# Tag Info

## New answers tagged homework-and-exercises

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According to Griffith, the integral over the volume, extending to beyond the volume, and effectively into all space, is just mathematical convenience- $\mid E\mid^2$ decreases rapidly as distance increases $(\propto\frac{1}{r^4})$. So, if we take all of space, the contributions of the integral beyond the volume we're concerned with is negligible, but that ...

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Take the kinematic relationship $$x_{end} - x_{start} = \frac{ v_{end}^2 - v_{start}^2 }{2 a }$$ which applies to constant acceleration and use it twice. First on the first interval to calculation the acceleration $a$ and then again on the second interval to find the distance traveled.

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Steve B is right. I just want to include a diagram. You can see that the surface $\rm S$ consisting of the conducting surfaces $\rm S_1, S_2, S_3,...., S_i$ enclosing the volume $\rm V$. Suppose $\rm S_1$ is at unit potential and all the others are at zero potential. If $\rm S_1$ has $\rm Q$ amount of charge then an equal amount of negative charge is ...

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For error propagation for one variable, it is best to use $$\delta f(x) = \left|\frac{d f(x)}{d x}\right| \delta x$$ which is to say that the uncertainty in the function should be weighted with the derivative(or how sensitive the function is to changing the variable) Now for your example, $$f(x) = x^n \Rightarrow \delta f(x) = \left|n x^{n-1} \right| ... 0 Approximation Slightly better and more elegant than the first. Let's make some assumptions. Let's assume that on impact, a fraction \lambda of the ball's Kinetic Energy is transferred to the liquid. Let's also assume that all of the energy transferred will go into shooting water up. This should give the upper bound on how high the water "tower" will be. ... 1 Your moment of inertia is incorrect. You must calculate is based on the individual masses and their distances from the pivot:$$\mathcal{I}=\Large\Sigma \large\left( m_ir_i^2\right).$$If you do this you should get an answer that agrees with what @ChrisDrost did, 2.47 s And you shouldn't assume that the center of mass is 1/3 of the way below the rotation ... 2 The simplest way is to exploit the symmetries here. So, instead of mindlessly going through the algebra, solving equations etc. you just use the fact that you are free to call any direction the x-direction, and set up a right handed coordinate system. In particular, this means that you are free to cyclically permute x, y, z in the equations. The problem of ... 1 Let me first do this the way that I know is correct: with Lagrangian mechanics. This says that all of the physics you need is contained in the Lagrangian, which is the kinetic energy minus the potential energy. Your three masses Left, Right, and Bottom make the kinetic energy K = \frac 12 m (v_L^2 + v_R^2 + v_B^2), where m = \text{1 kg}. Defining ... 0 Note that it is one dimensional motion. At the end of 2 sec, a force (break) is applied, causing a change of velocity of 4 units in 2 sec which means a decceleration of 2 m/s^2. Decceleration is constant since the Vx vs t graph is a st line. You can use use the appropriate equations to find the displacement in the first sec after the application of the ... 1 By simple calculus: Between t=2s and t=4s, v_x can be described by the function: v_x=8-2t. The definition of v_x is: v_x=\frac{dx}{dt}, so that dx=v_xdt. Integrating between t=2 and t=3 we get the distance travelled in that interval: x=\int_2^3(8-2t)dt=3m. Add to this the 8m travelled in the first 2s, so total distance is ... 1 The final position will be the initial position plus the area under the velocity versus time graph. That is the area between y =0 and the velocity function. I'm assuming you're not familiar with integral calculus, but if you look at the dimensions you arrive at by calculating this area you will find that it is meters. Good luck. 3 You can't "simply" calculate the average velocity from the velocity at the end points, unless the velocity graph is a straight line. Which it is between 2 and 3. But not between 0 and 3. So the approach you can take is this: What is the distance after 2 seconds of moving at 4 m/s? And what is the average velocity between t=2 and t=3 seconds (straight ... 0 Not a proof but: Because the potential only depends on r, it can be shown that the wave function is separable into \Psi=RY. (This might be shown by David Miller in his online quantum mechanics course at Stanford in his lectures in section 7 on the hydrogen atom.) Then we can say that the Y(\theta,\phi) portion is the spherical harmonics, same as for ... 0 Yes, the field outside the conducting surface is 0. If the net charge inside the conducting surface is 0, then the fields outside the surface are zero. It's basically the Faraday Cage effect. And the net charge inside the conducting surface is 0, because that's the definition of a capacitor. You might ask, "Um, well, but what if the net charge is not ... 0 OK so why is it possible to use the tracing trick he uses on p9 eqn 1.16 if the b_i are not orthoganol basis sets? 1 As @KyleKanos says, establishing a relationship with only two data points is dangerous. However, with that warning, if you want to test a relationship between quantities which you think might have some proportionality, you can take ratios of like quantities. Then test whether there is a relationship between various powers of the ratios. For example, if ... 0 For external points, cylinder behaves like a line of charge. So you just have to put the distance in the formula of electric field of a point charge as (radius + distance from the cylinders surface) 0 The rods will get attracted towards the plane due to polarisation of the rods because of the plane.the rodsbwill surely not attract each other.They will repell each other as the induced charges have the same sign(+ or-). 1 As a simplification, you can consider that you have a 2D viscous flow between two boundaries that approach each other. Assuming that the flow is symmetrical about the line (with the line along the Y direction), you can simplify this further to "no flow at x=0". What you are left with is a pressure distribution p(x,t) whose integral in x should equal the ... 0 Since this is a thin-walled tube, the cross-sectional area of the tube is (2\pi r)w where w is the tube thickness, and r is the tube radius. It follows that the stress is:$$ \sigma = \frac{T}{(2\pi r)w}$$where T is the force. If you double w, the stress is cut in half. Contrast with solid cylinder For a solid cylinder, the cross-sectional area ... 0 The premise of the question is false. \nu is never exactly 0. It tends toward 0 as \mu_c decreases. However, it is always non-zero and positive. Hence, the value of CV depends on the other variables in its definition. 0 Thank you so much for responding. I'm afraid I did not quite follow your response. Perhaps if I explain by means of the workings below. If I trace rho S+E in the picture here using the quick method of the formula 1.16 on p9 then lines three and four below quickly give the correct result of eqn 1.76. However I should get exactly the same result using my ... 1 Let I denote our integral:$$ I=\int{{e}^{x^2}(1+erfi(x))} dx $$Using IBP:$$ u=(1+erfi(x))\quad dv={e}^{x^2}dx\\ du=\frac{2}{\sqrt{\pi}} {e}^{x^2}dx \quad v=erfi(x)\frac{\sqrt{\pi}}{2} $$You get:$$ I=(1+erfi(x))erfi(x)\frac{\sqrt{\pi}}{2}-\int{erfi(x) {e}^{x^2}dx}\\ I=(1+erfi(x))erfi(x)\frac{\sqrt{\pi}}{2}-\int{erfi(x){e}^{x^2}-{e}^{x^2}dx} ...

