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Cry for help
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| Unanswered Cries for Help! Total So Far: 28 | | 0 | Click here to jump to the problem! | GR8677 #71
2010-02-18 08:22:12 | I like your idea about using common sense, and it works, however, in this case the answers B and C are very close, in the moment of the test, under pressure is very hard try to deicide about one of them. There is an alternative way to do some basic calculation to give the right answer, beteween B and C? |
| 6 | Click here to jump to the problem! | GR8677 #72
2009-06-03 13:03:06 | I had trouble thinking about why  . Both the Lagrangian and the standard force balancing solutions rely on the above equation, and it's not readily obvious to me as to why the bottom mass has an initial zero acceleration. Here is an intuitive way that might help:
When drawing force diagrams at equilibrium, one balances them so that  . So, for this system, before the string is cut, we have the spring trying to condense itself back to equilibrium, the two masses wanting to fall, and the string being the only thing standing between them and gravity. So (with $g>0$ defined) we have:
Top mass
 = 0)
Bottom mass
Now, after the string is cut, two things happen: tension goes to  and everything starts to move. However, let's just concentrate on the tension:
Let  . Since  is no longer a force, it has to go somewhere, i.e., it has morphed into motion, or  . Since there was no in the bottom equation, that doesn't change. Doing this allows us to write:
Top mass
 = 0)
Bottom mass
Now, we substitute the bottom mass into the top and obtain the desired solution:  .
I'm not sure whether this strategy of substitution works in general, and I would appreciate comments of any sort as to why it does or does not.
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| 11 | Click here to jump to the problem! | 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? |
| 12 | Click here to jump to the problem! | 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? |
| 13 | Click here to jump to the problem! | 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? |
| 14 | Click here to jump to the problem! | 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? |
| 18 | Click here to jump to the problem! | 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? |
| 20 | Click here to jump to the problem! | 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? |
| 21 | Click here to jump to the problem! | 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? |
| 22 | Click here to jump to the problem! | 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 |
| 23 | Click here to jump to the problem! | 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 |
| 24 | Click here to jump to the problem! | 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..." ? |
| 25 | Click here to jump to the problem! | 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? |
| 26 | Click here to jump to the problem! | 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. |
| 27 | Click here to jump to the problem! | 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. |
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