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## Hot answers tagged homework-and-exercises

14

The simplest formula for the centrifugal acceleration is $$a = r\omega^2$$ Here, $r$ is the radius which is 0.25 meters in your case. $\omega$ is the angular velocity which is $2\pi$ times the frequency $f$. Your $f$ is 1500 revolutions per minute which is $1500/60=25$ revolutions per second. In the SI units, we have $$a = 0.25\times 4\pi^2 \times 25^2 = ... 5 When two quantities of water (m_1 and m_2) at different temperatures (resp. T_1 and T_2) are mixed in adiabatic conditions (no heat loss and no external heating during mixing) the temperature T of the resulting mixture can be calculated from the heat balance (no heat is lost or added so the heat contained in both masses is found again in the ... 4 Well, actually you are looking for a one-parameter group of diffeomorphisms (or isometries if referring to the boost vector field). This group is obtained by solving the differential equation$$\frac{dx}{ds}= X(x(s))\tag{1}$$with a generic initial condition z at s=0 in the manifold M (Minkowski spacetime in your example). X is your vector field on ... 3 Calculating the sum of the interior angles precisely woud be a big task as we'd need to compute the trajectory of the light ray and there isn't a convenient analytic expression for this. However we can easily calculate an upper limit for the interior angles. The key fact we need to know is that the deflection angle \theta of a light ray in the ... 3 You meant the angular velocity vector \vec{\omega}, I think. But adding the velocity vector \vec{v} to the angular velocity vector \vec{\omega} would be like adding apples to oranges. Look even at the dimensions of the scalars of these vectors, for velocity that is \mathrm{m/s}, for angular momentum it is \mathrm{s^{-1}} (angles have no dimension ... 3 The ladder operators satisfy: \bf{a^{\dagger}}$$|n>=\sqrt{n+1}|n+1>\bf{a}$$|n>=\sqrt{n}|n-1> Taking into account <n|m>=\delta_{n,m} , you get the answer. 3 Take a trace of Einstein equations (trace of g_{\mu \nu} is D), you obtain$$R - \frac{D}{2} R + D \Lambda = 0$$Or$$R=\frac{D \Lambda}{D/2-1}$$Then substitute this expression for R into full Einstein equations and you obtain trivially$$R_{\mu \nu } = \frac{\Lambda}{D/2 - 1} g_{\mu \nu}$$2 It is actually possible, and not too difficult, to prove this without expanding the exponentials to first order only. What you are trying to prove is S^\dagger \gamma^0 = \gamma^0 S^{-1}, this is equivalent to$$ \gamma^0 S^\dagger \gamma^0 = S^{-1} $$because ( \gamma^0 )^2 = 1. Expand S^\dagger = \sum_n \frac{1}{n!} \left( \frac i 4 \omega_{\mu\nu} ... 2 It gets easier if you use the result from part 1. Then you also don't have to deal with the \mathcal O(\omega^2) (see my answer to your other question). In your calculation, you transformed \bar\psi and \psi, but not \gamma^\lambda. This is correct, as I will show in the end, but I will take another point of view which is really helpful here: ... 2 If you write the current as a function of time, I(t), then the root mean square current is:$$ I_\text{RMS}^2 = \frac{1}{\tau}\int_0^\tau I^2(t)dt $$where \tau is the period of the waveform. In this case I is always \pm 2 so I^2 is always 4 and the integral becomes:$$ I_\text{RMS}^2 = \frac{1}{\tau}\int_0^\tau 4dt = \frac{1}{\tau} 4\tau = 4 ...

2

I'm guessing you're misinterpreting this diagram: as meaning $\vec \omega$ is "along the plane of motion". That's not what the curved arrow is meant to denote, it is meant to denote the rotation around $\vec \omega$. The angular velocity $\vec \omega$ itself, since $\vec \omega\propto\vec r \times \vec v$, is always orthogonal to $\vec r$ and $\vec p$ and ...

1

If you accelerate your car with constant acceleration AND we then assume that the friction decelerates the car with constant (negative) acceleration, then you simply consider each situation for itself: $$x_{acc}=x_{0,acc}+v_{0,acc}t_{acc}+\frac12a_{acc}t_{acc}^2=\frac12a_{acc}t_{acc}^2$$ ...

