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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 ...

0

Your question has some inconsistencies and the first half of the question doesn't match up with the second half, consider then also want you to answer what happens if a→0 and a→R. I can assume that the question wants you to consider a shell and a cylinder with a thickness. (Also, they have given outer and inner radii, so that makes sense) You are expected to ...

0

I think there is a simpler approach: When you have pulse with velocity $v$, the value of the function has to be constant along lines given by equation $x = x_0 + vt$. In other words: $$f(x_0 + vt,t) = \rm{const}$$ Before we proceed further, we notice that the argument of exp can be simplified like this: $$f(x,t) = A \exp \biggl[-\biggl( \frac{ax-bt}{c} ... 1 Just solve the second order differential equation obtained from using Newton's Laws i.e.$$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 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 ... 0 a) consider a segment of the cable of length L. Calculate the charge contained on the inner cylindrical core of radius a. Then set the same charge to the outer cylinder of radius b. Calculate sigma from this charge and the area of cylinder with radius b and height L. b) for electrostatic case, the field for r < a must be zero if the inner core is a ... 0 Work is the change in kinetic energy. In both cases, the man starts on the ground at rest and end on the chair at rest. In both cases, the net work is zero. If you want to be more specific and describe the work due to his muscles, that positive work must exactly offset the negative work due to other sources. Gravity does negative work equal to minus the ... 0 I'll ask the question for completeness. As discussed in the comments, if you think about all the posible ways of dissipating energy, maybe you will doubt of the veracity of the assertion made in the exercice. But, the clue is not to think in if the teacher wanted to be "captious"... but that in this case all these ways of dissipating are really small ... 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 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 Newton's law of cooling is a corollary of Fourier's law of heat conduction:$$q=-\kappa \nabla T,$$where q is the heat flux, \kappa the heat conductivity and \nabla T the temperature gradient (in a single dimension \nabla T=\frac{dT}{dx}). In essence this law tells us that heat flows from hot to cold and that the heat flow is proportional to the ... 0 For c) You should think about what it means that an orbit is closed. It means that after some time t the particle will return to it's original position. For this to happen m_1\tau_{osc} and m_2\tau_{orb} have to be equal for some integer values of m_1 and m_2. This can only happen in your example if \sqrt(n+2) is rational. 0 Why don't you straight away solve this question, assuming our system to be a variable mass system. m\frac{dv}{dt}= F_{external}+ V_{relative}\frac{dm}{dt} Let the length of the left half of falling chain be x, and the length of the right half be y. Consider the left half as your system. If the initial length of chain was L, then that can be equated ... 1 using$$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 $$5 When two quantities of water (m_1 and m_2) at different temperatures (resp. T_1 and T_2) are mixed in adiabatic conditions (no heat loss and no external heating during mixing) the temperature T of the resulting mixture can be calculated from the heat balance (no heat is lost or added so the heat contained in both masses is found again in the ... 0 You arrived at almost correct answer. I get net horizontal accelerating force as:$$ F_x(\theta) = F_{pull}(\cos\theta+\mu\sin\theta) - \mu m g $$You are just missing the \mu before the sin and it also has a different sign. Double check your diagram, especially the orientation of forces. But nevertheless you still seem to arrive at almost correct ... 0 Did model rocketry for a while... from a practical standpoint this isn't feasible... unless you're talking about deploying something that drops at a rate you know at the time of the parachute deploying (like a weighted streamer. See the link below). Your chute deploys and you don't know the descent rate of it, so there's no way to find the vertical distance. ... 0 Drawing a free body diagram, the following equations can be found: F_y = F_\text{pull}\sin(\theta) + F_\text N - mg F_x = F_\text{pull}\cos(\theta) - F_\text R Stipulations on variables: F_\text N (normal force) must be positive; the ground cannot exert an attractive force. F_\text R (frictional force) must be less than or equal to \mu ... 1 You can decouple the horizontal and vertical motion of your rocket. In the vertical direction you have vertical thrust and gravity and horizontally you only have thrust (I ignore air resistance here). As you are interested in the altitude only, we only look at the vertical problem. All kinetic energy in the vertical direction is converted to potential energy ... 1 A short pendulum with a large rigid bob will run slower than expected, because the force required to rotate the bob is added to the force required to swing the bob, reducing the accelerations. As the pendulum length becomes "long" relative to the size of the bob, then this effect becomes "small" and you can better and better approximate the bob as a point. 0 For the system the Hamiltonian is : $$\notag H=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}+\frac{K}{2}\left(x_2-x_1\right)^2 \ .$$ You write down the Hamilton equation of motion: \notag \begin{cases} \dot{x_1}=\frac{p_1}{m_1}\\ \dot{x_2}=\frac{p_2}{m_2}\\ \dot{p_1}=k(x_2-x_1)\\ \dot{p_2}=k(x_1-x_2) \end{cases} ... 1 "The frequency doesn't change" is only true when the core is perfectly linear. For a real transformer, there will be some nonlinear effects (saturation) meaning that the sinusoidal input waveform will create harmonics in the output - second harmonics and higher frequencies will appear. But if you ignore those, then the flux change will vary sinusoidally at ... 0 Let \color{blue}{i=i_0\sin(\omega t)} be the input alternating current to the primary coil with a frequency \color{red}{\omega} then the voltage induced in the secondary coil of transformer is given as$$V_{in}=-M\frac{di}{dt}$$Where, M is the mutual inductance setting the value of i,$$V_{in}=-M\frac{d}{dt}(i_0\sin(\omega t))=-Mi_0\omega ...

