New answers tagged

1

That is because the potential looks like this: Here the potential is plotted over a 1D cross section. The outer ring is adding potential to the innermost area but since the potential at the innermost shell is given we know that it extends to the inside as well. The inner shell is adding 5V.


1

You need to consider the coordinates you are using and the matter density, in the first example you have polar cylindrical and in the second they are polar spherical coordinates. Thus Dirac's Delta Function should be written, respectively as $$ \delta(\vec{r} - \vec{r}_o) = \frac{1}{2\pi r} \delta(r - r_o)\delta(z - z_o), $$since de symmetry over $\phi,$ and ...


1

This is my own interpretation of what went wrong. I wanted to integrate from the point a to b along the current $I_1$ (I'm assuming that's how you choose the boundaries, integrate along the current). So my limits should reflect this direction of integration. This should be done by starting at $a_z=a$ and end at $b_z=a/2$. So I should swap the boundaries.


-1

I assume that you are working with the standard cosmological model. In this model you have $$H(z)=\sqrt{\Omega_{m}(1+z)^3+\Omega_{r}(1+z)^4+\Omega_{\Lambda}}$$ And it turns that the biggest contribution in the present is given by $\Omega_{\Lambda}$, so the universe is $\Lambda$ dominated as in your question. In this model you also have the equation for the ...


-2

Your approach is wrong , you have to use thrust force which is equal to m(-dv/dt)


-1

Your torque equation is incorrect. The torque produced by F2 is clockwise and that by F1 is counterclockwise. One of them should be negative. Also torque is force times radius.


0

The force balance on the piston at any time during the compression is given by: $$P_gA+F-mg-kx-P_{atm}A=0$$ where x is the upward displacement of the spring from its unextended length. If we multiply this by the differential (upward) displacement of the piston during the process $dx=\frac{1}{A}dV$, we obtain:$$P_gdV+Fdx-mgdx-kdx-P_{atm}dV=0$$Integrating ...


0

You have taken the wrong $\overset{\rightarrow}{dl}$, it is supposed to be $$\overset{\rightarrow}{dl}=dx \hat{x} - dz \hat{z} $$ This is because the current is moving along the opposite direction to $z\hat{z}$


0

Your system is the spring and the gas. When the first law is written in this form, $\Delta U= Q-W$, then $\Delta U$ is the change in internal energy of the system which includes the spring ($U_{\rm final} -U_{\rm initial})$, $Q$ is the heat input to the system (positive if into the system and negative if out of the system) and $W$ is the work done by the ...


0

Both the energies must be the same. Note that the odd condition that you have written can be combined with the even condition as the following: $$\frac{\sqrt{2mE}}{\hbar}=\frac{n\pi}{2}$$ For odd $n$ this gives you the cosine solution and for even $n$ it reduces to the sine solution. This means that for odd $n$ the wavefunction is pure cosine and for even $n$...


0

The answer you obtain by forcing the sine to vanish when you subtract both equations is correct, and you indeed get the second expression for the energy that you obtained at the beginning. However, as you correctly say, the answer should be exaclty the same as the answer you obtained for the well within $(0,L)$, so this means that you are missing some ...


0

A plane wave $Ae^{ikx}$ is not a normalizable wave function, as attempting to calculate the integral of its absolute value squared gives: $$ |A|^2\int_{-\infty}^{\infty}1dx, $$ and the integral is not finite. This in turn implies that a free particle of a definite momentum, which is represented by a plane wave, is not a valid solution to the system. What ...


2

Hint: You have to use the Fourier transform. You can use the definition: $$ \mathcal{F}\{f(x),k\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ikx}\,dx $$ So, for example, the first integral is: \begin{align} \int_{-\infty}^0 e^{i(k-k')x}\,dx &= \int_{-\infty}^\infty \theta(-x)e^{i(k-k')x}\,dx \\ &= \int_{-\infty}^\infty \theta(x)e^{i(k'-k)x}\,...


