# Tag Info

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 ...

4

Use kirchoff's loop and junction law. :)

3

My fallacy was in thinking that resultant translational motion varies by the distance ($d$) between the point of force application and the center of mass. It was not a fallacy or misconception. It is simply impossible that the linear velocity of a rod is the same if the point of application A (A-CM = d) varies. If A coincides with the ...

3

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: ...

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

No. Automobile tires to not expand radially to any great extent - the steel belts will keep that from happening. So, the tire radius still determines how far the car travels per rotation. Now, if your tires are slipping on the road, or are slipping with respect to the rims, than yes you have speedometer problems, but you have lots of other problems as ...

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 ...

2

If we treat the Earth as an isolated system then both its linear and angular momenta will remain constant. To answer your question you need only consider the angular momentum. The angular momentum is given by: $$L = I\omega$$ Since $L$ is a constant, if the moment of inertia changes from $I_1$ to $I_2$ then we have: $$I_1\omega_1 = I_2\omega_2$$ and ...

2

When we vary $F^{ab}F_{ab}$ with respect to the metric, we must also specify what we are holding fixed. Assuming that the context is that of electromagnetism, we consider the four-potential $A_b$ as an independent variable, and therefore under variations of other variables (such as the metric), it is held fixed, as is $F_{ab} = \partial_a A_b - \partial_b ... 2 It's worth drawing a diagram: The equations of motion are: $$x_1 = x_0 + \frac12 a_1 t^2$$ for$t_1 < \frac{v_1}{ a_1}$And $$x_1 = x_0 + \frac{v_1^2}{2 a_1} + v_1 (t - \frac{v_1}{a_1})$$ for the steady state. The same equations, with different suffixes, hold for the tiger. To solve, you start by considering all different orders of$t_1, t_2, ...

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

OK firstly operators act on kets as vectors, so for some unitary operator $\hat{U}$, the action of $\hat{U}$ sends $|\psi\rangle\rightarrow \hat{U}|\psi\rangle$ not $\hat{U}|\psi\rangle\hat{U}^\dagger$, which is normally not a well defined expression. For the rest of your question, since we are only dealing with a two dimensional space (or at least only a ...

2

You are sloppy with units, but the result is correct. To go from 25C to 3C is 22 cal/g. When you multiply by 300 g you have cal and your conversion to kJ is correct. Converting to W-hr is silly, but that is the unit of energy, not W/hr. You have 8.3 W-hr you want to remove. That chills the water assuming no new heat is added, so insulate the water. ...

2

Q = mc(t1-t2), Now, m = (density)(volume), Specific heat of water, c(in joule/gramCelsius) = 4.186, Hence, you can find the energy it would require for this conversion. . And the work you do can be a bit more pertaining to your efficiency.

2

Ok, so here are is my solution. I'll be happy is someone can provide something simpler. $a_1, v_1, t_1 - acceleratin, terminal\ velocity,\ time\ to\ reach\ it\ (for\ Tiger)$ $a_2, v_2, t_2 - acceleratin, terminal\ velocity,\ time\ to\ reach\ it\ (for\ You)$ $s_0 - starting\ distance$ Not let's consider three cases: 1) $a_1 < a_2 \wedge v_1 < v_2$ ...

2

As always, a diagram clears things up quickly: We are going to assume that there is no lift on the arrow (which is wrong - arrows do not fly like regular projectiles but that is not the point of this question). From thie diagram we can see the approach we need to take (I am not going to show the details of the steps, just give you some direction). the ...

2

Let's say you assume that the neutron star is spherically symmetric, e.g., ignore the effects of rotation. Then for a radial trajectory in the resulting Schwarzschild spacetime, the calculation is actually not quite wrong, although you must be careful in interpreting it. The reason is that orbits in a Schwarzschild spacetime have an effective potential that ...

2

By the very functioning of an open switch, there can be no current through the switch, and there is no constraint on the voltage across the terminals of the switch. This of course means that there cannot be a current through the entire branch containing R4 (and therefore the resistance R4 plays no role in this case). The current then flows through the rest ...

2

The equation $$\ddot{r} = -\mathbf{g},$$ is valid iff $\dfrac{h}{R_e} <<1$. The gravitational force is : $$\mathbf{F} = -m\dfrac{GM_e}{R^2}\mathbf{\hat{R}}.$$ Now one defines $r = R - R_e$ with $\dfrac{r}{R_e} <<1$. Then one has : \begin{align} \mathbf{F} &= -m\dfrac{GM_e}{R^2_e(r+R_e)^2}\mathbf{\hat{R}}\\ & = ...

2

You are specifically asking about a first order correction to the formula. Starting from $$F = \frac{GMm}{(R+h)^2}$$ for the projectile at height $h$, we can rearrange this as $$F = \frac{GMm}{R^2}\frac{1}{(1+\frac{h}{R})^2}$$ When $h\ll R$ we can use a first order Taylor expansion to write $$F = \frac{GMm}{R^2}\left(1-\frac{2h}{R}\right)$$ Finally, ...

1

For a given Bravais lattice you need to find the indices for which the structure factor $S_{hkl}$ doesn't vanish. For cubic lattices it's actually quite straightforward (e.g. for hexagonal ones can be already very tricky), knowing NaCl has a fcc structure, we know then the atomic positions in a unit-cell: Na $\rightarrow$ $[0,0,0]$, $[1/2,1/2,0]$, ...

1

Introducing: $a_ia_k^+=\delta_{ik}-a_k^+a_i$ Replacing 3 in 2 and solving: $\langle K|L\rangle=\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+|\rangle=\delta_{ik}\langle|a_ja_l^+|\rangle-\langle|a_ja_k^+a_ia_l^+|\rangle$ $\langle K|L\rangle=\langle|a_ja_ia_k^+a_l^+|\rangle$ $$=\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+\rangle$$ $$=\langle ... 1 Using the rule (3) you can sort the four operators in the order you prefer, by swapping terms. After each swapping, you obtain a piece with four operators, plus an additional one with two operators. And so on. In your case, by swapping the second and the third term,$$a_j a_i a_k^+ a_l^+ = - a_j a_k^+ a_i a_l^+ + \delta_{ki} a_j a_l^+ $$and by swapping ... 1 The square bracket transformation This is just the application of chain rule. The LHS means a derivative over the primed spacial coordinates while keeping unprimed spacial and time coordinates fixed.$$\nabla'[ \rho(\mathbf{x'},t')]_{ret} = \left(\sum_i \frac{\partial }{\partial x_i'} \hat{i}\right)[\rho(x_i',x_j',x_k',t')]_{ret}\\$$But the \rho is a ... 1 We start from the equation of time evolution:$$|\psi(t)\rangle = T\left(e^{-i\int_{-\infty}^{t}dt'H(t')}\right)|\psi\rangle$$Now, we need to evaluate the exponential with the following Hamiltonian:$$H(t) = H_0 + f(t)O_1$$while paying attention to the time ordering. To do this, for convenience, let us consider the (time-ordered) exponential part alone, ... 1 I'm guessing your questions all amount to whether general relativistic effects become important at the surface of a neutron star. To answer this we can compare the flat space metric (in polar coordinates):$$ ds^2 = -c^2dt^2 + dr^2 + r^2 d\Omega^2 \tag{1} $$with the Schwarzschild metric that describes the geometry outside a spherically symmetric mass:$$ ...

1

In an elastic collision involving two objects both momentum and energy are conserved. You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore one object will have the opposite momentum of the other. Since energy is the square of the momentum divided by twice the mass, this means ...

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