Hot answers tagged homework
21
The best way to solve it would be experimentally, by doing the run several times, with calibrated instrumentation by the roadside to measure your speed. The acceleration won't have been constant, so that's not an assumption we can use.
Knowing the 0-60 time capability won't really help; it could be different when accelerating up hill, compared to on the ...
16
This scientific problem – well, a more general one – has been solved in the following paper:
http://arxiv.org/abs/1204.0162
Because it's legal in my country to move backwards in time, I remember the future event – one minute from now – in which Andrew Gibson will mention that he has this paper hanging in his physics lounge. He will curse me. 11 minutes ...
13
Your calculation is incorrect.
$\text{Work} = \text{Force} \cdot \text{displacement} = F \cdot s$
The above product is a "dot" or "scalar" product, which means we only consider the displacement that occurs in the direction of the Force, which in the case of gravity is downwards. Can we set this vertical displacement to 0? No we cannot, and here is why:
...
10
Wavefunctions are found by solving the time-independent Schrödinger equation, which is simply an eigenvalue problem for a well-behaved operator:
$$ \hat{H} \psi = E \psi. $$
As such, we expect the solutions to be determined only up to scaling. Clearly if $\psi_n$ is a solution with eigenvalue $E_n$, then
$$ \hat{H} (A \psi_n) = A \hat{H} \psi_n = A E_n ...
9
Recall that a force perpendicular to the direction of motion does no work but simply changes the direction of the velocity vector.
The same thing happens here: Initially the ball's motion is perpendicular to the force of gravity and hence at this very moment, gravity does no work but slightly "rotates" this velocity vector towards the downward direction; as ...
8
You use the total amount of movement over time. So here that is|:
80km plus 60km equals 140km
Which gives you the correct answer.
Displacement, using Pythagoras, would be 100km, but you travelled 140km in that hour! You didn't travel along that hypoteneuse, so it is irrelevant here.
8
These kinds of proportionality questions are often best answered with dimensional analysis. You want to know a form a quantity with the units of time in terms of what you have.
You have a quantity $k$ with units $\frac{\text{Energy}}{\text{Distance}^3} = \frac{\text{Mass}}{\text{Distance} \times \text{Time}^2}$. You also have the mass $m$ (units of Mass) ...
7
To a very good approximation the transmission of a metal film falls exponentially with thickness i.e.:
$$ T = e^{-\alpha t}$$
where $\alpha$ is the absorption coefficient given on the web site Alexander mentioned, http://refractiveindex.info/?group=METALS&material=Copper, and at 500nm wavelength this gives $\alpha = 6.4297\times 10^5/cm$. So you just ...
7
The three capacitors are connected in parallel. There are only two nodes in this circuit. A series connection requires at least three. The equivalent capacitance is just the sum of the three capacitances.
UPDATE: The circuit can be redrawn such that the parallel connection is manifest.
7
If you solve for $t$ in Eq. (5.1), and plug that into equation (1.1), you'll see that the solution looks like $x_B \propto v_A^2 sin(\theta) cos(\theta)$. The function on the right is symmetric about $\pi/4$, thus, as long as $\theta$ doesn't equal $\pi/4$, there will be two solutions (symmetrically about $\pi/4$). Of course, in general, there could be ...
7
Alternatively I would look around the lab for an infrared thermometer. There exist in the market close focus ones that go down to 6mm in close focus option ( so as not to advertise, google space accurate infrared thermometers microscopes where I found the number in a one of the first hits).
I would choose a large ant, or attract more by a spot of honey ...
7
A common mistake when students begin the study of the quantum harmonic oscillator is to try to convert everything to integrals. The thing is, in most curricula, the QHO is also used as a way to secretly acquaint you with bra-ket notation, and all the conveniences it offers. In reality, you shouldn't need any integrals at all here.
$\lvert n \rangle$ is a ...
7
There is no 'only if' because it is not true:
\begin{align}
e^{A+B} = e^A e^B
\end{align}
does not necessarily imply $[A,B] = 0$.
One can easily find an example of this using matrices. Here's one:
\begin{align}
A=
\begin{pmatrix}
0 & 0 \\
0 & 2\pi i
\end{pmatrix},
B=\begin{pmatrix}
0 & 1 \\
0 & 2 \pi i
\end{pmatrix}.
\end{align}
$[A,B] \neq ...
7
Well, this certainly is an evil trick to play on first year students!
Escape velocity isn't actually a velocity at all. It's a speed, i.e., it's scalar quantity as opposed to a vector quantity. Note that when the escape "velocity" at r was calculated, the only assumption made was conservation of mechanical energy, and then magnitude of v is isolated from ...
6
Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metrix $\eta_{\mu\nu}$.
I) First recall the fact that
$SL(2,\mathbb{C})$ is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$.
This follows partly because:
There is a bijective isometry from the Minkowski space ...
6
If you want to prove that $\vec{L}=\vec{r}\times \vec{p}$ is constant with respect to time for a particle in a central force field $\vec F = \phi(r) \vec r$, just show that the angular momentum doesn't change with time, i.e. $\frac{d}{dt}\vec{L}=0$.
