to 500m. You may use a superior engine ground. The second, that's the most frequently encountered problem, is a weak ground in the computer system. Diagnosing an electrical short can be extremely tough and costly.
Diagram for pressure tank water treatment diagram hot water recirculating system diagram hot water system diagram home water well diagram water phase diagram diagram for soccer water particle diagram. | 1,440 | 6,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-22 | latest | en | 0.921724 |
http://www.mathacademy.com/pr/prime/browse.asp?LT=L&ANCHOR=cardinalarithmetic000000000000&TBM=Y&TAL=&TAN=&TBI=&TCA=&TCS=&TDI=&TEC=&TFO=Y&TGE=&TGR=&THI=&TNT=&TPH=&TST=&TTO=&TTR=&TAD=&LEV=B | 1,369,505,607,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706082529/warc/CC-MAIN-20130516120802-00004-ip-10-60-113-184.ec2.internal.warc.gz | 598,754,139 | 7,217 | BROWSE ALPHABETICALLY LEVEL: Elementary Advanced Both INCLUDE TOPICS: Basic Math Algebra Analysis Biography Calculus Comp Sci Discrete Economics Foundations Geometry Graph Thry History Number Thry Phys Sci Statistics Topology Trigonometry cardinal arithmetic – closure property cardinal arithmetic Let k and l denote cardinals and A and B denote sets such that k = |A| and l = |B|. Denote by BA the set of all functions from B to A, and by A × B the Cartesian product of A and B. The operations of cardinal arithmetic are then defined byk + l = |A union B|;k • l = |A × B|;kl = |BA|For finite cardinals, this corresponds to the operations of ordinary arithmetic, but is quite different in the case of transfinite cardinals. For instance, if k is not finite, then k + k = k.Cf. ordinal arithmetic. cardinality Two sets X and Y have the same cardinality if there exists a bijection between them. Specifically, the cardinality of X may be understood as a canonical object meant to represent the proper class of sets to which X is bijective. If the axiom of choice is assumed, then the cardinality of X is the unique cardinal number having the same cardinality as X Carroll’s Paradox ARTICLE Paradox published by Lewis Carroll as “What the Tortoise Said to Achilles.” See the article for a full exposition.Cf. Charles Lutwidge Dodgson. Cartesian plane The Cartesian product R2, represented graphically by two real number lines at right angles to one another, with the point (0,0) at the intersection.Used for graphing functions from the set of real numbers to itself. The quadrants, numbered I - IV as shown, indicate the regions of the plane where the x and y axes are positive and negative. Compare: Argand plane. Cartesian product For any collection {Ai}, i = 1, 2, 3, ..., n, of sets, the Cartesian productis the set of ordered n-tuples (a1,a2, ... ,an) with a1 an element of A1, a2 an element of A2, etc. The Cartesian product R2 of the set of real numbers is called the Cartesian plane, and in general n-dimensional real space is the Cartesian product Rn. The assertion that the Cartesian product of an infinite collection of non-empty sets is non-empty is equivalent to the axiom of choice. Cauchy sequence A sequence (x1, x2, x3, ... ) of elements of a metric space X with metric d(x, y) is Cauchy if for any e greater than zero there is some natural number N such thatIn other words, in a Cauchy sequence, the elements eventually become “arbitrarily close together.” If the metric space X is closed, this condition is equivalent to the sequence being convergent. CH See: continuum hypothesis. chain If X is a partially ordered set, then a subset Y of X is called a chain if it is totally ordered, that is, if for any two elements a, b of Y, either a b or b a.Cf. antichain. chain condition For a an infinite cardinal, a partial order P is said to have the a-chain condition if every antichain in P has cardinality not greater than a. If a = w, this is called the countable chain condition, or “c.c.c.” characteristic function Given a subset E of a space X, the characteristic function cE is defined by cE(x) = 1 if x is in E, and cE(x) = 0 otherwise. All properties of sets and set operations may be expressed by means of characteristic functions. choice, axiom of See: axiom of choice. circle In a plane, the locus of all points equidistant from a given point, called the center. The general equation for a circle in the Cartesian plane is given by (x - h) 2 + (y - k) 2 = r 2, where r is the radius of the circle (distance from the center to the locus of points), and (h, k) are the coordinates of the center.The interior of a circle is referred to as an open disk.A circle is also a conic section; a special case of an ellipse in which the foci coincide. Related article: Conics circumference Geometry: The distance around a circle in the plane, or around a great circle of a sphere.Graph Theory: The circumference of a graph G is defined as the length of the longest cycle of G. The circumference is ususally denoted by c(G), and is undefined if G has no cycles. class See proper class. closed General: A set is closed under an operation if applying the operation to its elements returns only elements in the set. For example, the set of integers is closed under addition, since adding two integers always gives another integer, but it is not closed under division, since dividing two integers may result in a non-integer.Geometry: A plane figure is closed if it consists of lines and/or curves that entirely enclose an area. Similarly, a figure in 3-dimensional space is called closed if it entirely encloses a volume.See the following listings for other uses of the word “closed” in mathematics.Topology: A set is topologically closed if it is not open. closed interval An interval of the real number line (or any other totally ordered set) which includes its endpoints. An interval containing only one of its endpoints is called half-open. Cf. open interval. closed set Topology: A subset E of a topological space X is closed if X - E (set difference) is open. In a metric space, E is closed if every convergent sequence in E converges in E; equivalently, if every accumulation point of E is in E.Set Theory: If a is a limit ordinal, then a set C contained in a is called closed if and only if for every limit ordinal b less than a, if C b is unbounded in b, then b C. C is called c.u.b. (“cub set” or “club set”) if and only if C is closed and unbounded in a.Cf. stationary set. closed set system If X is a set (or proper class) and F is a family of subsets of X, then F is called a closed set system providedX is a member of F, andF is closed under arbitrary intersections.Cf. filter. closure Topology: The closure of a subset E of a topological space is the smallest closed set containing E. It may also be expressed as the union of E with its accumulation points. If E is closed, then it is equal to its closure.Algebra: An algebraic closure of a field F is a field G containing F such that every polynomial with coefficients from F has a root in G. closure operator If X is a set, then a function C from P(X) into P(X) (i.e., a function on the power set of X) is called a closure operator providedY is contained in C(Y) for every subset Y of X, C(C(Y)) = C(Y) for every subset Y of X, andIf Y and Z are both subsets of X, with Y a subset of Z, then C(Y) is a subset of C(Z). Closure operators induce closed set systems. closure property A property of subsets of a set X is a closure property if X has the property and the intersection of any subsets of X having the property also has the property. cardinal arithmetic – closure property
HOME | ABOUT | CONTACT | AD INFO | PRIVACYCopyright © 1997-2013, Math Academy Online™ / Platonic Realms™. Except where otherwise prohibited, material on this site may be printed for personal classroom use without permission by students and instructors for non-profit, educational purposes only. All other reproduction in whole or in part, including electronic reproduction or redistribution, for any purpose, except by express written agreement is strictly prohibited. Please send comments, corrections, and enquiries using our contact page. | 1,715 | 7,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2013-20 | latest | en | 0.84243 |
http://math.hws.edu/original-website/TMCM/f95/quiz7.html | 1,542,226,110,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114213308-00248.warc.gz | 216,391,954 | 2,569 | ## CPSC 100, Fall 1995, Quiz #7
This is Quiz #7 from CPSC 100: Principles of Computer Science, Fall 1995. The answers given would receive full credit, but in many cases, there are variations or even completely different answers that would also receive full credit. The "Notes" are meant to clarify the answer or to provide some interesting extra information.
Question 1: Draw the picture that would be produced by the following xTurtle program:
``` fork(5)
turn( 45 * (ForkNumber - 1) )
forward(5)
turn(90)
forward(1)
back(2)
```
Note: There are five turtles. The turn statement then makes them fan out so that turtle 1 is facing 0 degrees, turtle 2 is facing 45 degrees, ..., and turtle 5 is facing 180 degrees. Then each turtle draws a "T" shaped figure.
Question 2: In xTurtle, the grab command is used to control access to shared data. Explain what this means and why it is necessary.
Answer: If a variable is declared before a fork statement, then all the turtles created by the fork will have access to the same location in memory. At some point in the program, it might be necessary to give one turtle exclusive access to that memory location. For example, if a turtle gets the value from that location, does a computation with it, and then stores the result back in memory, it is essential that no other turtle comes has come along and changed the value in memory while the first turtle was working. If all the turtles grab the variable before using it, then this problem will not occur since only one turtle at a time will have access to the variable.
Question 3: "Computers that have many CPU's and that use parallel processing can solve problems that can't be solved by ordinary computers." Is this true or false? Justify your answer.
Answer: This is false. An ordinary computer, with a single CPU, can use multitasking to simulate parallel processing and therefore can solve any problem that can be solved by a computer with multiple CPUs. Of course, it will take longer. (Multitasking means that the single CPU rapidly switches its attention from one process to another.)
Note: You can argue that a parallel processing computer could be so fast that it could solve problems that could not be solved by single-CPU computers in a reasonable amount of time. But that is a practical difference only, and not a difference in the problems that can (given enough time) be solved.
Question 4: What is the Internet ?
Answer: The Internet is a worldwide computer network consisting of thousands of smaller networks connected together (by routers, gateways, modems...). All the computers on this internet can communicate. The Internet provides a number of communication services to its users, including the World Wide Web, USENET news, and FTP file transfers.
Question 5: Explain what is meant by a protocol in the context of computer networks. Give an example of a protocol.
Answer: A protocol is a specification of a specific method of communication between computers on a network. For two computers to communicate, they must follow some mutually agreed protocol. An example is FTP, or File Transfer Protocol, which can be used to copy files from one computer to another.
Note: A protocol is a kind of black box, usually implemented by some subroutines. A process that wants to communicate simply calls the appropriate subroutines and all the details of communication will be handled inside the subroutine. For example, a process might call a subroutine named FTP_Send to send a file using FTP. The subroutine would probably send the file by calling on lower-level protocols such as TCP and IP, but the caller of the subroutine would not have to be aware of exactly how the file is being sent.
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# A mass m at the end of a spring vibrates with a frequency
A mass m at the end of a spring vibrates with a frequency of 0.832 Hz. When an additional 526 g mass is added to m, the frequency is 0.694 Hz. What is the value of m? | 72 | 252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-26 | latest | en | 0.910725 |
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# Serie 2 Electricidad y Magnetismo
1. How much charge is on each plate of a 4.00 F capacitor when it is connected to a 12.0 V battery? (b) If this same capacitor is connected to a 1.50 V battery, what charge is stored? 2. Two conductors having net charges of +10.0 C and -10.0 C have a potential difference of 10.0 V between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to +100 C and -100 C? 3. (a) If a drop of liquid has capacitance 1.00 pF, what is its radius? (b) If another drop has radius 2.00 mm, what is its capacitance? (c) What is the charge on the smaller drop if its potential is 100 V? 4. A 1-megabit computer memory chip contains many 60.0 f F capacitors. Each capacitor has a plate area of 21.010-12 m2. Determine the plate separation of such a capacitor (assume a parallel-plate configuration). The order of magnitude of the diameter of an atom is 10-10 m = 0.1 nm. Express the plate separation in nanometers. 5. When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates? 6. Four capacitors are connected as shown in Figure P26.21. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor if Vab = 15.0 V.
7. Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 F, C2 = 10.0 F, and C3 = 2.00 F.
8. Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 mm. 9. A parallel-plate capacitor in air has a plate separation of 1.50 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 250 V and disconnected from the source. The capacitor is then immersed in distilled water. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and (c) the change in energy of the capacitor. Assume the liquid is an insulator. 10. In the Bohr model of the hydrogen atom, an electron in the lowest energy state follows a circular path 5.2910-11 m from the proton. (a) Show that the speed of the electron is 2.19106 m/s. (b) What is the effective current associated with this orbiting electron? 11. An aluminum wire having a cross-sectional area of 4.0010-6 m2 carries a current of 5.00 A. Find the drift speed of the electrons in the wire. The density of aluminum is 2.70 g/cm3. Assume that one conduction electron is supplied by each atom. 12. A conductor of uniform radius 1.20 cm carries a current of 3.00 A produced by an electric field of 120 V/m. What is the resistivity of the material? 13. Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper. If the wire is to have a resistance of R = 0.500 , and if all of the copper is to be used, what will be (a) the length and (b) the diameter of this wire?
14. A toaster is rated at 600 W when connected to a 120 V source. What current does the toaster carry, and what is its resistance? 15. (a) What is the current in a 5.60 resistor connected to a battery that has a 0.200 internal resistance if the terminal voltage of the battery is 10.0 V? (b) What is the emf of the battery? 16. Two 1.50 V batteries with their positive terminals in the same direction are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.255 , the other an internal resistance of 0.153 . When the switch is closed, a current of 600 mA occurs in the lamp. (a) What is the lamps resistance? (b) What fraction of the chemical energy transformed appears as internal energy in the batteries? 17. The current in a loop circuit that has a resistance of R1 is 2.00 A. The current is reduced to 1.60 A when an additional resistor R2 = 3.00 is added in series with R1. What is the value of R1? 18. (a) Find the equivalent resistance between points a and b in Figure P28.6. (b) A potential difference of 34.0 V is applied between points a and b. Calculate the current in each resistor.
19. The ammeter shown in Figure P28.20 reads 2.00 A. Find I1, I2, and .
20. Determine the current in each branch of the circuit shown in Figure P28.21.
21. In Figure P28.21, show how to add just enough ammeters to measure every different current. Show how to add just enough voltmeters to measure the potential difference across each resistor and across each battery. | 1,244 | 4,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-30 | latest | en | 0.904149 |
https://byjus.com/rs-aggarwal-solutions-class-12-maths-chapter-13-methods-of-integration/ | 1,601,539,674,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00709.warc.gz | 288,484,288 | 123,990 | # RS Aggarwal Solutions for Class 12 Maths Chapter 13: Methods of Integration
The various methods of integrating the given set of questions are explained in brief under this chapter. Students who have difficulty in solving these problems can use the solutions PDF to clarify their doubts. It mainly improves problem-solving abilities among students in order to boost their exam preparation. The solutions are prepared by experts in a step by step manner based on CBSE exam pattern. For a better hold on the concepts covered under this chapter, RS Aggarwal Solutions for Class 12 Chapter 13 Methods of Integration PDF links are available below.
## RS Aggarwal Solutions for Class 12 Maths Chapter 13: Methods of Integration Download PDF
### Exercises of RS Aggarwal Solutions Class 12 Maths Chapter 13 – Methods of Integration
Exercise 13A Solutions
Exercise 13B Solutions
Exercise 13C Solutions
## Access RS Aggarwal Solutions for Class 12 Maths Chapter 13: Methods of Integration
.
