# Tag Info

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 ...

1

Your $E$ is potential energy in the rubber, which transforms to kinetic $K$. So your starting velocity $v$ will be: $E=1/2 mv^2$ From conservation of energy: $0-1/2 mv^2=0-mgh$ $h=v^2/2g$ and $v^2=2E/m$ Confirmed Interestingly enough the rubber does not obey Hook's law, and you need a lot more work if you want to find out what really ...

0

You may assume the origin to be at the the starting point of train A. Then find the time after which the two trains meet. Finally you can find the displacement of train A using the equation of motion $$S= vt + \frac{1}{2}at^2$$.

2

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 ...

0

Previous answers are great and explain the dynamics very well. I'd like to point out that this can be explained just as easily in a static situation. Imagine the weight the shaft has to carry. You don't even have to imagine a curve to note that the weight will automagically center (and lower the center of gravity of) the train in the first image. In ...

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 ...

9

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, ...

0

If the mass was simply attached at the current position and let free, your energy method is correct. If the mass was attached by hand and slowly allowed to stretch the spring until it remains fixed at a stable equilibrium point, the force method would have been correct. However, for your given situation I believe the question is asking you for the first ...

0

The two approaches do apply to different situations. Let's first look at the "energy method". Here you describe what happens to the different energy components as the mass starts to fall from $x=h$. The initial energy is indeed $mgh$. Now, after that the mass has fallen a distance $x$, the energy is composed of three components: the residual potential energy ...

0

Your force considerations are wrong. Note that there is only force on the block (i.e. $kx$); but there are actually two forces which are extending the spring. One is being applied by the block hanging(which you have considered). But you have not considered the force on the spring due to it's attachment to the ceiling. There will be an additional equal ...

0

The covariant derivative of a field have values in the Lorentz algebra in the representation of the field, spinorial representation in this case, I.e., $$\mathcal{J}^{ab} = - \frac{\imath}{4}[\gamma^a,\gamma^b].$$ Thus the covariant derivative for spinors is $$\nabla_\mu \epsilon = \partial_\mu \epsilon + \frac{\imath}{2} \omega_\mu{}^{ab}\mathcal{J}_{ab},$$ ...

-1

To see why the first configuration is used rather than the second, perform the following experiment: Hold a bowl in your hand and place a small ball inside. Move the bowl in circles at various speeds and observe the behavior of the ball. Now turn the bowl over and balance the ball on top. Again, move the bowl around and observe the ball. Which is more ...

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 ...

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

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 ...

-2

I can see a difference by thinking about the flanges... Given that 1) the flanges are on the inside of the wheels 2) the right handside of the diagrams is the outer part of the track where the flange will press against the rail... compare and where the red lines indicate the plane of the flange.... in the upper case the flange neatly pushes ...

0

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.

0

you can also take negative charges to define potential of a point.there would be electric field due to a -ve charge Q also and when you bring a point charge q(having -ve charge) near Q then you have to do work against the field.(because same charge repel each other )

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

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 ...

0

It appears to me that you did it right. Could there be a typo on the exam? I'm a high school physics teacher, and occasionally when I make up an exam and proof read it, an error slips through (which, of course, is embarrassing).

0

The force of the changing magnetic field on the electrons in the conductor can be deduced by imagining field lines traversing the conductor (from outside the ring to inside). This results in a force that is not necessarily parallel to the wire - but if the wire is very thin, then a tiny motion of the electrons will cause a polarization of the wire, after ...

0

I think I can answer in the case of the circular and square ring. Firstly, by cylindrical symmetry, we know that the E field is the same all the way around the circle. Why there is no radial component: Imagine rotating the system by 180 degrees about any diameter. At any location, the direction of the magnetic field would reverse, the direction of the ...

0

part i) answer is just $\int \int \frac{d^3p_1}{2E_1}\frac{d^3p_2}{2E_2} = \int\int \frac{E_1 dE_1 dcos(\theta_1) d\phi_1}{2}\frac{E_2 dE_2 dcos(\theta_2) d\phi_2}{2}$. Given that we have in our integrand a piece that only depends on relative angle between $\vec{p_1}$ and $\vec{p_2}$, if we orient our axes in the direction of $\vec{p_1}$ and integrate ...

0

Consider the net torque on the segment of string wrapped around the peg about the peg's centre. The reaction force the peg exerts on the string at each point is radial, and so does not contribute to this net torque. This net torque must therefore equal $\mathbf{T_1} \times \mathbf{r_1} - \mathbf{T_2} \times \mathbf{r_2}$, where $\mathbf{r_1}$ and ...