1

Inserting the expansion $$\psi=\int\frac{d^3p}{(2\pi)^32\omega_p}(a_pe^{-ipx}+b_p^\dagger e^{ipx})$$ into the expression for the Hamiltonian $$H=\int d^3x(\dot{\psi}^\dagger\dot{\psi}+\nabla\psi^\dagger\cdot\nabla\psi+m^2\psi^\dagger\psi)$$ we get H=\int d^3x\int\int\frac{d^3p}{(2\pi)^32\omega_p}\frac{d^3p^{\prime}}{(2\pi)^32\omega_p^{\prime}}(A+B+C) ... 1 The confusion lies with the definition of the transition probability. The transition amplitude between an H_0 eigenstate H_0 \left|m\right> = E_m \left|m\right> and another eigenstate H_0 \left|n\right> = E_n \left|n\right> due to a perturbation V after a time t is given by \begin{align} \left< n \right| U(t) \left| m \right> ... 1 The key is this: how is the force applied to the rider? Gravity pulls directly down on the passenger's mass, but what keeps the passenger from heading towards the center of the Earth? When you're in the passenger seat of a sports car, and the driver floors it, you feel pushed back into the seat. But what's actually happening is that the seat is pushing you ... 1 The rider feels "forced down" because the object to which they are attached is accelerating upwards. Because the acceleration is opposite to gravity, the normal force, \mathbf{F}_{N}, being exerted on the rider must increase in magnitude (relative to the "at rest" magnitude on a horizontal track) in order to produce a net upwards acceleration. Thus, it ... 1 "The frequency doesn't change" is only true when the core is perfectly linear. For a real transformer, there will be some nonlinear effects (saturation) meaning that the sinusoidal input waveform will create harmonics in the output - second harmonics and higher frequencies will appear. But if you ignore those, then the flux change will vary sinusoidally at ... 1 A short pendulum with a large rigid bob will run slower than expected, because the force required to rotate the bob is added to the force required to swing the bob, reducing the accelerations. As the pendulum length becomes "long" relative to the size of the bob, then this effect becomes "small" and you can better and better approximate the bob as a point. 1 You can decouple the horizontal and vertical motion of your rocket. In the vertical direction you have vertical thrust and gravity and horizontally you only have thrust (I ignore air resistance here). As you are interested in the altitude only, we only look at the vertical problem. All kinetic energy in the vertical direction is converted to potential energy ... 1 Remember Lenz's Law: as you change the flux through a coil, an e.m.f. is generated that opposes this change. Therefore, if I have two coils that are a certain distance apart, they will have a certain "shared" flux - flux due to A appearing in coil B, for example. Now if we bring A closer to B, we change the flux in B due to A, and will get a ... 1 Let the wavefunction bey(x,t)= A\sin\left[\frac{2\pi}{\lambda}(x- vt)\right].$$Now,$$\frac{\partial y}{\partial x}= \text{Rate of change of wavefunction when time is constant}\;,\\ \frac{\partial y}{\partial t}=\text{Rate of change of wavefunction when position is constant or transverse velocity}\; .$$Now differentiate y w.r.t. x keeping t ... 1 You should also match the derivatives at x=0 so that they took into account the \delta-function. If you take smooth well V_\epsilon(x) and consider small region near zero (-\epsilon,\epsilon) where the "meat" of the well is concentrated you may then integrate the Schrodinger equation at that region, ... 1 By using work energy theorem it can be solved. The velocity of the 2kg object till it reaches the 6kg object is given by$$\sqrt{2*g*5}=9.90m/s^2$$apply the conservation of momentum for plastic impact.$$m_1u_1+m_2u_2=m_1m_2V2*9.90+6*0=8*VV=2.475m/s^2$$work energy theorem$$\frac{1}{2}*8*2.475^2=\frac{1}{2}*72*(-x)^2+8*g*xx=1.801472656m$$... 1 using$$a^{\dagger} |n\rangle = \sqrt{n+1}| n+1 \rangle$$and$$ a |n\rangle = \sqrt{n} |n-1\rangle, $$apply these rules in order:$$ a(a^{\dagger}|n\rangle) = a\sqrt{n+1}|n+1\rangle = \sqrt{n+1}a|n+1\rangle = (\sqrt{n+1})^2 |n\rangle = (n+1)|n\rangle $$1 Newton's law of cooling is a corollary of Fourier's law of heat conduction:$$q=-\kappa \nabla T,$$where q is the heat flux, \kappa the heat conductivity and \nabla T the temperature gradient (in a single dimension \nabla T=\frac{dT}{dx}). In essence this law tells us that heat flows from hot to cold and that the heat flow is proportional to the ... 1 Yes , she is very correct. Consider motion of elevator. d= vt = 6*3 = 18 So, elevator travelled 18m. Now consider bolt h= ut + 1/2 gt^{2} -18 = 3u -10/2*9 u= 9m/s Now, v= u + at = 9 - 10*3 = -21 m/s so bolt falls with speed 21m/s. 1 This type of problem can be simplified if you use the frame of reference of the elevator. Now the bolt falls from rest, chasing the "starting point" which starts a distance h below and moves down at a constant 6  m/s. The bolt catches up in 3 seconds. Same problem, but the equations are simpler... 1 The summation indices on Z must match the summation indices on \bar{E}, because Z is the normalization constant for the total probabilities of being in every state. The n=0 state has zero wavenumber and doesn't exist. You're right to discard the analogy with para hydrogen. In that case, the peak is caused by interaction of the nuclear spins of the ... 1 Just solve the second order differential equation obtained from using Newton's Laws i.e.$$F=-\frac{k}{r^3}$$or$$m\frac{d^2r}{dt^2}=-\frac{k}{r^3} If you solve this differential equation, then your equation for the path will be of the radius as a function of time. The equation will be a non-central conic. HINT (TO SOLVE THE DIFFERENTIAL EQUATION): ...

1

The issue here is whether or not you have a sum over $l$. If you want $r_l^2$ to mean any of $r_1^2, r_2^2, r_3^2$, then when you write $r_l^2 = r_l r_l$ you should not be summing over $l$. So $[r_l^2,L_i] = 2i\hbar \epsilon_{ijl}r_jr_l$ is correct as long as you sum over $j$ but not over $l$. On the other hand, if in $r_lr_l$ you sum over $l$ you get the ...

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