Tag Info

Hot answers tagged

5

The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no $| 1 0 0 \rangle$ state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction ...


5

The reason is the samen as why the electric field inside a conductor is zero: if it isn't zero, the free electrons undergo a force and move (rearrange) untill they dont feel a force anymore. If the electrons don't feel a force, the electric field must be zero. At the surface of a conductor, the free electrons feel a force perpendicular to the surface, but ...


4

Let's figure it out. First let's figure out the force you would need to slide the object. In order to slide it you'd need a force that overcomes friction. $$ F_{\text{slide}} > \mu m g $$ where $\mu$ is the coefficient of friction. In order to tip it, you'd need to cause a torque that will cause the can to rotate about its far bottom corner. If you ...


4

You need to learn about the Eikonal equation and the equivalent ray path equation, which I talk about in my answer to the question Physics SE question "Ray tracing in a inhomogeneous media", and, if you need to know how it comes as the *slowly varying envelope approximation" from Maxwell's Equations, I talk about this in my answer to the question , "Optics: ...


4

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...


4

If you suppose that the scale works like a spring, which seems reasonable, then during standard use, the displacement $x$ of the scale is proportional to the mass $m$. The equilibrium relation is $$ mg=kx,\tag{1}$$ where $k$ is the stiffness of the spring. Assuming that, when you dropped a mass $M$ from a height $h$, all kinetic energy (which is equal to ...


3

If the rope is "radially directed" it means every point has the same angular velocity $\omega$. Assume a length $2\ell$, then you can integrate the force on the rope from $r-\ell$ to $r+\ell$ - gravitational force must equal centripetal force. This gives you an equation for $\ell$ as a function of $\omega$ and $r$. See if that gets you going.


3

The angular acceleration $\alpha$ is just $\alpha = a/r = 1/200\textrm{rad/s}^2$ where $a$ is the tangential acceleration and $r$ the radius of the circle. Here, we can see that $\alpha$ is constant, which allows us to use the constant acceleration equations in their angular form. For (b) you can use the angular constant acceleration equations. You have ...


3

If the wire used around the pulley is considered ideal and massless, tension in the wire is same at each point and hence vector A and vector C are equal in magnitude. This reduces one variable and you will be left with 3 equations and 3 variables.


3

First of all, the variation of the action produces the equations of motion. Inserting the solution of the equations of motion into the action simple provides you with the value of the action when evaluated on the solution. There's no problem with it being zero. However, when you do this, you should always worry about boundary terms, these are irrelevant for ...


2

You have that $$ E=2A_0\cos\left(\frac{\phi}{2}\right)\sin\left(\omega t + \frac{\phi}{2}\right) $$ which is correct. To get the intensity, you then square and time average this: \begin{align} I=\langle E^2\rangle&=4A_0^2\cos^2\left(\frac{\phi}{2}\right)\left\langle\sin^2\left(\omega t + \frac{\phi}{2}\right)\right\rangle\\ ...


2

Here's how. The ball gains a velocity $v$ due to gravity before hitting the ground. So each time it hits the ground its velocity is changed from $v$ to $-v$ (taking down as positive) during the collision, then returning again with $v$. The force $F_1= m\frac {dv}{dt}$ is experienced by the ball due to the collision, however this force is felt after the ...


2

How to prove it? EDIT: Consider a state $|\psi_n\rangle$, which is an eigenstate of the Hamiltonian $H$ having eigenvalue $E_n$. Then consider the action of $H$ on these two other states: $$ |\alpha\rangle \equiv AB|\psi_n\rangle $$ and $$ |\beta\rangle \equiv BA|\psi_n\rangle $$ EDIT: Ask yourself: Are these eigenstate of the Hamiltonian? If so, ...


2

According to Newtonian mechanics, it is true that the table exerts an equal but opposite force against gravity that results in $\Delta y = 0$, where $y$ is the up/down dimension. However, sliding block is clearly moving in the $x$ dimension (i.e. horizontally across the table). And it is also acted on by a force, namely friction. The block does not generate ...


2

No the point of contact is not at rest. It moves with the block. You are probably confusing with rolling motion in which the point of contact is always at rest. There the point of contact is rest because the lowest point on the disk has two contributions, one due to forward motion of disk as a whole (v) and one in the backward direction due to rotation ...