0

Your charged particle is subject to an electric force, q(Eo)r, directed radially outward from the origin where E is zero. As it gains speed, it will experience a magnetic force, qvB, at a right angle to its velocity. The particle will follow a curved path until it reaches a point where the radial velocity is zero. From that point, the curve back down will ...


1

Just a note on the third integral. $$\frac{1}{4\pi^2}\int_m^\infty dE\sqrt{E^2-m^2}e^{-iEt}.$$ If you don't want to explicitly do the calculation, as in Alex's answer, there is a plausibility argument. In the limit where $t$ is very large, the exponential oscillates very rapidly. The oscillations will cancel each other except in the region where $\sqrt{E^2-m^...


1

you just need to work it out using: $$T'^{\sigma\lambda}=\Lambda^{\sigma}_{\mu}\Lambda^{\lambda}_{\nu}T^{\mu\nu}$$ for a boost in the $z$-direction with only $T^{00}$ and $T^{zz}$: $${\bf T'}=\gamma^2\left[\begin{array}{cc}1 &-\beta\\-\beta &1\end{array}\right] \left[\begin{array}{cc}T^{00} &0\\0 &T^{zz}\end{array}\right]\left[\begin{array}{...


0

Hint : The (Lorentz) force acting on the particle is a linear function of the position $\:\mathbf r\:$ and the velocity $\:\mathbf v\boldsymbol{=}\dfrac{\mathrm d\mathbf r}{\mathrm dt}\boldsymbol{=}\mathbf{\dot{r}}\:$ of the particle. Equating this force to $\:m\mathbf a\boldsymbol{=}m\dfrac{\mathrm d\mathbf v}{\mathrm dt}\boldsymbol{=}m\mathbf{\dot{v}}\...


0

Look at the forces: •) $m_2$ has no acceleration along the $y-$direction, so $\vec F_{net}$ along the $y-$direction is $0$. Therefore, the force of tension in the string is: $$T = g m_2$$ •) Since $m_2$ is not accelerating along $y-$direction, the string is not rising or falling with acceleration; so the horizontal length of the string is not changing with ...


0

You may be looking for this $$v^2 = \dot r^2 + r^2 \dot \phi^2 ~.$$


0

Use the vector cross product to find the velocity vector $\vec{v}$ that results from the rotational velocity vector $\vec{\omega}$ at a location $\vec{r}$ relative to the rotation axis. $$ \vec{v} = \vec{\omega} \times \vec{r} $$ On a plane the above becomes $$ \pmatrix{v_x \\ v_y} = \pmatrix{-\omega\, y \\ \omega\, x} $$


0

Taking $M$ and $m_{2}$ as system, then if a force is applied surely both of this masses would accelerate with same acceleration Now the question states there shouldn't be any acceleration of $m_{2}$ in $y-$ axis then the string of pulley should not accelerate as well , which means there shouldn't be any acceleration of $m_{1}$ with respect to system of $M$ ...


0

Find the torque about the articulation, and set it to zero. I think your forces are as follows and therefore the torque about the pin is $$ \tau = -(L)\,N_1 + (2L) N_2 = 0 $$ Solve the above for $\omega$.


4

If you work out the problem, you will find that the acceleration on the end is just tangent to the walls. This is true for a solid body, or for two masses attached to a rigid link between them. $$ \begin{aligned} \vec{a}_{\rm K} & = \pmatrix{ 0 \\ -L \omega^2 \cos \varphi - L \dot \omega \sin \varphi } \\ \vec{a}_{\rm H} & = \pmatrix{L \dot \omega \...


-1

The rest-mass of the resultant particle is the same as the rest-mass of the original system. The momentum of the resultant particle is the same as the momentum of the original system. The velocity of the resultant particle can be calculated from its momentum and rest-mass. This may be useful for finding the rest-mass of the original system, and calculating ...


0

It seems to me that $F$ will apply a force to the string through the pulley and thus could act on $m_1$.


