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 3 Click here to jump to the problem! webtechcoupons2018-11-05 07:03:03 Aw, I thought this was a significant pleasant post. Godaddy Renewal coupons In idea, I should set up composing like this also – investing energy and real exertion to make an extraordinary article… yet precisely what do I say… I hesitate a lot through no methods seem to go did. Click here to jump to the problem!
 7 Click here to jump to the problem! TheBridge2018-10-27 02:12:16 But 1 + i - i is not equal to 0. The eigenvalues are actually 1, $-\\\\\\\\frac{1}{2} \\\\\\\\pm \\\\\\\\frac{\\\\\\\\sqrt{3}}{2}i$.Click here to jump to the problem!
 8 Click here to jump to the problem! TypT2018-10-26 22:10:37 Quick, dumb way I did it without having to figure out what exactly the question was asking:\r\n\r\nSpace time intervals have a Pythagorean relationship, the two numbers we are given are 3 and 5, to complete the Pythagorean triple we need a 4 so we get (C). It even has the trap answer of $4^2 = 16$ (E), to give a little more confidence in (C).Click here to jump to the problem!
 9 Click here to jump to the problem! The Bridge2018-10-26 19:15:17 That won\'t work. The formula on the left is for centripetal acceleration of someone standing on the Equator due to the rotation of the Earth, and works out to about .88 m/s2. This is in fact many times less than $g$, the acceleration due to gravity of something on the Earth\'s surface, which is the right side. If you knew the orbital period of an object at the Earth\'s surface off the top of your head (84 min), then this method would work, but it gives you $g\\approx$10 m/s2 in the table of information at the top of the test anyway, so the left side of the equation is completely unneeded.Click here to jump to the problem!
 10 Click here to jump to the problem! kronotsky2018-10-26 03:27:43 The relative differences in wavelengths is given by the factor (n-1), so this is what you are equating to 2e-4. Other than that, the situation happens to be exactly analogous, since increasing the index of refraction essentially increases the effective optical path length, which is the quantity used to derive the thin film equation.Click here to jump to the problem!
 11 Click here to jump to the problem! Ryry0132018-10-25 12:05:41 My half-BS solution: I tried to figure out which of the answers could reasonably come from 1.5. I thought 1.5 = 3/2, saw 2/3, and then picked it. A guessing strategy for if you know zero Physics.Click here to jump to the problem!
 13 Click here to jump to the problem! ryanmes2018-10-24 21:32:38 The easiest way to look at so many problems on this exam is limits. If d increases, then you need a smaller velocity to stay on the track. There is only one answer where this is the case (when d is in the denominator). And this is the answer.Click here to jump to the problem!
 14 Click here to jump to the problem! BruceLyu2018-10-24 07:35:28 This problem involves two plane waves (de-Broglie waves) in two regions. Initially, it is $e^{ikx}$ and after entering the region with potential energy V, the kinetic energy becomes $E-V_0$ and the corresponding wave is $e^{ik\'x}$. For a plane wave, the energy is $E=\\hbar^2 k^2/2m$ so \r\n$(\\frac{k}{k\'})^2 = \\frac{E}{E-V_0} = (\\frac{\\lambda\'}{\\lambda})^2$\r\nand solve it we get the final answer E.Click here to jump to the problem!
 20 Click here to jump to the problem! kronotsky2018-10-23 03:44:30 Right hand rule trick: if you put your thumb in the direction of I, and curl your fingers up into a fist, the fingers point along B. We want $q \\vec{v} \\times \\vec{B} = \\vec{I}$. Cyclic permutations and setting $q = -1$ gives $\\vec{I} \\times \\vec{B} = \\vec{v}$.Click here to jump to the problem!