Exercise 13a page: 621
Evaluate the following integrals:
1. ∫ (2x + 9) 5 dx
Solution:
Take 2x + 9 = t
So we get
2 dx = dt
It can be written as
2. ∫ (7 – 3x) 4 dx
Solution:
Take 7 – 3x = t
So we get
– 3 dx = dt
It can be written as
Solution:
Take 3x – 5 = t
So we get
3 dx = dt
It can be written as
Solution:
Take 4x + 3 = t
So we get
4 dx = dt
It can be written as
Solution:
Take 3 – 4x = t
So we get
– 4 dx = dt
It can be written as
Solution:
Take 2x – 3 = t
So we get
2 dx = dt
It can be written as
Solution:
Take 2x – 1 = t
So we get
2 dx = dt
It can be written as
Solution:
Take 1 – 3x = t
So we get
– 3 dx = dt
It can be written as
Solution:
Take 2 – 3x = t
So we get
– 3 dx = dt
It can be written as
Solution:
Take 3x = t
So we get
3 dx = dt
It can be written as
By integrating w.r.t. t
Solution:
Take 5 + 6x = t
So we get
6 dx = dt
It can be written as
Solution:
It is given that
By integrating w.r.t. t
Solution:
Take 2x + 5 = t
So we get
2 dx = dt
By integrating w.r.t. t
Solution:
Take sin x = t
So we get
cos x dx = dt
By integrating w.r.t. t
Solution:
Take sin x = t
So we get
cos x dx = dt
By integrating w.r.t. t
Solution:
Take cos x = t
So we get
– sin x dx = dt
By integrating w.r.t. t
Solution:
Take sin -1 x = t
So we get
Solution:
Take tan-1 x = t
So we get
Solution:
Take log x = t
So we get
1/x dx = dt
By integrating w.r.t. t
∫cos t dt = sin t + c
By substituting the value of t
= sin (log x) + c
Solution:
Take log x = t
So we get
1/x dx = dt
By integrating w.r.t. t
By substituting the value of t
= – cot (log x) + c
Solution:
Take log x = t
So we get
1/x dx = dt
By integrating w.r.t. t
By substituting the value of t
= log (log x) + c
Solution:
It is given that
Solution:
Take log x = t
So we get
1/x dx = dt
By integrating w.r.t. t
Solution:
Take √x = t
So we get
1/2√x dx = dt
By integrating w.r.t. t
∫cos t 2 dt = 2 sin t + c
By substituting the value of t
= 2 sin √x + c
Solution:
Take tan x = t
So we get
sec2 x dx = dt
By integrating w.r.t. t
∫et dt = et + c
By substituting the value of t
= e tan x + c
Solution:
Take cos2 x = t
So we get
– sin 2x dx = dt
By integrating w.r.t. t
∫- et dt = – et + c
By substituting the value of t
Solution:
Take ax + b = t
So we get
a dx = dt
It can be written as
Solution:
We know that
cos 3x = 4 cos3 x – 3 cos x
It can be written as
Solution:
Take -1/x = t
So we get
1/x2 dx = dt
By integrating w.r.t. t
∫et dt = et + c
By substituting the value of t
Solution:
Take -1/x = t
So we get
1/x2 dx = dt
By integrating w.r.t. t
∫cos (-t) dt = ∫cos t dt= sin t + c [Since, cos is an even function]
By substituting the value of t
= – sin 1/x + c
Solution:
It is given that
Solution:
Take e2x – 2 = t
So we get
2 e2x dx = dt
By integrating w.r.t. t
Solution:
Take log (sin x) = t
So we get
cos x/ sin x dx = dt
By cross multiplication
cot x dx = dt
By integrating w.r.t. t
Solution:
Take log (sin x) = t
So we get
cos x/ sin x dx = dt
By cross multiplication
cot x dx = dt
By integrating w.r.t. t
∫1/t dt = log t + c
By substituting the value of t
= log (log sin x) + c
Solution:
Take x2 + 1 = t
So we get
2x dx = dt
By integrating w.r.t. t
∫sin t dt = – cos t + c
By substituting the value of t
= – cos (x2 + 1) + c
Exercise 13B page: 658
Evaluate the following integrals:
1. (i) ∫sin2 x dx
(ii) ∫cos2 x dx
Solution:
(i) ∫sin2 x dx
Here, we know that 1 – cos 2x = 2 sin2 x
It can be written as
(ii) ∫cos2 x dx
Here, we know that 1 + cos 2x = 2 cos2 x
It can be written as
2. (i) cos2 (x/2) dx
(ii) cot2 (x/2) dx
Solution:
(i) cos2 (x/2) dx
Here, we know that 1 + cos x = 2 cos2 (x/2)
It can be written as
(ii) cot2 (x/2) dx
Here, we know that cosec2 x – cot2 x = 1
It can be written as
3. (i) ∫ sin2 nx dx
(ii) ∫ sin5 x dx
Solution:
(i) ∫ sin2 nx dx
Here, we know that 1 – cos 2nx = 2 sin2 nx
It can be written as
(ii) ∫ sin5 x dx
Here, we know that 1 – cos2 x = sin2 x
It can be written as
∫sin5 x dx = ∫(1 – cos2 x)2 sin x dx
Take cos x = t
So we get
– sin x dx = dt
By considering cos x = t we get
∫(1 – cos2 x)2 sin x dx = – ∫(1 – t2)2 dt
Here
– ∫(1 – t2)2 dt = – ∫(1 + t4 – 2t2) dt
We get
– ∫(1 – t2)2 dt = – ∫ dt + ∫ 2t2 dt – ∫ t4 dt
By integrating w.r.t t
4. ∫ cos3 (3x + 5) dx
Solution:
It is given that
∫ cos3 (3x + 5) dx
By substituting 3x + 5 = u
We get 3 dx = du
Here dx = du/3
So we get
Now by re-substituting the value of t = sin u and u = 3x + 5
5. ∫ sin7 (3 – 2x) dx
Solution:
We can write it as
= – ∫ sin7 (2x – 3) dx
By substituting the value of 2x – 3 = u
So we get
2 dx = du where dx = du/2
Solution:
Solution:
8. ∫ sin 3x cos 4x dx
Solution:
It is given that
∫ sin 3x cos 4x dx
Using the formula sin x × cos y=1/2(sin(x + y)-sin(y-x))
∫ sin 3x cos 4x dx = ½ ∫ (sin 7x – sin x) dx
On further simplification
∫ sin 3x cos 4x dx = ½ ∫ sin 7x dx – ½ ∫ sin x dx
By integrating w.r.t. x
∫ sin 3x cos 4x dx =
9. ∫ cos 4x cos 3x dx
Solution:
It is given that
∫ cos 4x cos 3x dx
Using the formula cos x × cos y=1/2(cos(x + y)+cos(x-y))
∫ cos 4x cos 3x dx = ½ ∫ (cos 7x + cos x) dx
On further simplification
∫ cos 4x cos 3x dx = ½ ∫ cos 7x dx + ½ ∫ cos x dx
By integrating w.r.t. x
∫ cos 4x cos 3x dx =
10. ∫ sin 4x sin 8x dx
Solution:
It is given that
∫ sin 4x sin 8x dx
Using the formula sin x × sin y=1/2(cos(y-x)-cos(y + x))
∫ sin 4x sin 8x dx = ½ ∫ (cos 4x – cos 12x) dx
On further simplification
∫ sin 4x sin 8x dx = ½ ∫ cos 4x dx – ½ ∫ cos 12x dx
By integrating w.r.t. x
∫ sin 4x sin 8x dx =
11. ∫ sin 6x cos x dx
Solution:
It is given that
∫ sin 6x cos x dx
Using the formula sin x × cos y=1/2(sin(y + x)-sin(y-x))
∫ sin 6x cos x dx = ½ ∫ (sin 7x – sin (-5x)) dx
On further simplification
∫ sin 6x cos x dx = ½ ∫ sin 7x dx + ½ ∫ sin 5x dx
By integrating w.r.t. x
∫ sin 6x cos x dx =
Solution:
We know that
1 + cos 2x = 2 cos2 x
So it can be written as
13. ∫ cos4 x dx
Solution:
14. ∫ cos 2x cos 4x cos 6x dx
Solution:
It is given that
∫ cos 2x cos 4x cos 6x dx
By multiplying and dividing by 2
= ½ ∫ 2 cos 2x. cos 4x. cos 6x dx
We can write it as
= ½ ∫ cos 2x (2 cos 4x cos 6x) dx
On further calculation
= ½ ∫ cos 2x [cos (4x + 6x) + cos (4x – 6x)] dx
So we get
= ½ ∫ cos 2x (cos 10x + cos 2x) dx
By further simplification
= ½ ∫ cos 2x cos 10 x dx + ½ ∫ cos2 2x dx
It can be written as
= ½. ½ ∫ 2 cos 2x. cos 10x dx + ½ ∫ (1 + cos 4x)/2 dx
By further multiplication
= ¼ ∫ [cos (2x + 10x) + cos (2x – 10x)] dx + ¼ ∫ (1 + cos 4x) dx
So we get
= ¼ ∫ [cos 12x + cos 8x] dx + ¼ [x + sin 4x/4]
By integrating w.r.t x
= ¼ [sin 12x/12 + sin 8x/8] + ¼x + sin 4x/16 + c
We get
= sin 12x/48 + sin 8x/32 + x/4 + sin 4x/16 + c
15. ∫ sin3 x cos x dx
Solution:
It is given that
∫ sin3 x cos x dx
By taking sin x = t
We get
cos x = dt/dx
It can be written as
cos x dx = dt
So we get
= ∫ t3 dt
By integrating w.r.t t
= t4/4 + c
By substituting the value of t
= sin4 x/4 + c
16. ∫ sec4 x dx
Solution:
It is given that
∫ sec4 x dx
We can write it as
= ∫ sec2 x sec2 x dx
So we get
= ∫ (1 + tan2 x) sec2 x dx
Take tan x = t
By differentiation we get
sec2 x dx = dt
It can be written as
= ∫ (1 + t2) dt
By integrating w.r.t. t
= t + t3/3 + c
By substituting the value of t
= tan x + tan3 x/3 + c
17. ∫ cos3 x sin4x dx
Solution:
It is given that
∫ cos3 x sin4x dx
We can write it as
= ∫ sin4 x cos2 x cos x dx
By taking sin4 x as common
= ∫ sin4 x (1 – sin2 x) cos x dx
Take sin x = t
We get cos x = dx/dt
Here cos x dx = dt
It can be written as
= ∫ t4 (1 – t2) dt
On further simplification
= ∫ t4 dt – ∫ t6 dt
By integration w.r.t. t
= t5/5 – t7/7 + c
By substituting the value of t
= sin5 x/5 – sin7 x/7 + c
Exercise 13C page: 678
Evaluate the following integrals:
1. ∫ x ex dx
Solution:
It is given that
∫ x ex dx
We can write it as
By integrating w.r.t x
= x ex – ∫ 1 ex dx
So we get
= x ex – ex + c
By taking ex as common
= ex (x – 1) + c
2. ∫ x cos x dx
Solution:
It is given that
∫ x cos x dx
We can write it as
By integrating w.r.t x
= x sin x – ∫ 1 sin x dx
So we get
= x sin x + cos x + c
3. ∫ x e2x dx
Solution:
It is given that
∫ x e2x dx
We can write it as
4. ∫ x sin 3x dx
Solution:
It is given that
∫ x sin 3x dx
We can write it as
5. ∫ x cos 2x dx
Solution:
It is given that
∫ x cos 2x dx
We can write it as
So we get
6. ∫ x log 2x dx
Solution:
It is given that
∫ x log 2x dx
We can write it as
7. ∫ x cosec2 x dx
Solution:
It is given that
∫ x cosec2 x dx
We can write it as
By integrating w.r.t x
= x (- cot x) – ∫ 1 (- cot x) dx
So we get
= – x cot x + ∫ cot x dx
Again by integration we get
= – x cot x + log |sin x| + c
8. ∫ x2 cos x dx
Solution:
It is given that
∫ x2 cos x dx
We can write it as
By integrating w.r.t x
= x2 sin x – ∫ 2x sin x dx
So we get
= x2 sin x – 2 [∫ x sin x dx]
Now apply by the part method
= x2 sin x – 2 ∫ x sin x dx
Again by integration
We get
= x2 sin x – 2 [x (- cos x) – ∫ 1. (- cos x) dx]
By integrating w.r.t x
= x2 sin x – 2 (- x cos x + sin x) + c
On further simplification
= x2 sin x + 2x cos x – 2 sin x + c
9. ∫ x sin2 x dx
Solution:
It is given that
∫ x sin2 x dx
We can write it as
sin2 x = (1 + cos 2x)/ 2
By integration
It can be written as
10. ∫ x tan2 x dx
Solution:
It is given that
∫ x tan2 x dx
We can write it as
tan2 x = sec2 x – 1
So we get
∫ x tan2 x dx = ∫ x ( sec2 x – 1) dx
By further simplification
= ∫ x sec2 x dx – ∫ x dx
Taking first function as x and second function as sec2 x
∫ x sec2 x dx – ∫ x dx
We get
By integration w.r.t x
= {x tan x – ∫ tan x dx} – x2/ 2
It can be written as
= x tan x – log |sec x| – x2/ 2 + c
So we get
= x tan x – log |1/cos x| – x2/ 2 + c
Here
= x tan x + log |cos x| – x2/ 2 + c
11. ∫ x2 ex dx
Solution:
It is given that
∫ x2 ex dx
By integrating w.r.t x
We get
= x2 ex – 2 [x ex – ∫1. ex dx]
By further simplification
= x2 ex – 2 [x ex – ex] + c
By multiplication
= x2 ex – 2x ex + 2 ex + c
By taking ex as common
= ex (x2 – 2x + 2) + c
13. ∫ x2 e3x dx
Solution:
It is given that
∫ x2 e3x dx
We can write it as
14. ∫ x2 sin2 x dx
Solution:
It is given that
∫ x2 sin2 x dx
We know that
Now by integrating the second part we get
15. ∫ x3 log 2x dx
Solution:
It is given that
∫ x3 log 2x dx
We can write it as
16. ∫ x log (x + 1) dx
Solution:
It is given that
Solution:
It is given that
Solution:
It is given that
We can write it as
= t et – ∫1. et dt
So we get
= t et – et + C
Now by replacing t with x2
19. ∫ x sin3 x dx
Solution:
It is given that
∫ x sin3 x dx
We know that
sin 3x = 3 sin x – 4 sin3 x
It can be written as
sin3 x = (3 sin x – sin 3x)/4
We get
So we get
20. ∫ x cos3 x dx
Solution:
It is given that
∫ x cos3 x dx
We know that
cos3 x = (cos 3x + 3 cos x)/4
It can be written as
21. ∫ x3 cos x2 dx
Solution:
It is given that
∫ x3 cos x2 dx
We can write it as
∫ x x2 cos x2 dx
Take x2 = t
So we get 2x dx = dt
x dx = dt/2
22. ∫ sin x log (cos x) dx
Solution:
It is given that
∫ sin x log (cos x) dx
Consider first function as log (cos x) and second function as sin x
So we get
= – cos x log (cos x) – ∫ sin x dx
Again by integrating the second term
= – cos x log (cos x) + cos x + c
23. ∫ x sin x cos x dx
Solution:
It is given that
∫ x sin x cos x dx
We know that
sin 2x = 2 sin x cos x
It can be written as
Consider first function as x and second function as sin 2x
24. ∫ cos √x dx
Solution:
It is given that
∫ cos √x dx
Take √x = t
So we get
1/2√x dx = dt
By cross multiplication
dx = 2 √x dt
Here dx = 2t dt
It can be written as
∫ cos √x dx = 2 ∫t cos t dt
Take first function as t and second function as cos t
By integrating w.r.t t
= 2 (t sin t – ∫ sin t dt)
We get
= 2t sin t + 2 cos t + c
Now by substituting t as √x
= 2√x sin √x + 2 cos √x + c
Taking 2 as common
= 2 (cos √x + √x sin √x) + c
25. ∫ cosec3 x dx
Solution:
It is given that
∫ cosec3 x dx
We can write it as
∫ cosec3 x dx = ∫ cosec x cosec2 x dx
So we get
By integration we get
= cosec x (- cot x) – ∫ (- cosec x cot x) (- cot x) dx
It can be written as
= – cosec x cot x – ∫ cosec x cot2 x dx
Here cot2 x = cosec2 x – 1
We get
= – cosec x cot x – ∫ cosec x (cosec2 x – 1) dx
By further simplification
= – cosec x cot x – ∫ cosec3 x dx + ∫ cosec x dx
We know that ∫ cosec3 x dx = 1
∫cosec3 x dx = – cosec x cot x – ∫ cosec3 x dx + ∫ cosec x dx
On further calculation
2 ∫cosec3 x dx = – cosec x cot x + ∫ cosec x dx
By integration we get
2 ∫cosec3 x dx = – cosec x cot x + log |tan x/2| + c
Here
∫cosec3 x dx = – ½ cosec x cot x + ½ log |tan x/2| + c
27. ∫ sin x log (cos x) dx
Solution:
It is given that
∫ sin x log (cos x) dx
Take cos x = t
So we get
– sin x dx = dt
It can be written as
∫ sin x log (cos x) dx = – ∫log t dt = – ∫1. log t dt
Consider first function as log t and second function as 1
By integrating w.r.t t
= – log t. t + ∫1/t. t dt
Again by integrating the second term
= – t log t + t + c
Now replace t as cos x
= t (- log t + 1) + c
We get
= cos x (1 – log (cos x)) + c
Solution:
Take log x = t
So we get
1/x dx = dt
Here
Again by integrating the second term we get
= t log t – t + c
Replace t with log x
= log x. log (log x) – log x + c
By taking log x as common
= log x (log (log x) – 1) + c
29. ∫ log (2 + x2) dx
Solution:
It is given that ∫ log (2 + x2) dx
We can write it as
= ∫1. log (2 + x2) dx
Consider first function as log (2 + x2) and second function as 1
So we get
Solution:
Consider log x = t
Where x = et
So we get
dx = et dt
We can write it as
32. ∫ e –x cos 2x cos 4x dx
Solution:
Using the formula
cos A cos B = ½ [cos (A + B) + cos (A – B)]
We get
cos 4x cos 2x = ½ [cos (4x + 2x) + cos (4x – 2x)] = ½ [cos 6x + cos 2x]
By applying in the original equation
∫ e –x cos 2x cos 4x dx = ∫ e –x (½ [cos 6x + cos 2x])
= ½ [∫e –x cos 6x dx + ∫e –x cos 2x dx]
Taking first function as cos 6x and cos 2x and second function as e –x
Now by solving both parts separately
33. ∫e √x dx
Solution:
Consider √x = t
We get
It can be written as
dx = 2 √x dt
where dx = 2t dt
We can replace it in the equation
∫e √x dx = ∫et 2t dt
So we get
= 2 ∫t et dt
Consider the first function as t and second function as et
By integration w.r.t. t
= 2 (t et – ∫1. et dt)
We get
= 2 (t et – et) + c
Taking et as common
= 2 et (t – 1) + c
Substituting the value of t we get
= 2 e √x (√x – 1) + c
34. ∫ e sin x sin 2x dx
Solution:
We know that
sin 2x = 2 sin x cos x
So we get
∫ e sin x sin 2x dx = 2 ∫ e sin x sin x cos x dx
Take sin x = t
So we get cos x dx = dt
It can be written as
2 ∫ e sin x sin x cos x dx = 2 ∫ e t t dt
Consider first function as t and second function as et
By integrating w.r.t. t
= 2 (t et – ∫1. et dt)
We get
= 2 (t et – et) + c
Here
= 2 et (t – 1) + c
By substituting the value of t
= 2 esin x (sin x – 1) + C
Solution:
Take sin -1 x = t
Here x = sin t
So we get
By integration we get
= t (- cos t) – ∫1. (- cos t) dt
We get
= – t cos t + sin t + c
It can be written as
cos t = √1 – sin2 t
Here we get
= – t (√1 – sin2 t) + sin t + c
By replacing t as sin-1 x
= – sin-1 x (√1 – x2) + x + c | 6,373 | 16,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2020-40 | latest | en | 0.929285 |
http://www.perlmonks.org/index.pl?node_id=1043713 | 1,524,210,257,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937161.15/warc/CC-MAIN-20180420061851-20180420081851-00043.warc.gz | 502,992,222 | 7,063 | Problems? Is your data what you think it is? PerlMonks
### Re: grouping numbers
by ww (Archbishop)
on Jul 11, 2013 at 12:44 UTC ( #1043713=note: print w/replies, xml ) Need Help??
Serious Error: "can't even get this to work" is not recognized as a valid error description.
Seriously, you'll need to tell us more about what you expected or want; why the output deviates from your heart's desire; and -- tho none appear when I test or execute your code -- error messages or warnings received from code you haven't shown.
If I've misconstrued your question or the logic needed to answer it, I offer my apologies to all those electrons which were inconvenienced by the creation of this post.
Replies are listed 'Best First'.
Re^2: grouping numbers
by ag4ve (Monk) on Jul 11, 2013 at 13:03 UTC
Seriously? The code runs fine for me.
The expected output of the code or what I'd like. Since I'm not sure if I'm on the right track with the code, what I'd like are three sets with 5,7,10,10, 50, and 70,72,75,79,80.
I'm not sure how else to explain this - between the code I'm stuck on, the use case, and abstract....?
What is significant about those groups of numbers?
Michael
Oh, they're 'close' together.
Not sure if this is any use but is does give the expected output. It loops through the sorted list working out the effect of adding each to 'current group average'. If change is greater than the value 'diff' it starts a new group
```#!perl
use strict;
my @data = (10,7,5,10,50,70,75,72,79,80);
my @arr = sort {\$a<=>\$b} @data;
my \$diff = 2; # group closeness
my @avg; #[count,sum]
my @grp;
my \$g=0;
for my \$i (0..\$#arr){
my \$val = \$arr[\$i];
#print "value \$val\n";
if (\$i == 0){
push @{\$grp[\$g]},\$val;
\$avg[\$g] = [1,\$val];
} else {
#work out new average with this element
my (\$n,\$sum) = @{\$avg[\$g]};
#print "count \$n sum \$sum\n";
my \$avg = \$sum/\$n;
my \$new_avg = (\$sum+\$val)/(\$n+1);
if (abs(\$new_avg - \$avg) < \$diff){
# join group
push @{\$grp[\$g]},\$val;
\$avg[\$g] = [\$n+1,\$sum+\$val];
} else {
# start new group
++\$g;
push @{\$grp[\$g]},\$val;
\$avg[\$g] = [1,\$val];
}
}
}
for (@grp) {
print join ',',@\$_,"\n";
}
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Notices? | 793 | 2,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-17 | latest | en | 0.899958 |
https://www.datasciencenotes.com/2015/10/ku-football-chase-for-winless-season.html | 1,642,534,516,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00109.warc.gz | 788,815,962 | 13,281 | ## Monday, October 5, 2015
### KU Football: The Chase for a Winless Season
Last week I posted on the probability of the University of Kansas Football team losing every game this season. Since then, the team lost again, thus increasing the probability of a winless season. I also found some great information on the topic of futility in college football.
