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Unanswered Cries for Help! Total So Far: 24

GR8677 #38

2008-11-05 15:41:44

I am confused about the order that the $\Delta t's$ are placed because in the official answer the first deduction corresponds EXACTLY to the answer choice A. Furthermore, the "correct" deduction has $\Delta t_1$ and $\Delta t_3$ reversed. Also, I have on my equation sheet $\Delta t_0 = \gamma \Delta t^\prime$ . Using this convention I was able to get the correct answer. Do I have this backwards, or is there a typo here?

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GR8677 #44 2008-11-02 14:36:44	I'm still confused. the equation $\(p_{i})$ =MV in no way shows the linear momentum of the particle contributing to the new angular momentum of the stick. Can someone clearly explain why angular momentum does not show up in the equation? Click here to jump to the problem!

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GR8677 #67 2008-10-28 11:10:05	So the absolute temperature is when T $\rightarrow \infty$ ? Is that what they mean by absolute temperature? Click here to jump to the problem!

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GR8677 #47 2008-10-27 09:47:20	kevglynn: Doesn't "a sealed and thermally insulated container" mean adiabatic ? Click here to jump to the problem!

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GR8677 #92 2008-10-14 10:56:46	Same question on choice (A)...the subject GRE test is coming soon, anyone care to explain? Click here to jump to the problem!

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GR8677 #93 2008-10-08 12:42:19	I am confused, why can't I use $mgsin(\theta)=T=$ force on string, so $a_{perp}=0$ , and $mgcos(\theta)=m(l\ddot{\theta})=ma_{para}$ , where $l$ =length of string. From this, $a_{para}=gcos(\theta)$ is correct, but $a_{perp}$ is wrong. Obviously the concept is incorrect here, but why? Click here to jump to the problem!

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GR8677 #22

2008-09-27 23:01:00

I take issue with this answer! This is the way I did it. 1. Given Rmin and Rmax you can sum them and divide in two to determine the semi-major axis, a. Thus choice E is eliminated 2. Given Rmin and the semi-major axis you just found, you can compute the eccentricity: Rmin=a(1-e) where a is the semi-major axis and e is the eccentricity 3. Given Vperi and Rmin you can calculate the mass of the planet: Vperi = sqrt{(GM(1+e))/Rmin} Thus choice B is eliminated 4. Similarly, since Rmax was also measured, you can now find the minimum velocity, which will occur at apastron: Vap= sqrt{(GM(1-e))/Rmax Thus choice C is eliminated Now we are left with the mass of the moon or the period of the orbit. Kepler's law is *actually* P^2= (4pi^2)a^3/(G(M+m)) Note the denominator contains the *sum* of the masses! So we truly have one equation left with two unknowns. You *cannot* calculate the *exact* period from here. You can only make an approximation where you neglect the mass of the moon, but that seems like a self-fulfilling prophecy or circular reasoning to me. Saying "a very small moon" is very subjective. What if the planet, too, is very small? They don't state the ratio of the planet to the moon, or mention that it is negligible in size compared to the planet. So how can we really use Kepler's law in approximate form in this case?

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GR8677 #82 2008-09-25 03:46:39	I don't understand. Why should we consider only reflections at the top and the bottom of the air gap? Why not the reflections at the top of the upper glass plat and the second glass plate? Click here to jump to the problem!

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GR8677 #94 2008-09-21 17:12:38	Wait, how are we in the clear??????? If gamma really is 5/4 then beta should be 3/5. Beta is clearly 3/4 = 0.75 So how is it a valid transformation? Click here to jump to the problem!

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GR8677 #97

2008-09-21 11:10:48

I am very confused... I had chosen D for my answer because that's the equation I had gotten to and I saw it in the answer choices... Now that I see that B is the right answer, I am having trouble seeing how B and D are not EXACTLY the same thing... the question DEFINES $v_{0}=\frac{1}{2}\omega_{0}R$ ...and if you just plug in this $\omega_{0}$ to choice B, you get choice D! What is the flaw in this logic? aren't $\omega_{0}, v_{0}$ , and R all constants? You yourself had $RMv_{0}$ before plugging in... did you just plug in for $v_{0}$ because you knew that the right answer was B or is there some reason why it is wrong to leave the answer like this? Is this possibly a mistake on the part of ETS?

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GR8677 #57 2008-09-21 08:48:59	Given the negative sign in Faraday's Law, wouldn't the induced emf in choice (A) be negative at t=0 rather than positive? Click here to jump to the problem!

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GR8677 #4 2008-08-18 01:43:44	I believe that choice E has the same dimensions as A&B. Click here to jump to the problem!

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GR8677 #4 2008-08-18 00:58:43	I believe that choice E has the same dimensions as A&B. Click here to jump to the problem!

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GR8677 #76 2008-08-01 13:01:11	Sorry, this should have been categorized under HELP :D Click here to jump to the problem!

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GR8677 #78 2007-10-21 21:28:08	I'm not really seeing why the tangential velocity of the two masses together is equal to the sum of their original linear velocities. Why can they be added this way? Click here to jump to the problem!

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GR8677 #16 2006-11-30 21:43:30	if you claim that $\overline{x}=2$ is the average and $\sigma=\sqrt{\overline{x}}$ is the standard deviation, then how do you go from there to get that $\sigma/\overline{x}=\sqrt{2C}/2C$ is the answer? I'm confused as to how the C's get in there Click here to jump to the problem!

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GR8677 #7 2006-11-01 10:44:16	Can you explain in greater detail? I can solve it, but I run into some messy algebra with quadratic equations. Can you explain how you so elegantly obtained these values? Click here to jump to the problem!

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GR8677 #41 2006-10-29 21:40:11	Can anyone direct me to a good explanation of the full spetroscopic notation and how it relates to the selection rules? More specifically, what part of this notation denotes m, the magnetic quantum number? Click here to jump to the problem!

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GR8677 #34 2006-10-27 15:15:32	hey i cant get enough practice problems for quant. i have got schaum seeries but it has only derivations any one has any material plz i want it Click here to jump to the problem!

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GR8677 #100

2006-10-27 14:07:24

hey watz up to begin iam a BE undergrad shifting to engineering physics.
any ways i just wanna know from where do i study error analysis. if any one has the notes on the comp plz mail them to me.
also if any one has practice problems (except for the four papers and two small other ones) specially for phy gre plz i want the soft copy.
bye, best of luck to everyone

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GR8677 #64 2006-10-19 13:49:09	What's this about a capacitor and an inductor? Problem 64 from 86 exams says "An alternating current electrical generator has a fixed internal impedance..." ? Click here to jump to the problem!

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GR8677 #57 2006-03-31 22:40:22	I don't understand. The problems gives the angle between the first minimum and the central max. Doesn't that correspond to m=1/2 which leads to d = 5 x 10^-5 which is B? Click here to jump to the problem!

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GR8677 #47 2005-11-10 21:33:24	Why is C the more correct answer than A? Your last comment in the introduction makes this seem confusing. Click here to jump to the problem!

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GR8677 #92 2005-11-08 21:14:27	If there is no selection rule for spin, then there would be no for j too. Recall that j=l+s. I was wondering this is because the photon emission obeys the conservation of spin momentum so that delta s is zero. Can anybody tell your opinion? Thanks. Click here to jump to the problem!

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