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# Princeton Review GMAT test - Accuracy

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Princeton Review GMAT test - Accuracy [#permalink]

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28 Jul 2013, 10:34
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Hello Everyone.

I just finished the Princeton review test (free) on their website and this is the score I got :

Verbal: 31 (31 out of 41 correct)
Quantitative: 39 (34 out of 37 correct)
Total (V & Q): 580

Do you guys know of any issues in scoring algorithm of PR ?

Thanks,
Gaurav

Aiming for 750.
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Re: Princeton Review GMAT test - Accuracy [#permalink]

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28 Jul 2013, 20:24
Why are you doubting it? Do you have a different score that's meaningfully different (less than 40 points is not meaningful even on the GMAT)
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Re: Princeton Review GMAT test - Accuracy [#permalink]

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28 Jul 2013, 21:09
I doubt it as the score is pretty low compared to the percentage of correct ansers. In Manhattan test I scored 680 with far many wrong answers. I think Princeton algo is faulty and assigning lower scores are a part of their marketing strategy.

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Re: Princeton Review GMAT test - Accuracy [#permalink]

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28 Jul 2013, 23:08
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gaurav780 wrote:
Hello Everyone.

I just finished the Princeton review test (free) on their website and this is the score I got :

Verbal: 31 (31 out of 41 correct)
Quantitative: 39 (34 out of 37 correct)
Total (V & Q): 580

Do you guys know of any issues in scoring algorithm of PR ?

Thanks,
Gaurav

Aiming for 750.

The score does seem a bit suspect to me (especially in Quant) if the numbers are what you have given above (any chance of misinterpretation or typo?). There are many factors (e.g. difficulty level of the questions) that decide your raw score but if you answer questions correctly, you are supposed to get harder questions. Since you missed out on only 3 questions, even if they are all in a row, they can't affect your score this much. Though I haven't heard of any major issues with the scoring algorithm of Princeton Review. Whatever the reason for your low score, I don't think Princeton Review will try to get a marketing boost using a faulty algorithm. If anything, it will have the opposite effect since today's clientele is discerning enough. It might be some temporary glitch somewhere. Just to be sure, I would suggest you to try out the GMAT prep test or our free test.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 07 Jul 2013 Posts: 3 Followers: 1 Kudos [?]: 1 [0], given: 0 Re: Princeton Review GMAT test - Accuracy [#permalink] ### Show Tags 06 Aug 2013, 10:48 Definitely not a typo. I tried a free test on gmat club - got 51Q and 31V - I think that rounds to 680. Anytips to improve verbal in a week ? Intern Joined: 09 Sep 2013 Posts: 8 Location: India GMAT 1: 710 Q50 V36 Followers: 1 Kudos [?]: 6 [0], given: 0 Re: Princeton Review GMAT test - Accuracy [#permalink] ### Show Tags 21 Sep 2013, 04:24 Even I had the same problem. I gave the free Princeton Test: Verbal 33 of 41 correct Quant 32 of 37 correct Score 670. I gave GMAT Prep test few days back...scored 710 with a lot many incorrect questions in there. Intern Joined: 12 Jul 2013 Posts: 9 Concentration: General Management, International Business Schools: HBS '17 GMAT Date: 06-12-2014 GPA: 3.66 WE: Information Technology (Other) Followers: 0 Kudos [?]: 6 [0], given: 9 Re: Princeton Review GMAT test - Accuracy [#permalink] ### Show Tags 24 May 2014, 04:53 I just took the Princeton Free GMAT Test and I became doubtful. Q 46(30 out of 37) and V41 (31 out of 41). The thing is I have never gotten about a 34 in Verbal. In fact the 34 I got was a retake on GMAT Prep 1. I started from a V25 (Manhattan GMAT) but I have been studying. The Verbal questions seemed easier. Could it be that I have improved sooooooo much? Wow!! I soooooo hope that is the case. I immediately Googled to see what others are saying about the Princeton Review Test. Does any one else feel the Verbal section was some what easier than normal GMAT Prep questions? _________________ Constantly sharpening the sword... Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7118 Location: Pune, India Followers: 2128 Kudos [?]: 13618 [0], given: 222 Re: Princeton Review GMAT test - Accuracy [#permalink] ### Show Tags 25 May 2014, 21:05 dakrownie wrote: I just took the Princeton Free GMAT Test and I became doubtful. Q 46(30 out of 37) and V41 (31 out of 41). The thing is I have never gotten about a 34 in Verbal. In fact the 34 I got was a retake on GMAT Prep 1. I started from a V25 (Manhattan GMAT) but I have been studying. The Verbal questions seemed easier. Could it be that I have improved sooooooo much? Wow!! I soooooo hope that is the case. I immediately Googled to see what others are saying about the Princeton Review Test. Does any one else feel the Verbal section was some what easier than normal GMAT Prep questions? Try an official GMAT Prep Test (mba.com) or Veritas Free Practice test (http://www.veritasprep.com/gmat/free-gm ... tice-test/). A V41 in these tests will validate your score. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Princeton Review GMAT test - Accuracy [#permalink]

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24 Jun 2014, 19:22
I recently took two Princeton Review tests:

Yesterday:
Q49 (31/37)
V37 (27/41)
690

Today:
Q43 (33/37)
V33 (31/41)
620

Even looking at which problems were experimental, it shouldn't cause this much of a difference. Some of the difficulty levels went down after correct answers.
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Re: Princeton Review GMAT test - Accuracy [#permalink]

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26 Jun 2014, 05:16
i have the same question..
Q47 (34 of 37 correct)
V35 (32 of 41 correct)
overall score - 660

while in gmat prep 1, i got overall score of 660 with 9 questions incorrect in Quant and 17 questions incorrect in verbal.. i don't think i would have improved so much in a week.. no matter how hard i study..

i felt that quants of princeton review was extreeeemmmeeeelllyyyyy easy and verbal was also a little easier (for ppl weak in verbal)

if anyone has analyzed then can you please throw some light on algo (which seems incorrect) of Princeton Review practice test and help us understand how do we predict our actual GMAT score based on this.. will be very helpful..
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Re: Princeton Review GMAT test - Accuracy [#permalink]

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19 Jul 2014, 11:49
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Hi,

First thing don't be scared of Princeton review scores

Yes there is some flaw in their algorithm.

I tried their diagnostic test with no preparation at all ( completed quant in 30 mins! )

Scored : 630
Quant : 33 out of 37
Verbal : 27 out of 41

Then after a preparation of 2 months with the following scores:

GMAT PREP 1 : 750
GMAT PREP 2: 720

KAPLAN 6 TESTS : 740,720,740,720,700 (This one sucks)
VERITAS : 720

I again tried princeton test:

Score : 640
Quant : 34 out of 37
Verbal 31 out of 41

Though Quant was easy 2-3 question are really hard no way (45 /51)
and verbal was of normal difficulty.

I know after reviewing carefully that I deserve greater than 640 for this attempt.
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Princeton Review GMAT test - Accuracy [#permalink]

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12 Oct 2014, 08:09
Something is wrong with PR mock. I just took PR mock and got 650(QA 48/VA32).
I got 5 wrong in QA (all are silly mistakes which I have myself to blame) but didn't agree with some of the OAs of VA (got 9 incorrect).

Out of 41 VA questions around 9 were experimental. If we see the difficulty level in VA, I started with 5 rating and ended up with 6. In between I got 8/9 level of experimental questions which surprisingly I got correct.