## BACKGROUND
I found some great information on horrible college football teams since I last posted, scouring the internet for infomation. Two great pieces:
• List of Winless Seasons. Winless seasons in college football are only pseudo-rare. By that I mean it happens fairly often (a couple teams each year), but it's still rare enough that we can keep a list of it. That list is interesting, and even includes seven (SEVEN!) winless seasons by my undergraduate alma-mater, Kansas State University. Also, the toilet bowl is an interesting clickhole I found myself in.
• The Bottom 25. This is a CBS attempt to rank the worst teams in college football. Guess who is considered the worst right now? The University of Kansas. An interesting insight from this analysis is that KU had a better chance to win its first Big 12 game this year more than any other Big 12 game. They already lost this game, so I need to change methodology to maintain an accurate probability estiamte.
## METHOD CHANGE
My prior methodology gives a good estimate of how likely it is that KU will go winless based on historical probabilities. This is especially true in the first week when the entire Big 12 season is laid out in front of us. But there is one bias to the estimate: KU is significantly more likely to win some games than other games. Complicating this, is that KU's statistically easiest game was the first game of the Big 12 season, after losing which, the probability of going winless increases dramatically.
While a team that only wins 6.8% of their games doesn't see a drastic difference in probabilities from game to game (wins are essentially a "fluke" and not as tied to opposition talent as competitive teams), we can still estimate the relative strengths of teams, using their individual probabilities to lose, and then weighting KU's probability by that. This methodology looks at the historical performance of each team, and adjusts the 6.8% by that performance. Here are the probabilities that each team will win their KU matchup this year:
The other worst team in the league (ISU) only has a 86.8 percent chance of beating KU, whereas the best team (Baylor) has a 98.3 percent chance to beat KU. Relatively speaking, this means KU is about 8 times more likely to beat ISU than Baylor.
Because KU played ISU first, their probability of a winless season increased from 53% to 61%. My initial model only had it at 57% following a conference week 1 loss, but that flat model didn't account for Iowa State's poor play. | 672 | 2,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-05 | latest | en | 0.959162 |
https://www.bogleheads.org/forum/viewtopic.php?f=10&t=265654&newpost=4247069 | 1,556,183,452,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578711882.85/warc/CC-MAIN-20190425074144-20190425100144-00225.warc.gz | 625,706,835 | 15,658 | ## Lognormal Inflation Model in Monte Carlo Retirement Planning
Discuss all general (i.e. non-personal) investing questions and issues, investing news, and theory.
Topic Author
K72
Posts: 35
Joined: Wed Dec 05, 2018 8:04 pm
### Lognormal Inflation Model in Monte Carlo Retirement Planning
New user, first post. From the posts I've read this looks like a great forum. Here goes. I've gained a lot of insight using Monte Carlo simulations for retirement planning, especially the impact of volatility. I've used one online tool and have had two different financial planners with access to commercial tools create financial plans. In every case I've been surprised (and disappointed) that for the most part, inflation is treated as a fixed value for the duration of the simulation. One was with a well known financial institution (when I inquired about how to include volatility with inflation all I was told was that I could request any fixed value I wanted for inflation). Another was with a private financial planner who had a fancy tool with flashing arrows showing movement of \$ from one bucket to another but he wasn't able to explain anything at all about what the tool actually does. The online tool used either historical inflation (but no detail about what historical meant) or you could input a mean and standard deviation (for a normal distribution most likely but it wasn't clear).
Needless to say I decided to try it on my own. Plotting CPI values for the past 80 years (maybe less I don't remember exactly), the histogram sure looked lognormal to me so I went ahead and made my own Monte Carlo model in Excel using a normal distribution for investment return and lognormal for inflation (parameters were determined via least squares). I used separate random values for each computation. Using a normal distribution for inflation doesn't make sense to me because in reality we haven't seen many instances of deflation. What I did was cap the upper limit at 16% (I remember too well the early 80's) with no possibility of deflation. I did correlation analysis between CPI and S&P 500 return and found very weak correlation at best.
Even though my inflation model may be overly pessimistic, I'm actually quite happy with the result but I want to be cautious about thinking I'm doing something sensible if I'm really not. Does this approach seem reasonable? I'd appreciate comments either way. Thanks!
kolea
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### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
No, it does not seem reasonable. Inflation is not an independent process from equities or bonds. They both respond to what inflation is doing, although somewhat differently. For a MC simulation, the key is not finding the right PDF model for inflation, it is finding the right correlation between inflation and bonds and equities. That is not to say that treating inflation as a constant is the right thing to do either, after all, the historic nominal PDF for equity returns has a built-in dependence upon inflation. If I were writing a MC simulation, I would keep inflation a constant but use a real PDF for equity returns and a real PDF for bond yields. But even that may be flawed since equities, bond yields, and inflation all have memory, i.e., they have a non-zero auto-correlation. We see this as momentum in equities. Similar for bonds. It is really tricky to create a MC simulation that correctly models reality. That is why I don't trust them.
Kolea (pron. ko-lay-uh). Golden plover.
AlohaJoe
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### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
Ken72 wrote:
Wed Dec 05, 2018 11:23 pm
Using a normal distribution for inflation doesn't make sense to me because in reality we haven't seen many instances of deflation.
I don't understand this sentence. If I have a normal process with mean of 15% and standard deviation of 2% it will never go below 0 but will still be a normal distribution. Whether a distribution is normal or not is unrelated to whether values ever go negative. That said, inflation has been repeatedly shown to fail tests for normality, since it has significant skewness & kurtosis. (Here is one of many entries on the subject: https://www.nber.org/papers/w4837.pdf) That said, pretending it is normally distributed is also probably okay-ish, depending on what you're trying to learn from the model.
What I did was cap the upper limit at 16% (I remember too well the early 80's) with no possibility of deflation.
Since there have repeatedly been periods of deflation (more common than high inflation, for that matter), it seems weird to eliminate the possibility.
Anyway, for a model it is fine-ish. It's not really any worse than assuming that equity returns are normal or lognormal. Or that there is no correlation between things. Or that there's no time-varying correlations. Or that there's no short-term momentum or long-term mean reversion. And all the other simplifying assumptions models make.
randomguy
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Joined: Wed Sep 17, 2014 9:00 am
### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
AlohaJoe wrote:
Thu Dec 06, 2018 12:36 am
Ken72 wrote:
Wed Dec 05, 2018 11:23 pm
What I did was cap the upper limit at 16% (I remember too well the early 80's) with no possibility of deflation.
Since there have repeatedly been periods of deflation (more common than high inflation, for that matter), it seems weird to eliminate the possibility.
Anyway, for a model it is fine-ish. It's not really any worse than assuming that equity returns are normal or lognormal. Or that there is no correlation between things. Or that there's no time-varying correlations. Or that there's no short-term momentum or long-term mean reversion. And all the other simplifying assumptions models make.
he remembers the 80s. Very unlikely anyone remembers the long deflationary periods in US history
I agree fixed inflation is bad. The question is this remotely better or are we just spewing out more numbers. I haven’t actually looked at I would expect a pretty strong link between bond returns and nominal inflation. Stocks to some extent also.
AlohaJoe
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### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
randomguy wrote:
Thu Dec 06, 2018 12:57 am
AlohaJoe wrote:
Thu Dec 06, 2018 12:36 am
Ken72 wrote:
Wed Dec 05, 2018 11:23 pm
What I did was cap the upper limit at 16% (I remember too well the early 80's) with no possibility of deflation.
Since there have repeatedly been periods of deflation (more common than high inflation, for that matter), it seems weird to eliminate the possibility.
Anyway, for a model it is fine-ish. It's not really any worse than assuming that equity returns are normal or lognormal. Or that there is no correlation between things. Or that there's no time-varying correlations. Or that there's no short-term momentum or long-term mean reversion. And all the other simplifying assumptions models make.
he remembers the 80s. Very unlikely anyone remembers the long deflationary periods in US history
But why is "long periods" and "US history" the important parts? Much of Europe & Japan has had deflationary periods for the past decade, as has been much discussed over the past years. Why pretend that can never ever possibly happen in the US when modeling a multi-decade period? Especially when we had deflation as recently as March 2018? There's also been deflation in December 2017, October 2017, July 2017, November 2016, July 2016. Is it so hard to imagine that those short periods might stretch from a few months to 12 months? Meanwhile, where are the occurrences of inflation being 16%? The highest annual inflation ever got in the US was 13%. So why allow inflation to go higher than it ever has but not allow deflation to even reach the levels it actually did?
It just doesn't seem like an honest, impartial decision. It seems like one that is "I'm scared of inflation so I'm going to bias my model towards it". Which is fine. Models are made for that kind of exploration. But then you start getting on tricky ground when ask on the internet about how realistic your model is.
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Joined: Thu Dec 06, 2018 1:59 am
### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
New member too. Great forum. Agree that zero correlation is very simplistic especially with respect to interest rates. I would assume in your model equity vol is much higher than inflation vol so pair correlation should matter less there.
Modeling real return as mentioned it’s a good alternative - (assuming normal distribution for real rates as they do go negative)
Regarding CPI distribution I would go for a shifted log normal to allow for sone negative inflation reading while maintaining a fatter tail on the right.
On the other hand constant inflation is even worse. Out of interest, what MonteCarlo tool are you using?
Valuethinker
Posts: 37626
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### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
Ken72 wrote:
Wed Dec 05, 2018 11:23 pm
New user, first post. From the posts I've read this looks like a great forum. Here goes. I've gained a lot of insight using Monte Carlo simulations for retirement planning, especially the impact of volatility. I've used one online tool and have had two different financial planners with access to commercial tools create financial plans. In every case I've been surprised (and disappointed) that for the most part, inflation is treated as a fixed value for the duration of the simulation. One was with a well known financial institution (when I inquired about how to include volatility with inflation all I was told was that I could request any fixed value I wanted for inflation). Another was with a private financial planner who had a fancy tool with flashing arrows showing movement of \$ from one bucket to another but he wasn't able to explain anything at all about what the tool actually does. The online tool used either historical inflation (but no detail about what historical meant) or you could input a mean and standard deviation (for a normal distribution most likely but it wasn't clear).
Needless to say I decided to try it on my own. Plotting CPI values for the past 80 years (maybe less I don't remember exactly), the histogram sure looked lognormal to me so I went ahead and made my own Monte Carlo model in Excel using a normal distribution for investment return and lognormal for inflation (parameters were determined via least squares). I used separate random values for each computation. Using a normal distribution for inflation doesn't make sense to me because in reality we haven't seen many instances of deflation. What I did was cap the upper limit at 16% (I remember too well the early 80's) with no possibility of deflation. I did correlation analysis between CPI and S&P 500 return and found very weak correlation at best.
Even though my inflation model may be overly pessimistic, I'm actually quite happy with the result but I want to be cautious about thinking I'm doing something sensible if I'm really not. Does this approach seem reasonable? I'd appreciate comments either way. Thanks!
Do you escape this problem by using real returns?
At least on an annual basis, CPI and S&P 500 would have very weak correlation. Dimson & Marsh in one of their Credit Suisse presentations (available online) has a slide on that. My guess would be that rising inflation actually has a negative correlation with S&P 500 (a guess).
(I think the intellectual confusion comes because of this: nominal bonds very clearly have a negative correlation with inflation, especially rising inflation - it's bad for bondholders; stocks pay high real returns but that does not mean inflation is good for stock returns (rising inflation in particular) because it tends to bring higher interest rates, which are generally bad for stocks (directly due to the alternative use of capital, indirectly due to the impact on the economy in general) also there is an "inflation drag" in corporate profits and taxes (inflation causes an inflation of accounting and taxable profits, but not in an economic sense).
It's very easy to confuse the fact that stocks pay high real returns with stocks as an inflation hedge. The important thing about stock returns is that they are *volatile* - that real return is paid for with a lot of risk. You can lose 25% of your portfolio in one day (fortunately in 1987 I did not have a lot of investments). And you can lose 50% in a very short period of time (in 1973-75 in the UK, you lost something like 80% in real terms). It's a well argued fact about stock markets that returns have been excessive (Mehra and Prescott 1986) -- the best guess I have seen is that investors are averse to short term losses in real wealth, thus safe assets are overvalued and risky assets are undervalued.
TIPS bonds should be strongly correlated with CPI, real estate a higher correlation than stocks (rents are linked to inflation, so are land prices in the long term). I am not sure the conditions which produced the general inflation of the 1970s following large price rises in commodities would be repeated - I am less sure a commodity is an inflation hedge.
Raybo
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Contact:
### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
I've written a couple Monte Carlo simulations of my retirement assets and market returns. Like you, I used a normal distribution for both bonds and stock returns. I chose to use a normal distribution for inflation, as well. I capped it a -2% but let the upside run, as high inflation is a real retirement worry. I made no attempt to correlate any of these. I just pulled a value from each distribution when I needed one.
Some thoughts:
I knew normal distributions wasn't close to accurate. But, when I did these simulations, I didn't have access to real market data going back decades. Now, I would likely gather the data on a monthly basis and simply put it into a customized step function for each distribution (stock, bond, inflation). I have been musing with also adding a modifying factor by analyzing the how one month's returns influences the next month's returns. I was never comfortable using normal distributions as my underlying data sources.
While investment returns and inflation are prime drivers of retirement survival, there are other important variables to consider. For example, I added a simplified federal tax system to my program. As I took money out of tax-deferred accounts to pay for retirement expenses, I calculated how much tax I'd owe on it. Taxes are not insignificant when funding retirements out of tax-deferred funds.
When I wrote my programs, I was still working. I added contributions to tax-deferred accounts into the model as well. If you are currently retired and not earning any income or putting money into IRAs, then this isn't important. But, what about Roth conversions?
I also added social security, both income and taxes on that income, to my calculations.
The data my model produced was a table of years I ran out of money and the (average) returns and inflation values that led to them (capped at 100 years of age). This way I not only got a graph of my possible outcomes, but I could also see what conditions would lead to the failure of my retirement funds.
I haven't revisited this simulation in many years as it was designed to help me plan my retirement. Since I retired in 2000, I haven't found it necessary. But, of late, the programming bug has bit me again and now that market return data is readily available and should the level of interest rise above my level of indolence, I might take another crack at it.
No matter how long the hill, if you keep pedaling you'll eventually get up to the top.
Topic Author
K72
Posts: 35
Joined: Wed Dec 05, 2018 8:04 pm
### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
I'd like to thank everyone that responded. It was helpful for pointing out that (1) I hadn't provided sufficient background/explanation as to what I was doing and (2) I'd made some assumptions w/o realizing I had. In event, I'd like to respond with clarification/comment:
1. I'm not retired but plan to be within a few months. My model includes estimates for expenses and includes income from pensions, Social Security, and RMD's from IRA's. I have no Roth.
2. I'm using inflation for the sole purpose of adjusting projected annual expenses. I really didn't want to use a fixed value because I don't think it sufficiently represents future reality.
3. For the inflation model, I used the U.S. CPI from about 1940 to present (no particular rationale for choosing 1940), which is how I got the lognormal looking distribution. I now realize I could go much further back (I did find some estimates of inflation back to 1800) and see the distribution is much more normal-ish with plenty of deflationary periods. This begs a question as to what period is reasonable to use for modeling inflation. I don't know the answer but I wanted to use something other than a fixed value.
4. I realize that a normal distribution of 15 with a sigma of 3 won't see much negative but I wouldn't think a model with this as inflation would be very useful. I would probably choose more like a mean of 4 with sigma of 2 or 3 but that would have plenty of negative. I just can't see using much negative inflation in a retirement model even though I know it could very well happen. I don't mind a bias toward higher inflation for the purpose of expense projection.
5. I made my own MC sim using Excel (so I could include the complexity of pensions and social security with cost of living assumptions, RMD's from IRA accounts and taxes). The online tool I used was https://www.portfoliovisualizer.com/mon ... simulation which has a variety of other tools that seem to be useful.
6. If I don't use MC then I'm not sure what to do for retirement planning to estimate if the money will last for 30 years (also an arbitrary choice on my part). I first tried fixed investment return and fixed inflation but quickly determined that all would be well as long as the investment return was higher than inflation by some margin. Also, I don't think I have the ability to predict investment returns or inflation for any future period of time. In the end I found MC to be a satisfying way to gain insight into retirement planning. Of course I realize whatever model is used is based on a lot of assumptions, but again, I'd rather do something rather than nothing and don't know what other approach to take.
Topic Author
K72
Posts: 35
Joined: Wed Dec 05, 2018 8:04 pm
### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
I also have a question. Several responses mentioned "real return". Can someone explain what this is? If it means investment return minus inflation, then I understand.
randomguy
Posts: 7411
Joined: Wed Sep 17, 2014 9:00 am
### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
AlohaJoe wrote:
Thu Dec 06, 2018 2:40 am
randomguy wrote:
Thu Dec 06, 2018 12:57 am
AlohaJoe wrote:
Thu Dec 06, 2018 12:36 am
Ken72 wrote:
Wed Dec 05, 2018 11:23 pm
What I did was cap the upper limit at 16% (I remember too well the early 80's) with no possibility of deflation.
Since there have repeatedly been periods of deflation (more common than high inflation, for that matter), it seems weird to eliminate the possibility.
Anyway, for a model it is fine-ish. It's not really any worse than assuming that equity returns are normal or lognormal. Or that there is no correlation between things. Or that there's no time-varying correlations. Or that there's no short-term momentum or long-term mean reversion. And all the other simplifying assumptions models make.
he remembers the 80s. Very unlikely anyone remembers the long deflationary periods in US history
But why is "long periods" and "US history" the important parts? Much of Europe & Japan has had deflationary periods for the past decade, as has been much discussed over the past years. Why pretend that can never ever possibly happen in the US when modeling a multi-decade period? Especially when we had deflation as recently as March 2018? There's also been deflation in December 2017, October 2017, July 2017, November 2016, July 2016. Is it so hard to imagine that those short periods might stretch from a few months to 12 months? Meanwhile, where are the occurrences of inflation being 16%? The highest annual inflation ever got in the US was 13%. So why allow inflation to go higher than it ever has but not allow deflation to even reach the levels it actually did?
It just doesn't seem like an honest, impartial decision. It seems like one that is "I'm scared of inflation so I'm going to bias my model towards it". Which is fine. Models are made for that kind of exploration. But then you start getting on tricky ground when ask on the internet about how realistic your model is.
Because you don't notice short periods of time and obviously the US is exceptional:) And again the deflation is at the point that nobody notices it as 4 months of 1% deflation (roughly what we had at the end of 2008) doesn't get much notice. It just isn't a big enough value. You could use international deflation but to fair then you should also end up using international inflation. And probably international bond and stock returns. And yes to be honest, you should use what the data is not what you wish it to be.
Personally I think all of this complexity is going to buy you basically nothing. You want to know what your SWR for a portfolio is for the next 30 or so years? About 5.5%+-2%. I don't think we can actually get much more accurate than that. Since you want to be conservative pick something down in the 3.5-4% range and plan on adjusting as life happens. One of the toughest things to accept is that you can't control everything.
AlohaJoe
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### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
Ken72 wrote:
Thu Dec 06, 2018 10:39 pm
6. If I don't use MC then I'm not sure what to do for retirement planning to estimate if the money will last for 30 years (also an arbitrary choice on my part).
