Hot answers tagged homework
4
I prefer to use Killing vectors and conservation laws to solve stuff like this, so let's analyze the problem using Killing vectors, and see if the results agree with your Euler-Lagrange equations.
Notice that the metric is invariant under translations of $v$. The associated killing vector is $\partial_v$ which in turn gives the following conserved ...
4
Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough
$$
E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)}
$$
For your assumption that the motion is strictly radial and downwards you have $v(t) = dr(t)/dt < 0$ so you can solve for $dr(t)/dt$ and get an ordinary first order ...
4
Since radioactivity is a random process, you'd expect some fluctuation in the number of decays, i.e. if you wait for the half life, there's no guarantee that exactly half of the nuclei will have decayed!
Based on your 3 estimates of the half life, you could just take the mean and go with that.
And how to find the isotope? There's lists for that. I just ...
3
The first resonant vibrational mode for a string clamped at both ends looks like:
You should be able to deduce the wavelength from that diagram. The second mode looks like:
Both of the images above are from http://www.clickandlearn.org/Physics/sph3u/Music/Music.htm and that site will spell it out in more detail for you.
If your string length is ...
3
Just some pointers to get you started: The side length allows you to calculate the flux through a benzene ring given a magnetic field, which should then give you the induced current, if you claim that the electrons in the benzene ring are metallic (not a bad approximation due to their delocalized nature).
Another thing you can look at is this article on ...
2
This problem has a recursive flavor that we'll not try to avoid.
Conservation of momentum tells us that
$$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$
Imposing the boundary condition $v(0)=0$ we find
$$v(n) = \frac{n}{n+p}v_0$$
as claimed.
Let $a_n$ be the time at which the $n$th bullet strike occurs.
We have $a_1=x_0/v_0$ and
$$v_0 (a_n - T) = v_0 ...
2
Any material between two nodes is displaced by the same direction. So the direction of B and C has to be the same as well as the direction of A and D due to symmetry. In addition, the direction of A must be the opposite of B since they are across from a node. Similarly the direction of C and D must be opposite.
So the two possible configurations are
...
2
why does the voltmeter not form a closed loop with the circuit and hence cause energy to be dissipated by internal resistance
It does, but the voltmeter probably has an impedance of 10,000,000 ohms - as most multimeters do - so the current flowing is negligibly small (0.6 microAmps) and therefore the voltage drop due to battery internal resistance is ...
2
The question you have formulated is not an easy one to answer (correctly). But the question you've formulated isn't quite the question that I see. The good news is that the text of the question you've posted implies a much simpler question; it's just asking for the energy change.
You can probably assume that the acceleration is dominated by the circular ...
2
The "correct" half-life is the one that best fits all the experimental data available.
Start with your original equation,$$N(t)=N_{0}e^{-\lambda t}$$ Take natural logarithms of both sides and obtain:$$\ln (N(t))=\ln(N_0)-\lambda \ t$$Note that $N(0)$ is not the same as $N_0$. The first is an experimental point, with the same sort of random and ...
2
Your answer would have been correct if, for example, the spheres were non-conducting and if the charges were distributed uniformly over their surfaces.
However, since the spheres are conducting, the surface charge distribution on each sphere will be altered because of the repulsion from the charges on the other sphere. In particular, the charges on each ...
2
Your angular velocity vector is
$$ \vec{\omega} = \Omega \frac{ \vec{r}_D - \vec{r}_A }{|\vec{r}_D - \vec{r}_A|} $$
where $\vec{r}_A = (0,0.2,0.12)$, $\vec{r}_D = (0.3,0,0)$, $\vec{r}_B = (0.3,0.2,0.12) $ in meters and $\Omega = 90\;{\rm rad/s}$.
Your velocity kinematics is
$$ \vec{v}_B = \vec{\omega} \times ( \vec{r}_B - \vec{r}_A ) $$
And acceleration ...
1
When imposing a periodic boundary condition, the amplitude of the wavefunction at coordinate $x$ must match that at coordinate $x+L$, so we have:
$$\Psi(x)=\Psi(x+L)$$
In your previous 'particle in a box' scenario, you mention that the general form of the wavefunction is given by a linear combination of sine and cosine with complex coeficients. It might be ...
1
Your approach ignores the body of the slinky and essentially describes two massive particles coupled by a very light spring, which is not allowed to oscillate or show any of the interesting dynamics a real slinky will exhibit.
Ideally, you should be using some sort of continuum-mechanics approach to this problem, e.g. treating the slinky as a very elastic ...
1
That line will get Coriolis acceleration $$\vec{a} = -2 \vec{\Omega} \times \vec{v}$$ ($\Omega$ is the angular speed of the earth's rotation, with a direction pointing into the ground from the view of the south pole). As it's going across the pole, there's a right angle between $\Omega$ and $v$ and the absolute value will be simply $$a = 2\Omega v$$ and the ...
