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21

Shift the upper configuration to the left a short distance at equilibrium. Result: the left wheel goes a little up, the right goes a little down, the train tilts clockwise, the center of mass is to the right of the centerline between the wheels, and therefore the center of mass provides a restorative force to push the train back to the right. Shift the ...


12

In both diagrams in the question, the left wheel has a smaller radius at the contact point than the right wheel. Because they're fixed to a common axle, in any given amount of time, the right wheel will travel a greater distance than the left, so the axle as a whole will rotate anti-clockwise (when viewed from above) about a vertical axis. As it does this, ...


7

There IS a potential, and all three bulbs will be on. First, you need a reference point to measure voltages. Let's take the wire right of the right battery, and define its potential as 0V. If both batteries generate a voltage of 5V, the wire between the batteries will have a potential of 5V, and the wire left of the batteries will have 10V: So, bulb 1 ...


5

A typical atom is roughly a few times $10^{-10} \text{m}$ wide. A piece of paper is say $(1/4) \text{m}$ wide. Therefore the ratio of the width of an atom to the width of a piece of paper is around $10^9$. A piece of paper is roughly the same width as a human, so $10^9$ is also a rough guess for the ratio of the width of a human to the width of an atom. The ...


4

For a 6 year old, you might want to focus on thickness instead of length, as the numbers get too big with length. A ream of paper (500 sheets) is a bit over an inch thick, say $3.5 \, \text{cm}$, so one sheet is $3.5/50 \, \text{mm}$, or $.07 \, \text{mm}$, which is $7 \times 10^{-5} \text{m}$. An atom has diameter $0.1 \, \text{nm}$ to $0.5 \, \text{nm}$ ...


3

For waves, reflection off of a fixed boundary causes an inverted reflected wave (a 180 deg phase shift), while reflection off of a free boundary causes no phase shift. Excellent examples of this can be found at PhET. See this. When light travels from a medium with a low index of refraction to a high-index material, it sees a "fixed" boundary and reflects ...


3

The gravitational mass of a neutron star is quite a lot less than its baryonic rest mass (plus the mass associated with the kinetic energy of its contents), because a bound neutron star, by definition, must have a total energy (the sum of its internal energy and gravitational potential energy) that is less than zero. In a “normal star” this is also true, ...


3

Yes, this is correct. Also there are field lines going out from the top of the top plate and the bottom of the bottom plate. And far away, the field lines should look like those of a single point charge, with charge equal to the total charge on the two plates.


3

Peskin & Schroeder, An Intro to QFT, are using that$^1$ $$i\Delta(x-y)~:=~\langle 0 | [\phi(x), \phi(y)] |0\rangle \tag{K} $$ vanishes for space-like vectors, see below eq. (2.53) on p. 28. In particular for equal times $x^0=y^0$, we have $$i\Delta(0,{\bf x}-{\bf y})~=~0.\tag{L}$$ Therefore at the physics level of rigor ...


3

The contact with the rail creates a kinematic center of rotation where the reaction forces meet. The rail car will tend to rotate about this center as a result of side loads. If the center is above the center of mass, the rail car acts like a hanging pendulum. A small deflection will cause a restoring torque opposing the swing. If the center is below the ...


2

This is a straightforward application of the divergence theorem. First, split the integral on the left into its vectorial components: $$ \int_{\partial \Omega}(P_0 - \rho g z)\left(-\mathrm d\mathbf{S}\right) = \sum_j \mathbf e_j\int_{\partial \Omega}(-P_0 + \rho g z)\mathbf e_j\cdot \mathrm d\mathbf{S} .$$ Then, apply the divergence theorem: \begin{align} ...


2

The mistake is in the second line, in the calculation of the differential mass element. The differential mass element in this case is a disc, of radius $r$ where $r = R \cos\theta$ as you have correctly used. However, the thickness of this differential disc is NOT $ R d\theta$ but $Rd\theta cos\theta$. Try to wrap your head around this. $Rd\theta$ is the ...


2

This answer uses Figures instead calculus as in the excellent Emilio Pisanty's answer. (1) Firstly : Horizontal hydrostatic pressure force cancels out Cut your body horizontally and take any section with infinitesimal height $\:dh_{1}\:$ as in Figure. Then \begin{align} \mathbf{F}_{\text{horizontal}}&=\sum_{m=1}^{m=N}\left(- ...


2

We can write total energy $E$ two ways: \begin{equation} E^2=p^2c^2+m^2c^4 \\ E=T+mc^2, \end{equation} where $T$ is kinetic energy. Eliminating $E$ from those two equations will give you the desired result.


2

What matters is the surface of the object (in this case sphere). On the sphere a charge would distribute uniformly. In the situation that you described what matters is the relative size of the surfaces of two spheres. Since they are equal the charge on both of the them is the same (and hence is the half of the original charge).