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The field of each quarter circle is evaluated separately about its symmetry axis: which is why the limits are taken the way the answer shows.

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Visualization The difference in height $h$ is always the same (here 10 m)! Remark This is of course only true if $g$ is constant, e. g. $h$ does not change "much". See also Wikipedia.

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Potential energy is given only as a difference of energies at different heights. So, if you want to know just how much does the person gain energy (or rather loose by friction in their muscles and joints) by walking down the hill, you might just use their height of their heels on the top of the hill and under the hill. But remember, you always have to use ...

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Theoretically, their center of mass is what you're looking for. It's somewhere near the stomach. High jumpers bend their body when they are jumping so that their center of mass can travel just above (or sometimes below) the bar which allows them to use the least energy to jump the highest: https://en.wikipedia.org/wiki/High_jump When a body is rotating or ...

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Incomplete answer: assuming the wall is a perfect reflector, we can remove it and ask at what point the sound reaches a second car driving away from the wall to the northwest, in a path reflected from the original car's through the line of the wall. We know the position $O$ of the first car when the driver pressed the horn, and at any given time $t$ we can ...

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Sound leaves the vehicle. When it strikes the wall, it reflects at a 45 degree angle and returns to the car. The triangle you will be using has legs that are length X, with a hypotenuse that is length Xsqrt2. That is for evey 1 foot of length in a leg, I will have 1.414 feet of length in the hypotenuse. What you want to solve for is time based on this ...

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The problem is arising since you are trying to take the trace in a non-orthonormal basis including both $|e_1\rangle$ and $|e_2\rangle$. If $A$ is an operator, the trace of the matrix $$A_{mn}\equiv \langle e_m|A|e_n\rangle$$ is not invariant under any basis $|e_m\rangle$. If this is the orthonormal eigenbasis, the trace is the sum of the eigenvalues. Any ...

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Your work is fine (depending on your units) and what you were asked to show is wrong. Though I do object to saying you have a force equal to $ma_c,$ I would just say that a net force orthogonal to the velocity makes it go in a circle of radius $r$ where $F=mv^2/r.$ And the problem is famous. The fact that the frequency doesn't depend on the velocity or the ...

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Solution using Laplace transforms Using the definition of the Laplace transform: $$\tilde{u}\left(y,s\right)=\int_{0}^{\infty}u\left(y,t\right)\exp\left(-st\right)dt$$ we can transform the PDE to an ODE: $$s\tilde{u} - u\left(y,0\right)=\nu \frac{d^2\tilde{u}}{dy^2} \rightarrow \frac{d^2\tilde{u}}{dy^2}-\frac{s}{\nu}\tilde{u}=0$$ with transformed boundary ...

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What happens if the particle is on the circular path? Hint. There must be something compensating the Lorentz force. EDIT: I think your identity for the angular velocity is wrong anyways. It should be $$\omega = \frac{qB}{m}$$

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A helpful starting point is to work backwards from the definition of the Christoffel symbols. Hope that helps.