Last 4 mocks:
VeritasPrep Mock 4 - 710(QA50/VA37)
VeritasPrep Mock 3 - 710(QA50/VA37)
VeritasPrep Mock 2 - 720(QA50/VA39)
GMATPrep Mock 3 - 760 (QA50/VA42)

Anyway, I will be completing the other mocks and let see how I stabilize.
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Re: Princeton Review GMAT test - Accuracy [#permalink]

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15 Feb 2015, 05:27
magatx wrote:
I recently took two Princeton Review tests:

Yesterday:
Q49 (31/37)
V37 (27/41)
690

Today:
Q43 (33/37)
V33 (31/41)
620

Even looking at which problems were experimental, it shouldn't cause this much of a difference. Some of the difficulty levels went down after correct answers.

I just had V29 with 26 of 41. This is not even possible right? If you scored a V37 with 27/41 !
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Re: Princeton Review GMAT test - Accuracy [#permalink]

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28 Feb 2015, 18:21
Verbal: 37 ( 32 correct out of 41)
Quantitative: 40 ( 28 correct out of 37)
Total (V & Q): 630

Is the scoring algorithm accurate..please advice?
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Re: Princeton Review GMAT test - Accuracy [#permalink]

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21 May 2015, 21:26
Hi Even my score have varied exponentially for Princeton review tests but not for Kaplan, I also am in doubt abt. the scoring system of Princeton

My Princeton Tests scores
Test 1 - 660 (QA - 32/37 , VA - 30/41)
Test 2 - 740 (QA - 28/37 , VA - 35/41)

Quant of Kaplan tests is balanced overall and i think close to actual GMAT questions
Verbal of Kaplan tests is little hard than actual GMAT
Test 1 - 740 (QA - 24/37 , VA - 23/41)
Test 2 - 760 (QA - 34/37 , VA - 30/41)
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Re: Princeton Review GMAT test - Accuracy [#permalink]

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22 May 2015, 10:23
Hi apoorv601,

The scoring algorithm on the Official GMAT is far more complicated than most people realize. It takes into account a number of different factors, including the relative difficulty of the question, whether you were expected to get it correct or not, the placement of the question, what's going on "around it", the "strings" of correct and incorrect answers, whether the question even counts or not (some questions are "experimental" and are worth 0), if you leave questions unanswered and incur a penalty, etc. As such, you shouldn't be spending time trying to figure it all out. You'd be better served working on building up your skills.

Since that algorithm is proprietary, no GMAT company has an exact match for it, thus CAT scores can vary a bit based on the 'biases' involved in their respective designs. Some are known to be more accurate, others less so.

With these CATs, did you take the ENTIRE CAT each time (including the Essay and IR sections)?
Have you taken any of the GMAC CATs?

GMAT assassins aren't born, they're made,
Rich
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# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Senior Manager Joined: 17 May 2015 Posts: 418 Followers: 8 Kudos [?]: 36 [0], given: 0 Re: Princeton Review GMAT test - Accuracy [#permalink] ### Show Tags 24 May 2015, 20:23 gaurav780 wrote: Hello Everyone. I just finished the Princeton review test (free) on their website and this is the score I got : Verbal: 31 (31 out of 41 correct) Quantitative: 39 (34 out of 37 correct) Total (V & Q): 580 Do you guys know of any issues in scoring algorithm of PR ? Thanks, Gaurav Aiming for 750. We can't comment for other tests but do note that no test prep company's algorithm will exactly match the real thing. The most we can do is get close. In terms of our scoring, many of our students report scores that are +/- 30 points from their actual GMAT test and so are a relatively good indicator of their performance. That said, no practice test can perfectly replicate the actual score algorithm, but this is quite close. You can always take a free GMAT PILL practice test here complete with detailed performance and timing analytics: http://www.gmatpill.com/gmat-practice-t ... ctice-test Intern Joined: 21 Apr 2015 Posts: 3 Followers: 0 Kudos [?]: 0 [0], given: 10 Re: Princeton Review GMAT test - Accuracy [#permalink] ### Show Tags 18 Apr 2016, 05:46 First Mock ever, 5 out of 6 DS errors are silly, in SC I'm clueless.. Verbal: 32 (29/41, CR 10/11, RC 11/13, SC 8/17) Quantitative: 48 (31/37, PS 19/19, DS 12/18) Total (V & Q): 650 Analytical Writing: * Integrated Reasoning: 5 (9/12) EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 8264 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 379 Kudos [?]: 2452 [0], given: 163 Re: Princeton Review GMAT test - Accuracy [#permalink] ### Show Tags 18 Apr 2016, 10:18 Hi bo2mac, A 650 is a strong first CAT score (the average score on the Official GMAT hovers around 540-550 most years). 1) How long have you been studying? 2) What is your goal score? 3) When are you planning to take the GMAT? GMAT assassins aren't born, they're made, Rich _________________ # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests

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Re: Princeton Review GMAT test - Accuracy [#permalink]

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18 Apr 2016, 11:02
EMPOWERgmatRichC wrote:
Hi bo2mac,

A 650 is a strong first CAT score (the average score on the Official GMAT hovers around 540-550 most years).

1) How long have you been studying?
2) What is your goal score?
3) When are you planning to take the GMAT?

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

1. Around a month
2. 720
3. July 2016

Still to take any GMATPrep mock
Re: Princeton Review GMAT test - Accuracy   [#permalink] 18 Apr 2016, 11:02

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# Latex formula not displaying correctly

#### sofiacardita

29 Aug, 2017 06:10 PM

This formula below displays correctly in https://stackedit.io/editor but not in Marked.

$$f_{L}(X)=β{0}+β{1}X_{1}+β{2}X{2}+...β{p} X{p}.$$

Is there anything I can do to fix it?

Thanks

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1. Support Staff Posted by Brett on 29 Aug, 2017 07:17 PM

I'm afraid I know so little about LaTeX and MathJax equations that I'm unsure exactly how this is supposed to work. However, I am seeing that escaping the underscores helps with either the MMD or GitHub processor…

$$f\_{L}(X)=β\_{0}+β\_{1}X\_{1}+β\_{2}X\_{2}+...β\_{p} X\_{p}.$$

In this case the Markdown processor isn't recognizing it as a math formula (because only the \$*formula* \$ syntax works at the processing point, so it's seeing emphasis tags and inserting <em>, breaking the MathJax rendering.

2. Posted by sofiacardita on 29 Aug, 2017 10:14 PM

Thank you! That was right!

3. sofiacardita closed this discussion on 29 Aug, 2017 11:07 PM.