2

Hint: $T = E - E_0 = m\gamma c^2 - mc^2 = mc^2(\gamma -1)$ and $p = |\vec p| = m\gamma |\vec v| = m\gamma v$


2

First a couple general results: given a function $f(t)$ and an interval $T = [t_1,t_2]$, the square of the mean of $f$ on $T$ is $$\langle f\rangle_T^2 = \biggl(\frac{1}{t_2 - t_1}\int_{t_1}^{t_2}f(t)\,\mathrm{d}t\biggr)^2$$ and the mean of the square of $f$ on $T$ is $$\langle f^2\rangle_T = \frac{1}{t_2 - ...


2

Hints: By mean/average Feynman means temporal mean/average defined as $$\tag{1} \langle f \rangle ~:=~\frac{ \int_{t_i}^{t_f}\! dt~ f(t)}{t_f-t_i} . $$ Inequality: The mean square is always greater than the square of the mean $$\tag{2}\langle f^2 \rangle ~\geq ~ \langle f \rangle^2.$$ There are several proofs of ineq. (2), e.g. the variance is always ...


2

Indeed, the $\vec{E}$ field in a parallel plate is independent of distance from the plate. This works because of the assumption $d \ll$ length of plate (thus, we can ignore side effects of the plate). And as Bort pointed out, it is the Voltage $V$ that scales linearly with respect to distance from the plate, while $\vec{E}$ will remain constant.


2

I am sorry to disappoint you, but there is no such formula that you can just apply. This is because it strongly depends on how and under what exact conditions and with wwhich tools you did the experiment. Think of this: If every methanol molecule reacts (burns) at once all at the same time, then the exact same amount of energy is spent, but it went really ...


2

Notice that you have implicitly chosen to measure angular momentum about the axle of the platform. That means that all the forces exerted by the axle on the platform are applied through the axis for rotation, meaning the torque they exert is $$\text{force} \times \text{lever arm} = F \times 0 = 0\,.$$ And there are no other forces present expect those ...


2

The operator $\delta(\gamma(z))$ is meant to be the operator dual to the state $|0\rangle_{NS}$ according to the state-operator correspondence. One could call it $E(z)$ or anything like that. But Polchinski uses the notation $\delta(\gamma(z))$ with this "nested" structure because the operator described in the previous paragraph may also be interpreted as ...


2

In fact you have been treating forces just like space vectors, meaning 'things' having a length, a direction and an orientation and that can be added and subtracted using the parallelogram rule. The fact that forces can be treated as vectors is a well-known property of forces.


1

The correct commutation relations for creation/annihilation operators are the following: $$ [a(k), a^{\dagger} (k')] = (2\pi)^3 \delta^{(3)} (k - k') $$ $$ [b(k), b^{\dagger} (k')] = (2\pi)^3 \delta^{(3)} (k - k') $$ Everything else commutes, including $[a, b^{\dagger}]$. From this, it is clear that $[H, a]$ can not depend on $b$.


1

It seems like the key problem is that you are conducting your integration over the wrong surface. The problem is, you have an infinitely extending sheet of charge on the surface $\rho=1.2$. Based on this, as you seem to have done, we work in cylindrical coordinates and consider a plane of constant $z$. Taking your expression: and converting it into ...


1

There are 4 standard kinematic equations from Newtonian mechanics, and you need what is usually considered to be the fourth equation. $\mathrm{(final\ velocity)^2 = (initial\ velocity)^2 + 2 \times acceleration \times distance}$ You know the initial velocity, final velocity, and distance. Solve for acceleration.


1

A few hints: The field at the middle wire is "indeterminate" since there is a singularity due to the current in the middle wire. If you sketch the field as a function of $x$ you would get something like this: (this is the plot of $\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}$ courtesy of Wolfram Alpha) The zeros in the field are easily seen as occurring ...


1

(a) You're about right on the calculation for the two points on $(-a, 0)$ and $(0, a)$ but the $2a-s$ thing looks weird. Is $s = x + a$ or something? On $(0, a)$ it should be $1/x + 1/(x + a) + 1/(x - a) = 0$ or $$(x + a)(x - a) + x (x - a) + x (x + a) = 0$$ simplifying to $3 x^2 = a^2$ and giving $x = \pm a \sqrt{1/3}$, which appears to be what you got by ...


1

Well, it's not energy, its power $P$. $~~~~~~~~~~~~P = \int F \cdot dv$ And since power is the derivative of energy $P = \dot E$, your world makes sense again ;). Regarding your problem I agree with Yanping Cai, the kinetic energy of the car $E_{kin}$ must be converted into potential energy of the spring $E_{spring}$. $~~~~~~~~~~~~\frac{1}{2} m_{max} ...


1

You'll find a one parameter family of solutions, because you have 4 independent quantities while in this problem you have the 3 independent units for mass, length and time. In your solution, you can see that the freedom to choose the parameter b comes from the fact that the density of air divided by the density if the object is dimensionless. To fix b ...



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