I agree with randomguy that there is limited value in trying to be too precise. I don't think there's anything wrong with using Monte Carlo. It is one tool among many.
There are tons of retirement calculators out there. By all means, build your own for fun. But most people are suited by just using http://cfiresim.com/ or https://www.i-orp.com or all the other ones you can Google. Most people use several and hope they all agree
kolea
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Location: Maui and Columbia River Gorge
### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
Ken72 wrote:
Thu Dec 06, 2018 11:37 pm
I also have a question. Several responses mentioned "real return". Can someone explain what this is? If it means investment return minus inflation, then I understand.
Yes, real return is the return that has been adjusted for inflationary effects. It is not strictly "subtracting" inflation since it is polynomial.
Kolea (pron. ko-lay-uh). Golden plover.
Valuethinker
Posts: 37626
Joined: Fri May 11, 2007 11:07 am
### Re: Lognormal Inflation Model in Monte Carlo Retirement Planning
Ken72 wrote:
Thu Dec 06, 2018 11:37 pm
I also have a question. Several responses mentioned "real return". Can someone explain what this is? If it means investment return minus inflation, then I understand.
Simple formula
nominal return = real return + inflation
Correct formula (the Fisher formula)
(1+nominal) = (1+ real) times (1+ inflation)
ie 7% inflation becomes 1+0.07 = 1.07 in the second term of that equation
or
real return= ((1+nominal)/(1+inflation))-1 answer will be a decimal (multiply by 100 to get a percentage)
The simple formula is fine for talking about it or thinking about it, but errors will creep in when compounded over a number of years. So in a spreadsheet, the Fisher formula is better. | 5,517 | 24,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-18 | latest | en | 0.963604 |
http://www.drumtom.com/q/what-are-the-ordered-pairs-f-x-2-x-1-squared-minus-5 | 1,488,268,140,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174154.34/warc/CC-MAIN-20170219104614-00050-ip-10-171-10-108.ec2.internal.warc.gz | 379,028,331 | 9,124 | # What are the ordered pairs? f(x)=2(x-1)squared minus 5?
### Answer this question
• What are the ordered pairs? f(x)=2(x-1)squared minus 5?
If we substitute the ordered pair ( x,y ) = (1 ,2) ... which of the given equations is a linear equation? 17 . y= 6 x2 +4, ... 2x−1 and y= 5 2x−2
Positive: 32 %
A summary of Ordered Pairs in 's Graphing Equations. ... or section of Graphing Equations and what it means. ... (x, y) = (5, 2): ...
Positive: 29 %
### More resources
Solutions of Equations in Two Variables 6.1 ... So the pairs x 5, y 0 and x 1, ... Which of the ordered pairs, (2, 0), (0, 2), (5, 2), (2, 5), and ...
Positive: 32 %
An equation or grouping of ordered pairs ... (x) = x2 + 5, find f(-2). f(x) = x2 + 5 2 f(-2) = (-2) + 5 ... m x 1 = -2 x 2 = 5 y 1 = 9 y 2
Positive: 27 %
Let function f be defined by the set of ordered pairs as follows: f = {(1,0),(4,5), ... f(x) = a*(b(x+c)) 2 + d ... both for f(x) and its inverse function ... | 348 | 953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-09 | longest | en | 0.763577 |
http://mathhelpforum.com/calculus/122605-coefficient-problem-print.html | 1,506,173,184,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689661.35/warc/CC-MAIN-20170923123156-20170923143156-00661.warc.gz | 218,017,152 | 2,663 | # Coefficient problem
• Jan 6th 2010, 01:14 AM
TomJerry
Coefficient problem
If any three successive coefficients in the expansion of $(1+x)^n$ are 36, 84 and 126. find n
I dont know how to begin this problem should i do a series of what ???
• Jan 6th 2010, 01:48 AM
Prove It
Quote:
Originally Posted by TomJerry
If any three successive coefficients in the expansion of $(1+x)^n$ are 36, 84 and 126. find n
I dont know how to begin this problem should i do a series of what ???
Draw Pascal's Triangle and see which row those three numbers appear in. | 158 | 553 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-39 | longest | en | 0.899185 |
https://socratic.org/questions/58f4d6e37c0149085c753cfd | 1,566,091,305,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00440.warc.gz | 675,181,563 | 5,819 | # Question #53cfd
Sep 26, 2017
${x}^{2} + {y}^{2} - 6 x - 10 y + 18 = 0$
#### Explanation:
Equation of a circle,given center coordinates and radius is
${\left(x - {x}_{1}\right)}^{2} + {\left(y - {y}_{1}\right)}^{2} = {r}^{2}$
Here ${x}_{1}$= 3 & ${y}_{1}$ = 5.
$\therefore$ Equation of the required circle is
${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = {4}^{2}$
${x}^{2} - 6 x + 9 + {y}^{2} - 10 y + 25 = 16$
${x}^{2} + {y}^{2} - 6 x - 10 y + 18 = 0$ | 221 | 467 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-35 | latest | en | 0.558491 |
http://clay6.com/qa/24140/if-three-numbers-are-in-gp-then-their-logarithms-will-be-in | 1,576,316,831,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540586560.45/warc/CC-MAIN-20191214094407-20191214122407-00220.warc.gz | 31,952,778 | 9,348 | Comment
Share
Q)
# If three numbers are in GP, then their logarithms will be in
$(a)\;AP\qquad(b)\;GP\qquad(c)\;HP\qquad(d)\;None\;of\;the\;above$
Explanation : a,b,c are in GP $\frac{b}{a}=\frac{c}{b}$
$b^2=ac$
$log\;(b^2)=\;log(ac)$
$2\;log\;b=log\;a+log\;c=log\;b+log\;b$
$log\;b-log\;a=log\;c-log\;b$
Therefore , $log\;a,log\;b,log\;c\;are\;in\;AP.$ | 172 | 356 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-51 | latest | en | 0.284189 |
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For more from Chase and Jason, check out their work at Football Perspective and The Big Lead.
# Pro Football Reference Blog
## How to fill out your brackets
Posted by Doug on March 12, 2007
So you're not much of a college basketball fan. You follow your alma mater, and possibly keep loose tabs on the rest of their conference, but that's about as far as it goes. But the tourney is good entertainment and, as is customary, you enter a bracket pool so you can have a rooting interest where none would otherwise exist. How do you maximize your chances winning the thing?
If you're like me, the first thing you do is you head someplace like this smorgasboard of computer ranking algorithms and check out a few of them to get a quick feel for which teams appear to be over- or under-seeded. Some of them even do the work for you by putting a specific probability estimate on each team's chances of advancing to each round.
Whatever the rules of your bracket pool, you probably get some sort of score associated with your entry. And the highest score wins. In most pools, you can use estimates like those above to compute (at least approximately) the expected score of each possible entry. Now simply find the entry with the highest expected score and turn it in.
That's what I used to do. Only recently did I realize that that's wrong. Maximizing your expected score is not the same as maximizing your chance of having the highest score. Your goal is the latter, not the former.
To see why they're not the same, imagine a simple pool where you are simply trying to pick the winner of the tournament. Let's say that in the very likely event of a tie, the winner will be selected randomly from among those who correctly picked the champion. You believe these are the probabilities of each team winning the tourney:
Ohio State: 25%
UCLA: 20%
Kansas: 15%
UNC: 15%
Florida: 10%
Texas A&M: 10%
Washington State: 5%
The "score" of your entry in this simple pool is either one or zero, depending on whether you pick the champ correctly or not. So the entry with the highest expected score is Ohio State. But Ohio State might or might not be the entry that maximizes your chance of winning the pool. It depends on who everyone else picked. If you were the only Buckeye-picker, then great. But if 90% of the other pool participants picked Ohio State, then you'd be better off picking Washington State.
So, while Ohio State is the "best" pick in some sense, it's also likely to be a "crowded" pick, and that's the problem. You may be better off going with a "worse" pick, if it's a pick that's less popular. That's a simple example, but the same issues are present in a real pool. Even if there aren't necessarily ties, the best picks are also going to be the most popular picks, and that's going to cause the same kind of crowding. If you pick the entry that you believe is most likely to occur, then there will be lots of other entries that look very similar to yours. This is problematic because you know you're going to miss on a lot of games. And if your entry is too centrist, it's likely that there will be an entry that looks just like it except that it got a few of the games you missed.
The other extreme is to pick an entry with Cinderellas and longshots aplenty. This avoids the crowding problem. With a wacky entry, even if you miss a lot of games, there are not likely to be many entries close to yours to capitalize on your mistakes. The problem here is that, if you turn in a wacky entry, you probably won't end up being even close. That's what makes it a wacky entry.
To make this a little more concrete, imagine two extreme strategies:
Strategy #1: pick a final four with two #1 seeds, a #2 seed and a #3 seed.
Strategy #2: pick a final four with two #9 seeds and two #7 seeds.
The upside of Strategy #1 you're very likely to hit at least a couple of the final four teams. The downside is that, if your final four hits, you're probably not the only one who has it.
The upside of Strategy #2 is that, even if you just get one or two of the final four teams correct, you're probably still doing better than everyone else. The downside is that you're not likely to hit even one.
And of course you don't have to be at one extreme or the other. There is a continuum of possibilities in between. So where do you want to position yourself? You can't answer that question unless you know what the other entries in your pool look like, and you're probably not going to know that. So you have to make some assumptions.
If it's a big contest with a mixture of hardcore and casual fans, I think it's reasonable to expect that the entries will generally cluster around the most likely outcomes, but that there will be some longshot entries mixed in. With that in mind, I'm going to make the following assumption:
Assume the entries in your pool are distributed the same as the distribution of actual outcomes of the tournament.
Roughly speaking, what this means is that, if you think Ohio State as a 25% chance of winning the tourney, then about 25% of the pool's participants will pick Ohio State to win it. If you think there is a 1% chance of a final four consisting of Florida, UCLA, Texas A&M, and Georgetown, then about 1% of the pool's entries will have that for a Final Four. If you think Virginia Tech has a 59% chance of beating Illinois in the first round, then around 59% of the entries will have Virginia Tech beating Illinois. And so on.
Is this a reasonable assumption? I think it's at least in the ballpark. Yahoo.com publishes the entries in its Tournament pick 'em contest and they match up reasonably well with objectively-generated probabilities (e.g. from Sagarin ratings and the like). Not perfectly, but reasonably. This shouldn't be too surprising. Sports gambling markets are often cited as an example of the wisdom of crowds and are generally believed to be pretty efficient.
So let's go back and apply this assumption to our drastically simplified pool, where we are only picking the champion. If these are the probabilities of each of these teams winning the title:
Ohio State: 25%
UCLA: 20%
Kansas: 15%
UNC: 15%
Florida: 10%
Texas A&M: 10%
Washington State: 5%
Then our assumption would imply that the above is also the distribution of entries. Twenty-five percent of the people would take Ohio State, 20% UCLA, and so on. If that's the case, then what is the best pick?
There is no best pick! Your chances of winning are the same no matter who you pick.
If there are 100 entries for example, then 25 of them took Ohio State. So if you are one of those 25 riding the Buckeyes, your chances of winning are 1%: a 25% chance they'll win, and then a 1-in-25 chance that you'll win the tiebreaker. If you take Washington State, you've also got a 1% chance of winning: 5% chance of the Cougars winning, then a 1-in-5 chance of winning the tiebreaker. Regardless of which team you look at, the analysis will turn out the same: you have a 1% chance of winning. One percent, of course, is one of a hundred, because you are one of a hundred people in the pool.
But that's an oversimplified situation. What happens in more complicated settings?
As many of you know, I teach math for a living. Last summer, I got a student and a colleague interested in investigating this question with me. Some very interesting (to us, anyway) mathematics arose from the investigation.
As an abstract model of the tournament prediction problem, we imagined the following game. Suppose that a random number, called the target, is to be chosen. Millions of participants will guess what the number will be, and whoever guesses closest is the winner. Let's say, just for example, that it is to come from the standard normal distribution. So there is about a 2/3 probability that the target will be between -1 and 1, a 95% chance that it will be between -2 and 2, a 99% chance that it will be between -3 and 3, and so on. Your job is to guess closer to the target than any other competitor, and let's assume that their guesses are distributed as independent standard normals as well. In other words, two-thirds of the guesses will be between -1 and 1, 95% between -2 and 2, and so on.
If you guess near zero, then you are likely to be close to the target. But you are also likely to be crowded out by the multitudes of other guesses that are in the same vicinity. If you make a guess far out in the tail, like say 3.4, then there aren't many guesses near yours, but the target isn't likely to be near your guess either. If you picture a standard bell curve, you can picture the choice as being between a tall skinny piece of the distribution (a guess near zero) or a short fat piece (a guess far from zero). Which gives you the better chance of winning?
As it turns out, it doesn't matter. Either is as good as the other. And anywhere in between is also just as good.
Even more interesting is that it does not matter that the distribution is standard normal. No matter what the distribution is (well, there are a few technical caveats, but I don't feel like I'm betraying the spirit of the results to say that it doesn't matter), as long as the distribution of entries is the same as the distribution of possible outcomes, and as long as the pool has a lot of entries, it doesn't matter what you guess.
So, at least to the extent that you believe our abstract game models your pool reasonably well, any guess is as good as any other. Fill out your bracket based on geography, uniform color, fierceness of mascot, or whatever other criteria you want. Your chances are as good as anyone else's.
If you're a casual follower of college hoops, you might find this liberating. While I haven't given you any actual advice on how to fill out your brackets, at least I've absolved you of any guilt you may have had about entering a contest where you have no idea what you're doing.
This entry was posted on Monday, March 12th, 2007 at 4:02 am and is filed under Non-football, Statgeekery. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed. | 2,373 | 10,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2013-20 | longest | en | 0.965038 |
https://studiegids.universiteitleiden.nl/en/courses/115405/advanced-measure-theory-bm | 1,723,669,669,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00093.warc.gz | 422,044,926 | 6,680 | Course
2022-2023
Students must have completed 'Inleiding Maattheorie’ (4082INMT3) or a comparable course on the fundamentals of measure and integration theory. Acquaintance with the basics of point set topology and metric spaces, and normed vector spaces (e.g. through the course ‘'Linear Analysis') is needed. The book by Cohn 'Measure Theory' provides ample background material.
## Description
The course starts by introducing and studying additional structures on the set of finite measures. For example, they constitute a convex cone that can be embedded in a vector space: the signed measures. This is an ordered vector space with natural norm(s) defined on it. This order structure relates to the so-called Hahn-Jordan decomposition of signed measures. Absolute continuity of measures and the Radon-Nikodym Theorem are discussed.
The core of the course considers Borel measures on topological spaces, mainly locally compact Hausdorff or separable complete metric spaces (Polish spaces). Various regularity concepts for (signed) measures are introduced. The Riesz Representation Theorem is proven, that identifies the dual space of continuous functions on a locally compact Hausdorff space, vanishing at infinity., with the particular class of signed Radon measures.
Considering Borel measures on non-locally compact base spaces leads to various mathematical complications. In the course we focus on the case when the underlying space is Polish (i.e. metrisable, becoming a separable complete metric space), which is a common assumption in Analysis and Probability Theory. We discuss weak convergence of measures and the associated Dudley metric, which is defined by a norm on the signed measures. This introduces a weaker norm (and topology) than that related to the order structure. It is a highly useful concept, e.g. in Probability Theory. Important are relative compactness results for sets of measures: uniform tightness of measures and the Prokhorov Theorem.
The topological structure enables discussion of dynamics in spaces of measures. We provide examples of those defined by so-called Markov operators and one-parameter semigroups of such operators. Important concepts are: invariant (probability) measures and ergodic measures, the existence (Krylov-Bogolyubov Theorem), possible uniqueness and stability of invariant measures and conditions for that.
## Course objectives
The course introduces students to more advanced topics in measure and integration theory, such as norms and weak (vector space) topologies on the vector space of signed measures. Understanding of these concepts allows her/him to consider applications to Dynamical Systems and Markov processes. This provides a good starting point for further study, either in the direction of Analysis (e.g. equations in spaces of measures) or Probability Theory (e.g. Markov processes)
## Timetable
You will find the timetables for all courses and degree programmes of Leiden University in the tool MyTimetable (login). Any teaching activities that you have sucessfully registered for in MyStudyMap will automatically be displayed in MyTimeTable. Any timetables that you add manually, will be saved and automatically displayed the next time you sign in.
MyTimetable allows you to integrate your timetable with your calendar apps such as Outlook, Google Calendar, Apple Calendar and other calendar apps on your smartphone. Any timetable changes will be automatically synced with your calendar. If you wish, you can also receive an email notification of the change. You can turn notifications on in ‘Settings’ (after login).
For more information, watch the video or go the the 'help-page' in MyTimetable. Please note: Joint Degree students Leiden/Delft have to merge their two different timetables into one. This video explains how to do this.
## Mode of instruction
• Lectures (2 hours per week)
• Three take-home assignments with exercises, organized per topic discussed
## Assessment method
The final grade of the course is computed by weighted average from two components:
1. three take-home individual assignments (practicals, equally weighted average; 25%)
2. written exam (75%)
A retake exam is oral, over a selection of topics from the course material. The final grade in case of a retake exam is simply the mark for the retake exam (100%).
The course combines well-established results with those that are recent developments in the field of Analysis and Probability Theory. Thus, not a single book can and will be used. Detailed Lecture Notes will be provided with ample references to the literature. Recommended books (but not mandatory):
• On fundamentals of measure theory: Donald L. Cohn, Measure Theory ISBN: 978-1-4614-6955-1 (Print) 978-1-4614-6956-8 (Online) (available as e-book via Leiden University Library).
• Encyclopaedic, on topics of the course and beyond: V.I. Bogachev, Measure Theory, Volume 1 and 2, Berlin: Springer-Verlag, 2007
See further the references in the Lecture Notes (made available through Brightspace).
## Registration
From the academic year 2022-2023 on every student has to register for courses with the new enrollment tool MyStudyMap. There are two registration periods per year: registration for the fall semester opens in July and registration for the spring semester opens in December. Please see this page for more information.
Please note that it is compulsory to both preregister and confirm your participation for every exam and retake. Not being registered for a course means that you are not allowed to participate in the final exam of the course. Confirming your exam participation is possible until ten days before the exam.
Extensive FAQ's on MyStudymap can be found here.
## Contact
Lecturer: Dr. S.C. Hille (shille[at]math.leidenuniv.nl)
Teaching assistants: see Brightspace pages of the course. | 1,225 | 5,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.903395 |
https://web2.0calc.com/questions/bullet-question | 1,532,227,994,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593004.92/warc/CC-MAIN-20180722022235-20180722042235-00280.warc.gz | 803,716,090 | 5,237 | +0
# Bullet question!
0
166
5
A bullet travels at 850 m/s. How long will it take a bullet to go 1 Km?
Guest Feb 16, 2017
#1
+4
0
1.17 seconds i think
LAX18 Feb 16, 2017
#2
0
850 meters / 1,000 =0.85 km/sec.
