# Tag Info

8

Let $|n'\rangle$ be a basis of the Hilbert space, then $$\textrm{tr}\Big[|\alpha\rangle\langle\alpha|A\Big]=\sum_{n'}\langle n'|\alpha\rangle\langle\alpha|A|n'\rangle=\sum_{n'}\langle\alpha|A|n'\rangle\langle n'|\alpha\rangle = \langle\alpha|A\left(\sum_{n'}|n'\rangle \langle n'|\right)|\alpha\rangle=\langle\alpha|A|\alpha\rangle$$

3

My equation is $$\frac{x^2}{a^2} + \frac{x^2}{b^2} = 0$$ That's not the equation you want for a satellite. That equation describes an ellipse with its center at the origin. You want an ellipse with the origin at one of the foci: $$r = \frac{a(1-e^2)}{1+e\cos\theta}$$ where $r$ is the distance from the origin to a point on the ellipse. $a$ is the ...

3

I think that you're making this problem more complicated than it has to be in order to simply determine if the assembly will tip over or not. You don't really need the spatial distribution of the forces being exerted by the table or ground on the assembly. All you need to note is that if the pivot point is at x=D1 then the ground will exert whatever ...

2

If there is no torque, then $$\sum \tau = \mathbf{r_1}\times\mathbf{F_M}+\mathbf{r_2}\times\mathbf{F_m}=0$$ Therefore, $$\mathbf{r_1}\times M\mathbf{g}+\mathbf{r_2}\times m\mathbf{g}=0\tag{1}$$ where $\mathbf{r_i}$ denotes the position of the center of mass of the combined system relative to the force applied. If we give the box dimensions $h$ and $l$, the ...

2

Another way to see this is to observe that any state $|\psi⟩\in\mathcal H$ can be extended to an orthonormal basis of the Hilbert space, and in that basis the trace $\operatorname{Tr}\left(|\psi⟩⟨\psi|\hat A\right)$ is exactly $⟨\psi|\hat A|\psi⟩$. More explicitly, for any $|\psi⟩\in\mathcal H$ there exists a sequence ...

2

Perhaps the easiest way to see that there can't be a potential difference between $A$ & $B$ is a symmetry argument. You're tempted to say that $A$ is at a higher potential than $B$ so that current will flow from $A$ to $B$. But continuing along the loop, I find that current must also flow from $B$ to $A$, which would lead me to conclude that $B$ is at a ...

2

In my (ancient) copy of Hecht and Zajac (1980), the answer is found in figure 10.18. It shows that for slit spacing $a$ and slit width $b$, peaks in the diffraction pattern are spaced $\lambda/d$ while the first zero due to the finite width is at $\lambda/b$. In the figure, $a = 3b$ and the third peak is suppressed:

2

The EMF created by a changing magnetic field is not considered to arise from a potential. This can easily be seen because when there is an emf, a charge can move around in a complete circle and dissipate energy the whole way around, but a potential cannot drive a charge around in a circle, because potentials are conservative. The two pieces of the electric ...

2

A free body diagram on the $2m$ mass would have $2mg$ down and $T$ up. This would give a Newton's 2nd Law equation, assuming up to be the positive vertical direction, of $$T-2mg=2ma_{2v}$$. The $m$ mass free-body diagram would yield two downward forces, $T$ and $mg$ with a Newton's 2nd Law equation of $$-T-mg=ma_{1v},$$ assuming the tension magnitude in the ...

2

Just use snell's law, that is, $\mu \,\sin\theta$=constant, where $\mu$ denotes the refractive index and $\theta$ is the angle between the ray and the normal between a generic point and the point of incidence.The rest is math, you need to express $\sin\theta$ in terms of the slope at that point and solve the resulting differential equation.

2

Which is right? So let's first establish what is correct. Suppose we have a circle of radius $r$ and mass $m$ centered on a circle of radius $\bar R$ (not quite what you have defined; in your case $\bar R = R + r$). We'll say that the position of the $r$-circle on the $\bar R$-circle is angle $\phi$ and, from the angle $\phi$, we will measure an angle ...

2

Torque $\vec{\tau}(\vec{r})$ is an vector multiplication of radius-vector $\vec{r}$ on applied force vector $\vec{F}$, i.e. $\vec{\tau}=[\vec{r}\times\vec{F}]$ Here the radius-vector (or position vector) $\vec{r}$ is the vector from the point where the torque is defined to the point where the force is applied (see image). On the picture you shown ...

2

So let's start from the relations you gave and transform one of them from ket to bra. $$\left|i\right> = \mathcal{ CPT}\left | \bar{i}\right>$$ $$\left<f\right| = \left< \bar{f}\right| (\mathcal{ CPT})^{\dagger}$$ Using the CPT invariance condition, $\left(\mathcal{ CPT} \right)T \left(\mathcal{ CPT}\right)^{-1}= T^{\dagger}$, It is easy ...

2

It depends on what exactly you are asking. Suppose we take 64g of copper i.e. one mole of copper. Each copper atom contributes one conduction electron, so our chunk of copper contains $6.023 \times 10^{23}$ (Avagadro's number) conduction electrons with a total charge of 96488 coulombs. John's answer involves removing those electrons by a chemical reaction. ...

