New answers tagged

0

I don't know where you have trouble, but it's straightforward if you just assume $R(x,t)$ and $S(x,t)$ in the form given for $\psi$ and then expand out the operations, where: $$\begin{align} \psi &\equiv R(x,t)e^{i S(x,t)\hbar} \\ \partial_t \psi &\rightarrow \Big(\frac{\partial_t R}{R} + \frac{i \partial_t S}{\hbar} \Big)\psi \\ \nabla \psi &\...


0

I think some errors crept into your calculation. And at any rate - your confusion is around $a_0$ - which is the quantity you need to solve for (the initial acceleration needed to reach the ball at just the right moment). The player has to cover a distance $d$ in the time that the ball covers a distance $\sqrt{2} d$. The velocity of the ball is $v_b$, so ...


7

The gravitational potential field can be found by a full volumetric integration from the overall volume, or planet, or whatever: $$ \Phi(\mathbf r) = -G\int_V\frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|}dV $$ This just comes from having a distribution for $M$ in the potential formula: $$ \Phi = \frac{GM}{r} $$ Also, its easy to see that, if you are ...


2

$x_{ps}=PS-l$ and $x_{qs}=QS-l$ not the way you had it. The stretched length of the strings is $l$, so if the stretched length is $PS$ then the extension has to be $PS-l$.


4

Use the Baker-Campbell-Haussdorf formula to commute the displacement operators, $$ e^{\alpha a^\dagger} e^{-\alpha^* a} = e^{|\alpha|^2} e^{-\alpha^* a} e^{\alpha a^\dagger} $$ and rearrange the matrix element as $$ \langle n| e^{\alpha a^\dagger} e^{-\alpha^* a} |m \rangle = \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \langle 0 | a^n e^{-\alpha^* a} e^{\alpha a^\...


2

Assuming you have it set up that $\langle n\mid m\rangle=\delta_{nm}$ Then taking the sum: $$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{n!}\sqrt{m!}}{\sqrt{(n-l)!}\sqrt{(m-k)!}}\langle n-l|m-k\rangle$$ This simplifies to: $$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{n!}\sqrt{m!}}{...


1

It is at least possible to simplify your second expression. Note that (if I've not miscalculated) $$\langle m|e^{{\alpha}a^{\dagger}}e^{-\alpha^*{a}}|n\rangle = \left( \langle n|e^{{-\alpha}a^{\dagger}}e^{\alpha^*{a}}|m\rangle \right)^{*}$$ so we can without loss of generality assume $n > m$. Now, in your second sum, we can use that $$\langle n-l|m-k\...


0

It seems to depend mainly on the spin. See the comsol.com simulation at https://www.comsol.com/blogs/physics-behind-baseball-pitches/ For the slider, in essence a lateral motion, Just spin it laterally, right to left if you are righty, and it'll curve to the left. The axis of rotation would be vertical, if righty the direction of the angular momentum vector ...


3

The force the fluid does on the bottom piece does not depend on the height of the water column of the reservoir. It does depend on the height $h$ of the water column in the plate. This can be easily seen by the fact the water is static so the pressure at any horizontal plane is the same. The pressures in $a$ and $b$ are the same. The force of the water on ...


1

I could give you some intermediate steps, as I suspect the purpose of the exercise is to familiarize you with Poisson Brackets. Absorb mw into q and Q to spare yourself complication. You wish to compute $\{ Q,P\}$ and check whether it is =1. the PBs are linear differential operators on each of their arguments, so they obey Leibniz's chain rule. So you ...


16

Rigorously speaking, the probability to find the electron at a distance exactly equal to $r$ from the nucleus is $0$. On the other hand, we can define the probability to find the electron in a volume $d \mathbf{r}$ as $$P(d \mathbf r) =| \Psi(\mathbf r)|^2 d \mathbf r = |\Psi(r,\theta,\phi)|^2 r^2 \sin \theta \ dr \ d\theta \ d \phi$$ where I have ...