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Less nuclear waste: The fusion reactors will not produce high-level nuclear wastes like their fission counterparts, so disposal will be less of a problem. They have also been used on earth for uncontrolled release of large quantities of energy in the thermonuclear or ‘hydrogen’ bombs. I try putting 3He2 --> 4He2 + 1H1 and the site says can;t give useful feedback. The problem is "The fusion of helium-3 to form helium-4 and hydrogen-1." Figure 1: The equation for nuclear fission. These fragments, or fission products, are about equal to half the original mass. Nuclear Fission. 1 a neutron inside the nucleus of an atom breaks down changing into a proton. Nuclear Fusion Nuclear Physics From A Level Physics Tutor . The energy from the Sun - both heat and light energy - originates from a nuclear fusion process that is occurring inside the core of the Sun.The specific type of fusion that occurs inside of the Sun is known as proton-proton fusion.. La fusion nucléaire est plus difficile à réaliser que la fission car ici, il faut rapprocher des atomes si près l'un de l'autre qu'ils vont se coller. and I need to create a nuclear fusion equation to represent it. Power Production Nuclear. 2 Answers. The released energy reaches Earth in the form of electromagnetic radiation. For example, an alpha particle is a helium nucleus (He) with a charge of +2 and a mass number of 4, so it is symbolized $_2^4\text{He}$. Nuclear fusion is a process in which atomic nuclei are fused together to form heavier nuclei. Extremely high temperatures (on the order of 1.5 x 10 7 °C) can force nuclei together so the strong nuclear force can bond them. The isotopes of hydrogen needed for the hydrogen bomb fusion reaction were placed around an ordinary fission bomb. A 4 1H = 2He + 2 β + Mass. Fission Vs Fusion Nuclear fission and Nuclear Fusion are the Types of Nuclear Reactions. This process also fuses four protons into a Helium nucleus, by using Carbon (C), Nitrogen (N) and Oxygen (O) nuclei as catalysts. If light nuclei are forced together, they will fuse with a yield of energy because the mass of the combination will be less than the sum of the masses of the individual nuclei. 1 decade ago. e noyau 235U absorbe le neutron et, sous l’effet des forces, se rompt brutalement en deux. Already have an account? Light nuclei are combined in order to release excess binding energy and they form a heavier nucleus. Here is a typical fusion equation: Nuclear fusion is essentially the antithesis of the fission process. The basic difference between nuclear fission and fusion is that in fission reaction, a heavy nucleus splits into two daughter nuclei with the emission of energy, while in nuclear fusion two … How To Solve Quadratic Equations Using Scientific Calculator Casio Fx 82ms . 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# Relativistic functions (plasmapy.formulary.relativity)¶

Functions for calculating relativistic quantities ($$v \to c$$).

## Functions¶

 Return the Lorentz factor. Calculate the relativistic energy (in joules) of an object of mass m and velocity v.


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# Limit cos

Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.

Now 1-cos(θ) is just the negative of cos(θ)-1, the numerator of our limit. At very small angles this distance approaches zero because the cosine function is fairly flat and close to 1 in the vicinity of θ = 0. Example 1. Find the limit: $$\lim_{x \rightarrow 0} \, \frac{sin(5x)}{x}$$ Nov 10, 2020 Mar 27, 2020 This video works through the limit of (cos x - 1)/sin x. This limit is commonly found in a Calculus 1 class.#mathematics #calculus #limits***** Jun 11, 2018 So that new limit does not exist! And so L'Hôpita l's Rule is not usable in this case.

But when x goes to 0 from the negative side 1/x goes instead to negative infinity. This means that the limit as x goes to 0 for Cos (x)/x is undefined as the left and right limits do not agree. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Free limit calculator - solve limits step-by-step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy. Solution for Find the limit. cos(70) – 1 sin(90) - lim Nov 18, 2020 · Limits and Derivatives Class 11 MCQs Questions with Answers.

## Join the Action Alerts PLUS Community today!

The calculator will use the best method available so try out a lot of different types of problems. You can also get a better visual and understanding of the function by using our graphing tool. Limits, a foundational tool in calculus, are used to determine whether a function or sequence approaches a fixed value as its argument or index approaches a given point. Limits can be defined for discrete sequences, functions of one or more real-valued arguments or complex-valued functions.

### If you look at the formal definition of the limit of a function, you'll see that for the limit to exist, (cos x)/x has to approach some real number L as x approaches 0. The fact that (cos x)/x is unbounded as you approach 0 from either side is enough to say that the limit doesn't exist.

And so L'Hôpita l's Rule is not usable in this case. BUT we can do this: limx→∞ x+cos(x)x = limx→∞ (1 + cos(x)x) As x goes to infinity then cos(x)x tends to between −1∞ and +1∞, and both tend to zero.

It is possible to calculate the limit at a of a function where a represents a real : If the limit exists and that the … Find the limit lim x→0 x 2 cos(1/x) Solution to Example 1: As x approaches 0, 1 / x becomes very large in absolute value and cos(1 / x) becomes highly oscillatory. However cos(1 / x) takes values within the interval [-1,1] which is the range of cos x, hence-1 ≤ cos (1/x) ≤ 1 Multiply all terms of the above inequality by x 2 (x not equal to 0) Oct 03, 2016 (1 − cos(x)) = 0. The left and the right limits are equal, thus lim x→0 sin(x) = 0, lim x→0 (1 − cos(x)) = 0. Similarly, lim x→0 Showing that the limit of (1-cos (x))/x as x approaches 0 is equal to 0. This will be useful for proving the derivative of sin (x). Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.

Answer. Answer: (d) e-1/2 Hint: See full list on mathdoubts.com As per cos double angle formula, the term $\cos{2x}$ can be expanded in cosine or sine but there is no limit rule in cosine. Hence, it is better to expand the cosine double angle function in sine. $=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{ ormalsize \dfrac{1-(1-2\sin^2{x})}{x^2}}$ Feb 14, 2006 · what is the limit for cos (n pi) and sin (n pi) for n>1 also??

Find the limit: $$\lim_{x \rightarrow 0} \, \frac{sin(5x)}{x}$$ Nov 10, 2020 Mar 27, 2020 This video works through the limit of (cos x - 1)/sin x. This limit is commonly found in a Calculus 1 class.#mathematics #calculus #limits***** Jun 11, 2018 So that new limit does not exist! And so L'Hôpita l's Rule is not usable in this case. BUT we can do this: limx→∞ x+cos(x)x = limx→∞ (1 + cos(x)x) As x goes to infinity then cos(x)x tends to between −1∞ and +1∞, and both tend to zero. And we are left with just the "1", so: limx→∞ x+cos(x)x = limx→∞ (1 + cos… This free calculator will find the limit (two-sided or one-sided, including left and right) of the given function at the given point (including infinity). Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x.

It is possible to calculate the limit at a of a function where a represents a real : If the limit exists and that the … Find the limit lim x→0 x 2 cos(1/x) Solution to Example 1: As x approaches 0, 1 / x becomes very large in absolute value and cos(1 / x) becomes highly oscillatory. However cos(1 / x) takes values within the interval [-1,1] which is the range of cos x, hence-1 ≤ cos (1/x) ≤ 1 Multiply all terms of the above inequality by x 2 (x not equal to 0) Oct 03, 2016 (1 − cos(x)) = 0. The left and the right limits are equal, thus lim x→0 sin(x) = 0, lim x→0 (1 − cos(x)) = 0. Similarly, lim x→0 Showing that the limit of (1-cos (x))/x as x approaches 0 is equal to 0. This will be useful for proving the derivative of sin (x). Solve your math problems using our free math solver with step-by-step solutions.

Complex functions The trigonometric functions, including cosine, are usually viewed as functions that take real number Scope and limitations are two terms that address the details of a research project. The term scope refers to the problem or issue that the researcher wants to study with the project. Limitations is the term used for constraints that impact Join the Action Alerts PLUS Community today! Very few people can claim that they have achieved all that they are capable of. Chief of Product Management at Lifehack Read full profile Very few people can claim that they have achieved all that they are capable of. In the Western world m Reacting to robots, redefining our relationship with animals and revealing the things that are killing us now — a special series As we learn more about our similarities, the battle for non-human rights grows Why it’s in our nature to leave Before discussing the study implications, the limitations of this study must be acknowledged. First, this report encompasses the perspectives of key stakeholders in five states.

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The specific heat of an ideal gas is

1.  Proportional to $\mathrm{T}$

2.  Proportional to ${\mathrm{T}}^{2}$

3.  Proportional to ${\mathrm{T}}^{3}$

4.  Independent of $\mathrm{T}$

Explanation is a part of a Paid Course. To view Explanation Please buy the course.