1 / 0.85 =1.1765..... seconds to travel 1 km.
Guest Feb 16, 2017
#3
0
Guest Feb 16, 2017
edited by Guest Feb 16, 2017
#4
0
yes
Guest Feb 21, 2017
#5
0 | 171 | 371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-30 | latest | en | 0.836673 |
http://wikiomni.com/pages/Archimedes%27_principle | 1,534,539,654,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212910.25/warc/CC-MAIN-20180817202237-20180817222237-00483.warc.gz | 476,110,208 | 20,215 | Language
• #### Related Blog
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# Archimedes' principle
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the center of mass of the displaced fluid[1]. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.[2]
## Explanation
A floating ship's weight Fp and its buoyancy Fa must be equal in size.
In On Floating Bodies, Archimedes suggested that (c. 250 BC):
Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.
Practically, Archimedes' principle allows the buoyancy of an object partially or fully immersed in a liquid to be calculated. The downward force on the object is simply its weight. The upward, or buoyant, force on the object is that stated by Archimedes' principle, above. Thus, the net force on the object is the difference between the magnitudes of the buoyant force and its weight. If this net force is positive, the object rises; if negative, the object sinks; and if zero, the object is neutrally buoyant - that is, it remains in place without either rising or sinking. In simple words, Archimedes' principle states that, when a body is partially or completely immersed in a fluid, it experiences an apparent loss in weight that is equal to the weight of the fluid displaced by the immersed part of the body.
## Formula
Consider a cube immersed in a fluid, with its sides parallel to the net direction of gravity. The fluid will exert a normal force on each face, and therefore only the forces on the top and bottom faces will contribute to buoyancy. The pressure difference between the bottom and the top face is directly proportional to the height (difference in depth). Multiplying the pressure difference by the area of a face gives the net force on the cube – the buoyancy, or the weight of the fluid displaced. By extending this reasoning to irregular shapes, we can see that, whatever the shape of the submerged body, the buoyant force is equal to the weight of the fluid displaced.
${\displaystyle {\text{ weight of displaced fluid}}={\text{weight of object in vacuum}}-{\text{weight of object in fluid}}\,}$
The weight of the displaced fluid is directly proportional to the volume of the displaced fluid (if the surrounding fluid is of uniform density). The weight of the object in the fluid is reduced, because of the force acting on it, which is called upthrust. In simple terms, the principle states that the buoyant force (Fb) on an object is equal to the weight of the fluid displaced by the object, or the density (ρ) of the fluid multiplied by the submerged volume (V) times the gravity (g) or Fb = ρ x g x V[3]. Thus, among completely submerged objects with equal masses, objects with greater volume have greater buoyancy.
Suppose a rock's weight is measured as 10 newtons when suspended by a string in a vacuum with gravity acting on it. Suppose that, when the rock is lowered into water, it displaces water of weight 3 newtons. The force it then exerts on the string from which it hangs would be 10 newtons minus the 3 newtons of buoyant force: 10 − 3 = 7 newtons. Buoyancy reduces the apparent weight of objects that have sunk completely to the sea floor. It is generally easier to lift an object up through the water than it is to pull it out of the water.
For a fully submerged object, Archimedes' principle can be reformulated as follows:
${\displaystyle {\text{apparent immersed weight}}={\text{weight of object}}-{\text{weight of displaced fluid}}\,}$
then inserted into the quotient of weights, which has been expanded by the mutual volume
${\displaystyle {\frac {\text{density of object}}{\text{density of fluid}}}={\frac {\text{weight}}{\text{weight of displaced fluid}}}}$
yields the formula below. The density of the immersed object relative to the density of the fluid can easily be calculated without measuring any volume is
${\displaystyle {\frac {\text{density of object}}{\text{density of fluid}}}={\frac {\text{weight}}{{\text{weight}}-{\text{apparent immersed weight}}}}.\,}$
(This formula is used for example in describing the measuring principle of a dasymeter and of hydrostatic weighing.)
Example: If you drop wood into water, buoyancy will keep it afloat.
Example: A helium balloon in a moving car. When increasing speed or driving in a curve, the air moves in the opposite direction to the car's acceleration. However, due to buoyancy, the balloon is pushed "out of the way" by the air, and will actually drift in the same direction as the car's acceleration.
When an object is immersed in a liquid, the liquid exerts an upward force, which is known as the buoyant force, that is proportional to the weight of the displaced liquid. The sum force acting on the object, then, is equal to the difference between the weight of the object ('down' force) and the weight of displaced liquid ('up' force). Equilibrium, or neutral buoyancy, is achieved when these two weights (and thus forces) are equal.
## Refinements
Archimedes' principle does not consider the surface tension (capillarity) acting on the body.[4] Moreover, Archimedes' principle has been found to break down in complex fluids.[5]
## Principle of flotation
Archimedes' principle shows buoyant force and displacement of fluid. However, the concept of Archimedes' principle can be applied when considering why objects float. Proposition 5 of Archimedes' treatise On Floating Bodies states that:
Any floating object displaces its own weight of fluid.
In other words, for an object floating on a liquid surface (like a boat) or floating submerged in a fluid (like a submarine in water or dirigible in air) the weight of the displaced liquid equals the weight of the object. Thus, only in the special case of floating does the buoyant force acting on an object equal the objects weight. Consider a 1-ton block of solid iron. As iron is nearly eight times as dense as water, it displaces only 1/8 ton of water when submerged, which is not enough to keep it afloat. Suppose the same iron block is reshaped into a bowl. It still weighs 1 ton, but when it is put in water, it displaces a greater volume of water than when it was a block. The deeper the iron bowl is immersed, the more water it displaces, and the greater the buoyant force acting on it. When the buoyant force equals 1 ton, it will sink no farther.
When any boat displaces a weight of water equal to its own weight, it floats. This is often called the "principle of flotation": A floating object displaces a weight of fluid equal to its own weight. Every ship, submarine, and dirigible must be designed to displace a weight of fluid at least equal to its own weight. A 10,000-ton ship's hull must be built wide enough, long enough and deep enough to displace 10,000 tons of water and still have some hull above the water to prevent it from sinking. It needs extra hull to fight waves that would otherwise fill it and, by increasing its mass, cause it to submerge. The same is true for vessels in air: a dirigible that weighs 100 tons needs to displace 100 tons of air. If it displaces more, it rises; if it displaces less, it falls. If the dirigible displaces exactly its weight, it hovers at a constant altitude.
While they are related to it, the principle of flotation and the concept that a submerged object displaces a volume of fluid equal to its own volume are not Archimedes' principle. Archimedes' principle, as stated above, equates the buoyant force to the weight of the fluid displaced.
One common point of confusion[by whom?] regarding Archimedes' principle is the meaning of displaced volume. Common demonstrations involve measuring the rise in water level when an object floats on the surface in order to calculate the displaced water. This measurement approach fails with a buoyant submerged object because the rise in the water level is directly related to the volume of the object and not the mass (except if the effective density of the object equals exactly the fluid density)[7]. | 1,852 | 8,275 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-34 | latest | en | 0.925985 |
http://hubpages.com/art/Learn-to-Draw-Shapes-Freehand | 1,487,687,990,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170741.49/warc/CC-MAIN-20170219104610-00614-ip-10-171-10-108.ec2.internal.warc.gz | 126,746,277 | 15,921 | # Learn to Draw Shapes Freehand
Source
## Practice Makes Perfect When Drawing Freehand
Drawing shapes freehand may seem just like simple doodling for most folks, however, these shapes actually have a purpose. You're actually training your hand to develop a finer dexterity when you're drawing these freehand shapes.
The connection between your eye/brain and hand - known as "hand-eye coordination" - is also being developed to a greater degree. When you just think you're doodling or scribbling mindlessly, you're actually teaching yourself to be a better artist.
Did you know that every person has a touch of the artists deep inside them? With some folks, that "inner artist" is very close to the surface and is easily accessed. With other folks, the inner artists needs to be coaxed and persuaded to come out.
SO, if you want to learn to draw shapes freehand, you're actually going to learn to be a better artists - even though you didn't know it!
If you've never drawn freehand shapes before, then this tutorial is for you.
Let's begin...
.
## Step 1: Draw a Simple Square
OK, so we all know what a square is, right?
Four straight lines of equal length which are perpendicular to one another.
Simple enough. Let's begin by drawing a single, straight line about one inch in length, as shown to the right.
.
## Step 2: Draw a Second Line at a Right Angle
Now, let's add another line the same length as the first but horizontal.
When you finish this step, your drawing should look like the letter "L."
At this point in your drawing, focus on making the lines as straight as possible.
.
## Step 3: Complete the Square
Now you're basically going to draw two more lines, each one parallel to the one across from it.
Remember, a square has sides of equal length so make sure you are drawing lines that are the same size, otherwise your square may resemble a parallelogram instead.
That's a basic square.
Let's move on to the next shape.
.
## Step 4: Draw a Triangle
OK, so you have a square, now let's try the triangle. Start by drawing a shape that looks like the number "7."
Again, let's try to draw this triangle with 3 equal sides.
I know these seem like very basic shapes, but there is so much going on between your brain, eyes and hand when you're drawing freehand like this.
Remember, these simple shapes train your brain, eyes and hand to work together.
.
## Step 5: Finish the Triangle
This is so easy, just draw a third line and connect the two ends of your number 7 shape.
When you first start out drawing freehand like this, you may notice your lines are not exactly straight - that's OK. You can't expect to draw straight lines right from the start.
Like most things, getting better takes practice.
.
## Step 6: Draw a Circle
Sounds so simple, right? Just draw a circle - right!
Perhaps one of the hardest of the basic shapes to draw freehand is the circle.
To get this shape just right, you must train your brain, eyes and hand to work closely and exactly. Your first attempts may look somewhat irregular - that is OK.
Remember, drawing is a learning experience. Each attempt teaches us something and we become better.
.
## So Here Are My Freehand Shapes
Here, you can see my three shapes drawn freehand.
How did I do? How do your shapes look?
If your shapes don't look as good as you would like, all you need to do is practice your freehand drawing. Next time you'r on the phone, waiting for a train or just bored - start practicing your freehand drawing of shapes.
I have been drawing since before I can remember, about 50 plus years. I have drawn almost every day over that time and have developed a lot of skills. SO many, that I have made a living as an artist for all of my life, even today.
If you're interested in drawing and becoming more proficient, all you need to do is look around and start drawing the things you see.
If you practice just a few minutes each and every day, you will begin to get better and better gradually.
If you want to get started drawing, take a look at some of the drawing books and other products at Amazon.com (see box below).
Good luck!
## When do you typically find yourself drawing freehand shapes?
See results without voting
## More by this Author
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If you want to learn how to draw chrome, then this is the step-by-step drawing tutorial for you. Learning to draw chrome is actually quite easy if you know the basic drawing techniques. More here ...
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DDE 3 years ago from Dubrovnik, Croatia
Awesome and so simple to learning on how to draw shapes.
MKayo 3 years ago from Texas Author
Thx, DDE! | 1,122 | 5,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-09 | latest | en | 0.961143 |
http://www.slideserve.com/liam/chapter-6-arrays-and-vectors | 1,503,348,877,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00126.warc.gz | 675,382,439 | 17,561 | 1 / 27
# Chapter 6: Arrays and Vectors - PowerPoint PPT Presentation
Chapter 6: Arrays and Vectors . Presentation slides for Java Software Solutions Foundations of Program Design Second Edition by John Lewis and William Loftus Java Software Solutions is published by Addison-Wesley
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### Chapter 6: Arrays and Vectors
Presentation slides for
Java Software Solutions
Foundations of Program Design
Second Edition
by John Lewis and William Loftus
Instructors using the textbook may use and modify these slides for pedagogical purposes.
• Arrays and vectors are objects that help us organize large amounts of information
• Chapter 6 focuses on:
• array declaration and use
• arrays of objects
• sorting elements in an array
• multidimensional arrays
• the Vector class
• using arrays to manage graphics
has a single name
Each value has a numeric index
scores
79 87 94 82 67 98 87 81 74 91
Arrays
• An array is an ordered list of values
0 1 2 3 4 5 6 7 8 9
An array of size N is indexed from zero to N-1
This array holds 10 values that are indexed from 0 to 9
• A particular value in an array is referenced using the array name followed by the index in brackets
• For example, the expression
scores[2]
refers to the value 94 (which is the 3rd value in the array)
• That expression represents a place to store a single integer, and can be used wherever an integer variable can
• For example, it can be assigned a value, printed, or used in a calculation
• An array stores multiple values of the same type
• That type can be primitive types or objects
• Therefore, we can create an array of integers, or an array of characters, or an array of String objects, etc.
• In Java, the array itself is an object
• Therefore the name of the array is a object reference variable, and the array itself is instantiated separately
• The scores array could be declared as follows:
int[] scores = new int[10];
• Note that the type of the array does not specify its size, but each object of that type has a specific size
• The type of the variable scores is int[] (an array of integers)
• It is set to a new array object that can hold 10 integers
• See BasicArray.java(page 270)
int[] list = new int[LIMIT];
// Initialize the array values
for (int index = 0; index < LIMIT; index++)
list[index] = index * MULTIPLE;
list[5] = 999; // change one array value
for (int index = 0; index < LIMIT; index++)
System.out.print (list[index] + " ");
• Some examples of array declarations:
float[] prices = new float[500];
boolean[] flags;
flags = new boolean[20];
char[] codes = new char[1750];
• Once an array is created, it has a fixed size
• An index used in an array reference must specify a valid element
• That is, the index value must be in bounds (0 to N-1)
• The Java interpreter will throw an exception if an array index is out of bounds
• This is called automatic bounds checking
Bounds Checking
• For example, if the array codes can hold 100 values, it can only be indexed using the numbers 0 to 99
• If count has the value 100, then the following reference will cause an ArrayOutOfBoundsException:
System.out.println (codes[count]);
• It’s common to introduce off-by-one errors when using arrays
for (int index=0; index <= 100; index++)
codes[index] = index*50 + epsilon;
• Each array object has a public constant called length that stores the size of the array
• It is referenced using the array name (just like any other object):
scores.length
• Note that length holds the number of elements, not the largest index
• See ReverseNumbers.java(page 272)
• See LetterCount.java(page 274)
double[] numbers = new double[10];
System.out.println ("The size of the array: " + numbers.length);
for (int index = 0; index < numbers.length; index++)
{
System.out.print ("Enter number " + (index+1) + ": ");
}
System.out.println ("The numbers in reverse:");
for (int index = numbers.length-1; index >= 0; index--)
System.out.print (numbers[index] + " ");
int[] upper = new int[NUMCHARS];
int[] lower = new int[NUMCHARS];
char current; // the current character being processed
int other = 0; // counter for non-alphabetics
System.out.println ("Enter a sentence:");
// Count the number of each letter occurance
for (int ch = 0; ch < line.length(); ch++) {
current = line.charAt(ch);
if (current >= 'A' && current <= 'Z')
upper[current-'A']++;
else
if (current >= 'a' && current <= 'z')
lower[current-'a']++;
else
other++;
}
// Print the results
System.out.println ();
for (int letter=0; letter < upper.length; letter++)
{
System.out.print ( (char) (letter + 'A') );
System.out.print (": " + upper[letter]);
System.out.print ("\t\t" + (char) (letter + 'a') );
System.out.println (": " + lower[letter]);
}
System.out.println ();
System.out.println ("Non-alphabetic characters: " + other);
• An initializer list can be used to instantiate and initialize an array in one step
• The values are delimited by braces and separated by commas
• Examples:
int[] units = {147, 323, 89, 933, 540,
269, 97, 114, 298, 476};
char[] letterGrades = {'A', 'B', 'C', 'D', 'F'};
• Note that when an initializer list is used:
• the new operator is not used
• no size value is specified
• The size of the array is determined by the number of items in the initializer list
• An initializer list can only be used in the declaration of an array
• SeePrimes.java(page 278)
int[] primes = {2, 3, 5, 7, 11, 13, 17, 19};
System.out.println ("Array length: " + primes.length);
System.out.println ("The first few prime numbers are:");
for (int prime = 0; prime < primes.length; prime++)
System.out.print (primes[prime] + " ");
• An entire array can be passed to a method as a parameter
• Like any other object, the reference to the array is passed, making the formal and actual parameters aliases of each other
• Changing an array element in the method changes the original
• An array element can be passed to a method as well, and will follow the parameter passing rules of that element's type
• Sorting is the process of arranging a list of items into a particular order
• There must be some value on which the order is based
• There are many algorithms for sorting a list of items
• These algorithms vary in efficiency
• We will examine two specific algorithms:
• Selection Sort
• Insertion Sort
Given an array numbers of size length,
1 For index = 0 to length - 1 do
1.1 find min, index of smallest element from index to length - 1
1.2 switch elements at locations index and min
expand 1.1
1.1.1 min = index
1.1.2 for scan = index+1 to length do
1.1.2.1 if (numbers[scan] < numbers[min])
1.1.2.1.1 min = scan;
• An example:
original: 3 9 6 1 2
smallest is 1: 1 9 6 3 2
smallest is 2: 1 2 6 3 9
smallest is 3: 1 2 3 6 9
smallest is 6: 1 2 3 6 9
• See Sorts.java(page 290) -- the selectionSort method
int[] grades = {89, 94, 69, 80, 97, 85, 73, 91, 77, 85, 93};
for (int index = 0; index < grades.length; index++)
public class Sorts {
// Sorts the specified array of integers using the selection
// sort algorithm.
public static void selectionSort (int[] numbers) {
int min, temp;
for (int index = 0; index < numbers.length-1; index++) {
min = index;
for (int scan = index+1; scan < numbers.length; scan++)
if (numbers[scan] < numbers[min])
min = scan;
// Swap the values
temp = numbers[min];
numbers[min] = numbers[index];
numbers[index] = temp;
}
}
• The approach of Insertion Sort:
• Pick any item and insert it into its proper place in a sorted sublist
• repeat until all items have been inserted
• In more detail:
• consider the first item to be a sorted sublist (of one item)
• insert the second item into the sorted sublist, shifting items as necessary to make room to insert the new addition
• insert the third item into the sorted sublist (of two items), shifting as necessary
• repeat until all values are inserted into their proper position
• An example:
original: 3 9 6 1 2
insert 9: 3 9 6 1 2
insert 6: 3 6 9 1 2
insert 1: 1 3 6 9 2
insert 2: 1 2 3 6 9
• See Sorts.java (page 290) -- the insertionSort method
public static void insertionSort (int[] numbers) {
for (int index = 1; index < numbers.length; index++) {
int key = numbers[index];
int position = index;
// shift larger values to the right
while (position > 0 && numbers[position-1] > key)
{
numbers[position] = numbers[position-1];
position--;
}
numbers[position] = key;
}
}
• Both Selection and Insertion sorts are similar in efficiency
• The both have outer loops that scan all elements, and inner loops that compare the value of the outer loop with almost all values in the list
• Therefore approximately n2 number of comparisons are made to sort a list of size n
• We therefore say that these sorts are of order n2
• Other sorts are more efficient: order n log2 n | 2,487 | 9,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-34 | longest | en | 0.826688 |
https://www.convertunits.com/from/technical+atmosphere/to/dyne/square+centimetre | 1,638,151,881,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00154.warc.gz | 819,745,161 | 12,884 | ››Convert technical atmosphere to dyne/square centimetre
technical atmosphere dyne/square centimetre
How many technical atmosphere in 1 dyne/square centimetre? The answer is 1.0197162129779E-6.