0

Take it one step at a time Time to reach $v_1$ under acceleration $a_1$ $$t_1 = \frac{v_1}{a_1}$$ Distance traveled to reach $v_1$ $$x_1 = \frac{1}{2} a_1 t_1^2$$ Total distance traveled during cruise (time $t_1$ to $t_2$) $$ x_2 = x_1 + v_1 (t_2-t_1) $$ Total time to stop from $v_1$ under deceleration $a_3$ $$ t_3 = t_2 + \frac{v_1}{a_3}$$ Total distance ...


1

So the vacuum works by a pressure differential (Difference) $\Delta P \vec{A}_{lid}= \vec{F}_{vacuum}$ where $\Delta P$ is the difference between the atmospheric pressure and the pressure of the air inside the container (which probably wont be zero, depending on your Vacuum pump, which i know nothing about) $\vec{F}_{vacuum}$ is the force holding the lid ...


0

Neglecting the effects of rotation on the 'apparent' strength of gravity, the gravitational field strength is similar to the electric field strength around a uniformly charged circular disk (or cylinder). This is because gravitational and electrostatic forces both obey $1\over r^2$ laws. The field strength along the axis of the disk is quite easy to derive (...


0

You are right in all you have written. But a more fruitful sentence could be: Friction always tries to prevent sliding. (Kinetic friction) If two boxes are sliding over one another, the friction on the top block will pull in the lower block's direction, and friction on the lower block will pull in the top block's direction. Friction tries to keep them ...


6

First you should understand why gravity does not change as you move around on Earth's surface. The easiest way to explain this is using symmetry. The Earth is roughly a sphere. On a sphere there are no special points so gravity must act the same everywhere. A flat planet can actually be made to have constant gravity. If you had a flat planet that extended ...


1

Thanks to Ján Lalinský for getting me to double check my faulty memory. My assumption of $\langle\chi_n|[H_0,z]|\chi_m\rangle = 0$ was wrong. Here is the solution after double checking my work. $$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\chi_n|H_0\,\,z|\chi_m\rangle-\langle\chi_n|z\,\,H_0|\chi_m\rangle$$ $$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\...


0

You are orbiting at the surface of the earth, so $r$ is fixed. Since $r$ is fixed you are not free to change $a$; it is fixed by Newton's Law of gravitation $$ a = \frac{F}{m} = G\frac{m_\mathrm{earth}}{r_\mathrm{earth}^2} = 9.8\;\;\mathrm{m/s}^2$$ Since $a$ is fixed and $r$ is fixed, $v$ is determined.


1

Assuming no air resistance, the only force acting on the satellite is the force of gravity, $\bf{F_g}$. The stated circular orbit also means that centripetal force, $\bf{F_c}$ is involved, which causes the satellite to follow the circular path. Due to this, the centripetal acceleration can be directly equated to g, which allows you to immediately solve for ...


1

Volume is not a meaningful measure of quantity, for the reason you hint at in your question. You can say how many moles (or grams) of water you drank - more useful if you want to know about the impact on your body chemistry. This is related to my answer about scales measuring in grams rather than Newtons. Can you see how?


0

Your mass density must have units kg/m${}^3$. The simplest way to achieve this is to write the density $$\rho({\bf r}) = \frac{M}{V}\left(\frac{r}{r_0}\right)^2$$ where $r_0$ is a constant having units of distance, and $M$ is the total mass, and $V$ is the volume of the cylinder. We choose $r_0$ by integrating to get the total mass $$ M = \int \rho({\bf r}...


0

As an example of how silly this is let's say you like to eat hot food (at least, hot enough not to die of food poisoning assuming you eat meat, and you cook it with a microwave which consumes 1,000 watts, which you run for ten minutes a day (three meals, three minutes per meal). I'm assuming that a microwave is the most efficient practical way of cooking ...


0

Some idea of how wildly ambitious these claims are can be gleaned from a BBC TV programme that was broadcast about six years ago. In this show, they put a family of four in a house for a day and powered it through pedal power; they needed scores of cyclists. The show is no longer available directly from the BBC, but this clip gives you some idea. ...


1

For the horizontal portion, there is no acceleration, initial velocity is 12 m/s and final, I have no idea. See Newton's first law an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Your rock moving at 12 m/s horizontally has momentum in that direction. That momentum is ...