Difficulty Level:


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# Notion of Independent Random Variables

Given a probability space $(\Omega,\Sigma,P)$. A random variable $X$ is a function mapping $\Omega \to \mathbf R$. On the other hand, we know that sum of two random variables is still a random variable.

Now, consider two random variables $X_1, X_2$ which are independent and identical. For a $\omega \in \Omega$, $(X_1+X_2)(\omega)=X_1(\omega)+X_2(\omega)$. Since $X_1$ and $X_2$ are identical, the equation would become $(X_1+X_2)(\omega)=2X_1(\omega)$. This result is obviously wrong! Can any one point out the mistake in the reasoning?

This problem appears when I am studying sum of i.i.d. random variables. For a sequence of i.i.d. random variables {$A_i$}. In the textbook, there is an event {$\omega\in\Omega\mid\sum_{i=0}^\infty A_i is finite.$}, but I think that it must be {$\omega\in\Omega^n\mid......$} or something like that. However, then there comes another problem why sum of random variables with domain $\Omega$ euals a random variable with domain $\Omega^n$.

• By "identical," you probably mean identically distributed. Identically distributed does not mean that the random variables are equal. That is, $X_1 \overset{d}{=} X_2$ does NOT mean $X_1 = X_2$. – Clarinetist Mar 25 '16 at 13:22

You should be careful with the phrase "identical". i.i.d means "independent and identically distributed". So $X_1$ and $X_2$ are not the same functions, they share the same distribution function.


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# Calculus

I'm a little confused with this integration problem: If the definite integral from 0 to 2 of (e^(x^2)) is first approximated by using two inscribed rectangles of equal width and then approximated by using the trapezoidal rule with n=2, the difference between the two approximations is what?

1. 👍 0
2. 👎 0
3. 👁 1,085
1. x f(x)
0 1
1 e
2 e^4

So, if there are 2 rectangles of width 1, then the area, using left-sides is

1*1 + 1*e = e+1 = 3.718

using right-sides, it's

1*e + 1*e^4 = 57.316

Using the trapezoidal rule, we have

1(1+e)/2 + 1(e+e^4)/2 = 30.517

Kind of a coarse approximation.

1. 👍 0
2. 👎 1
2. Thank you!

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2. 👎 0

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# A2 Physics, Electrical and Magnetic Fields, Capacitance, Electromagnetic Induction and Alternating Current, 5-8

## Electric Fields

### Coulomb’s Law

#### Static Electricity

The electrostatic effect is a result of static electricity.

• Rubbing two materials together can move electrons by friction

• Leaves a build up of charge

• Eg. A Van der Graaf generator will remove belts from the metal dome via the belt

• Approach a material of polarised molecules will cause the to realign

• Eg. Comb and paper or water

#### Charge

• Measured in coulombs

• An electron has a charge of $$-1.6\times 10^{-19}C$$

Materials can be charged by induction.

• A negatively charged object is brought close to the object

• The object is momentarily earthed

• The electrons are repelled, and flow through the earth

• This will leave a net positive charge

#### Coulomb’s Law

Coulomb’s Law states that the force between two point charges separated by a distance $$r$$ in a vacuum is directly to the product of the two charges and inversely proportionate to the square of their separation.

$$F=\frac{Qq}{4\pi\epsilon_{0}r^{2}},\nonumber \\$$

where $$\epsilon_{0}$$ is the permittivity of free space, $$8.85\times 10^{-12}Fm^{-1}$$.

• This assumes a vacuum

• Negligible difference in air

• Also assumes uniform point charges

Over microscopic distances, this force is much more important than the gravitational force, due to the small masses in comparison to charges. However, over macroscopic distances, the larger masses become more important as gravity is only attractive, whereas electrostatic forces can be attractive or repulsive.

## Electric Fields

A charged object will create an electric field around itself. Further to this, a charged particle will experience a force in this region.
Electric field lines show the force that would act on a positive test charge at that point. The proximity of the lines denotes the strength of the field. The test charge must be sufficiently small in size and charge so as not to distort the electric field.
The electric field strength, $$E$$, is the force per unit charge, in $$NC^{-1}$$.

$$E=\frac{F}{Q}\nonumber \\$$

In a radial field, field strength can be determined by using coulomb’s law. This is given by

$$E=\frac{Q}{4\pi\epsilon_{0}r^{2}}\nonumber \\$$

## Generating Electric Fields

A Van der Graaf generator can be used to generate an electric field. A Van der Graaf will normally be positively charged.

## Uniform Fields and Potential Difference

A uniform field can be demonstrated using a metallised polystyrene ball on a nylon thread between two oppositely charged plates. The uncharged ball will be attracted to the nearest plate due to its metal paint. Upon contact, the ball will be charged and repelled, and this will be repeated.
The work done, $$W$$, by an electric field in moving a particle from one plate to another (in a uniform field) is given $$W=Fd$$, with $$d$$ being the separation. Electric potential difference, $$\Delta V$$, is the work done per unit charge. Thus,

\begin{gather*} \Delta V=\frac{W}{Q} \\ \therefore Fd=Q\Delta V \\ \therefore E=\frac{\Delta V}{d} \\ \end{gather*}

As such, electric field strength can be measured in $$Vm^{-1}$$ or $$NC^{-1}$$.

## The Electron Gun

An electric field can be used to accelerate electrons.

• Electrons emitted from a heated cathode by thermionic emission

• High potential difference between cathode and anode

• Electrons accelerated towards anode

In this case

$$E=q\Delta V,\nonumber \\$$

where $$E$$ represents the energy transferred, and $$\Delta V$$ is the potential difference between the two electrodes.

## Charged Particles in an Electric Field

In a uniform field, the force is constant. Therefore, the acceleration must also be constant. This will give the charged particle a parabolic path, which can be treated using SUVAT.


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# Just an integral

1. Jul 20, 2006

### scavok

I had to reverse a double integral and evaluate, and I was able to do the reversal, but I was stumped on how to integrate this. I imagine there's a way to simplify it since I can't see how any of the integration techniques I've learned will help, but nothing comes to mind.

Last edited: Jul 20, 2006
2. Jul 20, 2006

### TMFKAN64

3. Jul 20, 2006

### Office_Shredder

Staff Emeritus
hehehe.... did you actually try it?

I don't think that's the answer he's looking for


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# Signal distortion under non-linear channel

The channel is given to have these characteristics, and its required to study the affect on Frequency Modulated signal and its demodulation.
Since there are no sharp corners (like during clipping) will there be a significant difference in the frequencies? Can a regular high-pass filter be used still to demodulate the FM signal.

• Hi Tanya- What is the purpose of your question (this looks like a homework problem, so getting the context---we don't bypass the benefit and purpose of homework but we can certainly help you with where your stuck) – Dan Boschen Jul 29 '20 at 2:54
• Yes it was part of additional problems to think on. But I am not familiar with non-linear transformations hence asked here. – Tanya Choudhary Jul 29 '20 at 3:56


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# 23. Field of a Charged Conducting Disc

## Charge Density on the Disc

This is a classic problem, solved in the early nineteenth century by Green (and is Jackson problem 3.3!)  It turns out that the capacitance of a disc of radius $a$ is $C=8{\epsilon }_{0}a.$  A sphere of radius $a$ has capacitance $4\pi {\epsilon }_{0}a$, a factor $\pi /2$ greater.  Cavendish measured the ratio sphere/disc experimentally as 1.57, probably in the 1770's.  Very impressive.