2

The state you have given is not normalisable as a consequence of the results of the calculations you have done. Even if the first state (with coefficient $A$) was not present it would not be normalised. To normalise what you have given, another constant needs to multiply everything through (so that the relative proportions are unchanged


2

Whenever you want to prove something, you need to get your hypotheses straight. I understand that you want to prove that a mass $m$ on a spring with constant $k$ moves harmonically with frequency $\sqrt{k/m}$, but you seem to be assuming that result in your "proof": First off, writing $E = \frac12 k A^2$ already implies that the motion is harmonic with ...


2

The geodesic equation can be derived by extremizing the length ("proper time" in the case of general relativity) of a path connecting two fixed points. One requires that, after introducing a parameter $\lambda$ so that for the geodesic $x^{\mu} = x^{\mu}(\lambda)$ connecting points $A$ and $B$: ...


2

In this case the wave is a sound wave i.e. a compression wave moving through air. The velocity of the wave is determined by the elastic properties of the air. Specifically it is given by: $$ v = \sqrt{\gamma\frac{P}{\rho}} $$ where $P$ is the air pressure, $\rho$ is the air density and $\gamma$ is a constant called the adiabatic index. So for any given ...


2

This particular case is easy, because we're working in the $xz$ plane so the sums are straightforward. The general case for two arbitrary vectors would be a lot harder. Anyhow, let's take the particular example you describe: The angle (1 0 a) makes with (1 0 0) is just tan$^{-1}$(a) and the angle (1 0 b) makes is tan$^{-1}$(b). You want the $a$ angle to ...


2

On the elevator two forces act.One is the gravitational force and other is the normal force. so from F=ma we can say- F(normal)-F(Gravitational)=ma so,F(n)=f(g)+ma so,F(n)=m(g+a) You just need to put the values in these equations and solve for acceleration. And as for the direction know one thing-When the body moves upwards the elevator shows a greater ...


1

Pay close attention to how you defined $x$: it's not the coordinate of the desired point on the $x$ axis, it's how far left of the positive charge that point is. So when your equation tells you $x = 4.83$, that means the desired point is $4.83$ units left of the positive charge. But, presumably the question you've been asked wants the coordinate of the ...


1

In my opinion, your solution is correct. The book's solution is correct as well, but you would need to draw a different diagram for it to make sense. Their assumption is to take by default, r as lying on the positive x axis. In this way, the distance between the $Q_2$ and the required point is the mod of $r-2$. Since this expression is being squared, it ...


1

Gauss law is useful only in the cases of high symmetry systems like sphere, infinitely long (or very long and thin) cylinder, or infinite plane. You can't even apply it to curved sides of "real", short cylinder. You need to show somehow that gravitational field is the same everywhere on you Gauss surface, so that integral in the Gauss law turns out to be a ...


1

Why not try to draw the Feynman diagrams first and characterize them by different topologies; then count the possible contractions for each cases?


1

The wave came back to him after 3.5s passed. That gives a clue as to what kind of signal this was (and wasn't). It wasn't electromagnetic radiation (e.g. light). It was sound. Sound is a wave carried by a medium, in this case, air. The velocity of the source of a sound has no effect on the velocity of the propagating sound wave. This gives you the ...


1

In that case, $\hat{\sigma}$ here refers to a vector formed by $\hat{\sigma}_x$, $\hat{\sigma}_y$ and $\hat{\sigma}_z$ as its Cartesian components. The individual components of the expectation value of the magnetic moment vector would then be obtained using the corresponding components of the Pauli spin operators.


1

The problem is translationally symmetric along one spatial dimension, which means that it is effectively a two-dimensional problem. You have a weird understanding of what "the dimension of a potential" is - there's really no such concept. But if you define the dimension $n$ by matching a radially symmetric $\phi$ to either $v(r)=b\log r +c$ when $n=2$ or ...


1

In this question the force F has to be calculated using coulomb's law.In coulomb's law it clearly states that both the 02 bodies should be charged.So in this question one will gain charge up to 6 uC and the other will be zero.So there will be no force between them. The question has not mentioned whether the two balls are insulators or conductors.It should ...


1

Looks like textbook hybridization problem, did you check the usual suspects, or e.g. this one?



Only top voted, non community-wiki answers of a minimum length are eligible