0

When ice is heated from 0 to 4 degrees C, it actually contracts. The water molecules get closer together and the water occupies less volume. However, above 4 degrees C water expands(i.e. in your case). This is explained by Charles's law but any way this change in volume is not very drastic. This calculator might be helpful


4

$1$ litre of water will remain almost $1$ litre as long as it is in the liquid state, no matter what the temperature is. The following formula gives you an order of magnitude estimate of the expansion: $$\Delta V=V_0\ \Delta T \ \beta$$ where $\beta$ is the coefficient of thermal expansion and $V_0$ is the initial volume. For water, $\beta \approx 10^{-...


1

What you want is $\rho(\textbf{r})=Cr^2$, where $C$ is a constant. Find $C$ by integrating this function over cylinder volume, and equating the result to total mass $m$.


0

As you know that there are two geometries of x ray diffraction namely Bragg geometry and Laue geometry. In Bragg geometry the crystal is in reflective mode. I suppose that crystal is in that mode only. Since it is not mentioned anywhere I assume that it is a single crystal i.e. only one plane is reflecting. As you know that Bragg diffraction condition yields ...


1

Write $$ U = \frac{1}{2}(I_1 + Z_1) \otimes U'_{00} + \frac{1}{2}(I_1 - Z_1) \otimes U'_{11} + (X_1 + iY_1) \otimes U'_{01} + (X_1 - iY_1) \otimes U'_{10} = \left(\begin{array}{cc} U'_{00} &U'_{01} \\U'_{10}& U'_{11}\end{array}\right) $$ where $U'_{ij} \in G_n$. In terms of the latter, $$ U' |\Psi\rangle = \sqrt{2} \;\langle 0 | U | 0\otimes \Psi \...


0

The answers are correct, the justification has some major problems. You have a ramp, so the slope is not 0. That means that the normal force (which is perpendicular to the ramp) and the gravity (vertical) are not collinear, so they cannot cancel each other. In a reference frame with an axis parallel to the ramp, the gravity has a component perpendicular to ...


2

The radius of the wavepacket is unrelated to the radius of the particle. For instance you could have a particle of zero size (as the electron is suspected) but whose probability of being found somewhere has a certain size. For instance a wave packet: The wave packet only gives you the probability of finding a particle at that place (that is, more likely ...


8

A human can produce about 100 W power continuously. That is equivalent to a couple of lightbulbs. So, while pedalling continuously you can keep one room reasonably lit. In one hour, that means you can generate a total energy of 0.1 kWh. A top athlete will be able to do better, so let's say it is possible to generate 0.2 kWh in 1 hour. That is still nowhere ...


0

Assume all masses are the same and we have a perfectly symmetrical collision. It seems a logical assumption doesn't it? Then, try to understand the following conservation of energy equation. Use it to solve for $v_{af}$: $$\frac{1}{2}mv_{ai}^2 = \frac{1}{2}mv_{af}^2 + \frac{1}{2}mv_{b}^2 +\frac{1}{2}mv_{c}^2$$ Once you've done that, go back and solve the ...


0

You will agree that the velocity of A depends on the elasticity of the collision. Now, momentum is conserved regardless of the elasticity of the collision. Therefore, you cannot expect momentum conservation alone to determine the outgoing velocity of A. In detail, the total outgoing momentum depends on two unknowns, i.e. A's velocity and B and C's common ...


0

You are right in both respects : This is a very poor question, expecting you to guess what might have gone wrong. There could be several explanations : bulbs with different resistances, or different ways of connecting them. The question is wrong. Your logic is correct in both cases. So either (a) there is some other valid reason in the question-setter'...


1

The above isn't drawn correctly. And it is far too complicated in my opinion. I think I was able to solve this using high school physics: This bugged me a long time. I am surprised no one ever posted an answer on the internet. The length of his rope was 60.8 meters, assuming a drop above the target building of about 12 meters. The difference in height ...


