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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 ...


5

This answer uses Figures instead calculus as in the excellent Emilio Pisanty's answer. ($\:h\:$ = depth of immersed horizontal surface from the rest open surface of the fluid) (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 ...


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

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

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 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 ...


3

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

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

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

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

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

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.


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.


1

There are two observations that can be made about this problem. 1) If T1 is not equal to T2, the string will slip on the peg, which is frictionless. 2) If the reaction force from the peg, force R1, is not perpendicular to the peg's surface, there will be a component of R1 that is parallel to the surface of the peg, and the peg will rotate. For a ...


1

If $H$ commutes with $A_1$, then it will indeed share an eigenbasis with it. Your mistake is in supposing that it will share the same eigenbasis with both $A_i$s. Examples are easy to provide: On the trivial side, if $H=E_0\mathbf 1$ is trivial, then it shares an eigenbasis with $A_1=x$ and it shares an eigenbasis $A_2=p$, but it cannot share an ...


1

Assuming that the angle theta is measured relative to the vertical (e.g., the position of the string when the pendulum is at rest), a careful free body analysis indicates that the acceleration of the pendulum is g * sin(theta). This means that the acceleration of the pendulum continuously varies as it swings. This is relevant because the kinematic ...


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

The coefficient of friction is the ratio of the frictional resistance force to the normal force. The coefficient of static friction is the ratio of the maximum amount of friction that must be overcome to start an object moving, to the normal force. The coefficient of kinetic friction is the ratio of the amount of friction that must be overcome to keep an ...


1

You know that $$m\vec a=q\vec v \times \vec B,$$ So in particular, since $\vec B = B\hat z,$ we have $$ma_x=qv_yB,\text{ and } ma_y=-qv_xB. $$ And in our case $B=-\beta x$ so we have $$m\ddot x=-q\dot y\beta x\text{ and } m\ddot y=q\dot x\beta x. $$ You can take the time derivative of the left equation and get $$m\dddot x=-q\ddot y\beta x-q\dot y\beta ...


1

Bulb H is connected to second battery directly. As long as this bulb is not short circuited, there will be some applied potential difference across its terminals and hence it will be ON


1

I think your limiting procedure should work. For the homogeneous Maxwell equations, when we restrict to a surface of constant $x$, the pull back of the field strength looks something like $$\iota^* F = E^\perp_i dt \wedge dx^i + B_x dy \wedge dz,$$ where $E^\perp$ is the component of $E$ in the $yz$-plane. Now your timelike pillbox integral will have ...



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