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 0 Click here to jump to the problem! deneb2018-10-15 23:29:30 My reasoning is the breaking happens continuously. It\'s not quantized like an electron jumping energy levels. You don\'t go from 10 m/s to 5 m/s without slowing down to 9, 8, etc. Click here to jump to the problem!
 5 Click here to jump to the problem! deneb2018-10-13 21:26:52 I\'m confused what the answer is describing. Is the s=1 referring to the entire helium atom? This particular helium atom has two electrons which each have s=1/2, so together s=1? And then since it\'s in the S orbital, l=0, then j=1? Click here to jump to the problem!
 6 Click here to jump to the problem! Hassan00132018-10-12 15:08:27 I think the better way is to just use the \\\\\r\n\r\n$\\Delta z=z_2 - z_1 = 2\\pi / k$ \\\\\r\n\r\nand again combine both 2 waves to make the vector for the electric field.\r\n\r\nnow if we calculate the vector\'s magnitude we easily find the solution for choice (A)\r\n\r\nthis is recombination after portions : \\\\\r\n\r\n$\\vec{E} =E_1 e^{i(k z_1 - \\omega t)} + E_2 e^{i(k z_1 + 2\\pi k - \\omega t + \\pi)}$ \\\\\r\n\r\n\r\nnow it needs just to calculate the magnitude of E to get the answer...\r\nClick here to jump to the problem!
 7 Click here to jump to the problem! Hassan00132018-10-12 15:05:09 I think the better way is to just use the $\\Delta z=z_2 - z_1 = 2\\pi / k$\r\nand again combine both 2 waves to make the vector for the electric field.\r\n\r\nnow if we calculate the vector\'s magnitude we easily find the solution for choice (A)\r\n\r\nthis is recombination after portions : \r\n$\\vec{E} =E_1 e^{i(k z_1 - \\omega t)} + E_2 e^{i(k z_1 + 2\\pi k - \\omega t + \\pi)}$\r\n\r\nnow it needs just to calculate the magnitude of E to get the answer...\r\nClick here to jump to the problem!
 8 Click here to jump to the problem! deneb2018-10-10 05:14:29 Agreed, just a simple ratio problem. \r\n\r\nT1 = 4/3 hours\r\nT2 = 24 hours\r\n\r\nR1 = 1 Re\r\nR2 = a * Re\r\n\r\nsolve for a\r\n\r\n18^2 = a^3\r\n\r\na = 18^(2/3)\r\n\r\nwe can approximate 18^(1/3) to be around 2.5, which squared is 6.25. Answer is B. Click here to jump to the problem!
 11 Click here to jump to the problem! deneb2018-10-08 04:12:51 You have to divide both sides by $v^2_x$, and since $v_x$ is just the original velocity given in the problem, that\\\\\\\'s how you get $v^2$ in the denominatorClick here to jump to the problem!
 12 Click here to jump to the problem! cquinone2018-10-04 19:26:08 The area estimations I\'m seeing here and in other solutions are wrong, you can easily calculate the range of V to be 2-5, so area should be estimated as 1/2(b)*(h) = 1/2(5-2)(500-200)= 1/2(900) = 450. This is equidistant to 300 and 600, but you know that your estimate was slightly high as the isotherm BC dips down. So you pick -300.Click here to jump to the problem!
 13 Click here to jump to the problem! Batman2018-09-13 09:06:44 Rotation of electron in its orbital creates a magnetic moment that interacts with the inhomogeneous magnetic field. The magnetic moment of a proton is very small compared to that of electron and hence negligible in this case. \\\\r\\\\nHowever, we have methods to work with the nuclear magnetic moment too.Click here to jump to the problem!