Your first thought might be: doesn't the charge all migrate to the edge, just as charge in a solid goes to the surface? The answer is no$—$and it's worth thinking about why not. Consider first a uniformly charged sphere. Pick an arbitrary point P inside (but not the center$—$that's obviously trivial). We know there's no field, but how exactly does the cancellation from all the surface elements work? Take one element of surface, subtending a narrow cone at P, and continue the lines of the cone through P to trace out another surface element on the opposite side.  These two surface elements won't have the same area: they subtend the same solid angle at P, so their areas are in the ratios of the squares of their distances from P.  But the electrostatic field at P goes as the inverse square of the distance, so the two forces exactly balance.

Now back to the disc: it's got circular symmetry, but if the charge is around the edge, this cancellation no longer works, because the corresponding small segments of the circle scale linearly with distance from an interior point, but the inverse square law of force still holds, so the nearer charge is more important.  In other words, if we start with a large number of equal charges arranged equally spaced around the edge, then move one of them inwards and allow those still on the edge to rearrange slightly to restore equal spacing, the one moved inwards will move further away from the edge, towards the middle. Evidently, if all the charges are free to move, the equilibrium arrangement will have a nonzero density over the whole disc.

OK, so what about the charge distribution on the disc?   Finding the field is Jackson problem 3.3, but Jackson takes the charge density as given.  Zangwill says there is no truly simple way to calculate it.  I beg to differ$—$Purcell gives the simple explanation presented below, but Kelvin was almost certainly aware of it a century earlier.

Here’s the analysis: we’re looking for an (obviously circularly symmetric) charge distribution such that everywhere on the disc the electric field has zero horizontal (plane of the disc) component.  Otherwise, it's not electrostatics. We know, of course, the field will have a vertical component, opposite on the two sides, proportional to the charge density.

The key is to think about the argument for a uniformly charged sphere.  Zangwill actually states at the end of his analysis (based on taking the disc as the limiting case of an oblate ellipsoid, that the charge distribution equals the projection of a uniformly charged hemisphere onto its equatorial plane.  This is the clue!

On the uniformly (surface) charged sphere, the forces from the two elements of charge $\sigma d{A}_{1},\sigma d{A}_{2}$ as felt at P are equal and opposite, as we've argued. But now suppose we project those charges on to the equatorial plane (see figure). The ratio of their distances from P is unaltered, as is clear from the second diagram. Therefore, the forces from them at P are still equal and opposite. Doing this for the entire sphere, we see that a disc with charge density equal to that projected to the equatorial plane from a uniformly charged sphere will generate zero electric field parallel to the disc, and so will be in electrostatic equilibrium.

So what is the charge distribution?  We use $\rho$ for the radial variable on the disc, following standard cylindrical coordinate notation.

The ratio of the projected surface area to the area on the sphere is $\mathrm{sin}\theta =\sqrt{{a}^{2}-{\rho }^{2}}/a,$ so the appropriate charge density on the disc is proportional to the inverse of this.  The integral to get the total charge $Q$ is simple, the result is a surface charge density on the disc

$\sigma \left(\rho \right)=\frac{Q}{2\pi a\sqrt{{a}^{2}-{\rho }^{2}}}.$

## Finding the Field

Jackson's (problem 3.3) suggested method of solution, using spherical harmonics, is a bad idea, for reasons that will become evident.

The good news is that the problem can be solved in elegant fashion using oblate spheroidal coordinates (a nineteenth century favorite) and the correct solution makes clear why ordinary spherical harmonics lead to difficulties.  You can get the first part, the distant potential, correctly with ordinary spherical harmonics (as suggested by Jackson), but you cannot match it to a solution close to the disc.  The reason is simple. To match the angular spherical harmonics, you must match an odd power of distance outside to an even power inside. However, the correct solution of the problem is an odd power of distance along the $z\text{-}$ axis at all distances. The potential has a singularity at the edge of the disc, so the usual matching conditions at the crucial radius $a$ are invalid. There are ways around this: the inner region can be broken into two, above and below the disc, with separate expansions in each, either even or odd Legendre polynomials form a complete set if restricted to half the angular range.  This is correct, but technically difficult and not very transparent.  Several “solutions” posted on the web are simply wrong.

(Note: If you’re not familiar with oblate spherical harmonics, you can get comfortable with them in less time that you’ll otherwise spend on this Jackson problem.)

## Oblate Spherical Harmonics

The natural way to solve this problem is to use oblate spheroidal harmonics $\left(\mu ,\nu ,\phi \right).$   This becomes evident on looking at the lines of constant $\mu$ in the coordinate diagram here (from Wikipedia).  They look just like the equipotentials we would intuitively expect from a charged disc, and in particular the $\mu =0$ line corresponds to a thin disc in 3D, just what we expect for a conductor.  Of course, we haven’t proved anything yet!  To prove our guess correct, we need to establish that there is some function of $\mu$ only which satisfies the three-dimensional ${\nabla }^{2}\phi =0$ everywhere except at $\mu =0.$

The coordinates look like this in cross section, in three dimensions they are rotationally symmetric about the central line $\nu =\pi /2.$

The Cartesian coordinates are:

$\begin{array}{l}x=a\text{ }\mathrm{cosh}\mu \text{ }\mathrm{cos}\nu \text{ }\mathrm{cos}\varphi ,\\ y=a\text{ }\mathrm{cosh}\mu \text{ }\mathrm{cos}\nu \text{ }\mathrm{sin}\varphi ,\\ z=a\text{ }\mathrm{sinh}\mu \text{ }\mathrm{sin}\nu .\end{array}$

(This is not as bad as it looks: the azimuthal $\varphi$ dependence is trivial, from the corresponding rotational symmetry, so we're left with, essentially, a function of $\rho =\sqrt{{x}^{2}+{y}^{2}}$ and $z.$ We have to be careful with signs, though, since $\rho$ is defined as positive, so to get the right picture it's better to plot an $xz$ cross section as shown above.

(In this cross section$—$azimuthal $\varphi =0$ $—$the equations are equivalent to a complex variable equation, $x+iz=\mathrm{cosh}\left(\mu +i\nu \right),$ using the addition formula for cosh, recall these were the equipotentials we found in 2D for a charged strip. In that 2D case, the potential was simply $\mu .$ That can’t be the case here, though, because ${\nabla }^{2}$ is different.)

The constant $\mu$ surfaces are ellipsoids (here oblate spheroids):

$\frac{{x}^{2}+{y}^{2}}{{\mathrm{cosh}}^{2}\mu }+\frac{{z}^{2}}{{\mathrm{sinh}}^{2}\mu }={a}^{2}.$

Note from the figure that we take $0\le \mu <\infty .$

This is often written in cylindrical coordinates, corresponding to a constant $\phi$ surface:

$\frac{{\rho }^{2}}{{\mathrm{cosh}}^{2}\mu }+\frac{{z}^{2}}{{\mathrm{sinh}}^{2}\mu }={a}^{2}.$

The foci of all the ellipses are at $a,-a,$ the ellipses are said to be "confocal". The lengths of the major and minor axes are $2a\mathrm{cosh}\mu ,\text{ }\text{ }2a\mathrm{sinh}\mu$ respectively.

Constant $\nu$ surfaces are:

$\frac{{\rho }^{2}}{{\mathrm{cos}}^{2}\nu }-\frac{{z}^{2}}{{\mathrm{sin}}^{2}\nu }={a}^{2}.$

A tricky point: we parametrize the whole plane with positive $\mu ,$ so from $z=a\mathrm{sinh}\mu \mathrm{sin}\nu ,$ $\nu$ has the same sign as $z.$ Therefore these surfaces are half-hyperbolas, divided at $z=0,$ with foci also at $a,-a,$ and they are field lines. That is, $\nu$ is discontinuous at $\mu =0,$ across the (infinitely thin) disc. (In the next lecture, we look at the field for a hole in a charged conducting sheet. We can use the same coordinate system, but must change the parametrization so that the discontinuity only happens across the charged sheet, the opposite situation to the disc.)