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The problem is that you have not solved the question yet. What you have found is not the friction between the boxes. It is something else. As you actually state yourself, you have instead found the maximum [static] friction. This is just the maximum possible value and not at all necessarily equal to the actual friction. Static friction can be anything from ...

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If it is static friction, then the two blocks are stuck together and both have the same acceleration. In that case, the top block has a net force of 40 N (100 N pull - 60 N friction) and the bottom block has 60 N (just from the friction). Since 60N < 80N max friction, then the ansatz that this is static friction is consistent.

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You don't need angular velocity to find the linear (translational) speed. If you have answered question d.ii then you have an equation like $$T_{cen}=ma_{cen}$$ where $a_{cen}$ is the central (radial) acceleration, and $T_{cen}$ is central (radial) force towards the center. You now need to know the following formula for a circular orbit: ...

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Use Fresnel coordinates: $$ma_n=m\frac{v^2}{r}=F_n\\ ma_t=\frac{dv}{dt}=F_t\\$$ Now $F_n$ denotes the total forces in the radial direction, and using a diagram you find: $$F_n=P_n+T_n=T\sin{(30)}$$ Now for the tangential direction do the same thing to find: $$F_t=P_t+T_t=-mg+T\cos{(30)}$$ Note that the speed is constant so $dv/dt=0$: $$F_t=0\\ ... 1 While not necessary to solve this problem, I want you to know that... Concept # 0: the angular velocity of circular motion is directly proportional to the linear velocity of motion,$$ v = \omega r $$where v is the linear velocity, \omega is the angular frequency, and r is the radius of circular motion. Concept # 1: whenever an object exhibits ... 0 The electron doesn't absorb the photons, it scatters them. The energy absorbed by an interaction depends on the scattering angle - this can be determined using the Compton formula for the wavelength of the scattered photon. And the electron tends to scatter more at different angles (proportional to the Thomson differential cross section). From this I found ... 0 To solve these, it's enough that you know:$$ v = v_0 + at  x = x_0 + v_0 t + \frac{1}{2} a t^2 $$Fill in the values you know, then solve for the value you want. 0 The carbon dioxide panic is so widespread that these days, virtually all carmakers make the data available for the CO2 emissions per kilometer directly. For example, this is the table for all Škoda models https://www.fleetnews.co.uk/cars/co2-emissions-calculator/skoda/ Replace "skoda" in the URL by "audi" or another brand to get the numbers you're ... 1 As the value of b increases the resistance between the outer and the inner shells will converge to 1/4 \pi \sigma a. If we consider the outer shell to be at the "infinity", the resistance between the "infinity" and the inner shell will be 1/4 \pi\sigma a. We can think of the situation in which there are two shells in the infinite sea of poorly ... 1 yep, think of  \xi  as a unit vector and replace all instances of it with  \epsilon \xi where  \epsilon  is some small number. Then you will see that those two terms are second order in  \epsilon . 0 The suggestion of Kevin Zhou, on other hand, has the advantage that we could try to control all the poles of the original well and the halved one. The polology of the square well was studied in Nussenzveig "The poles of the S-matrix of a rectangular potential well or barrier" Nuclear Physics, 11:409–521, 1959. and revisited in his book "Causality and ... 0 I'd say that the non adiabatic problem is ill-defined. Lets see. All the initial solutions \Psi_n(x) = A_n \sin(n \pi x / 2a) have a probability current \Psi'^* \Psi - \Psi^* \Psi' equal to zero everywhere, so in principle you can cut them at any point without having a leak of probability. The problem is that the boundary conditions for the domain of ... 0 This is more geometry than physics, you might be better on Mathstackexchange, but a few hints: This is simple resolving of vectors. Think about what you WOULD do if you knew the angle theta and the magnitude of Fb in order to find the magnitude of F. By my reckoning you turn out with a simultaneous equation for theta and the magnitude F. Apologies if this ... 1 If this is a correct description of what happens, can we conclude that g does same work on P and on P'? Yes. This is correct. If g acts perpendicularly to the velocity, it performs work of magnitude zero. This is also correct. The reason the two statements above are not contradictory is that the work done by the gravity changes the direction of ... 1 Taking the Hermitian conjugate reverses the order of the \psi's. You have$$ L_M^\dagger = \left( \bar{\psi}\psi\right)^\dagger = \left( \psi^\dagger \gamma^0\psi\right)^\dagger = \psi^\dagger{\gamma^0}^\dagger \psi = \psi^\dagger\gamma^0\psi = \bar{\psi}\psi = L_M \ ,  where we use that $\gamma^0$ is Hermitian.

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I will not answer your question directly; only give you some tools that should help you answer the question (in practice) yourself. To focus the attention, find below a typical heating/cooling diagram for a frozen pure substance. The vertical axis marked $T$ represents temperature (in degrees Celsius). Three significant temperatures are indicated on the ...

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