1

Yes it is true. The only DC magnets that use "no" power are superconducting magnets (like in MRI systems). Of course for those, there is significant power needed to keep the windings at superconducting temperature... and the cooling system will typically use several kW. "How much power does a junkyard magnet use" is not an easy thing to answer: but ...

1

Yes, in electrical circuits of only passive elements (Resistors, Capacitors, and Inductors) only the Resistors dissipate power (as heat). Active elements like transistors can also dissipate power, and if the currents in the circuit are changing with time, then power can be radiated away in electromagnetic waves. Therefore, the electrical resistance of the ...

1

If the first measurement yields the value $A_1$ with certainty, this means the initial state has collapsed into $u_1$ after the first observation. In particular one has, inverting the above back: $$|u_1\rangle =\frac{\sqrt{3}}{2}|v_1\rangle + \frac{1}{2}|v_2\rangle.$$ Now a measurement of the observable $B$ must be performed and then one more measurement ...

1

The point of natural units is to rescale your units so that $c = 1$ and $\hbar = 1$ and $k_B = 1$. This is technically a type error because the quantities on both sides have different dimension, but it means "in the dimensions that give this the appropriate size." So this means that you have a $\text{cm time}$ unit, for example, which is the time it takes ...

1

Let's do natural units the other way around. Suppose that we've always worked with natural units, we measure time and distances in the same units and then some crazy physicist comes along who puts in factors of c in equations, e.g. $$ds^2 = dt^2 - dx^2 - dy^2 - dz^2 \longrightarrow c^2 dt^2 - dx^2 - dy^2 - dz^2$$ He then defines a meter and a second such ...

1

"As accurate as possible" is a fuzzy concept. Given that you ask this question, I expect that a few simplifying assumptions are justified. For an object in a circular orbit of constant radius $R$, orbiting a perfectly spherical earth of constant density, the kinetic and potential energy can be calculated. Their relationship is beautifully simple, as derived ...

1

Your answer is perfectly fine. As you can see one can choose an abritrary phase $\exp(i\phi)$ for $c_n$ in the equation $$E_n + \frac{\hbar \omega}{2} = |c_n|^2$$ and it will still hold. This relates to the fact that you can always choose an arbitrary phase for the eigenfunctions $\psi_n$. All physical observables (e.g. $A_{nn} =\langle ... 1 Recall that states or wave-functions are only defined up to an overall phase, i.e.$\psi(x)$and$e^{i \alpha(x)} \psi(x)$are both wave-functions that describe the same state. The wave-function generically is a complex function of the form$\psi = f(x) e^{i h(x)}$where$f(x)$and$h(x)$are real functions. It is then often convenient to make a choice of ... 1 In the rules of quantum mechanics, every state$|\psi\rangle$is a "vector" which has a "dual", which is usually a complex conjugate$\langle \psi|$and every measurement in some state is described by an average$\langle A\rangle$and an operator$\hat A$which is its own conjugate transpose: together these say that in state$|\psi\rangle$the average ... 1 I'll risk moderatorial opprobium with a partial answer because you have come so close. You correctly use the SUVAT equation$v^2 = u^2 + 2as$to find that the velocity of the ball just before it strikes the ground is$v_i = -7$m/s (using the sign convention that upwards is positive). So far so good. Now you know the ball rises back up to a height of 1.8m, ... 1 Your expression for the acceleration due to the kinetic friction is incorrect. Remember, $$f_k = \mu_kn,$$ where$n$is the normal force. To find the normal force, you have to use what you know about the centripetal acceleration. Draw a free-body diagram, and label the weight of the car and the normal force, and then you know that ... 1 An accelerating object has a changing velocity. Obviously so since the object starts with zero velocity and the velocity increases with time according to the SUVAT equation: $$v = u + at$$ So your equation 1.1 is no use here. It calculates the average velocity. This could actually be used to calculate the acceleration, but the working is a bit involved ... 1 Both are right. The first approach gives the compression where the net force on the object is zero. The second approach gives the compression when the velocity of the object is zero. When the block falls on the spring, it oscillates between$x=\frac{2mg}{k}$and$x = 0$. Since the spring is ideal and the air resistance is negligible, this oscillation does ... 1 Your equation 1.1 can be used with constant velocity. Here you have to use the$2^{\text{nd}}$equation. ie$a = 2d/(t^2)$. So, the answer is$118.4 \, \text{cm}/s^2$. 1 The work-energy theorem is certainly the easiest way to do the problem, but you can also solve it by calculating the force. In any situation where you need to calculate the response of an object to a force you use Newton's second law. This tells us (after a minor rearrangement): $$\frac{d^2x}{dt^2} = \frac{F}{m} \tag{1}$$ In this case the force on the ... 1 Time of flight is determined only by the vertical component of velocity - it is the time interval between when the projectile was released ($y=y_0$) and when it reaches the ground ($y_t=0\$). As the collision is with a vertical wall it acts in the horizontal direction (assuming no friction during the short duration of the collision) and so has no effect on ...

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