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8

I think the colloquial term for that type of plot is "spaghetti diagram" because you have a bunch of lines running across it. It's really the mass fraction as a function of interior mass. From our stellar structure equations, we have that $$ \frac{dm}{dr}=4\pi r^2\rho, $$ which is derived from the mass-continuity equation, so you can relate the radius, $r$ ...


3

In order to solve this question, we need to determine the distance from the eyes of the observer to the horizon, as this will be the relevant distance the car travels within those 2 seconds. Let's define a few variables: $d =$ distance between observer and horizon $v =$ velocity of the car $t =$ time for car to travel from horizon to observer $h =$ distance ...


3

The phase space has points $x^{i}\in V_{2}$. The standard inner product is $x^{i}\delta_{ij}x^{j}$ with $\delta_{ij}$ an invariant tensor under the action of the isometry group SO(2). If $R^{i}_{j}$ are the group matrices, \begin{equation} \delta_{ij}=[R^{-T}]_{i}^{\ k}[R^{-T}]_{j}^{\ l}\delta_{kl} \ . \end{equation} Set up a one-parameter subroup ...


2

As Nemis L. pointed out, the expectation value $\langle H\rangle$ is constant, because of Ehrenfest's theorem: $$\frac{d}{dt} \langle H \rangle = \frac{1}{i \hbar} \langle [ H,H ] \rangle = 0.$$ The other way of seeing this is that the state can be written as a superposition of orthogonal energy eigenstates. Obligatory image: Goldberg, Schey, and ...


2

The thing you don't appear to have is a direct means to measure your altitude. But you have enough tools to do the job. First, weight your sandbag "on the ground". Let's say the weight is $W$. Now climb to a very high altitude (you said you could go to any altitude... go high). Find an altitude at which the weight of the sandbag is noticeably reduced, and ...


2

From conservation of angular momentum we can derive angular velocity as a function of height: $$\omega(h) = \omega_0 \left(\frac{R + h_0}{R+h}\right)^2$$ where $\omega_o$ is the angular velocity of the earth, $R$ is the radius of the earth, $h$ is the current height and $h_0$ is the initial height. The horizontal velocity (in the frame of reference of the ...


2

Just a back-of-the-envelope calculation.. $a_{\rm centrifugal} = R \Omega^2 = 6.4 \times 10^6 \times \left( 2\pi/86400 \right)^2 \sim 0.03 \;\rm{m}\, \rm{s^{-2}}$ is about 0.3% of the gravity of Earth. Of course, it would depend on the latitude you live at.


2

The height $h$ is probably the vertical displacement pointing downwards. Therefore: $$ h = \left(-\mathbf{\hat j}\right)\cdot\mathbf s = -|\mathbf{\hat j}||\mathbf s|\cos\alpha = -s\cos\alpha $$ Now we can derive: $$ \frac{dh}{ds} = -\frac{d}{ds}\left(s\cos\alpha\right) = -\cos\alpha \quad\Longrightarrow\quad \frac{dh}{ds} = -\cos\alpha $$ Therefore, ...


2

The frictional force would be in the forward direction if the object is a rotating object. Let us say the object we are talking about is a wheel. At a point of time, the frictional force between the wheel and the surface will be only at the point of contact of the wheel with the surface i.e the bottom most point of the wheel, say A. With respect to the ...


2

Actually, there is a reversible process that will allow you to end up with a larger temperature, and this does not violate the second law. To see this, you can couple your system to Carnot engine, which will extract and "store" work until both the reservoir and the case are at the same temperature. Now use the stored energy to heat the case. Actually your ...


2

As you have already worked out, $(xp)^\dagger = p^\dagger x^\dagger = px \neq xp$, so that the product of two self-adjoint operators is not needfully self-adjoint, and indeed from this equation we see that the product is self adjoint if and only if $p$ and $x$ commute. To complete your thinking, how indeed do we calculate the product of the two ...


2

You do not need to spend any energy to keep an object stationary in a gravitational field, but you do need a force that is opposite to gravity, so that the object can be at rest. For instance, you can put the object on a table, or "hang" it, and the object will stay there without any energy requirement. So the answer is zero, you do not need to do any ...


1

Partial answers: For skew-symmetric $A$, then a possible action is $$\tag{1}S[q]~=~\int \!dt~ L, \qquad L~=~\frac{1}{2}\dot{q}^T\dot{q}+\frac{1}{2}q^TA\dot{q}$$ If one is allowed to use Lagrange multipliers $\lambda_i$, then a possible action is $$\tag{2}S[q,\lambda]~=~\int \!dt~ L, \qquad ...


