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| Latest 25 Comments | | 0 | Click here to jump to the problem! | kimseonta
2009-01-07 01:18:00 | Sorry about the one that I posted below...
it is totally incorrect...so I am now going to post the solution again...
The rocket is in a free space -> No external forces
the momentum must be conserved.
Pi=Pf
mv=(m+delta m)(v+ delta v)+(V+v)( - delta m)
, where the delta m should be negative when considering a rocket propulsion. And V is the velocity [Not speed] of the rocket's exhaust from the rocket's frame.
[m + (delta m)](delta v) - V(delta m) = 0
take /(delta t) and limit[delta t -> 0]
Then,
mdv/dt - Vdm/dt = 0
Here, of course, V is negative[vector]. Let u be the speed[scalar] of the rocket's exhaust from the rocket's frame.
 u = -V
mdv/dt + udm/dt = 0: Here is the answer. |
| 1 | Click here to jump to the problem! | kimseonta
2009-01-06 20:13:36 | This is my own way to make the typo (?) clear...
The same idea goes: the conservation of the momentum.
Pi = Pf
mv = (m + dm)(v+dv) + (dm)V, where V <0 of course.
= mv+mdv+vdm+dmdv+(dm)v
Now there exist two appoximations that one can make.
- V >> v : one can simply assume that the speed of the exaust from a rocket should be much faster than the one of the rocket.
-dmvm is almost zero relative to the other parts of the equation.
so the simplified version will be here...
mdv+Vdm=0
m(dv/dt) + V(dm/dt) = 0 |
| 3 | Click here to jump to the problem! | engageengage
2009-01-05 14:33:55 | adding onto my earlier comment, to get that same result, you have to throw away second order powers of  , which are tiny anyways. |
| 6 | Click here to jump to the problem! | Walter
2008-12-28 14:52:36 | There are four possible spin states for the two electron system that are either symmetric or antisymmetric. There is only one that is antisymmetric, (hence singlet), but three that are symmetric, (hence triplet).
You need to be able to pick out the symmetric spin eigenfunctions. These will look the same after you exchange the labels for the electrons. This gives I and III.
Relabeling II makes it change sign - it's the antisymmetric singlet. |
| 7 | Click here to jump to the problem! | sonnb
2008-11-08 19:58:35 | The free particle has an inital total energy of E, which is equal to its kinetic energy. When it enters the region of potential energy V, its total energy remains the same (by conservation of energy) so that its new kinetic energy is E-V. |
| 12 | Click here to jump to the problem! | carl_the_sagan
2008-11-07 22:05:17 | I'm not sure if this is a valid way to tackle the problem but it's how my messed up line of thinking went:
The block B needs to be pushed up equal to the force gravity exerts on it to stay in place. Take gravity to be 10m/s^2 for convenience and it needs 40N.
However, the coefficient of friction is only .5, so you have to push twice as hard, so 80N. This means you need to accelerate B by 80N/4kg = 20m/s^2 horizontally to produce enough "lift" to balance things. Since the entire system is 20kg, F = 20kg*20m/s^2 = 400N |
| 14 | Click here to jump to the problem! | Poop Loops
2008-11-07 20:35:25 | This is such a load of crap. In the question they say the loop rotates with angular speed  and then they ask you for the angular speed of the loop. What the hell does that mean??? |
| 15 | Click here to jump to the problem! | carl_the_sagan
2008-11-07 20:29:54 | This is just a glorified rate problem like you see on the general GRE. Gamma gets you half the work done in 24 minutes, Beta gets you half the work done in 36 minutes. Working together you have:
 = 14.4
Granted they are really decays, but the same idea applies, and you didn't have to mess with logs and any other nonsense. |
| 16 | Click here to jump to the problem! | eliasds@yahoo.com
2008-11-07 20:03:23 | FYI, the questions on OSU's website are just taken from the 4 practice tests. I recommend taking the real practice tests before going back over OSU's "homework".
PS: Good luck to everyone taking the exam tomorrow!!!!!!!! |
| 17 | Click here to jump to the problem! | carl_the_sagan
2008-11-07 19:48:26 | A really cheesy way is to totally disregard the question and just look at the answers. Typically on a question like this they are going to have to repeat the correct value for each column a few times.
so "m" is repeated in L, "1/k" is repeated in C, and "x" is repeated in Q. The only answer which lines all these multiple-occurring rows is B.
Granted this has nothing to do with physics, but this isn't about physics, its about getting the right answer. |
| 18 | Click here to jump to the problem! | carl_the_sagan
2008-11-07 19:42:52 | Also did it this way, if you can see a quick trick like this and arrived at a supplied answer, it's almost always a good thing. |
| 19 | Click here to jump to the problem! | carl_the_sagan
2008-11-07 19:32:58 | This explanation makes by far the most sense to me. Though I never took a solid state course, I recall this from the bit we got to in modern and condensed matter lab. |
| 22 | Click here to jump to the problem! | flux
2008-11-07 18:16:24 | Because the surface is a conductor, the E field must go to zero. This is similar to a rope with a fixed end. The E field will thus flip over and head the other direction. This leaves us with choices A and C. Using the right-hand-rule brings us to the conclusion that the B field holds its direction. Choice C it is! |
| 23 | Click here to jump to the problem! | askewchan
2008-11-07 16:57:57 | There's actually a very cool plot of this at the very bottom of the article:
http://en.wikipedia.org/wiki/Binding_energy |
| 24 | Click here to jump to the problem! | askewchan
2008-11-07 16:55:59 | Actually, binding energy goes down only after iron or so. Binding energy is lowest in Hydrogen, rises to a maximum around iron, then it goes down.
This is why energy is released by fusion for light elements, but by the opposite (fission) for heavy elements. The ultimate endpoint of both of these processes is iron, from which you can no longer milk any binding energy. |
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