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varsha
2017-02-21 11:20:22
Either the question is wrong or there is a typo in the answer, it is v0 = 1/2 Rw0 and not v0 = 1/2 R2 w02.
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varsha
2017-02-21 11:15:10
Either the question is wrong or there is a typo in the answer, it is v0 = 1/2 Rw0 and not v0 = 1/2 R2 w02.
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NAJUM UDDIN
2017-01-20 04:49:49
THANKS \r\n
3 Click here to jump to the problem!
varsha
2017-01-17 11:10:51
How did epselon become >> than kT??....please explain,
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varshakr
2016-12-27 10:46:29
depends on units
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bluejay27
2016-12-25 22:22:53
of Young and Freedman
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bluejay27
2016-12-25 22:22:27
Check page 606 and 607. D is the correct answer.
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bluejay27
2016-12-22 22:12:03
Air is obviously less dense than the atomic radius 10^{-10}?? What does this mean?
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DownWithETS
2016-10-30 14:36:23
I used this method also, but it doesn\\\\\\\'t really work. X rays range from wavelength=10 nm-.1 nm. If you assume the high-end energy limit of .1 nm, then it works, but using either 1 or 10 nm values gets us a different order of magnitude, which are both possible choices. Is there some reason you know to choose the highest energy x-rays?
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GrahamS
2016-10-29 04:34:12
Dimensional analysis won\'t work here because they are asking about a proportionality. That implies that there could be constants that they are leaving out, constand that could have totally unknown units. Anyways, I think somebody who took basic E&M shouldn\'t ever forget the formula for the magnetic field outside of a wire of current, which makes this problem extremely quick, even wuicker than those that can be done with dimensional analysis.
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GrahamS
2016-10-28 17:37:58
That doesn\'t give answer E. Answer E has a b^4 in it, your method (which is the same thing that I did originally) has a b^2.
11 Click here to jump to the problem!
GrahamS
2016-10-28 17:37:14
That doesn\'t give answer E. Answer E has a b^4 in it, your method (which is the same thing that I did originally) has a b^2.
12 Click here to jump to the problem!
junksneeze
2016-10-27 18:46:32
For finding the radius of mars, I found it easiest to think this way: the surface of mars is a circle described by r^2 = x^2 + y^2. Starting at x=0 for example, y = r. We are told that when x = 3600 meters, y = r - 2. So r^2 = 3600^2 + (r - 2)^2. Solve for r.
13 Click here to jump to the problem!
SarlCagan
2016-10-27 16:53:54
If the centre particle B isn\'t moving, then you can think of it as a wall. Then A and C don\'t know about each other, and will each independently oscillate as if they were simple oscillators, with frequency \\omega = \\sqrt{k/m}\r\n\r\nSo the answer is choice B (which in this case is the only one that has particle B at rest)
14 Click here to jump to the problem!
Dudeleb
2016-10-27 15:16:34
here is a fun little rhyme to help you remember the resistance of a wire. Resistance = RHO RHO RHO YOUR BOAT L OVER A! RESISTANCE RESISTANCE RESISTANCE LIFE IS BUT A DREAM (of an electron traveling down some conducting material). The rest is math. Plug and chug. Like a beer. Am I RIGHT!? Ah man. This test is making me crazy. Cheers all.
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psitae
2016-10-27 14:27:54
probably very close to the speed of light -->  \\frac{2}{\\sqrt 5} c
16 Click here to jump to the problem!
benjamin_DW
2016-10-27 14:18:09
What\'s not clear to me is why the substance **must** cool at the same rate that we heated it. Just because we heated it at a rate of 100W does not mean it will necessarily cool at a rate of 100W. In fact, by Newton\'s law of cooling, the temperature \\Phi(t) is given by: \r\n\r\n \\frac{d\\phi}{dt} = -\\kappa (\\Phi - \\Phi_{0})\r\n\r\nWhere  \\Phi(0) is the background temperature, and \\kappa is a constant dependent on the geometry of the material.[and even this of course, is an approximation]
17 Click here to jump to the problem!
psitae
2016-10-27 04:07:49
very good! you used the fact that water doesn\'t compress to realize that the two water levels have the same distance to the 20 cm point!