0

From a point on the sphere exactly one half of the space is visible. This one is created be the plane touches the sphere in this given point. This is the average horizon in practice. So we can use Pythagoras' theorem to calculate the distance to an object at a certain height, when it rises above the horizon: $$(\textrm{Earths Radius})^2 + (\textrm{ground ...


3

No you cannot assume that. The initial rotation is about the major axis, and it will continue to be so (in the absence of torque, and since you were already rotating about the major axis). Instead, since $\omega_2=\omega_3=0$, your equations for the evolution of the angular momentum don't require the moments of inertia to be the same.


1

There is extensive (100s of articles) lit. on the web regarding atmospheric drag and drag in general. A spherical bob and a very thin rod makes the problem very simple. One may use this graph: http://eis.bris.ac.uk/~memag/Teaching/Multi/dragcurve.pdf , and then calculated the Reynolds number to find the drag force using the formula given by the first ...


0

Your first expression should have a additional factor in the integrand called a density of states (DOS) and is usually denoted $\rho(E)$, i.e., $$U(t)=\sum\int\rho(E)dE|E,\alpha\rangle\langle E,\alpha|e^{-iEt/\hbar}.$$ The density of states is the number of states in the energy range $(E,E+dE)$. Look at the parabolic dispersion section of the density of ...


3

All you need to do is calculate the perigee distance $r_p$ that is the distance of closest approach. Then if $r_p < R_A$ your satellite will crash and burn. Once again we start from the vis-viva equation: $$ v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right) \tag{1} $$ The parameter $a$ is the semi-major axis of the ellipse, and it is related to the ...


2

As drawn the disc's angular speed $\omega$ is too fast as related to the velocity of the centre of mass of the disc $v$ for the no slipping condition ($v = R \omega$, with $R$ the radius of the disc) to be satisfied. You can think if the frictional force as trying to accelerate the centre of mass whilst at the same time the frictional force applies a torque ...


0

The problem was imperfectly posed, which leaves me trying to answer multiple issues; I'll try to answer all, but the order of answering is completely arbitrary. An ideal chain has inelastic links, but this would not affect a stretched coiled chain. Nevertheless, a stretched coiled chain would not give the answer you get from simple energy conservation ...


1

Lagranges equations are: ${\partial L \over \partial q_i} = {d \over dt }{ \partial L \over \partial q_i'}$ where $q_i'={dq_i \over dt}$ You can find constants of motion using lagranges equations and Hamiltons equations. You already know that the Hamiltonian is conserved-time is not explicit. (The energy is not always equal to the Hamiltonian) You can ...


8

If you're trying to simulate a 2D solution of the Laplace equation (which is the only unambiguous reading of your post as currently stated; if that's not what you're doing then you should clarify your question with exactly what it is you're doing and how), then your code is wrong. The reason is that your results don't obey the maximum principle: a harmonic ...


0

You have only counted the energy stored in one spring. There are two springs in the diagram. The springs have elastic potential energy initially as well as finally. Look at the formula you have used : you have calculated the increase in elastic energy stored in each spring: $\frac12 k (x_2^2-x_1^2)$. There was elastic energy stored in the springs before ...


0

I agree with @Matt S's comment that you didn't state the problem because we can't see what is supposed to be happening in your diagram. Anyway, for every reasonable assumption I can make about what happens, the time of ascent is different to the time of descent. I assume your diagram shows a projectile in a friction-less parabolic trajectory that ends in an ...


0

You need use the hint twice. The last equation in your statement should be an inequality. In order to obtain this inequality you need use the hint again.


1

Here is how to answer this question (rather than the specific answer, which I don't think you're after). Firstly they tell you that the two bits of the path are the same length ('the first half of the path') and are straight. Call the length $l$. They also give you the angles, $\alpha_1$ and $\alpha_2$ of the two bits of the path to some reference line. ...


0

If both the ropes are of identical material, then the fact that they have equal tension in them means, by symmetry, that they make equal angles with horizontal, call it $\theta$. Then the only equation you have is $mg=2Tsin\theta$. So unless $\theta$ is given you cannot solve the problem. If you make, say, $\theta$ small so that the hanging ropes are close ...



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