1

Hints: Given the Maxwell equation: $$ \nabla\times\mathbf h=\frac{4\pi}{c}\mathbf J $$ Take the curl of both sides, what do you get? How can (1.4) be used here? Vector calculus tells you $$ \nabla\times\nabla\times\mathbf a=\nabla\left(\nabla\cdot\mathbf a\right)-\nabla^2\mathbf a $$ What assumptions must you make to get the London equation?


1

Since you want the expectation value on a state with definite energy, say $|n\rangle$, then only $aa^\dagger$ and $a^\dagger a$ will give you some meaningful contributions (of course you can discard $a^\dagger a$ if $n=0$). So since it is an excited state you're interested in I will assume that $n\geq0$. Your expectation value is then $$\langle ...


1

I think your solution is correct. The width $\Delta x(T)$ of the wave packet at any time $T$ can be obtained from $$\Delta x(T)=\Delta x(T=0)+\Delta v_g T\text{ ,}$$ where $\Delta v_g$ is the uncertainty in the group velocity $v_g$, which causes the widening. You have implicitly used $$v_g=\frac{p}{m_e}\Longrightarrow \Delta v_g=\frac{\Delta ...


1

Your problem here is using the wrong version of Gauss's law, or at least one that is not applicable in this situation. $$ \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$$ describes Gauss's law where $Q$ is the entire charge enclosed within the surface. If we introduce a dielectric material, such that $\vec{D} = \epsilon_r \epsilon_0 \vec{E}$ and a ...


1

This is a very interesting problem. Probably the first thing that is going to happen is that the ball will roll to the right, so that it will be in contact with both the 'floor' and the right-hand wall. Presumably, what will cause the ball to start lifting will be a combination of two things: The ball may start spinning and try to lift on the right-hand ...


1

I see that you already have the solution with you. So, I will try to help you with the tension in the chain zero part. The tension here will be zero even if the chain is acceelerating because the acceleration is caused by the gravitational force which is pulling it down and this force is acting on every single atom and molecule of the chain. In other words, ...


1

In principle you could get the displacement from accelerometer measurements, if you also had an estimate of the orientation of the phone at all times. You would need to use the phone orientation to convert each instantaneous acceleration measurement into the same coordinate frame, and then subtract off a constant component representing gravity, then ...


1

In order to perform the (possibly singular) Legendre transformation, it is necessary to have information about pertinent rank conditions of the structure constants $\rho$, $\omega$, $\Sigma_{ij,k\ell}$, $\epsilon_{ij}$ and $P_{ijk}$. In this answer, we will sketch how the (possibly singular) Legendre transformation is performed in principle: We will use ...


1

Is it really a mistake? Let me see using QM. For calculating $-i \hbar \int dx \ x \ \psi^* \frac {∂ \psi}{∂x},$ we can integrate by parts, use the fact that the solutions vanish at $+\infty$ and $-\infty$, and obtain $i \hbar \int dx \frac {∂ (x\ \psi ^*)}{∂x}\psi$ = $i \hbar \int dx \ \psi ^* \psi$ + $i \hbar \int dx \ x \frac {∂\psi ^*}{∂x}\psi ...


1

Let's break up the process into two steps, which necessarily must happen in sequence--the first step is one in which the reservoir heats the body up from a temperature of 273 K to its own temperature of 363 K, and the second step is one in which the reservoir heats the body from 363 K to 373 K. The first step involves a positive entropy change, so it can ...


1

From the potential application of the question, I assume that question is to provide design parameters for a water slide. Because the depth of the water will be reduced from acceleration, the real question is how to add resistance to the canal. Resistance can be provided through the disruption through the use of twists and turns. I would generate ...


1

Here is how I would do it: From conservation of energy you find $$mgh = \frac12 m v^2 + \frac12 I \omega^2$$ And the relationship between $v$ and $\omega$: $v=\omega R$. The result then follows.


1

I see your update but I don't understand the equality, more exactly the right hand side. I get there $2(x_1 −x_0) +(x_3 −x_2 )$. This equality is also directly obvious. From it one gets indeed your relation between accelerations $2a_2 +a_1 −a_{P_{smallest}} =0$


1

The Fresnel transmission coefficients at the Brewster angle between two media of $n=1$ and $n=1.5$ are not 0.86 and 1. I haven't checked the Maths, but looking at this: https://www.geogebratube.org/student/m325541 I would say the transmission coefficients are $t_s=0.6$ and $t_p=0.66$. The reflection coefficients $r_s=-0.4$ and $r_p=0$. The transmission ...



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