Using the product rule we get two terms:
$\frac{d}{dt}\vec{L}=\frac{d}{dt}(\vec{r}\times \vec{p}) = ...
6
I) OP is given a problem of the form
$$\tag{1} \dot{q}~=~f(q,p), \qquad \dot{p}~=~g(q,p), $$
where $f$ and $g$ are two given smooth functions. OP is asked to derive the integrability condition for the eqs. (1) to be Hamilton's eqs.
$$\tag{2} \dot{q}~=~\frac{\partial H}{\partial p}, \qquad
\dot{p}~=~-\frac{\partial H}{\partial q}.$$
OP correctly ...
6
The formula you wanted to use gives you the magnitude of the average velocity, not the average speed.
To get the magnitude of the average velocity, you take the total displacement (which is a vector!), divide by the total time, and find the magnitude of that vector. What you get is:
$$\text{Magnitude of Average Velocity}= \biggl| \frac{\sum_i \vec{d}_i} ...
6
think about this with an example: the sine and cosine functions. They both average individually to zero over an interval. You can multiply those averages and still obtain zero. But if you multiply sin by itself and then average, you get a very distinct non-zero result.
When the functions are arbitrary, the average of the product quantifies statistical ...
6
Actually the conducting disk problem is solved very easily in the so-called oblate spheroidal coordinates.
First, alter the coordinates so that your disc is centered at the origin and is orthogonal to the $z$-direction. I will follow the notation of the Wiki article:
$$
x=a\cosh\mu\cos\nu\cos\phi\\
y=a\cosh\mu\cos\nu\sin\phi\\
z=a\sinh\mu\sin\nu
$$
where ...
5
First choose a direction for u, along the z-axis. Then the integral is
$$ I = \int {1\over (x^2 + y^2 + A z^2 + B)^{5/2} } dx dy dz $$
Rescale z by $\sqrt{A}$ to get rid of A and restore rotational invariance.
$$ I = {1\over \sqrt{A}} \int {1\over (x^2 + y^2 + z^2 + B)^{2.5}} dx dy dz $$
Now you do find the B dependence immediately from rescaling x y and ...
5
You mistake is that you use the absolute value "of the spatial components" (your words) of the velocity only. Picking spatial components only is clearly not a Lorentz-covariant procedure, so it cannot calculate the invariant "feelings of the astronauts".
Instead, the right condition is given by the same inequality but $|d u^\mu / d \tau|$ is the length of ...
5
From where did you get the idea that one can get a spin zero representation? The product of an even/odd number of Fermion representations always gives a Boson/Fermion representation.
In your particular case, repeated use of $$1/2 \otimes s = (s-1/2) \oplus (s+1/2)$$ gives $$1/2 \otimes 1/2 \otimes 1/2=(0\oplus 1) \otimes 1/2
= (0 \otimes 1/2) \oplus (1 ...
5
I'm not sure I entirely understand the question. This is what I think you're asking; please ignore the rest of this answer if I've misunderstood you.
If the plane is stationary (and I assume there is no friction) then a block on the plane will feel a force down the plane of $mg \space \sin\theta$, so it will accelerate down the plane. If we push the plane ...
5
For the partition sum, you have so sum $e^{-E}$ ($T=1$) over all possible eigenstates of the system where $E$ is the energy of the corresponding state.
Two bosons can be in the states 10
$|kl\rangle$, with $1\leq k \leq l \leq 4$ where we accounted for the degeneracy by introducing an additional state with $E_4 =2E$. The corresponding partition sum reads ...
5
The question is meant to remind the student that the temperature relevant to physics is the absolute temperature in Kelvin, and not the temperature in relation to the triple point of water (in degrees Celsius).
That is, the zero point of absolute temperature is well-defined as the state without any thermal energy, and doubling absolute temperature is hence ...
5
The nature (and glory) of the dirac delta function is that the volume integral
$$ \int_{\Delta V} dV' \delta ( \boldsymbol{r-r'} )
= \left\{
\begin{array}{cc}
1 & \text{if } \Delta V \text{ contains } \boldsymbol{r}\\
0 & \text{if } \Delta V \text{ does not contain } \boldsymbol{r}
\end{array} \right. $$
Using this function, you can write the ...
5
You are confusing in the index, such calculations must be carried out very carefully.
I would start with your difention.
$$M_i=\epsilon _{ijk} q_j p_k$$
$$M_p=\epsilon _{pnm} q_n p_m$$
$$\{M_i, M_p\}=\sum_l \left(\frac{\partial M_i}{\partial q_l}\frac{\partial M_p}{\partial p_l}-\frac{\partial M_i}{\partial p_l}\frac{\partial M_p}{\partial q_l}\right)$$
...
5
No, you are wrong. Particular for the following statement:
If I integrate the square of this equation between $r=0$ and $r=x$, am I right in assuming I am calculating the probability of finding the electron in a sphere radius x?
The probability density at any points is given by $|\psi(r,\theta,\phi)|^2$. Certainly, the probability is for any region $V$ ...
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