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Unanswered Cries for Help!      Total So Far: 45
 0 Click here to jump to the problem! GR8677 #842018-04-05 03:13:54 well I got down to D and E as well. But instead of taking m to infinity, I took it to zero. In that way D gives an infinitely large frequency. I can\'t make sense out of it. I think a massless m2 simply makes itself disappeared so the system is just a pendulum with a meaningless spring attached it oscillating with f = sqrt (l/g). \r\nThere is something wrong with this reasoning. I hope somebody can help me eventually.\r\nClick here to jump to the problem!
 2 Click here to jump to the problem! GR8677 #392017-04-08 00:55:51 I\'m confused. Choice (E) resembles the function y = $\\frac{1}{x^2}$ which has positive concavity, not negative like the graph shows.Click here to jump to the problem!
 4 Click here to jump to the problem! GR8677 #632015-12-19 21:09:12 I am really very happy after getting the unlimited roblox robux generator which allowed me to acquire the whole world in playing roblox with much ease after following the given link.Through it http://www.robloxfreerobuxgenerator.com/ i was able to access the whole world in roblox and enjoyed every little bit of it.Click here to jump to the problem!
 5 Click here to jump to the problem! GR8677 #982015-09-04 09:22:09 I tried to solve this problem in an alternative way, which is less simple and convenient than the official solution on the site, but unfortunately failed to get the correct answer. I can\'t find my mistake and I am quite baffled. I shall deeply appreciate if someone would like to offer me help soon!\r\n\r\nMy solution is to integrate the intensity of electric field E at the point P first, and then integrate the electric potential at the same point. \r\n\r\nAssume the linear electric density $\\lambda$,\r\nQ=$\\lambda$L, dQ=$\\lambda$dx,\r\n\r\ndE=$\\frac{kdQ}{(l+x)^2}$ =$\\frac{k \\lambda}{(l+x)^2}$dx, \r\n\r\nE=$\\int_0^l$$\\frac{k\\lambda}{(l+x)^2}$dx,\r\n\r\nApplying the substitution method, set u=x+l, du=dx, so\r\nE=k$\\lambda$$\\int_l^{2l}\\mathrm{u}^{-2}\\,\\mathrm{d}u$=k$\\lambda$[$\\frac{-1}{u}$]=$\\frac{kQ}{2l^2}$.\r\n\r\nHence, V=$\\int_l^\\infty\\vec{E}\\mathrm{d}\\vec{l}$=$\\int_l^\\infty$$\\frac{kQ}{2l^2}$$\\mathrm{d}l$=$\\frac{kQ}{2}$[$\\frac{-1}{l}$]=$\\frac{kQ}{2l}$.\r\n\r\nSo my answer is the choice B. \r\n\r\nCan someone tell me where I did wrong? Click here to jump to the problem!
 6 Click here to jump to the problem! GR8677 #982015-09-04 08:42:18 I tried to solve this problem in an alternative way, which is less simple and convenient than the official solution on the site, but unfortunately failed to get the correct answer. I can\'t find my mistake and I am quite baffled. I shall deeply appreciate if someone would like to offer me help soon!\r\n\r\nMy solution is to integrate the intensity of electric field E at the point P first, and then integrate the electric potential at the same point. \r\n\r\nAssume the linear electric density $\\lambda$,\r\nQ=$\\lambda$L, dQ=$\\lambda$dx,\r\ndE=$\\frac{kdQ}{(l+x)^2}$\r\n =$\\frac{k \\lambda}{(l+x)^2}$dx, \r\nE=$\\int_0^l$ $\\frac{k\\lambda}{(l+x)^2}$dx,\r\nApplying the substitution method, set u=x+l, du=dx, so\r\nE=k$\\lambda$$\\int_l^{2l}\\mathrm{u}^{-2}\\,\\mathrm{d}u$=k$\\lambda$[$\\frac{-1}{u}$]=$\\frac{kQ}{2l^2}$.\r\n\r\nHence V=$\\int_l^\\infty\\vec{E}\\mathrm{d}\\vec{l}$\r\n =$\\frac{kQ}{2l^2}$$\\int_l^\\infty$$\\mathrm{d}x$.\r\nFinally, there comes out infinity of the electric potential. \r\n\r\nCan someone tell me where I did wrong? \r\nClick here to jump to the problem!
 8 Click here to jump to the problem! GR8677 #582015-03-23 12:27:14 I'm a little confused about notation in chemistry. In Quantum Mechanics, the notation of (2s+1)^L_(j) is used where L is a place holder for the total angular momentum quantum number (l=0 ->S, l=1->P, l=2 ->D, etc). How does 2s+1 relate to the proton number, in this case Z=11. The solution says to sum up the superscripts to see which add up to 11 but why?Click here to jump to the problem!
 19 Click here to jump to the problem! GR8677 #102011-02-28 16:53:23 Why is the direction of the -q and -2q equal to $\hat{x}$ and not -$\hat{x}$? I understand that the forces add up from drawing out the Forces... but mathematically my magnitude sum is not as I expected it to be since i used -$\hat{x}$ as the direction of the force on q for -q and -2q.Click here to jump to the problem!
 22 Click here to jump to the problem! GR8677 #672010-04-09 19:57:04 Just think of it as the superposition of two waves one that is polarized and one that is partly polarized add these together (think a circle and a ellipse) and you will get the combined wave.Click here to jump to the problem!
 24 Click here to jump to the problem! GR8677 #412010-03-27 13:14:14 Some texts include the spin of an odd nucleon in the total Russell-Saunders coupling sum of spins. I don't know why it is sometimes included and sometimes not. Perhaps that is why ETS has designated the state as S=1, because they are including the spin of the proton. The problem with this method is that all atoms would have a total integer spin, zero for even Z, and one for odd Z. Does anyone know when to include the odd nucleon and when not to in the total angular momentum? It seems that it should always be included. I don't think that electronic shielding in larger atoms would change the total angular momentum.Click here to jump to the problem!
 27 Click here to jump to the problem! GR8677 #992009-04-01 18:35:24 Please excuse my ignorance. How did you get this formula for $E_{f}$?Click here to jump to the problem!
 28 Click here to jump to the problem! GR8677 #672008-10-28 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!
 31 Click here to jump to the problem! GR8677 #222008-09-27 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?Click here to jump to the problem!
 32 Click here to jump to the problem! GR8677 #822008-09-25 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!
 33 Click here to jump to the problem! GR8677 #942008-09-21 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!
 38 Click here to jump to the problem! GR8677 #162006-11-30 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 thereClick here to jump to the problem!
 40 Click here to jump to the problem! GR8677 #412006-10-29 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!
 41 Click here to jump to the problem! GR8677 #342006-10-27 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!
 42 Click here to jump to the problem! GR8677 #1002006-10-27 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 everyoneClick here to jump to the problem!