We assume you are converting between technical atmosphere and dyne/square centimetre.
You can view more details on each measurement unit:
technical atmosphere or dyne/square centimetre
The SI derived unit for pressure is the pascal.
1 pascal is equal to 1.0197162129779E-5 technical atmosphere, or 10 dyne/square centimetre.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between technical atmospheres and dynes/square centimeter.
Type in your own numbers in the form to convert the units!
››Quick conversion chart of technical atmosphere to dyne/square centimetre
1 technical atmosphere to dyne/square centimetre = 980665 dyne/square centimetre
2 technical atmosphere to dyne/square centimetre = 1961330 dyne/square centimetre
3 technical atmosphere to dyne/square centimetre = 2941995 dyne/square centimetre
4 technical atmosphere to dyne/square centimetre = 3922660 dyne/square centimetre
5 technical atmosphere to dyne/square centimetre = 4903325 dyne/square centimetre
6 technical atmosphere to dyne/square centimetre = 5883990 dyne/square centimetre
7 technical atmosphere to dyne/square centimetre = 6864655 dyne/square centimetre
8 technical atmosphere to dyne/square centimetre = 7845320 dyne/square centimetre
9 technical atmosphere to dyne/square centimetre = 8825985 dyne/square centimetre
10 technical atmosphere to dyne/square centimetre = 9806650 dyne/square centimetre
››Want other units?
You can do the reverse unit conversion from dyne/square centimetre to technical atmosphere, or enter any two units below:
Enter two units to convert
From: To:
››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 593 | 2,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.706149 |
http://www.ocw.titech.ac.jp/index.php?module=General&action=T0300&GakubuCD=2&GakkaCD=321700&KeiCD=17&KougiCD=202101978&Nendo=2021&vid=03&lang=EN | 1,670,662,741,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710421.14/warc/CC-MAIN-20221210074242-20221210104242-00467.warc.gz | 83,208,333 | 19,208 | ### 2021 Applied Probability and Statistical Theory
Font size SML
Undergraduate major in Electrical and Electronic Engineering
Instructor(s)
Kajikawa Kotaro Sakaguchi Kei
Class Format
Lecture
Media-enhanced courses
Day/Period(Room No.)
Tue1-2(H101) Fri1-2(H101)
Group
-
Course number
EEE.M231
Credits
2
2021
Offered quarter
3Q
Syllabus updated
2021/5/10
Lecture notes updated
-
Language used
Japanese
Access Index
### Course description and aims
It is vital to acquire knowledge and skills in statistics and probability in various fields related to electrical engineering and information communication engineering. By combining lectures and exercises, the course enables students to understand and learn the fundamentals of mean, variance, characteristic function, etc. in the first part (probability) and those of unbiased estimation, maximum likelihood estimation, Bayesian inference, hypothesis testing, machine learning, etc. in the second part (statistics).
The ability to derive statistically significant information will be very useful in the real world.
### Student learning outcomes
Students will be able to learn how to analyze data in various fields related to electrical engineering and information communication engineering by using probablistic methods and statistical techniques. The course provides specific examples in engineering, which will give a deeper understanding. Many practical exercises and exams will enable students to acquire knowledge effectively.
Corresponding educational goals are:
(1) Specialist skills Fundamental specialist skills
(6) Firm fundamental specialist skills on electrical and electronic engineering, including areas such as electromagnetism, circuits, linear systems, and applied mathematics
### Keywords
Probability, Mean, Variance, Maximum likelihood estimation, Bayesian inference, Hypothesis testing, Machine learning
### Competencies that will be developed
✔ Specialist skills Intercultural skills Communication skills Critical thinking skills Practical and/or problem-solving skills ✔ ・Fundamental specialist skills on EEE
### Class flow
To cultivate practical ability, students are given many exercise problems, which are related to a previous class and a class on the day.
### Course schedule/Required learning
Course schedule Required learning
Class 1 Permutations Understand the Permutations
Class 2 Conditional probability Calculate conditional probability.
Class 3 Mean and variance Calculate mean and variance
Class 4 Characteristic function Understand characteristic function
Class 5 Random variable and distribution I Understand random variable and distribution
Class 6 Random variable and distribution II Understand random variable and distribution
Class 7 Test level of understanding with exercise problems Test level of understanding for classes 1–6
Class 8 Statistical inference Understand data analysis by statistical inference
Class 9 Unbiased estimation Understand unbiased estimation
Class 10 Maximum likelihood estimation and Bayesian inference Learn how to perform maximum likelihood estimation and Bayesian inference
Class 11 Hypothesis testing, Classification, Machine learning Learn how to perform hypothesis test and classification
Class 12 Regression, Prediction, Machine learning Learn how to perform regression and prediction
Class 13 Stochastic process Understand stochastic process and method of data analysis
Class 14 Test level of understanding with exercise problems Test level of understanding for classes 8–13
### Out-of-Class Study Time (Preparation and Review)
To enhance effective learning, students are encouraged to spend approximately 100 minutes preparing for class and another 100 minutes reviewing class content afterwards (including assignments) for each class.
They should do so by referring to textbooks and other course material.
N/A
### Reference books, course materials, etc.
Watanabe, Sumio, Noboru, Murata, Probability and statistics: a bridge to information science, Corona-sha; ISBN: 9784339060775 (in Japanese)
Ogura, Hisanao, stochastic process for physics and engineering, Corona-sha; ISBN: 9784339004229, 9784339004236 (in Japanese)
Shibata, Fumiaki, Probability and statistics, Iwanami-shoten; ISBN: 9784000079778 (in Japanese)
Baba, Noriyuki, Kuchii, Shigeru, Campus seminar: statistics, mathema-shuppan, ISBN: 9784907165314 (in Japanese)
Suyama, Atsushi, Machine learning based on Bayesian inference, Kodansha; ISBN: 9784061538320 (in Japanese)
### Assessment criteria and methods
Students' knowledge of probability and statistical, and their ability to apply them to problems will be assessed.
Exam 50%, exercise problems 50%.
### Related courses
• LAS.M101 ： Calculus I / Recitation
• LAS.M105 ： Calculus II
• EEE.M211 ： Fourier Transform and Laplace Transform
• EEE.S341 ： Communication Theory (Electrical and Electronic Engineering)
### Prerequisites (i.e., required knowledge, skills, courses, etc.)
Students must have successfully completed both Calculus I and Calculus II. | 1,020 | 5,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-49 | latest | en | 0.857554 |
https://fr.slideserve.com/georgio/332-437-lecture-4-variable-entered-karnaugh-maps-and-mixed-logic-notation | 1,628,101,286,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00073.warc.gz | 281,484,562 | 22,786 | 332:437 Lecture 4 Variable-Entered Karnaugh Maps and Mixed-Logic Notation
# 332:437 Lecture 4 Variable-Entered Karnaugh Maps and Mixed-Logic Notation
## 332:437 Lecture 4 Variable-Entered Karnaugh Maps and Mixed-Logic Notation
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##### Presentation Transcript
1. 332:437 Lecture 4Variable-Entered Karnaugh Maps and Mixed-Logic Notation • Variable-Entered Karnaugh Maps • MUX-based and Indirect-Addressed MUX Design • ROM-based Logic design • Buffers and drivers • Mixed-Logic notation • Summary Material from An Engineering Approach to Digital Design, by W. I. Fletcher, Prentice-Hall Bushnell: Digital Systems Design Lecture 4
2. Variable-Entered Karnaugh Maps (VEM) • Allow 8 to 16 variable maps to be represented as 3 to 5 variable maps • Needed because typical Boolean design problem involves 8 or more Boolean variables Bushnell: Digital Systems Design Lecture 4
3. AB CD 00 01 11 10 00 1 00 1 00 1 00 1 11 1 1 11 11 1 1 11 01 01 1 1 1 1 01 1 1 1 1 01 10 10 10 1 1 10 1 1 AB CD 00 01 11 10 AB CD 00 01 11 10 Six-Variable K-Map AB CD 00 01 11 10 EF = 00 EF = 01 EF = 10 EF = 11 Bushnell: Digital Systems Design Lecture 4
4. Variable-Entered Map (VEM) • Conventional logic minimization: • Time consuming • Error-prone • VEM Key idea: • Represent values of function in terms of its variables (called map-entered variables) within Karnaugh map framework • Group like variables in Karnaugh map cells Bushnell: Digital Systems Design Lecture 4
5. Map Entry 0 c c c + c or 1 X Situation No minterms with this condition f = 1 when c = 0 f = 1 when c = 1 f = 1 when c = 0 or 1 f = X (don’t care) Example VEM f (a, b, c) = a b c + a b c + a b c + a b c Arbitrarily choose c as map entered variable Bushnell: Digital Systems Design Lecture 4
6. ab c 0 1 00 1 0 01 0 1 11 0 1 10 1 0 XOR/XNOR Gate Grouping • New way of grouping map variables • For this example: • f = c b Bushnell: Digital Systems Design Lecture 4
7. a b 0 1 c 0 c c 1 c + c 0 f b a Example Result • VEM technique reduced an ordinary 3-variable K-map to a 2-variable map • Must group only like minterms in VEM • f = c ( b ) + c (a b) Bushnell: Digital Systems Design Lecture 4
8. ab c 0 1 00 1 0 01 0 1 11 0 0 10 1 1 a b f c Conventional K-Map Method • f = (c b) a + a b Conventional Way VEM Way # 2-input gates 4 4 # inverters 2 0 Bushnell: Digital Systems Design Lecture 4
9. Use of DeMorgan’s Theorems to Transform Logic Gates Bushnell: Digital Systems Design Lecture 4
10. VEM Techniques • Use as many map-entered variables as you like • Works well for partially-minimized functions – use remaining variables as map-entered variables. • Example • f (w, x, y, z) = Sm (3, 4, 6, 7, 10) + Sd (9, 11, 12, 14, 15) • Choose w, x, y as ordinary K-map variables • Make z the map-entered variable • Only appears inside K-map entries Bushnell: Digital Systems Design Lecture 4
11. Example (continued) Map Entry 0 z z 1 z z + z X z X z X + z X w 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 x 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 f 0 0 0 1 1 0 1 1 0 X 1 X X 0 X X • Truth table Bushnell: Digital Systems Design Lecture 4
12. wx y 0 1 00 0 z 01 z z + z 11 z X z X + zX 10 zX z + zX Example (concluded) • Variable-Entered K-Map • f = y z + x z + w y z Bushnell: Digital Systems Design Lecture 4
13. MUX-Based Logic Design • Use circuit inputs to select from a variety of Boolean functions for the circuit output • Functions wired to MUX inputs • Often more efficient than SOP or POS design • Problem: Sometimes you need a 5-input MUX (32 different input function selections), but there are really only 5 distinctly different input functions • Wastes chip area with a large MUX • Solution: Indirect-addressed MUX Bushnell: Digital Systems Design Lecture 4
14. b X X X X X 0 0 1 c X 0 0 1 1 0 1 X All else e 1 0 0 0 0 X X X a 0 0 0 0 0 1 1 1 d X 0 1 0 1 X X X Z f g f g g h h f g 1 X Example Indirect-Addressed MUX-Based Design • Example truth table: Substitute ROM for MUX Bushnell: Digital Systems Design Lecture 4
15. A0 A1 A2 MUX Select f g f g g h h g 1 0 0 1 2 3 4 5 6 7 Z ROM and MUX Implementation ROM 000 001 001 010 010 011 011 100 100 001 101 110 000 A0 A1 A2 Word Address a b c d e 00001 to 01111 (odd only) 00000 01000 00010 01010 00100 01100 00110 01110 10000 to 10011 10100 to 10111 11000 to 11111 All others Bushnell: Digital Systems Design Lecture 4
16. ROM 1 1 0 1 1 1 1 1 a b c a 0 0 0 0 1 1 1 1 b 0 0 1 1 0 0 1 1 c 0 1 0 1 0 1 0 1 F 1 1 0 1 1 1 1 1 0 1 2 3 4 5 6 7 F Straight ROM Logic Implementation • Sometimes the best design method Bushnell: Digital Systems Design Lecture 4
17. a b F c Alternate Discrete Logic Implementation • May very well use as much chip area as ROM implementation – each ROM cell needs only 1 transistor Bushnell: Digital Systems Design Lecture 4
18. Logic Gate Choices • Extra unused inputs • Terminate properly by connecting to non-controlling logic value • OR/NOR/XOR – connect to 0 (Vss) • AND/NAND – connect to 1 (VDD through pullup resistor) • EQUIVALENCE – connect to 1 • Unconnected extra inputs act like radio antennas • Collect electrical noise & randomly fluctuate between logic 0 & 1 Bushnell: Digital Systems Design Lecture 4
19. Choice of Logic Realization • CMOS, nMOS, & TTL logic families • Fewer transistors in NAND/NOR gates than in AND/OR gates • NAND/NOR also faster than AND/OR • Gate substitution: Use 3-input AND gate instead of cascaded 2-input AND’s (faster) Bushnell: Digital Systems Design Lecture 4
20. Form Substitution • All 3 realizations are exactly the same, but the single NOR gate realization is better Bushnell: Digital Systems Design Lecture 4
21. t1 … t2 Buffers/Drivers • Buffers have more driving current than ordinary CMOS gates – used for large fanout gates (more than 8-10). • Do not load up an ordinary CMOS gate with more than 8-10 gates Delay = t1 + n t2 Delay = t1 + t2 Bushnell: Digital Systems Design Lecture 4
22. Collector Base Emitter Buffers/Drivers (continued) • Drivers – adapted for higher current & voltage levels than buffers • Examples: TTL CMOS, CMOS TTL conversions • Reason: TTL output I-V specification does not match analog circuit specification • Examples: relays, analog switches, fluorescent displays, appliance controls • High-Voltage Drivers • Usually have open-collector Bipolar outputs Bushnell: Digital Systems Design Lecture 4
23. Enable Buffers/Drivers (continued) • Line drivers – provide (source) large I or accept (sink) large I • Source: 40 mA with 50W load • Sink: 60 mA • For interface lines (long) between digital systems • Bus driver – Line Driver with tri-state output When disabled, goes into high impedance state at output Bushnell: Digital Systems Design Lecture 4
24. Buffers/Drivers (continued) • Line Receiver – actually a driver – used at receiving end of bus interface to latch & amplify signals sent over bus • Load on bus is single load – but signal amplified & sent many places Bushnell: Digital Systems Design Lecture 4
25. Logic Types • Positive Logic (active high) • 0 = 0 V., 1 = 1.0 V. • Negative Logic (active low) • 0 = 1.0 V., 1 = 0 V. • Mixed-Logic – logic with: • Negative (positive) logic inputs • Positive (negative) logic outputs • Arbitrary mixture of positive & negative logic • Most systems designed this way • Can be confusing Bushnell: Digital Systems Design Lecture 4
27. Multiple Gate Interpretations • Positive logic: Negative logic: Bushnell: Digital Systems Design Lecture 4
28. Why Do We Need Negative Logic? • Easier to hold unused inputs of circuit at high voltage level rather than at low voltage level • Hold many unused inputs high – only active ones need to be held low • Easier to label signals in terms of their functions whether or not data is positive or negative • Example: IN / OUT • Often get better noise immunity with negative logic than with positive logic • Particularly true on buses • Low hardware noise margin often better than high hardware noise margin Bushnell: Digital Systems Design Lecture 4
29. Mixed-Logic Design Rules • If an input (output) is negative, LABEL IT with / and make sure that it flows into (from) a bubble • If an input (output) is positive, DO NOT LABEL IT with / and make sure that it does not flow into (from a bubble) • Not always possible to follow these rules in every part of the circuit due to Boolean function Bushnell: Digital Systems Design Lecture 4
30. /A /C /B C A B Example • If A, B, C are negative logic, better notation: • Meaning: If A is turned on or B is turned on, then C is turned on • Negative logic has the OR operation, so the negative logic picture corresponds to that Bushnell: Digital Systems Design Lecture 4
31. Real-Life Circuit Design • Mixed-logic occurs most frequently • 2-level SOP or POS forms may use way too much hardware • Always true for very large circuits • Instead, use multi-level logic Bushnell: Digital Systems Design Lecture 4
32. /b /b a a f f c c d d Transformations • Can always add bubbles to both ends of a wire without changing circuit function • Turn AND-OR into NAND-NAND form Bushnell: Digital Systems Design Lecture 4
33. Summary • Variable-Entered Karnaugh Maps • MUX-based and Indirect-Addressed MUX Design • ROM-based Logic design • Buffers and drivers • Mixed-Logic notation Bushnell: Digital Systems Design Lecture 4 | 2,842 | 9,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-31 | latest | en | 0.561484 |
https://edurev.in/course/quiz/attempt/12087_RRB-ALP-Technician-Mock-Test--English--13/9220a5f2-36c8-4230-8b64-b47176ad7050 | 1,660,033,426,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00546.warc.gz | 231,304,262 | 60,584 | # RRB ALP & Technician Mock Test (English) - 13
## 75 Questions MCQ Test RRB ALP & Technician Exam (Group C ) - Mock Tests | RRB ALP & Technician Mock Test (English) - 13
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Attempt RRB ALP & Technician Mock Test (English) - 13 | 75 questions in 60 minutes | Mock test for Railways preparation | Free important questions MCQ to study RRB ALP & Technician Exam (Group C ) - Mock Tests for Railways Exam | Download free PDF with solutions
QUESTION: 1
Solution:
QUESTION: 2
### Present ages of X and Y are in the ratio 5:6 respectively. Seven years hence this ratio will become 6:7 respectively. What is X's present age in years?
Solution:
Let the present ages of X and Y be 5x years and 6x years respectively.
Then, 5x + 7/6x + 7 = 6/7
⇒ 7(5x + 7) = 6(6x + 7)
⇒ x = 7
∴ X's present age = 5x = 35 years
QUESTION: 3
### Find the same relationship among the four alternative numbers given under it. Given set : (6, 13, 22)
Solution:
QUESTION: 4
OABC is a rhombus whose vertices A, B and C lie on a circle with centre at O. If radius of circle is 10cm. The area of the rhombus
Solution:
If A, B, and C lie on the circle, O is the center of the circle, and the radius is 10 cm, then OA, OB, OC, AB, and BC must all measure 10 cm because OA, OB, and OC are all radii, and AB and BC are congruent because of congruent sides of a rhombus.
Hence, the area of the rhombus is twice the area of an equilateral triangle with sides that measure 10 cm.
The area of an equilateral triangle of side s is :
Hence the area of the rhombus is
QUESTION: 5
Graphite is a soft lubricant extremely difficult to melt. The reason for this anomalous behaviour is that graphite
Solution:
QUESTION: 6
An example of phylum Echinodermata is
Solution:
QUESTION: 7
Find the correct combination
Anil, introducing a girl in a party, said, " She is the wife of the grandson of my mother". How is Anil related to the girl?