Recalling that the definition of an ellipse is that the sum of its distances from two points (the foci) is a constant (in fact the length of the major axis), and for a hyperbola the difference is constant (also the length of its major axis, but be careful with our notation, in most places the lengths of these major axes are $2a,$ but in these coordinates, $2a$ is the distance between the foci).

We define the (always positive!) distances (switching now from $\rho z$ to just the $xz$ plane) by

${d}_{1}^{2}={\left(x+a\right)}^{2}+{z}^{2},\text{ }{d}_{2}^{2}={\left(x-a\right)}^{2}+{z}^{2}.$

It follows that

$\mathrm{cosh}\mu =\frac{{d}_{1}+{d}_{2}}{2a}\text{ }\mathrm{cos}\nu =\frac{{d}_{1}-{d}_{2}}{2a}.$

(The equipotential ellipse major axis has length $2a\mathrm{cosh}\mu$.)  We can also see from this that the curves cut at right angles (think of moving incrementally so that the sum/difference from two fixed points is constant).

### Electrostatics of a Charged Conducting Disc

Obviously, from the pictures, the natural choice for potential is some function of the coordinate $\mu .$

So we need to solve ${\nabla }^{2}\Phi =0$ for this spheroidal system, that is (Wikipedia):

$\begin{array}{l}{\nabla }^{2}\Phi =\\ \frac{1}{{a}^{2}\left({\mathrm{sinh}}^{2}\mu +{\mathrm{sin}}^{2}\nu \right)}\left[\frac{1}{\mathrm{cosh}\mu }\frac{\partial }{\partial \mu }\left(\mathrm{cosh}\mu \frac{\partial \Phi }{\partial \mu }\right)+\frac{1}{\mathrm{cos}\nu }\frac{\partial }{\partial \nu }\left(\mathrm{cos}\nu \frac{\partial \Phi }{\partial \nu }\right)\right]+\frac{1}{{a}^{2}{\mathrm{cosh}}^{2}\mu {\mathrm{cos}}^{2}\nu }\frac{{\partial }^{2}\Phi }{\partial {\varphi }^{2}}\\ =0.\end{array}$

But, thankfully, we are looking for a potential that is a function of $\mu$ only, that is, a solution of:

The solution is (with  )

$\Phi \left(\mu \right)={\Phi }_{0}\left(1-\frac{2}{\pi }\mathrm{arctan}\mathrm{sinh}\mu \right)=\frac{2}{\pi }{\Phi }_{0}\mathrm{arcsin}\frac{1}{\mathrm{cosh}\mu }.$

Now

$\mathrm{cosh}\mu =\frac{\sqrt{{\left(\rho +a\right)}^{2}+{z}^{2}}+\sqrt{{\left(\rho -a\right)}^{2}+{z}^{2}}}{2a}$

So on the $z\text{-}$ axis,

$\Phi \left(\mu \right)=\frac{2}{\pi }{\Phi }_{0}\mathrm{arcsin}\frac{a}{\sqrt{{a}^{2}+{z}^{2}}}=\frac{2}{\pi }{\Phi }_{0}\mathrm{arctan}\frac{a}{\left|z\right|}.$

And, for the far region $z>a$, we can use the standard expansion, to get

$\mathrm{arctan}\left(a/z\right)=a/z-\frac{{\left(a/z\right)}^{3}}{3}+\frac{{\left(a/z\right)}^{5}}{5}-\frac{{\left(a/z\right)}^{7}}{7}+\cdots =\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}{\left(a/z\right)}^{2n+1}}{2n+1};\text{ }z\ge a,$

and this agrees with Jackson’s result for $z>a.$

For $z obviously the series diverges, but we can just use

$\mathrm{arctan}\left(a/z\right)=\pi /2-\mathrm{arctan}\left(z/a\right).$

Therefore, our solution for the potential along the $z\text{-}$ axis is

$\begin{array}{c}\Phi \left(\mu \right)=\frac{2}{\pi }{\Phi }_{0}\mathrm{arctan}\frac{a}{\left|z\right|}\\ =\frac{2}{\pi }{\Phi }_{0}\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}{\left(a/z\right)}^{2n+1}}{2n+1},\text{ }z\ge a,\\ ={\Phi }_{0}-\frac{2}{\pi }{\Phi }_{0}\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}{\left(z/a\right)}^{2n+1}}{2n+1},\text{ }0\le z\le a.\end{array}$

Notice we can't get this result by matching Legendre polynomial expansions at $r=a,$ because ${z}^{-\left(2n+1\right)}$ would have to match with ${z}^{2n}.$

This is just on the axis, of course: let’s check that our expression for the potential is in fact constant over the whole disc in the $\left(x,y\right)$ plane.  We have

$\Phi \left(\mu \right)=\frac{2}{\pi }{\Phi }_{0}\mathrm{arcsin}\frac{1}{\mathrm{cosh}\mu }=\frac{2}{\pi }{\Phi }_{0}\mathrm{arcsin}\frac{2a}{\sqrt{{\left(\rho +a\right)}^{2}+{z}^{2}}+\sqrt{{\left(\rho -a\right)}^{2}+{z}^{2}}}.$

We know that the equipotentials are ellipsoids (here oblate spheroids), and this is clear from the formula, remembering that an ellipse is defined as the set of curves such that the sum of distances from two fixed points, the foci, is constant.  This is exactly what we see in the denominator.  The foci are all at $a,-a.$ The flat disc itself is a degenerate ellipsoid, with foci of a cross section, such as the $xz$ plane, coinciding with the ends of the major axis.  Therefore, the denominator is the sum of distances from the two ends, and for a point on the disc, that's just $2a.$ This establishes that the potential is indeed constant on the disc.  (Note: at least one online "solution" takes positive square roots for $z=0$ in this expression, so gets the denominator equal to $2\rho ,$ which is nonsense.)

Exercise:  For $z=0$ but outside the disc, the potential is nonzero, it's worth looking at it for small distances away from the disc, but in the plane.

### Charge Density on Disc from Potential

The charge density is of course proportional to the strength of the perpendicular electric field strength.

The potential near the origin on the $z\text{-}$axis is ${\Phi }_{0}-\frac{2}{\pi }{\Phi }_{0}\frac{z}{a},$ so the field strength is $2{\Phi }_{0}/\pi a.$

To find the normal field strength, hence the charge density, at other points on the disc, consider an equipotential very close to the plate, in cross section an ellipse with major semi-axis we can take to be $a$ and minor semi-axis $b\ll a.$

Then the equation of the ellipse is very close to  $\frac{{\rho }^{2}}{{a}^{2}}+\frac{{z}^{2}}{{b}^{{}^{2}}}=1,$ and in particular the height of the equipotential above the disc at distance $\rho$ from the center is $z=\left(b/a\right)\sqrt{{a}^{2}-{\rho }^{2}}.$

Taking an equipotential sufficiently close, the field strength must vary across the disc as the inverse of the distance between this equipotential and the (equipotential) disc, that is, it must be

$E\left(\rho \right)=\frac{2{\Phi }_{0}}{\pi \sqrt{{a}^{2}-{\rho }^{2}}},$

and the charge density is

$\sigma \left(\rho \right)=\frac{4{\epsilon }_{0}\text{ }{\Phi }_{0}}{\pi \sqrt{{a}^{2}-{\rho }^{2}}},$

confirming the argument at the beginning of the lecture.