1

You are correct that the resistance will increase. If you take the resistance of one element to be $R$, then in parallel in the effective resistance is $\frac{1}{2}R$, and ins series it is $2R$, so quantitatively, it increases by a factor of four. Now, does this increased resistance mean that more power will be dissipated? Well this depends on your type of ...


6

This is a good question, and highlights a common misconception in Newtonian mechanics. For concreteness, let's work in two dimensions. It's true that at every moment, you can write the velocity as a rotation about an instantaneous point $\mathbf{r}_0$ of rotation, which means that the velocity $\mathbf{v}$ of any point in the body satisfies $$\mathbf{v} = \...


1

I don't remember the details of the Kronig Penney model. However, what the author says makes sense to me, I will explain you why. There are usually two kind of approximations which one use in band theory. One is assuming parabolic dispersion $$ E(k)=\frac{\hbar^2k^2}{2m_e^*} + E_0 $$ Neglecting the offset energy $E_0$, this is equivalent to the dispersion of ...


4

Isospin. It looks like a homework problem, with a summary answer in Okun's book, p 63. You need quark diagrams like a hole in the head. This strangeness-changing decay violates isospin by 1/2, so assuming the $\Delta I=1/2$ piece of the hamiltonian dominates, and, since Λ is an isosinglet (so irrelevant to the isospin amps), you just consider the addition ...


0

The field they are asking for is, B, at point, P. The formula you are starting with is reasonably accurate if the length of the solenoid is large compared with the diameter. In this case, start by finding the B field at a point on the axis of symmetry of a ring of current, (Ni/L)(dL). Then add the contributions (at point P) from all such rings by ...


0

It's purely trigonometric. A pair of sound wave interference can be modeled as: $$ \cos(\omega_1\,t)+\cos(\omega_2\,t) = 2\,\cos\left(\frac{\omega_1+\omega_2}{2}\,t\right)\cos\left(\frac{\omega_1-\omega_2}{2}\,t\right) $$ In right side of equation first frequency term is modulation angular frequency and second term is beat frequency.


2

It doesn't work for a cylindrical capacitor. The capacitance per unit length for the cylindrical capacitor is: $$ C = \frac{2\pi\epsilon}{\ln(b/a)} $$ The way we work out the capacitance is to put a charge $Q$ on the capacitor and calculate the electric field between the plates. Then we integrate the field to get the potential difference $V$ and the ...


1

The beat frequency, mathematically, is indeed $\frac{f - f_0}{2}$, as can easily be shown using the factor formula in trigonometry. However, the perceived beat frequency is twice of that, which is $f - f_0$. This is because the beat frequency modulates the amplitude of the sound wave. In one full cycle of $2\pi$, the amplitude goes both positive and negative,...


2

In a spherical capacitor, you have two conductive concentric spherical shells. Since a spherically distributed charge can be modeled as a point charge (and you can prove this by using Gauss's law with a spherical gaussian surface assuming spherical symmetry) the potential between the two shells is: $$\Delta V = k_eQ[\frac{1}{R_2} - \frac{1}{R_1} ]$$ Since ...


1

your system has one generalized coordinate $\varphi$ thus: $$x=L\sin \left( \varphi \right)$$ $$y=L\cos \left( \varphi \right) $$ with: $$T=\frac{m}{2}(\dot{x}^2+\dot{y}^2)$$ and $$U=m\,g\,y$$ you get: $$\ddot{\varphi}=\frac{g}{L}\,\sin(\varphi)\tag 1$$ the velocity vector is : $$\vec{v}= \begin{bmatrix} \dot{ \varphi}\,y \\ -\dot{\varphi}\,x \\ \end{...


0

You are confusing distance with coordinates. The correct equation to use would be $y$. Why? Because the variable $y$ itself can be either positive or negative, and this sign is automatically built into the definition of $y$. It would be extraneous (and incorrect) to add an extra negative sign in front of it. The same thing applies if you choose to orient ...


