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 8 Click here to jump to the problem! DownWithETS2016-10-30 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?Click here to jump to the problem!
 9 Click here to jump to the problem! GrahamS2016-10-29 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.Click here to jump to the problem!
 10 Click here to jump to the problem! GrahamS2016-10-28 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 $b^2$.Click here to jump to the problem!
 11 Click here to jump to the problem! GrahamS2016-10-28 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 $b^2$.Click here to jump to the problem!
 12 Click here to jump to the problem! junksneeze2016-10-27 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.Click here to jump to the problem!
 13 Click here to jump to the problem! SarlCagan2016-10-27 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)Click here to jump to the problem!
 15 Click here to jump to the problem! psitae2016-10-27 14:27:54 probably very close to the speed of light --> $\\frac{2}{\\sqrt 5} c$Click here to jump to the problem!
 16 Click here to jump to the problem! benjamin_DW2016-10-27 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] Click here to jump to the problem!
 20 Click here to jump to the problem! anGRE2016-10-26 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\nClick here to jump to the problem!
 23 Click here to jump to the problem! dc7719572016-10-24 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 + E2^2Click here to jump to the problem!