GR9277 #72



Alternate Solutions 
jmason86 20090903 19:19:22  I did this one by limits and general test taking strategy.
Limits: Q > finite value as T > . Eliminates (D) and (E)
ETS will generally force you to choose between very similar answers. Eliminates (C)
With only (A) and (B) left, it is a good idea to guess. But if you can get a hunch, that is even better. It's probably totally flawed thinking, but I saw the lack of an exponential in the numerator for (B), which made it look similar to a BoseEinstein distribution. The full expression in (A) reminded me of the MaxwellBoltzmann distribution that comes from the problem statement. Choose (A)
 

Comments 
ernest21 20190930 05:00:43  Cool! I\'ll surely be coming back for the next posts from you. You\'re an incredibly engaging writer that I can freely recommend this article to my college students. clash royale best decks   BerkeleyEric 20100917 17:04:59  If you remember that , then immediately you can see that there will have to be the squared sum in the denominator, so only A and B remain. And there still has to be the exponential in the numerator (a derivative won't get rid of that), so the answer must be A.
flyboy621 20101022 16:11:48 
+1

  jmason86 20090903 19:19:22  I did this one by limits and general test taking strategy.
Limits: Q > finite value as T > . Eliminates (D) and (E)
ETS will generally force you to choose between very similar answers. Eliminates (C)
With only (A) and (B) left, it is a good idea to guess. But if you can get a hunch, that is even better. It's probably totally flawed thinking, but I saw the lack of an exponential in the numerator for (B), which made it look similar to a BoseEinstein distribution. The full expression in (A) reminded me of the MaxwellBoltzmann distribution that comes from the problem statement. Choose (A)
  RebeccaJK42 20070323 10:27:19  Where does the k in the numerator come from?
alpha 20070330 22:37:37 
k is Boltzmann constant.. The k on the numerator is actually canceled out by the in the denominator

Richard 20070914 00:05:03 
The fact of the matter is, you have a factor of
introduced by the derivative.
To make it look pretty, they multiplied the top and bottom by giving (with the extra ) the factor . Then of course, you have the .

  Andresito 20060329 10:05:29  I could not obtain T^2 in the denominator. Is there an error in the solution provided by the ETS?
radicaltyro 20061031 14:19:25 
Hi Andresito,
Review your calculus and try again. The answer is correct.

Blue Quark 20071101 08:00:11 
Andresito you made the same mistake I did. The T is in the denominator, not the numerator. Thus when you differentiate you get a (T^(2)) in front of the exponential in addition to the normal e/k

drizzo01 20121108 07:40:44 
am I the only one who doesn't see a T in the denominator of the second term? I appreciate that T^2 would be in the denominator if there was a T to being with, but as far as I can see, the only T's are within the expression for the exponential, which don't come out upon taking a derivative.

drizzo01 20121108 07:46:58 
wait nvm, got it.

 




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