1
For a free particle, the energy/momentum eigenstates are of the form $e^{i k x}$. Going over to that basis is essentially doing a Fourier transform. Once you do that, you'll have the wavefunction in the momentum basis. After that, time-evolving that should be simple.
Hint: The fourier transform of a Gaussian is another Gaussian, but the width inverts, in ...
1
For pushing it up, we have to overcome friction(act downwards) as well as the $mg\sin\theta$. So, $$3N=f+mg\sin\theta$$
Now the block is just slipping , so friction is acting upwards, and so does the force applied externally.So,
$$N+f=mg\sin\theta$$
Eliminate $N$ and use $f=\mu mg\cos\theta$.
Solve for $\mu$ you get your answer.
1
The ground will provide all of the static friction. Imagine what would happen if the upper block contributed even a tiny amount to the static friction: It would have to move forward due to the reaction force. Having M2 inch along you pull M1 (which stays stationary) would be very strange indeed.
Static friction always acts to prevent relative motion. It ...
1
You are halfway there, because you already wrote down the Hadamard gates. The remaining part, $I-2\left|0\right>\left<0\right|$ (I negated it from what you have, since this gives a simpler solution), is diagonal in the computational basis. Write down these diagonal entries and say out loud to yourself what this operator "does".
1
The voltmeter does indeed draw current, but very small. Its internal resistance should be high, unless it's a very cheap voltmeter.
Putting the voltmeter's internal resistance, let us say 1MOhm just to make up a value, in series with the battery's internal resistance of a fraction of one Ohm, makes a voltage divider providing practically the entire ...
1
Stevin's loop (or the "epitaph of Stevinus") gives us the answer to the first part. That is, the loop will not move if its two sides span the same vertical distance down the triangle.
So its center of mass lies at the $y$-axis point one-half of the way down the vertical chain length (as well as one-half of the way down the slanted chain length -- the ...
1
Focus on the integral
$$ I_{ij}(k) = \int k_i k_j\ \mathrm{d}\Omega_k.$$
This is a rank 2 symmetric tensor which can only depend on $\vec{k}$ through its magnitude $k^2$, since the direction has been integrated over. So the only possibility is that $I_{ij}$ is proportional to the unit tensor (Kronecker delta):
$$ I_{ij}(k) = f(k^2) \delta_{ij},$$
where ...
1
You can prove this by induction. I'll drop the operator hats as they're a pain to write.
First step:
Suppose $\hat H = \hat q_k$. Then $[ H, p_i] = [q_k, p_i] = i\hbar \delta_{ik} = -\frac{\hbar}{i} \frac{\partial H}{\partial q_i}$
So in the special case that the Hamiltonian has this form, the claim is true!
Now suppose that the claim is true for $H = ...
1
Assuming you want to place the $5$ $kg$ mass at a point such that entire COM coincides with $(1,2,3)$.
Then we can assume a $6$ $kg$ mass at $(1,2,3)$ and a $5$ $kg$ point mass lying at $(-1,3,-2)$. And then the new $5$ $kg$ mass should be placed such that $COM$ of these two masses is lying at $(1,2,3)$ .
So the new $5$ $kg$ mass should be placed at ...
1
A standing wave is a wave that has nodes. The points of the wave go up and down in some places, and remain at zero at others (the nodes). The general form of a standing wave is a sine curve that remains at a fixed position, but its amplitude changes in time between $+A_0$ and $-A_0$. Specifially, there is a time where the wave form is completely flat.
...
1
NOTE: This comment was too long so I'll make it an answer.
I would assume that the amount that the spring contracts is negligible compared to the distance that the mass has fallen.
But anyway, regarding your amplitude problem: consider that when you have a mass on a level surface connected to the spring, there is an interplay between kinetic and spring ...
1
There are a number of related ways of thinking about this. The answer of webb can be put on a slightly more explicit ground. In the "spin coherent states" path integral for the quantum Heisenberg model, solutions of the classical Heisenberg model are extrema (or saddle-points).
You could also, more prosaically, perform a Holstein-Primakoff transformation to ...
1
Irrespective of the dimensions, Poisson equation is always true. That is, if $\phi$ is the electric potential and $\rho$ is the charge density then, $\nabla^2\phi = \rho/\epsilon_0$. The Green's function of this equation satisfies $\nabla^2 G(\vec{x},\vec{x}^\prime) = \delta(\vec{x} - \vec{x}^\prime)$.
A Fourier transform of this equation is $k^2G(\vec{k}) ...
1
Hey you're getting wrong, the equivalent emf(voltage) of system will be 16V-8V because both have opposite poles facing each other, so their will be net flor of current (-ve to +ve) according to the cell of greater emf(16V cell).Then your total resistance is $$5+1.6+1.4 = 8 \Omega$$(all are in series) . $$I(Current) = E(e.m.f or Voltage)/R(Resistance) = 8/8 = ...
Only top voted, non community-wiki answers of a minimum length are eligible