Solution:
QUESTION: 8
The number of leap years from 1799 AD to 1902 AD is
Solution: Both 1800 AD and 1900 AD are not leap year. The number of leap years from 1804 AD to 1896 AD is 24
QUESTION: 9
The idea 'omnis cellula e cellula'which means that all living cells arise from pre-existing cells,was given by
Solution:
QUESTION: 10
Choose the word which is least like the other words in the group.
Solution:
QUESTION: 11
A white solid which is yellow when hot but changes to white again on cooling is
Solution:
QUESTION: 12
If CABLE = 96372 and RISK - 8415, what word is made by 37265 ?
Solution:
QUESTION: 13
If 'ish lto inm' stands for 'neat and tidy'; 'qpr inm sen' stands for 'small but neat' and 'hsm sen rso' stands for 'good but erratic', what would 'but' stand for ?
Solution:
QUESTION: 14
Choose the word which is least like the other words in the group .
Solution:
QUESTION: 15
Complete the analogous pair.
Export : Earning :: Import : ?
Solution:
QUESTION: 16
Which of the following statements about Polar vortices is/are correct?
Solution:
QUESTION: 17
What will be the compound interest acquired on a sum of Rs 12000 for 3 years at the rate of 10% p.a?
Solution:
QUESTION: 18
Which among the following statements with reference to Aadhaar is INCORRECT?
Solution:
QUESTION: 19
The United Nations Framework Conventions on Climate Change (UNFCCC) is an international treaty drawn at:
Solution:
QUESTION: 20
Name the newly appointed director general of the National Investigation Agency (NIA).
Solution:
QUESTION: 21
The 11th Association of Southeast Asian Nations (ASEAN) Defense Ministers Meeting was held in which country?
Solution:
QUESTION: 22
Which insurance firm has launched the first ever micro-insurance plan ‘insurance khata’ targeting the under-served social sections with seasonal incomes?
Solution:
QUESTION: 23
Which day of the last week did Satish meet Kapil at Kapil's residence ?
I. Kapil was out of town from Monday to Wednesday. He returned on Thursday morning.
II. On Friday night Satish telephoned his friend to inform that only yesterday he had got approval of Kapil after personally explaining to him all the details.
The question given above has a problem and two statements giving certain information. You have to decide if the information given in the statements is sufficient for answering the problem. Indicate your answer as
Solution:
QUESTION: 24
A square field ABCD of side 90 m is so located that its diagonal AC is from north to south and the corner B is to the west of D. Rohan and Rahul start walking along the sides from B and C respectively in the clockwise and anticlockwise directions with speeds of 8 km/hr. Where shall they cross each other the second time ?
Solution:
QUESTION: 25
Who has recently been appointed to the post of Economic Relations Secretary in the Ministry of External Affairs?
Solution:
QUESTION: 26
If (x + 2) (x - 5) (x - 6) (x - 1) = 144, then the values of x are:
Solution:
[ x − 6 x + 2 ] [ x − 5 x + 1 ] = 144
(x2 − 4 x − 12) (x2 − 4 x − 5) = 144
Let x2 − 4 x = y
y − 12 y − 5 = 144
QUESTION: 27
That the earth is almost spherical in shape was first stated by
Solution:
QUESTION: 28
The value of the expression tan 1o . tan 2o . tan 3o … … tan 87o . tan 88o . tan 89o , is
Solution:
QUESTION: 29
Solution:
QUESTION: 30
Which of the following is not a mosquito borne disease?
Solution:
QUESTION: 31
When a body is accelerated, then
Solution:
QUESTION: 32
Three-fourth of 68 is less than two-third of 114 by
Solution:
QUESTION: 33
Which of the following is true in the given figure, where AD is altitude to the hypotenuse of a right angled ∆ABC?
1. ∆ABD and ∆CAD are similar triangles.
2. ∆ABD and ∆CDA are congruent triangles.
3. ∆ADB and ∆CAB are similar triangles.
Select the correct answer using the codes given below :
Solution:
QUESTION: 34
If ABCD is a cyclic quadrilateral in which ∠DAC=27o, ∠DBA=50o and ∠ADB=33o, then ∠CAB is equal to :
Solution:
QUESTION: 35
The acceleration due to gravity of a body moving against gravity is
Solution:
QUESTION: 36
The longest thread like structure attached to the nucleus is called :
Solution:
QUESTION: 37
Three numbers are in the ratio 1:2:3 and their H.C.F. is 12. The numbers are :
Solution:
Given ratio of numbers = 1 : 2 : 3
HCF = 12
So Numbers are
(12 x 1), (12 x 2) and (12 x 3)
= 12 , 24, 36
QUESTION: 38
Which of the following best fits in the blank?
‘Apple, Application,_______, Approval, Apricot, April’
Solution:
QUESTION: 39
How much amount of financial assistance has been extended by the Government of India for the upgradation of Kankesanthurai (KKS) Harbour in Sri Lanka?
Solution:
QUESTION: 40
Proteolytic enzyme of pancreatic juice is
Solution:
QUESTION: 41
The relation between u, v and f in a spherical mirror is :
Solution:
QUESTION: 42
If Film Actors are represented by a large circle on the right, TV Actors are represented by a large circle on the left, which combination will best represent Stage Actor?
Solution: is (C) because some TV Actors can be Film Actors and Vice-versa. Stage Actors can also be either Film Actors or TV Actors or both.
QUESTION: 43
In questions below, equations have become wrong due to wrong orders of signs. Choose the correct order of signs from the alternative given.
5×6÷3-12=13-6
Solution:
QUESTION: 44
Which of the following is not a true solution?
Solution:
QUESTION: 45
Which of the following is not an ore of aluminium ?
Solution:
QUESTION: 46
The start-up incubation centre named Centre for Entrepreneurship Opportunities and Learning (CEOL) has recently been inaugurated in which state by Nirmala Sitharaman?
Solution:
QUESTION: 47
A body moving along a straight line at 20 m/s undergoes an acceleration of 4 m/s2. After two seconds its speed will be ________.
Solution:
QUESTION: 48
Choose the correct alternative from the given ones that will complete the series.
3, 4, 12, 16, 48, ___
Solution:
QUESTION: 49
The difference of the sum and the difference of two numbers is 24.Find the smaller one.
Solution:
QUESTION: 50
Three-fifth of one-third of a number is 30. What is 20% of that number?
Solution:
QUESTION: 51
Who has been named to be the next Foreign Secretary of India to succeed S. Jaishankar who retires on 28 January?
Solution:
QUESTION: 52
2 pipes A and B can fill a cistern in 20 minutes and 25 minutes respectively, Both are opened together but at the end of 5 minutes B is turned off. Total time taken to fill the cisten is
Solution:
25(1 - t/20 ) = 5
∴ t = 16 minutes
QUESTION: 53
A man gains 10% by selling a certain article for a certain price. If he sells it at double the price, then the profit made is
Solution:
QUESTION: 54
A sum of Rs 7000 is to be divided among A, B and C so that A shall receive 2/5 as much as B and C together and C receives 3/4 of what A and B receive together. Share of B is
Solution:
QUESTION: 55
Which one of the following is the surgical method of contraception is employed in human male :
Solution:
QUESTION: 56
Amit borrowed a sum of money with simple interest as per the following rate structure (a) 6% per annum for the first three years (b) 8% per annum for the next five years (c) 12% per annum for the next four years. If he paid a total of Rs.5040 as interest at the end of twelve years, how much money did he borrow ?
Solution:
QUESTION: 57
The simplified value of [{(1/3)÷(1/3)×(1/3)}/{(1/3)÷(1/3)of(1/3)}]-(1/9) is :
Solution:
QUESTION: 58
What should come in the place of the question mark (?) in the following equation? 1500 of 45% + 1700 of 35% = 3175 of ?%
Solution:
QUESTION: 59
Which among the following countries has won the 2018 Blind Cricket World Cup?
Solution:
QUESTION: 60
Solution:
QUESTION: 61
Below is given statement followed by two arguments numbered I and II. You have to consider the statement and the following arguments and decide which of the arguments is strong in the statement.
Statement :
Should higher qualification be the only criteria for internal promotions in any organisation?
Arguments :
I. Yes, why not? In fact only higher qualification is more important than other factors.
II. No, quality of performance and other factors are more important than mere higher qualification in case of internal promotion.
Solution:
QUESTION: 62
Below is given statement followed by two assumptions numbered I and II. You have to consider the statement and the following assumptions and decide which of the assumptions is implicit in the statement :
Statement:
The multinational fast food chains are opening up a large number of Plus Coffee Shops with piped modern music in different cities of India and these are serving various snacks with coffee.
Assumption:
I. A large number of persons may become regular customers of these coffee shops.
II. The people will like to enjoy the comfortable environment while drinking coffee with snacks.
Solution:
QUESTION: 63
Below is given statement followed by three conclusions numbered I, II and III. You have to consider the statement and the following conclusions and decide which of the conclusions is follows in the statement :
Statements :a. Some caps are umbrellas.
b. Some umbrellas are raincoats.
c. All raincoats are trousers.
d. All trousers are jackets.
Conclusions:I. Some raincoats are caps.
II. Some trousers are umbrellas.
III. All raincoats are jackets.
Solution:
QUESTION: 64
The mean of marks in statistics of 100 students in class was 72. The mean of marks of boys was 75, while their number was 70. The mean marks of girls in the class are
Solution:
QUESTION: 65
Mean of a set of 22 readings is 20. If each reading is first multiplied by 4 and then 10 is added to it, the new mean is
Solution:
QUESTION: 66
Below are the statements followed by four conclusions numbered I,II,III and IV. You have to consider the statements and the following conclusions and decide which of the conclusion(s) follows the statement(s).
Statements :
a. All towns are villages.
b. No village is forest.
c. Some forests are rivers.
Conclusions :
I. Some forests are villages.
II. Some forests are not villages.
III. Some rivers are not villages.
IV. All villages are towns.
Solution:
QUESTION: 67
Sound travels at 330 metres a second. How many km away is a thunder cloud when its sound follows the flash after 10 seconds?
Solution:
QUESTION: 68
If 6 men and 8 boys can do a piece of work in 10 days and 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys to do the same type of work will be
Solution:
QUESTION: 69
For plant metabolism, calcium, magnesium and sulphur are
Solution:
QUESTION: 70
The length, breadth and height of a cuboid are in the ratio 1 : 2 : 3. The length, breadth and height of the cuboid are increased by 100%, 200% and 300% respectively. Then the increase in the volume of the cuboid is
Solution:
Suppose length, breadth and height of cuboid are respectively x, 2x, 3x
∴ Volume of cuboid = x x 2x x 3x = 6x3
Now, increased length, breadth, height of cuboid
= x x 200/100 , 2x x 300/100 , 3x x 300 /100 = 2x, 6x, 9x
∴ Volume of new cuboid = 2x x 6x 9x = 108x3
Increase in volume of cuboid = (108 - 6)x3 = 102 x3 = 6x3 x 17
= 17 times the original volume.
QUESTION: 71
The minimum distance between the source and the reflector, so that an echo is heard is approximately equal to ______.
Solution:
QUESTION: 72
If by arranging the letters of the word NABMODINT, the name of the game is formed, what are the first and the last letters of the word so formed ?
Solution:
QUESTION: 73
The type of energy possessed by a simple pendulum, when it is at the mean position is
Solution:
QUESTION: 74
Five business leaders, Bill Gates, Donald Trump, Jack Welch, Laxmi Mittal, and Warren Buffet, were being judged for the award of Business leader of the year. The selection committee ranked them on five different parameter viz. Empathy, Vision, Focus, Creativity, and Intuition. The five leaders were given points on these five parameters. The topper gets 5 and the least gets 1 point. There are no ties. The person with highest total gets the award. Bills Gates gets the award by scoring 24. Warren Buffet gets 5 points in Creativity and 3 points in Intuition. Jack Welch got consistent score in 4 of the parameters. Their final standings were in alphabetical order of their names.
Q. What is the overal score of Jack Welch?
Solution:
QUESTION: 75
Five business leaders, Bill Gates, Donald Trump, Jack Welch, Laxmi Mittal, and Warren Buffet, were being judged for the award of Business leader of the year. The selection committee ranked them on five different parameter viz. Empathy, Vision, Focus, Creativity, and Intuition. The five leaders were given points on these five parameters. The topper gets 5 and the least gets 1 point. There are no ties. The person with highest total gets the award. Bills Gates gets the award by scoring 24. Warren Buffet gets 5 points in Creativity and 3 points in Intuition. Jack Welch got consistent score in 4 of the parameters. Their final standings were in alphabetical order of their names.
Q. What is the rank of Donald Trump in Creativity?
Solution:
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code | 3,958 | 15,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-33 | latest | en | 0.891066 |
http://www.talkstats.com/showthread.php/48545-Tricky-problem-solution | 1,508,383,411,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823220.45/warc/CC-MAIN-20171019031425-20171019051425-00594.warc.gz | 573,099,709 | 13,167 | 1. ## Tricky problem solution
Hi,
A soccer team plays matches.
-- If in a previous match it won, it has a 0.6 chance to win in the next match (if it failed, the chances to win in the next match are 0.51).
-- If in a match before the previous one it had won, the chances of winning next time are 0.55, but if it had failed, the chances of winning are 0.49.
So the outcome of the next match is statistically dependant on two past events, and the outcome of a previous match depends on the match before too, respectively.
If the team have won both previous matches, what are the chances it will win the next time?
Could you please indicate possible way of solving this problem?
Alexey
2. ## Re: Tricky problem solution
Hello,
So, I am interested in multivariate conditional probability. I have never read any literature on it. May I aks you to advise me on the authors to read or even specific articles?
Alexey
3. ## Re: Tricky problem solution
Originally Posted by alexeymosc
Hello,
So, I am interested in multivariate conditional probability. I have never read any literature on it. May I aks you to advise me on the authors to read or even specific articles?
Alexey
It just seems to me to be an ill-defined problem.
4. ## Re: Tricky problem solution
Originally Posted by Dason
It just seems to me to be an ill-defined problem.
I am not a pro in probability theory (I had a basic course in the university only). However, I am interested in analysis of time series with hidden dependancies and what some calls the long memory. It is just a hobby. I will try to explain what my diea is.
I have recently came across a time series of high frequency currency quotes which made me think. I transformed the series into returns and if the return has a positive sign, let's denote it as a 1, and if the sign is negative, then 0.
below are probabilities of occuring 1 given 1 or 0 at previous step(s):
lag 1
0---0,56
1---0,44
lag 2
0---0,45
1---0,55
lags 1 and 2
0---0---0,51
0---1---0,41
1---0 ---0,59
1---1---0,49
So the conditional probability when analyzing 2 events is different from that based on 1 previous event, which means according to my understanding that the uncertainty about the latest event becomes less.
So, given only lag 1 and lag 2 pair-wise probabilities, is it possible to come at the 3-event conditional probability wihout having additional knowledge about the process?
5. ## Re: Tricky problem solution
I have already discussed this problem at a probability forum in Russia (where I am from ), and I was told by a probability associate professor that the problem that I formulated cannot be solved using only pair-wise probabilities. I also experimented with frequencies on the real data and came to that conclusion as well. That means if I know conditional probabilities between A & B, and between B & C, I still cannot figure out a conditional probability of A given C. It was intriguing to me, but seems unresolvable. Kolmogorov's axiomatization of conditional probability involves knowing the frequency of a joint event A & C (in the following equation they are A&B): . There is no other way to get the figure. However, I think that a thought giant in probability - who I am not - can have a ground to dig this problem and look for possible ways to estimate this figure.
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http://bedtimemath.org/category/daily-math/sports/page/2/ | 1,519,161,071,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813109.36/warc/CC-MAIN-20180220204917-20180220224917-00195.warc.gz | 38,646,116 | 15,134 | # ‘Sports’
##### Speedy Scooter Dog
October 14, 2017 4:00 pm
We’ve seen dogs ride skateboards and surfboards, which takes great balance. But for a dog to ride a scooter is a different story. Well, Norman the French sheepdog can do just that, and has set the doggie scooter-riding world record. When you think about all the hours of training and scooter rides it took for him to do this, it really adds up and makes his feat that much more impressive!
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https://www.jiskha.com/questions/1002324/lim-x-to-1-of-sin-1-cosx-x | 1,575,853,870,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540515344.59/warc/CC-MAIN-20191208230118-20191209014118-00233.warc.gz | 768,935,868 | 5,340 | # Calculus
lim x to 1 of (sin(1-cosx))/x
1. 👍 0
2. 👎 0
3. 👁 206
1. Are you sure x is not approaching zero?
then you have
cos x = 1 - x^2/2 ......
and
sin (x^2/2)/x
= x/2
= 0
1. 👍 0
2. 👎 0
posted by Damon
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https://techcommunity.microsoft.com/t5/excel/more-then-64-levels-of-nesting-need-a-simpler-formula/td-p/3035669 | 1,642,692,367,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00210.warc.gz | 612,989,109 | 75,461 | SOLVED
# More then 64 levels of nesting need a simpler formula
Occasional Contributor
# More then 64 levels of nesting need a simpler formula
Hello All,
I'm new to excel and just bumped into the max nesting . Can anyone please help me to simplify the formula or help me find a way around it please?
I need the IF Formula to run a search value from A2 to A36 and from C2 to C36 and if true assign value on the cell next to it. I've done the formula here but u cant use it since is too big.
I can I work around this?
