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## Hot answers tagged homework-and-exercises

4

a. Immediately after the switch is closed, are either or both bulbs glowing? Explain. They will both glow as some current passes through them as the capacitor is charging. b. If both bulbs are glowing, which is brighter? Or are they equally bright? Explain. They are both equally bright, because an equal and opposite charge is flowing on to ...

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The way to do this is using the Wigner-Eckart theorem. The way it is applied to your problem is as follows: $$\left\langle nlm |\vec{r}| n'l'm'\right\rangle = \left\langle nl ||\vec{r}|| n'l'\right\rangle \left\langle l' m' 1 q | l m\right\rangle$$ where the second factor is a Clebsch-Gordan coefficient and $q=-1,0,1$ indicates the type of transition. For ...

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Think about this from the perspective of a person in the elevator. No windows, they can't look outside. As far as they are concerned, they live on a small box-like planet where the acceleration due to gravity is 9.8 + 1.2 = 11 m/s$^2$. In a system where the acceleration due to gravity appears to be 11 m/s$^2$, a bolt drops 2.7 m. How long does it take to ...

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At the instantaneous moment shown in the diagram, we can write: $$2R\alpha_{ring}=a_{disc}$$ as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration $=2R\alpha_{ring}=2a_{ring}$ so we find that: $$a_{disc}=2a_{ring}$$ Note that when the string moves to another position this will not be true, ...

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123hoedjevan gives you a wrong answer. The principle of least action states that the physical configuration of the system of fields realizes a minimum of the action with respect to compactly supported variations of the fields which, by the very definition of compactly supportedness, must then vanish on the boundary of the support itself. This in turn means ...

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A crucial hypothesis is missed in your construction. Each $\phi_i$ must also satisfy $\phi_i \not \perp \psi_i$, otherwise $\langle \psi_i |E_i \psi_i\rangle >0$ is false. This point provides an answer to your last question as well. If $\psi$ is an added further vector, linearly dependent on the vectors $\psi_i$, the construction you made cannot be ...

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Remember that the variation of the angular momentum equals the external torque. If there are no external torque (as in your case), the angular momentum is conserved.

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The principle of conservation of angular momentum says that angular momentum remains conserved unless an external torque acts on it. The net torque on a body is defined as: $$\vec{\tau\,}=\dfrac{\mathrm d\vec{L\,}}{\mathrm dt}$$ We can clearly see from this definition that since external torque on the body is zero, the angular momentum is going to remain ...

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In your scenario, angular momentum $m v r$ is preserved (because your pulling force is radial, with no tangential component). So if you reduce $r$ by half, $v$ must double, and since $\omega = v/r$, it increases by a factor of four. Note this means in a small amount of time that the area swept out by the string is proportional to $v$ and $r$. Since they ...

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When deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to $T$ (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it. If we take a segment small enough that we can consider it as a straight line, then the ...

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a) Assume that light has to travel 1 m to reach the origin of the reference frame I, so it'll take $t=1/c$ seconds to reach. Now, using trigonometry (the sine rule in the first equation), we get: $$\frac{\sin{\Delta\theta}}{\omega t} = \frac{\sin{\theta}}{ct}$$ $$ct\cos{\Delta\theta} = 1-{\omega t}\cos{\theta}$$ Dividing the first equation with the ...

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The external force acts only for the small time when the cue has been struck. Once it moves, there is no force. This means that the ball is moving with zero external force, which means according to Newton's second law, the velocity of the ball is the same. here the act of friction is of less importance as it requires in a billiard play. So the center of mass ...

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The name "Gaussian noise" actually has to do with the higher order correlations in the noise, such as: $$\langle \eta(t) \eta(t+\tau_1) \eta(t+\tau_2) \rangle,$$ $$\langle \eta(t) \eta(t+\tau_1) \eta(t+\tau_2) \eta(t+\tau_3) \rangle,$$ and so on. If the noise is Gaussian then all of these higher order correlations can be rewritten in terms of the two-term ...

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Hint for (a) : Use the Figure above to prove that : $$\tan\left(\phi^{\prime}-\phi\right)=\dfrac{w\sin\phi}{c-w\cos\phi}=\dfrac{w\sin\theta}{c+w\cos\theta}\;=\;-\tan\left(\theta^{\prime}-\theta\right) \tag{a-01}$$ This explains why I find the last equality instead of the first as I post in one of my comments : we refer to ...

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You don't. Two given spacetimes can have their metrics written in the same way but may have different coordinate ranges. A simple example is just a spacetime with spatial coordinate identified , such as the cylinder spacetime : $$ds^2 = -dt^2 + d\theta$$ Identical to Minkowski space, which is its universal cover. Of course, two things to watch out for : ...