The total charge is therefore

$Q=\underset{0}{\overset{a}{\int }}2\pi \rho \sigma \left(\rho \right)d\rho =\underset{0}{\overset{a}{\int }}2\pi \rho \frac{4{\epsilon }_{0}\text{ }{\Phi }_{0}}{\pi \sqrt{{a}^{2}-{\rho }^{2}}}d\rho =8{\epsilon }_{0}\text{ }{\Phi }_{0}\underset{0}{\overset{a}{\int }}\frac{\rho }{\sqrt{{a}^{2}-{\rho }^{2}}}d\rho =8{\epsilon }_{0}\text{ }{\Phi }_{0}a.$

A sphere of radius $a$ has capacitance $4\pi {\epsilon }_{0}a$, a factor $\pi /2$ greater.

## What About a Disc of Finite Thickness?

Glancing at the long thin ellipse pictured above, it’s at least plausible that the capacitance of a disc of thickness $2b$ would be well approximated by that of this ellipsoid.  Assuming $b\ll a,$ the potential along the $z$ axis

$\Phi \left(z=b\right)\cong {\Phi }_{0}\left(1-\frac{2}{\pi }\frac{b}{a}\right).$

Think about the electric field lines emanating from charge $Q$ on the infinitely thin disc. They all pass through this thin ellipsoid too, and beyond it are identical from that same charge actually on the ellipsoid. But the potential on the ellipsoid is down by the factor in the equation above, so it follows that the capacitance of the ellipsoid is increased by the inverse, to leading order:

${C}_{b}\cong 8{\epsilon }_{0}a\left(1+\frac{2}{\pi }\frac{b}{a}\right).$

In fact, a lot of work has been done on this problem, using a more accurate shape for the disc (a thin cylinder, like a coin) beginning with Maxwell in 1877.  Even Thomson (Lord Kelvin) found Maxwell’s paper hard to follow, so we won’t try.  A more accessible presentation has been given by Kirk McDonald (online) updated in 2019, carefully analyzing charge distribution on a coin-like cylinder.  For the infinitely thin disc case as discussed above, the charge density diverges as the inverse square root of distance from the edge, but for the thin cylinder there are upper and lower edges, and a cylindrical surface between, and the charge density near those edges diverges only as the inverse cube root of distance.  (This follows from our earlier 2D discussion of charge near corners.)  A reasonable assumption is that this cube root behavior prevails for distances from the edge of order the thickness, but away from that the charge distribution is as for the infinitely thin disc.  Taking the crossover to be at $b,$ and matching the charge densities there, McDonald finds the charge on the cylindrical surface makes a big enough contribution that the finite thickness correction becomes of order $\sqrt{b/a}$ rather than $b/a.$ For details, see his paper.


text:
# Line Emission Sources¶

LineSourceModel assumes that the emission is occuring at a single energy, and that it may or may not be broadened by thermal or other motions. In the former case, the emission is a delta function at a single rest-frame energy $$E_0$$:

$\varepsilon(E) = A\delta(E-E_0)$

In the latter case, the emission is represented by a Gaussian with mean $$E_0$$ and standard deviation $$\sigma_E$$:

$\varepsilon(E) = \frac{A}{\sigma_E\sqrt{2\pi}}e^{-\frac{(E-E_0)^2}{2\sigma_E^2}}$

When creating a LineSourceModel, it is initialized with the line rest-frame energy e0 and an emission_field field specification that represents the normalization $$A$$ in the equations above, which must be in units of counts/s. Optionally, the line may be broadened by passing in a sigma parameter, which can be a field specification or YTQuantity, corresponding to either a spatially varying field or a single constant value. In either case, sigma may have units of energy or velocity; if the latter, it will be converted to a broadening in energy units via $$\sigma_E = \sigma_v\frac{E_0}{c}$$.

Note

In most cases, you will want velocity broadening of lines to be handled by the inputted velocity fields instead of by the sigma parameter. This parameter is designed for thermal or other sources of “intrinsic” broadening.

## Examples¶

An example of an unbroadened line:

e0 = YTQuantity(5.0, "keV") # Rest-frame line energy
emission_field = ("gas", "line_emission") # Line emission field (normalization)
line_model = pyxsim.LineSourceModel(e0, line_emission)

An example of a line with a constant broadening in km/s:

e0 = YTQuantity(6.0, "keV")
emission_field = ("gas", "line_emission") # Line emission field (normalization)
sigma = (500., "km/s")
line_model = pyxsim.LineSourceModel(e0, line_emission, sigma=sigma)

An example of a line with a spatially varying broadening field:

e0 = YTQuantity(6.0, "keV")
emission_field = ("gas", "line_emission") # Line emission field (normalization)
sigma = "dark_matter_velocity_dispersion" # Has dimensions of velocity
line_model = pyxsim.LineSourceModel(e0, line_emission, sigma=sigma)


text:
In isosceles triangle the length of a leg is 17 cm, and the base is 16 cm. find the length of the altitude

In isosceles triangle the length of a leg is 17 cm, and the base is 16 cm. find the length of the altitude to the base. answer under 10 minutes.

Why would a government have to be very careful when it decides which goods should have a tariff? How

1. alexcuevaz90 says:

15 cm

Explanation:

use the Pythagorean Theorem to solve this problem

2. Expert says:

The correct answer is clerk, i did the quiz

3. Expert says:

Increased profits reduced health risks

4. olivya2005d says:

Hey there,

I have 2 minutes left.

Your questions states: In isosceles triangle the length of a leg is 17 cm, and the base is 16 cm. Find the length of the altitude to the base. Your correct answer would be 15cm.

Hope this helps.
$In isosceles triangle the length of a leg is 17 cm, and the base is 16 cm. find the length of the al$


text:
# Electric force vs magnetic force, physics class-12

SHARING INCREASES KNOWLEDGE

# ELECTRIC FORCE VS MAGNETIC FORCE

In this article, we are going to understand the basic difference between electric and magnetic forces and we will also compare the effects of the electric and the magnetic fields on the charged particle, so let’s get started…

Whenever you talk about the electric and magnetic forces, you should keep in mind that whether it is an electric force or magnetic force they will influence the motion of charged particles.[latexpage]

However, the resulting change to the charged particles’ trajectory will qualitatively differ for both forces. Below we will briefly review the two types of force and compare their effects on a charged particle.

## ELECTROSTATIC FORCE AND MAGNETIC FORCE ON THE CHARGED PARTICLES

### IN ELECTROSTATIC

Recall the electrostatic chapter and see what you have learned, if the electric field E is unchanged then the electrostatic force F on any charged particle q is given as-
$$F=qE$$
Where F is the force vector, q is the charge, and E is the electric field vector. Note that the direction of electric force F is identical to electric field E if the charge is positive, but if the charged particle is negative then the direction of electric force is in opposite direction. This electric field may be produced by a larger charge, Q, acting on the smaller charge q over a distance r so that:
$$E=\frac{F}{q}=K|\frac{Qq}{qr^2}|=K|\frac{Q}{r^2}|$$
It should be noted that the electric force F acts parallel to the electric field E. The curl of the electric force will be zero, i.e.
$$\nabla \times E = 0$$
From this, one can conclude that the electric field can work and a charged particle in a pure electric field follows the tangent of an electric field line.