18 Click here to jump to the problem!
TubbyBeef
2016-10-26 22:44:59
You are correct. A fluid with density half that of steel would not float freely in water. \r\n\r\nThe notion that they are imiscible and this prevents the more dense fluid from sinking relies on the tube being narrow.. narrow enough that surface tension effects can dominate over buoyancy effects and the Rayleigh–Taylor instability present. \r\n\r\nThere is a reason we don\'t use capillary tubes as manometers. :-/
19 Click here to jump to the problem!
anGRE
2016-10-26 12:04:05
Wow, the formatting totally got fucked. All of the forward slashes just disappeared. Rough.\r\n\r\nWell I spent enough time on this. Just ignore it.
20 Click here to jump to the problem!
anGRE
2016-10-26 12:01:38
This problem is just misleading. The observer measures the wavelengths to a precision of four significant digits, but the answer choices are only stated to two significant digits. It should say MOST NEARLY somewhere in the problem statement. I failed to solve the problem under testing conditions because of the poor phrasing. \r\n\r\nIn case anyone is interested in my method:\r\n\r\nThe ratio \\frac{\\lambda}{\\lambda_0} = \\frac{607.5}{121.5} = 5, exactly. So I was expecting nice numbers to fall out of the calculation.\r\n\r\nNoting that this is a redshift, and that in general stuff doesn\'t move faster than c, I narrowed to choices (C) and (D).\r\n\r\nInstead of solving the equation for beta, which I worry might end in a nasty quadratic, I plug these two choices in as fractions of c to the equation for the relativistic doppler shift:\r\n\r\nFor (C), sqrt{\\frac{1+(12/15)}{1-(12/15)}} = 3\r\nFor (D), sqrt{\\frac{1+(14/15)}{1-(14/15)}} = sqrt{29}\r\n\r\nRemember, I expect to see a nice round 5. Obviously root 29 is closer, but given the precision to which the wavelengths are stated, I expect to land on one of the answer choices exactly. So I conclude I either have made an algebra error, or I misremembered the doppler shift formula. I mark no answer choice.\r\n\r\nIt\'s crazy that 32% of people who took the exam got this question correct. Arriving at the correct answer involves an unwarranted approximation at best and a math error at worst.\r\n\r\n
21 Click here to jump to the problem!
dc771957
2016-10-24 22:53:42
If the particle were initially at the origin, or somewhere else along the x-axis, at rest, it would follow motion similar to E only it would be bouncing along the x-axis, not the y-axis. Initially v=0 so the force due to the magnetic field is zero, as the particle begins to accelerate in the +y direction due to the force from the electric field, it begins to feel a force from the magnetic field, which will initially make the particle accelerate in the +x direction. For this setup, the magnetic force will always be trying to force the particle to make a right turn. As the particle begins to move in the +x direction, the magnetic field will start to pull the particle in the -y direction, keep in mind that the electric field is still acting in the +y direction. As the particle begins to return to the x-axis, with a velocity in the -y direction, following a similar path shape from answer E, it will slow down because the electric field is pulling it upwards. The particle will have enough kinetic energy to resist the pull of the electric field until it returns to the x-axis, where it will briefly stop, before repeating this process. \r\nIf you take a look at answer E, you will see that the magnetic and electric fields will be pulling the particle in the +y direction when it intersects with the y-axis; the electric field always acts in +y, and the particle will have a velocity in the -x direction with magnetic field still in the +z direction. So, no, E is not the correct answer, but that didnt stop me from picking when I first looked at this problem.
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dc771957
2016-10-24 17:42:22
GODDAMNIT I HATE LATEX. Im so glad im only taking the physics GRE and not trying to learn latex right now
23 Click here to jump to the problem!
dc771957
2016-10-24 17:41:17
Yo fuck latex and whatnot, heres my solution. Intensity a.k.a. irradiance is going to be proportional to the superposition of E1 and E2 dotted with itself. So we will have \\left( E_1+E_2 \\right) dotted with itself. This will give us \\left( E_1 dot E_1 + 2 E_1 dot E_2 + E_2 dot E_2 \\right). E1 dot E2 will be zero, since the two waves are orthogonal and polarized perpendicularly to each other. So the final intensity will be proportional to E1^2 + E2^2
24 Click here to jump to the problem!
dc771957
2016-10-24 15:04:54
So what I did was take the real parts of the original wave. You see that x will look like cos and y will look like -cos, as many people have stated. So we know that when x=0, y=0. Then you can see that when x=1, y=-1 and vice versa. Plotting just these three points you see that the answer is B.
 
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