0

Given enough string at the top, so that neither spring hits the pulley, then they do act as one. Add answer A.


0

I found the solution in--- Siegfried Flugge's Springer book "Practical Quantum Mechanics",2nd reprint(1994),page-178 , in addition to solutions of problems involvin a bunch of other similar potentials,which are treated in relatively less books.


3

I'd like to add to what 'PM 2Ring' wrote. The observer will measure the rocket to have a constant length no matter where it is in the observer's frame of reference (assuming it is moving at a constant velocity -- in which case it will be length contracted). However, the observer will observe the rocket to be longer when it moves towards him and shorter ...


2

The difference between measurement and observation is crucial in relativity. When we observe the rocket, the finite speed of light affects our observation. In general, light from the head and the tail of the rocket will take a different amount of time to travel to the observer. When we measure the rocket, we compensate for time delays caused by the finite ...


0

The straight forward way it to round the error to the most significant digit, and keep the measured value to the accuracy of the error, thus the value is $0.0002 \pm 0.0004$. See Significant figures


2

We take the Wigner function $$W(\alpha)=\frac{1}{\pi^2}\int \text{e}^{\alpha \beta^*-\alpha^*\beta}\text{Tr}\left(\hat \rho \hat D(\beta) \right) \text{d}^2\beta,$$ and write the displacement operator as $\hat D(\beta)=e^{\beta\hat a^\dagger-\beta^*\hat a}=e^{-\beta^*\hat a}e^{\beta\hat a^\dagger}e^{\frac 1 2|\beta|^2}$ using the BCH formula such that $$...


0

It you measure the period directly, then you do not need formula - the error is determined by your measurement device + the reaction time, if you need to switch it on/off (like a chronometer). On the other hand, if you measure the length of the pendulum, then it is necessary to propagate the error using the equation: $$\Delta T = \left|\frac{\partial T(l,g)...


1

The details of the acceleration when the ball hits the walls depends on how squishy the ball and walls are. If the ball and the walls are infinitely rigid then the bounce takes zero time and the acceleration is infinite. The trouble is that as you say you get the change in velocity is $\Delta v = \infty \times 0$ and that is undefined. In practise the ball ...


0

Too many questions for one post. I'll answer the first and the last since the second is a little bit involved and you can find the solution looking to the book Lie Algebras in particle physics by Georgi. Take the $SU(N)$ Lie algebra. Being a complex Lie group you'll need at least $2N^2$ parameters, $N^2$ for the real part and $N^2$ for the imaginary. If ...


-1

I think your metric is wrong. The $dtd{\phi}$ component in particular. I just tried it with the metric from my notes and it's quite simple. Just find $\frac{d\vec{R}}{d\lambda}$ and use the metric to find its magnitude


1

I think this phrasing is a bit misleading if isolated. If this was true as written for any state $\psi$, then the tensor product of two qubits would have dimension $2$ instead of $4$. But if you look at the next sentence it is clear what the author actually means Here $|ii〉 ≡ |i〉_A⊗|i〉_B$, both local bases $\{|i〉\}_{A,B}$ depend on the state $|\Psi\rangle$...


1

The answer provided by roshoka is correct. I only want to briefly discuss why the time dependence can be "ignored" in this case. As the question asks for the probability of obtaining a particular energy eigenvalue at a later time, then you must first understand what the wave function looks like at a later time t. For a system in which the potential is time ...


0

See, you can think of it in two ways the stopping voltage is always opposite to the electrons velocity hence it should also be given a negative sign .....second one is here the electrons works against an external field and loses its energy so work should be positive


0

A block sliding without friction would behave itself in exactly the same way as a pendulum, the normal force being provided by the bowl rather than a string. Its period can be calculated from energy conservation, without linearization: $$\frac{m}{2}\left(\frac{dx}{dt}\right)^2+V(x)=E,$$ so that $$\frac{dt}{dx}=\sqrt{\frac{m}{2}}\frac{1}{\sqrt{E-V(x)}},\\ \...


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