0

This is a mechanical advantage problem. For fixed pulleys, only the direction of motion is changed, and there is no mechanical advantage. A 1 N force directed downward on one side of the fixed pulleys (the small ones) produces a 1 N force directed upward on the rope on the other side of the fixed pulleys. For a movable pulley (the large one), there are ...

-3

Twin paradox modified for a periodic universe I do wish people would keep speculations out of physics lessons. The Planck mission found no evidence of any kind of toroidal topology. There's no evidence at all that we live in "spatially periodic" universe. Besides, the old game was Asteroids, not Pac Man. Can part (b) be solved only using special ...

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

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

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

I think an easier way would be to do $$\frac{u \left(\sin(53^\circ) - \sin(33.7^\circ)\right)}{2g}$$

0

Update: The answer is 120 degrees using the angles between all the axis x = cos^-1 (+/- sqrt(1-cos^2(y)-cos^2(z)) x = cos^-1(+/- 0.493) x = 60 or 120

0

The angles given give you 2 out of 3 of the direction cosines, namely $\gamma_y$ and $\gamma_z$. What relationship do the three direction cosines $\gamma_x$, $\gamma_y$ and $\gamma_z$ fulfill? Recall that they are the components of a unit vector, since they are the lengths of the projection onto the three axes of a unit vector pointing along the boom.

-1

Use the trigonometric table for finding the values ( if you want an easy way out) or try doing a bit of adjustments here or there to suit your needs.

0

It would certainly seem that Alice is younger, assuming that Bob is the one in the inertial reference frame described at the beginning of the problem. The logic would bethat as measured in Bob's frame, $\Delta x_B = 0$, $\Delta x_A = R$ ($R$ here being the "radius" of the Universe; in other words, Alice has "come back" to Bob's location), and $\Delta t_A = ... 0 If you take away mass in Newtonian mechanics, then any force leads to an infinite response. The spring force then promises an infinite negative-feedback response to any deviation from equilibrium, and depending on how you take the limits involved, you either get infinite sinusoidal motion with a period of 0, or perfect rigidity with no motion and a period of ... 1 Assuming that the angle theta is measured relative to the vertical (e.g., the position of the string when the pendulum is at rest), a careful free body analysis indicates that the acceleration of the pendulum is g * sin(theta). This means that the acceleration of the pendulum continuously varies as it swings. This is relevant because the kinematic ... 0 If$[A,B]=0$then there is a unitary transformation$U_{A,B}$that diagonalises both$A$and$B$simultaneously. This transformation depends on the pair of commuting operators$(A,B)$, so that for a different pair there could be a different unitary. Assume that all the eigenvalues of$H$have multiplicity 1. Then there exists a unique (up to a phase factor) ... 1 If$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$, but it cannot share an ... 1 There are two observations that can be made about this problem. 1) If T1 is not equal to T2, the string will slip on the peg, which is frictionless. 2) If the reaction force from the peg, force R1, is not perpendicular to the peg's surface, there will be a component of R1 that is parallel to the surface of the peg, and the peg will rotate. For a ... 0 You have$x$and$y$(the displacements along those axes) represented as functions of some variable. Velocity is defined the instantaneous rate of change of the displacement, with respect to time. (i.e. its time derivative) The velocity in the$x$and$y$directions therefore, would be$\frac{dx}{dt}$and$\frac{dy}{dt}$respectively. (Because projection ... 0 Each of the components of$A^i$depends on$(x,y,z,t)$thus you must use the chain rule and evaluate$(dA^i/dt) + (dA^i/dj * dj/dt)$where the j's are summed over x, y, and z. 0$T = \sqrt{\frac{m}{k}}$from solving the differential equation$m\ddot{x}+kx = 0$However the mass doesn't change when you go to the moon only weight changes. so the period should be the same. 0 Assume T1>T2 and T1 and T2 acting in the same direction. The difference T1-T2 will just lead to a slipping of the string over the frictionless peg. This will not cause a reaction force. The 'common' component, i.e. the part of the force that is felt on either side of the peg T2 causes the reaction force R. So R is equal to the minimum of T1 and T2: ... 2 Indeed, the$\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}$will remain constant. 0 I do not know whether I understood exactly your question. In any case, I do not think that you need to use Helmholtz decomposition to establish that$E=-\nabla\phi-\partial_t A$(using$c=1$). Particularly, the fields do not have to fulfill the hypotheses of Helmholtz decomposition in all space for this identity to be true. Maybe we could work in ... 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 ... 0 I'd say the problem doesn't make much sense as it stands. Without terms that include a time derivative, the system would just immediately jump the the configuration with the smallest$L\$, as the speed with which it does so is not "punished" (does not increase the action). In other words, the solution with the smallest action is the one with ...

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 ...

0

It's s not always 0, but it is safe to assume that it is the relative difference between the plates that is important, not the absolute value.

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