### IN MAGNETISM

But in contrast to the magnetism, the magnetic force on a charged particle (when it is moving with v velocity) is orthogonal to the magnetic field, and it is given as-
$$F=q(v\times B) =qvB\sin\theta$$
where B is the magnetic field vector, v is the velocity of the particle and θ is the angle between the magnetic field and the particle velocity. The direction of magnetic force can be easily determined by the use of the right-hand rule.

If the particle velocity by any means, aligned parallel to the magnetic field, or it becomes is zero, then the magnetic force will also be zero. This is different from an electric field, where the speed of the particle at any given point in time does not affect the magnitude or direction of the electric force.

The dependency on the angle of the fi magnetic field also causes has the effect that the charged particles move in a circular or helical fashion perpendicular to the magnetic field lines, while a particle in an electric field moves in a straight line along an electric field line.

Another difference between the electric and magnetic force is that the work done by the magnetic force on the charged particle is zero because the particle moves in a circular motion and ends up at the same initial point. We can express this mathematically as follows:
$$\oint B\cdot dr=0$$

## KEY POINTS BETWEEN ELECTRIC FORCE AND MAGNETIC FORCE

 In the case of positive charge, the force on a charged particle due to an electric field is directed parallel to the electric field vector and anti-parallel in a negative charge. It does not depend on the velocity of the particle. But the magnetic force on a charged particle is perpendicular to the magnetic field vector, and it depends on the velocity of the particle. The right-hand rule can be used to determine the direction of the force. The electric field may do work on the charged particle but the magnetic field does not work.

Stay tuned with Laws Of Nature for more interesting and useful content.

SHARING INCREASES KNOWLEDGE


text:
Mathematics
OpenStudy (anonymous):

Solve by completing the square: $3x^2 - 6x - 5 = 0$

OpenStudy (anonymous):

OK

OpenStudy (anonymous):

$3x^2-6x=5$ $x^2-2x=5/3$ $(x-1)^2=5/3 +1$ $x-1=\pm \sqrt{(8/3)}$ $x=1\pm \sqrt{8/3}$

OpenStudy (anonymous):

you failed to complete the square...

OpenStudy (anonymous):

why?

OpenStudy (anonymous):

completing the square means that you take b/2 and square it to get c...

OpenStudy (anonymous):

I am pretty sure that I did what the question asked

OpenStudy (anonymous):

completing the square is not shown in your work. you moved the 5 to the other side and divided by three. this is not completing the square

OpenStudy (anonymous):

completing the square would start like: ${-6 \over 2}, -3^2 = 9$


text:
Football plays and drills
Category
Name
27-Dec-2017
Running
Playname: Side line strip drill

Running back starts with ball in right hand (using the right side-line).

Bags are set-up as shown progressively getting closer to the sideline as you get closer to the goal.  Bags are shown with Triangles.

As Running back approaches the bags they are given a pop from the defender pushing them toward the sidelines.

Running back holds ball to outside.

As a second progression of the drill have defenders standing on the sideline trying to strip the ball out of the right hand of the running back.

One swipe, then eventually whatever your need.


text:
# In how many ways can you choose three distinct numbers … [closed]

In how many ways can you choose three distinct numbers from the set of {1,2,3,...,19,20} such that their product is divisible by 4 ?

## closed as off-topic by Namaste, Hayden, Gina, user147263, Davide GiraudoJul 21 '14 at 15:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Hayden, Gina, Community, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.

• It might be easier to find number of triples which are not divisible by 4 – Jack Yoon Jul 21 '14 at 14:08

Strategy: There are $\binom{20}{3}$ possible choices. Let us see how many are bad.

We could choose all odd. Easy to count.

We could choose $2$ odd, and the other divisible by $2$ but not by $4$, that is, one of the numbers $2,6,10,14,18$. Again, it should not be hard to count these.

You can break it down like this. First we ask how many numbers are divisible by four in the set $\{1,...,20\}$ that is $4,8,12,16,20$. If exactly one is chosen from this list we have $$5\cdot {15 \choose 2}$$ ways of doing this. If two are chosen we have $${5 \choose 2} \cdot 15$$ ways of doing that. And finally there are $${5 \choose 3}$$ ways of picking three of the numbers divisible by four.

Then ask how many ways are there to choose two numbers divisible by only two? This includes $2,6,10,14,18$ and then to avoid over counting we can only choose a number not divisible by four from the rest. If we choose it to be an odd number we find: $${5 \choose 2} \cdot 10,$$ and if we choose three even numbers (not divisible by four) we find: $${5 \choose 3}$$ ways of doing this.

The sum of all of these is the answer.

• (I think you may have to correct for over-counting in the first step, since you could choose 4, 5, 8 or 8, 5, 4.) – user84413 Jul 21 '14 at 14:36
• You are right. I think it just needs to be divided by 2... – Joel Jul 21 '14 at 14:39
• It might be easier to break the first step into cases, depending on the number of multiples of 4 chosen. (I believe $5\binom{19}{2}$ is odd.) – user84413 Jul 21 '14 at 15:01
• I should never do combinatorics without my coffee. I think this fixes it. – Joel Jul 21 '14 at 15:13
• This looks almost right, but I think maybe there is slight over-counting in the second step. (For example, you could have 2, 6, 10 or 10, 2, 6.) – user84413 Jul 21 '14 at 15:30
1. We need three numbers, one of which is divisible by 4, and any other two

In this set, the divisible by 4 are only 5 $\{4,8,12,16,20\}$. And we don't care what the other 2 will be, as they all will do the job. =>$5\cdot19\cdot18\div2 = 855$

2. We need three numbers, two of which are divisible by 2, but not by 4, and one undivisible by 4

In the set, the divisible by 2, but not by 4, are only 5$\{2,6,10,14,18\}$. We need 2 of this set, and 1 from the set, exlcuding only divisible by 4 $\{1,2,3,5,...,19\}$, which are 15, but since we used 2 of them, there are only 13 left. => $5\cdot4\cdot13\div2=130$

Giving us a total of $985$ ways

• This is not correct. Do you see why? – Jack Yoon Jul 21 '14 at 14:20
• Yep, thanks, already updated it. – Peter Jul 21 '14 at 14:42
• Still not true. Following Andre's method I get 795. Firstly dividing by two in case 1 is not enough. – Jack Yoon Jul 21 '14 at 15:04


text:
# American Institute of Mathematical Sciences

2018, 25: 1-7. doi: 10.3934/era.2018.25.001

## Zermelo deformation of finsler metrics by killing vector fields

 1 Centre International de Rencontres Mathématiques-CIRM, 163 avenue de Luminy, Case 916, F-13288 Marseille -Cedex 9, France 2 Institut für Mathematik, Fakultät für Mathematik und Informatik, Friedrich-Schiller-Universität Jena, 07737 Jena, Germany

Received  October 10, 2017 Published  March 2018

Fund Project: The authors thank Sergei Ivanov for useful comments. V. M. was partially supported by the University of Jena and by the DFG grant MA 2565/4.

We show how geodesics, Jacobi vector fields, and flag curvature of a Finsler metric behave under Zermelo deformation with respect to a Killing vector field. We also show that Zermelo deformation with respect to a Killing vector field of a locally symmetric Finsler metric is also locally symmetric.

Citation: Patrick Foulon, Vladimir S. Matveev. Zermelo deformation of finsler metrics by killing vector fields. Electronic Research Announcements, 2018, 25: 1-7. doi: 10.3934/era.2018.25.001
##### References:

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The unit ball of $\tilde F$ (dashed line) is the $v$-translation of that of $F$ (bold line). If a vector $J$ is tangent to the unit ball of $F$ at $\xi$, it is tangent to the unit ball of $\tilde F$ at $\xi + v$
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