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Forget anything about a capacitor and just consider the resistance of the conducting liquid. Think of the liquid as made up of thin $dr$ concentric shells of radius $r$ and find the resistance of a shell in terms of the resistivity, radius and thickness. Then do the integration to find the resistance of all the liquid.

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I want to charge a 12v 100ah battery which need 20 amp of current. You can charge the battery at any current proving it is not too high and it might be that the 20 amp is the maximum charging current? If the adapter gives a constant (regulated) 12 volts then you will not be able to charge a battery of the lead-acid type it will require more than 12 V ...

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The concept of 'straight' is a bit ill defined in GR and has no real definition. In fact in a sense the geodesics themselves be seen as 'straight' lines; they are the shortest paths connecting 2 points (this is what in normal Euclidean space would be a 'straight line') In the LC connection they are the integral curves of some vector field $V$ with $... 1 Your mistake is in your conservation energy equation. The way you wrote it is valid only when falling from infinity, from rest. The correct is: $$dE=dK+dU=0,$$ that is $$mvdv=-\frac{K}{x^2}dx,$$ where$K\equiv Gm_1m_2$. Integrating from$(x_i,v_i)$to$(x,v)$we get $$\frac 12m(v^2-v_i^2)=K\left(\frac{1}{x}-\frac{1}{x_i} \right).$$ This is the correct ... 1 I can do it by considering$m_1$and$m_2$to be a system, which would give me$a=F/(m_1+m_2)$. How can I use a free-body diagram instead to calculate the acceleration? But this is by the use of a free-body diagram. Otherwise, how would you know that it is the force$F$you should include? Because, your acceleration expression comes from Newton's 2nd ... 1 Hint: first find the angular velocity of the disc by conservation of energy,$\frac{1}{2}I\omega^2 = mgh$,$\omega$= angular velocity of the pulley, I = moment of inertia of the pulley. It is mentioned that the pendulum cover a horizontal distance L in t seconds. And we also know that $$v_{linear-velocity} = R\omega$$ So time t can be found ... 1 I) For the record, here is the operator calculation that OP wants to avoid. The benefit of the calculation is that the operators are not sandwich with any bra/ket representation, and hence we do not have to worry about whether the bra/ket representation is faithful. Let us put$\hbar=1$for simplicity. The starting point is the CCR $$[x^i, ... 1 You should take moments about the points r1 and r2, which then you will get two seperate equations, which is what you have done, and that's fine. Take another look at both of your equations, r1 and r2 are forces and since you are taking moments about both those points, you need to multiply r1 and r2 by their distance. What you have essentially done is equate ... 1 Hints: The starting point is the 2-point relation$$T(\phi(x)\phi(y)) ~-~:\phi(x)\phi(y): ~=~ C(x,y)~{\bf 1}, \qquad C(x,y)~\equiv~\langle 0 | T(\phi(x)\phi(y))|0\rangle,\tag{1} $$cf. this Phys.SE post. The relevant Wick's theorem is a nested Wick's theorem$$ T(:\phi(x)^n::\phi(y)^m:)~=~\exp\left( ... 1 There are two mistakes. As AccidentalFourierTransform pointed out, the coefficient$7.181\times 10^{-16}$, when converted from MeV to eV, should give$7.181\times 10^{-46}$. Mega means a million, and it to the fifth power gives$10^{30}$, not just$10^{15}$. In this way, the OP has to add a$10^{-15}$factor to his result. That makes his result$10^{-3}$... 1 On page 68 the paragraph headed Calculating displacement given a time and acceleration includes the text: Assume that you’re on your traditional weekend physics data-gathering expedition. Walking around with your clipboard and white lab coat, you happen upon a football game. Very interesting, you think. In a certain situation, you observe that the ... 1 You will need to use two free body diagrams, one for each mass, one FBD will include the force of A on B and the other one the force of B on A. Both mass have the same acceleration (do they), so you end up with a little system of three equations and three unknowns. 1 In general, the change in angular momentum resulting from a change in moment of inertia depends on how the change is implemented, and to some extent your perspective. In physics, you can think of global conservation laws as constraints that feed into your interpretation of a system. Consider the simple problem of determining the change in linear momentum ... 1 First, let's clarify the expression of$|\phi\rangle$. The kets$|+\rangle_z$and$|-\rangle_z$are eigenvectors of$\hat{S}_z$such that $$\hat{S}_z |+\rangle_z = +\frac{\hbar}{2}|+\rangle_z \\ \hat{S}_z |-\rangle_z = -\frac{\hbar}{2}|-\rangle_z$$ This means that in the$\{|+\rangle_z = ...

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The derivation of ideal gas equation from Hamilton's equations will take the same procedure as what you have seen in Wikipedia. Since you said you haven't understood the way in which the equation is derived I will give you a step by step explanation on it. So we have a system of perfect gas molecules. Of course they are non-interacting. We are going to ...

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