=IF(B2>0;IF(ISNUMBER(SEARCH(Recap!\$A\$2;N2));Recap!\$B\$2;IF(ISNUMBER(SEARCH(Recap!\$A\$3;N2));Recap!\$B\$3;IF(ISNUMBER(SEARCH(Recap!\$A\$4;N2));Recap!\$B\$4;IF(ISNUMBER(SEARCH(Recap!\$A\$5;N2));Recap!\$B\$5;IF(ISNUMBER(SEARCH(Recap!\$A\$6;N2));Recap!\$B\$6;IF(ISNUMBER(SEARCH(Recap!\$A\$7;N2));Recap!\$B\$7;IF(ISNUMBER(SEARCH(Recap!\$A\$8;N2));Recap!\$B\$8;IF(ISNUMBER(SEARCH(Recap!\$A\$9;N2));Recap!\$B\$9;IF(ISNUMBER(SEARCH(Recap!\$A\$10;N2));Recap!\$B\$10;IF(ISNUMBER(SEARCH(Recap!\$A\$11;N2));Recap!\$B\$11;IF(ISNUMBER(SEARCH(Recap!\$A\$12;N2));Recap!\$B\$12;IF(ISNUMBER(SEARCH(Recap!\$A\$13;N2));Recap!\$B\$13;IF(ISNUMBER(SEARCH(Recap!\$A\$14;N2));Recap!\$B\$14;IF(ISNUMBER(SEARCH(Recap!\$A\$15;N2));Recap!\$B\$15;IF(ISNUMBER(SEARCH(Recap!\$A\$16;N2));Recap!\$B\$16;IF(ISNUMBER(SEARCH(Recap!\$A\$17;N2));Recap!\$B\$17;IF(ISNUMBER(SEARCH(Recap!\$A\$18;N2));Recap!\$B\$18;IF(ISNUMBER(SEARCH(Recap!\$A\$19;N2));Recap!\$B\$19;IF(ISNUMBER(SEARCH(Recap!\$A\$20;N2));Recap!\$B\$20;IF(ISNUMBER(SEARCH(Recap!\$A\$21;N2));Recap!\$B\$21;IF(ISNUMBER(SEARCH(Recap!\$A\$22;N2));Recap!\$B\$22;IF(ISNUMBER(SEARCH(Recap!\$A\$23;N2));Recap!\$B\$23;IF(ISNUMBER(SEARCH(Recap!\$A\$24;N2));Recap!\$B\$24;IF(ISNUMBER(SEARCH(Recap!\$A\$25;N2));Recap!\$B\$25;IF(ISNUMBER(SEARCH(Recap!\$A\$26;N2));Recap!\$B\$26;IF(ISNUMBER(SEARCH(Recap!\$A\$27;N2));Recap!\$B\$27;IF(ISNUMBER(SEARCH(Recap!\$A\$28;N2));Recap!\$B\$28;IF(ISNUMBER(SEARCH(Recap!\$A\$29;N2));Recap!\$B\$29;IF(ISNUMBER(SEARCH(Recap!\$A\$30;N2));Recap!\$B\$30;IF(ISNUMBER(SEARCH(Recap!\$A\$31;N2));Recap!\$B\$31;IF(ISNUMBER(SEARCH(Recap!\$A\$32;N2));Recap!\$B\$32;IF(ISNUMBER(SEARCH(Recap!\$A\$33;N2));Recap!\$B\$33;IF(ISNUMBER(SEARCH(Recap!\$A\$34;N2));Recap!\$B\$34;IF(ISNUMBER(SEARCH(Recap!\$A\$35;N2));Recap!\$B\$35;IF(ISNUMBER(SEARCH(Recap!\$A\$36;N2));Recap!\$B\$36;IF(ISNUMBER(SEARCH(Recap!\$C\$2;N2));Recap!\$D\$2;IF(ISNUMBER(SEARCH(Recap!\$A\$3;N2));Recap!\$D\$3;IF(ISNUMBER(SEARCH(Recap!\$C\$4;N2));Recap!\$D\$4;IF(ISNUMBER(SEARCH(Recap!\$C\$5;N2));Recap!\$D\$5;IF(ISNUMBER(SEARCH(Recap!\$C\$6;N2));Recap!\$D\$6;IF(ISNUMBER(SEARCH(Recap!\$C\$7;N2));Recap!\$D\$7;IF(ISNUMBER(SEARCH(Recap!\$C\$8;N2));Recap!\$D\$8;IF(ISNUMBER(SEARCH(Recap!\$C\$9;N2));Recap!\$D\$9;IF(ISNUMBER(SEARCH(Recap!\$C\$10;N2));Recap!\$D\$10;IF(ISNUMBER(SEARCH(Recap!\$C\$11;N2));Recap!\$D\$11;IF(ISNUMBER(SEARCH(Recap!\$C\$12;N2));Recap!\$D\$12;IF(ISNUMBER(SEARCH(Recap!\$A\$13;N2));Recap!\$D\$13;IF(ISNUMBER(SEARCH(Recap!\$C\$14;N2));Recap!\$D\$14;IF(ISNUMBER(SEARCH(Recap!\$C\$15;N2));Recap!\$D\$15;IF(ISNUMBER(SEARCH(Recap!\$C\$16;N2));Recap!\$D\$16;IF(ISNUMBER(SEARCH(Recap!\$C\$17;N2));Recap!\$D\$17;IF(ISNUMBER(SEARCH(Recap!\$C\$18;N2));Recap!\$D\$18;IF(ISNUMBER(SEARCH(Recap!\$C\$19;N2));Recap!\$D\$19;IF(ISNUMBER(SEARCH(Recap!\$C\$20;N2));Recap!\$D\$20;IF(ISNUMBER(SEARCH(Recap!\$C\$21;N2));Recap!\$D\$21;IF(ISNUMBER(SEARCH(Recap!\$C\$22;N2));Recap!\$D\$22;IF(ISNUMBER(SEARCH(Recap!\$C\$23;N2));Recap!\$D\$23;IF(ISNUMBER(SEARCH(Recap!\$C\$24;N2));Recap!\$D\$24;IF(ISNUMBER(SEARCH(Recap!\$C\$25;N2));Recap!\$D\$25;IF(ISNUMBER(SEARCH(Recap!\$C\$26;N2));Recap!\$D\$26;IF(ISNUMBER(SEARCH(Recap!\$C\$27;N2));Recap!\$D\$27;IF(ISNUMBER(SEARCH(Recap!\$C\$28;N2));Recap!\$D\$28;IF(ISNUMBER(SEARCH(Recap!\$C\$29;N2));Recap!\$D\$29;IF(ISNUMBER(SEARCH(Recap!\$C\$30;N2));Recap!\$D\$30;IF(ISNUMBER(SEARCH(Recap!\$C\$31;N2));Recap!\$D\$31;IF(ISNUMBER(SEARCH(Recap!\$C\$32;N2));Recap!\$D\$32;IF(ISNUMBER(SEARCH(Recap!\$C\$33;N2));Recap!\$D\$33;IF(ISNUMBER(SEARCH(Recap!\$C\$34;N2));Recap!\$D\$34;IF(ISNUMBER(SEARCH(Recap!\$C\$35;N2));Recap!\$D\$35;IF(ISNUMBER(SEARCH(Recap!\$C\$36;N2));Recap!\$D\$36; « False »)
6 Replies
# Re: More then 64 levels of nesting need a simpler formula
``````=INDEX(ColumnContainingDesiredValues,
MATCH(YourCriteria,ColumnContainingDataAsPerYourCriteria,0),1)``````
# Re: More then 64 levels of nesting need a simpler formula
Click on the blue button in cell H1 of the attached sheet. Maybe this is what you want to do.
# Re: More then 64 levels of nesting need a simpler formula
Everything in Excel could be done by several ways. If at first you explain your logic, illustrate it with small sample (with manually added desired result if possible), mention on which Excel version/platform you are that will be much easier and faster to make concrete suggestion.
In general you did such way, but it's not clear you search for entire value and text as part in another text, are all values texts or some numbers, which Excel do you use.
Above as a comment, I vote for @Juliano-Petrukio suggestion. It could be more concrete with more concrete sample.
# Re: More then 64 levels of nesting need a simpler formula
Hello Sergei, Thank you for this precisions:
The Version I use is Excel 2016 in Windows 10 (1607)
The logic I'm searching is:
I have a table of values that associate with certain users - eg If "RED" user "John"
After that I have an extract that I need to run through for this data
eg. the value color comes out in column "N" and I must assign user in column "S" but the assignment needs to respect the row
If "N1" the cell contains the word "RED" the cell "S1" needs to be populated with "John" .
The suggestion of Juliano, can I keep the matching row condition?
I've redone a new sample file
Hope this is more clear.
Thank you for you help and time
best response confirmed by KuroDesu (Occasional Contributor)
Solution
# Re: More then 64 levels of nesting need a simpler formula
``````=IFERROR( | 2,333 | 5,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-05 | latest | en | 0.593232 |
https://solvedlib.com/n/question-122-ptsconsider-the-laplace-equation-defined-in,11386071 | 1,695,805,431,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00269.warc.gz | 578,124,784 | 20,579 | # Question 122 ptsConsider the Laplace equation;defined in the inside of the circle, z? +y? < The boundary condition given by
###### Question:
Question 12 2 pts Consider the Laplace equation; defined in the inside of the circle, z? +y? < The boundary condition given by u (1,y) 312 { for (T,y) € C,where C is the circle y = Mark all values which the function u(r,y) CANNOT attain inside the circle:
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##### Question 21ptsIndicate whether the descriptions below refer to systolic or diastolic more than once_ pressure. Answers may be choseThis is the pressure from the blood being pumped out of the heart when the ventricles [Choose contract Choose systolic pressurc This is the pressure excrted on the blood diastolic pressure when the ventricles relax and the arterial Choose walls recoil (snap back in place like a rubber band)This is the higher of the two blood pressure readings[Choose _The average of t
Question 2 1pts Indicate whether the descriptions below refer to systolic or diastolic more than once_ pressure. Answers may be chose This is the pressure from the blood being pumped out of the heart when the ventricles [Choose contract Choose systolic pressurc This is the pressure excrted on the bl... | 2,545 | 9,384 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.84089 |
https://cs.stackexchange.com/questions/39669/monotone-frameworks-transfer-functions-for-flow-edges-instead-of-labels | 1,719,107,558,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00683.warc.gz | 171,821,836 | 41,229 | # Monotone Frameworks: Transfer functions for flow edges instead of labels
So, in generic program analysis, we have a lattice $L$ with a join operation $\sqcup$, program with statements labelled, and for each label $b$, a transfer function $F_b : L \rightarrow L$.
The goal is to find, for each label $b$, a fix-point $x$ such that $F_b(x)=x$. This is a common approach used in dataflow analysis,
In "Principles of Program Analysis" by Nielson, Nielson and Hankin, they reference that "Some formulation of monotone frameworks associate transfer functions with edges (or flows) rather than nodes (or labels)." However, in the textbook, they simply reference an exercise where the reader is asked to implement such a framework.
I'm wondering if anyone has a reference to either a book or paper where such a formulation is given. In particular, I'm wondering how a typical worklist algorithm for finding a fixed-point would be formulated.
Your formulation is not stated in the way I typically think of it. Traditionally rather than thinking of looking for a fixed point at each node, we're looking for a solution to a system of simultaneous equations:
$$x_b = F_b \left( \bigsqcup_{p \in \mathtt{pred}[b]}x_p \right) \; \mathrm{for\:all\:nodes}\:b$$
If $F_b$ has the distributive property over $\sqcup$ then you can simply rewrite this as
$$x_b = \bigsqcup_{p \in \mathtt{pred}[b]} F_b(x_p) \; \mathrm{for\:all\:nodes}\:b$$
and since you have a function evaluation for every edge $p \rightarrow b$ you might as well just rewrite it as
$$x_b = \bigsqcup_{p \in \mathtt{pred}[b]} F_{p \rightarrow b}(x_p) \; \mathrm{for\:all\:nodes}\:b$$
If you don't like that handwaving, then here's some more: think of it as taking $F_b$ for every node with one predecessor ($p$) and pushing $F_b$ onto the edge making it $F_{p \rightarrow b}$. For nodes $b$ with more than one predecessor split them into two nodes, $b$ and $b'$. The predecessors of $b$ are still the same, then there's an edge $b \rightarrow b'$ and then the successors of the original $b$ become the successors of $b$. Now the original label $F_b$ becomes $F_{b\rightarrow b'}$. The join operation still happens on the nodes.
Note that while transfer functions are associated with edges, data flow facts are still associated with nodes. So the worklist algorithm doesn't really change. Actually, with the normal (transfer functions on nodes) formulation, the worklist algorithm is really a worklist of edges, not nodes, so nothing changes. (For an example of this take a look at these slides that came up when I googled for "dominance frontier". http://www.seas.harvard.edu/courses/cs252/2011sp/slides/Lec04-SSA.pdf. They give the worklist algorithm on page 7, and it's based on edges.)
A paper using the edge formulation is:
G. Ramalingam: Data Flow Frequency Analysis. PLDI 1996: 267-277.
(I haven't read the paper beyond the first couple of pages to see that they were using the edge formulation, it's what came up after a bit of Googling.)
I wasn't familiar with the edge transfer function approach until you asked this question. But I also know of at least two cases where it is important to take edges into account.
The first is when dealing with (post)dominance frontiers. The natural definition of the dominance frontier of a node is a set of edges, but the original (more complicated) formulation was in terms of nodes. See for example the (brief) discussion of this in
Pingali, Keshav; Bilardi, Gianfranco: Optimal Control Dependence Computation and the Roman Chariots Problem, ACM ToPLaS, 19(3):462-491, 1997.
Another example is in the classic
Knoop, J; Rüthing, O; Steffen, B: Lazy Code Motion, PLDI, 1992.
In Lazy Code Motion you sometimes need to move computations onto edges, but in the original formulation in the paper they were trying to move those computations to different nodes, which gave them trouble with branch nodes that target join nodes. They called these critical edges and had to add a node along the edge. This is explained in the following lecture notes by Cooper and Torczon: https://www.cs.utexas.edu/~pingali/CS380C/2013/lectures/LazyCodeMotion.pdf
• Thanks for this! This is quite helpful. Part of my confusion was that, in NNH, they distinguish between solving for $x_{b \circ}$, the context value, and $x_{b\ldotp}$, the effect value, where the context was equal to the join of all its predecessors effecct, and $x_{b\ldotp} = F_b(x_{b\circ})$. So in an edge-based version, would there be a join operation in computing the effect as well? Commented Feb 23, 2015 at 10:43
• Sorry, messed up the latex, should be $x_{b\bullet}$ and $x_{b\circ}$ Commented Feb 23, 2015 at 10:51
• Something like what's described on slides 38-44 of this presentation Commented Feb 23, 2015 at 10:52
• Yes, in an edge based version there is a join. It is still on the nodes. But since, more often than not, you are working with monotone functions you can do the join incrementally. See slide 84 of the presentation you linked. The line $A[l'] \leftarrow A[l'] \sqcup f_l(A[l])$ Commented Feb 23, 2015 at 12:25
• That is: the calculation is the same: you evaluate $F_{p\rightarrow b}(x_p)$ and then join the result into $x_b$. Commented Feb 23, 2015 at 12:36 | 1,377 | 5,246 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-26 | latest | en | 0.90225 |
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Stefani Joanne Angelina Germanotta is Lady Gaga's real name.
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A quadratic equation is one in which the highest power of the unknown is two. Its general form is ax^2+bx+c=0.
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The quotient of 6x^2 + 5x - 1 and 2x + 1 is 3x + 1 with a remainder of -2. To solve this. Divide 6x^2 and 2x. Multiply 3x and 2x + 1, you'll get 6x^2 + 3x. Subtract 6x^2 + 3x from 6x^2 + 5x, you'll get 2x, then bring down -1. Divide 2x - 1 and 2x + 1, you'll get 1 with a remainder of - 2. So the final result is 3x + 1 with a remainder of -2.
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,332 | 3,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-34 | latest | en | 0.872875 |
https://brainly.com/question/72708 | 1,484,666,872,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279923.28/warc/CC-MAIN-20170116095119-00386-ip-10-171-10-70.ec2.internal.warc.gz | 795,870,873 | 9,490 | 2014-06-10T14:34:16-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
X + y = 34.5
xy = 297
From the first one . . . y = 34.5 - x
Plug that into the second one . . . x(34.5 - x) = 297
Eliminate parentheses . . . 34.5x - x² = 297
Bend that around and tidy it up . . . x² - 34.5x + 297 = 0
Apply the quadratic formula to that, and find . . .
x =18
x = 16.5
2014-06-10T21:35:26-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. | 269 | 952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-04 | latest | en | 0.920716 |
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PROPORTION
If two ratios w : x and y : z are equal, that is w : x = y : z then the four quantities w, x, y & z are said to be in proportion. This is also written as w : x : : y : z, the quantities w, x, y, z which are called the terms of the proportion, need not all be of the same kind. However the terms of the first ratio must be of the same kind, and the terms of the second ratio must also be of the same kind.
For your better understanding, we will provide you some example –
Example.1) Show the numbers such as 12, 36, 72, 216 are in proportion
Ans.) As per the given condition 12, 36, 72, 216 are in proportion because –
12 1
----------- = ---------- = 1 : 2
24 2
72 1
And ---------- = --------- = 1 : 2
216 2
So, numbers such as 12, 36, 72, 216 are in proportion.
Example.2) Show the quantities 4 h, 12 h, 16 L, 48 L are in proportion.
Ans.) As per the given condition 4 h, 12 h, 16 L, 48 L are in proportion because –
4 h 1
----------- = --------- = 1 : 3
12 h 3
16 L 1
------------- = ---------- = 1 : 3
48 L 3
So, the quantity such as 4 h, 12 h, 16 L, 48 L are in proportion.
The first and the fourth terms of a proportion are called the “EXTREMES”, while the second and the third terms are called the “MEANS”. The fourth term is also called the “FOURTH PROPORTION” to the other three terms.
w y
If, w : x = y : z, then -------- = -------- or wz = xy
x z
hence we can conclude that, the product of the "Extremes" = the product of the "Means"
above rule is known as the “CROSS PRODUCT RULE”. | 508 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.83872 |
https://www.physicsforums.com/threads/work-energy-problem.729799/ | 1,527,130,271,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034502-00398.warc.gz | 808,017,035 | 16,896 | # Homework Help: Work-Energy Problem
1. Dec 24, 2013
### Dumbledore211
1. The problem statement, all variables and given/known data
A stone is thrown keeping target to a mango hanging in the branch of a mango tree. The velocity of the stone while hitting the mango is 9.8ms^-1. If the boy uses half of the energy used before the stone can reach the same height of the mango. Mass of the mango is 250gm.
2. Relevant equations
v^2=u^2-2gh
3. The attempt at a solution
According to the question the final velocity of the stone hitting the mango should be 9.8ms^-1. The velocity with which the boy projected the stone vertically should be greater than 9.8ms^-1 since it should be enough to counteract the gravitational acceleration. So we can say that the kinetic energy the the stone gains is half the kinetic energy the boy uses to project the stone. The workout is shown below and if I make any silly mistakes or misinterpret the question please point out my mistake and please see if the answer I got is reasonable enough
Let the velocity with which the boy projects the stone be v1
final velocity of the stone v2=9.8ms^-1
According to first condition
Ek1/2=Ek2
v1^2/2= v^2
v1= √2 v2
Hence, v1= 13.86ms^-1≈
v2^2=v1^2-2gh or, (9.8)^2= (13.86)^2-19.6h or, 96.04-192.1= -19.6h or, h= 96.06/19.6
h= 4.9meters
Sorry, I don't have the answer to this prob in my book so I have to make sure if this is the correct one..
2. Dec 24, 2013
### CWatters
The question doesn't make much sense. Can you check the wording.
3. Dec 24, 2013
### Dumbledore211
That is the part which I found bewildering to understand but I think what the question meant is that the final kinetic energy of the stone hitting the mango becomes half the kinetic energy of the boy that he uses to project the stone vertically . Then the whole problem makes sense.........
Last edited: Dec 24, 2013
4. Dec 24, 2013
### CWatters
...if you assume that the weight of the Mango is given just to confuse you.
Assuming you are correct then another way to solve the problem is to apply conservation of energy... the PE gain is where the other half of the initial energy went so.
mgh = 0.5 * 0.5mV12
then substitute V1 with √2 V2 | 613 | 2,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-22 | latest | en | 0.899858 |
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