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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 $0.5 \, \text{nm}$ ...


4

This paper is interesting. It uses the method of calculating the number of nucleons in the neutron star, $N$, based on the radius, $r$, the number density as a function of radius, $n(r)$, and the metric function $\lambda$, which comes from the equations of general relativity: $$N=\int_0^R 4\pi r^2e^{\lambda/2}n(r)dr=\int_o^R4\pi r^2 ...


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 ...


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 ...


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

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

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

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

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

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 ...


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 ...


1

The way I like to phrase pretty much all of dimensional analysis is, "you can only take an arbitrary mathematical function of dimensionless parameters: mathematics doesn't directly deal in any other sorts of functions." When you see $[[R]] = \text m, ~~[[M]] = \text {kg},~~[[v]] = \text{m/s},~~[[\rho]] = \text{kg}/{\text m^3}$ your first question needs to ...


1

Since drag is the resistance force of an object moving through a fluid (a fluid friction term), then you can make the physical argument for the value of $b$. This frictional force exists only at the boundary between the object and the fluid itself (a surface force), this means that the drag force must be independent of the mass of the object, thus $b=0$. ...


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

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 ...



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