0

If the question is viewed correctly the batteries are not connected to each other, therefore it is an open circuit. Or if it is connected the batteries are facing each other and the values are not mentioned and so let us assume they are equal. Then the net voltage is zero which is the same as no voltage so V across the 4 ohm resistance is zero.

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if you imagine correctly, the magnetization and demagnetization is happening at the frequency of the AC supply. Therefore when primary side changes flux in one direction there is the same change in the secondary side and the same time interval. The same way the opposite direction in another same interval of time. So the frequency the Primary AC had will be ...

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For an object to move in a constant velocity the zero acceleration then the Forces should be zero. The $$\mu=\frac{65*g*\sin 30}{65*g*\cos30}=0.57735$$ Your answer is perfectly true. $$\mu<1$$ if your question is why does the object still move even when force is zero. Example is Terminal Velocity.in air. The reason the object comes down is that there is ...

0

Transformer work on mutual induction. Due to this frequency always be constant. Read mutual induction carefully you will get your answer. V=L(di/dt)

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 $$R_{\mu \nu } = \frac{\Lambda}{D/2 - 1} g_{\mu \nu}$$

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, ...

0

Let $\theta$ is the slope angle. The normal force is $F_{N} = mg\cos{\theta}$, which you have calculated. Now there are three important points to consider. The object is in constant speed, so $F_{net} = 0$. There are three forces acting on the object $F_{friction}, ~F_{gravity} ~and ~ -F_{normal }$ whose net sum is zero to justify (1). $F_{friction}$ by ...

0

in both cases the direction of vectors take if one vector which in angular velocity and take another vector which in linear velocity in that cases first perform rotational motion and another perform linear motion so resultant vector have no perfect direction.

1

Let the wavefunction be $$y(x,t)= A\sin\left[\frac{2\pi}{\lambda}(x- vt)\right].$$ 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 $t$ ...

0

You cannot add angular velocity to horizontal velocity...they are two diff things

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 ...

0

For eg if certain amount of clockwise current is sent through a coil nd then a second coil is brought closer....as we bring it closer the repulsion increases and the repulsion is of same magnitude as of the experienced current and hence the current is said to be increased

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.

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. $$m_1u_1+m_2u_2=m_1m_2V$$ $$2*9.90+6*0=8*V$$ $$V=2.475m/s^2$$ work energy theorem $$\frac{1}{2}*8*2.475^2=\frac{1}{2}*72*(-x)^2+8*g*x$$ $$x=1.801472656m$$ ...

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 ...

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 ...

0

After reviewing the problem further, it looks like there is an error in the text. In fact, transforming to polar coordinates should be $P_\rho(x,y,P_x,P_y)=\dfrac{x*P_x+y*P_x}{(x^2+y^2)^{1/2}}$ and NOT $P_\rho(x,y,P_x,P_y)=\dfrac{x*P_x-y*P_x}{(x^2+y^2)^{1/2}}$ which was given.

0

When the piston is moved suddenly, then the pressure on the piston is greater than the average pressure in the chamber: the process is not reversible and some energy is lost to fluid as opposed to an adiabatic reversible process. According to the first principle, and because the heat transfer to the system is 0 (insulating cylinder) the internal energy of ...

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 ...

0

As pointed out above, this is a standard ODE in which u are normally to assume a solution to the DE as Psi(x)=Aexp(ax) [where A and a are both constants] If u then substitute for the assumed solution in the Schrodinger equation, u have; a^2+k^2=0 or a^2=-(k^2) or a=+/-(ik) since u now have two values for a (ik and -ik), it means you should have two ...

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 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 ...

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Yes, the ratio is for example $4/5$ and they are asking for minimum distance $n_1=4k$; $n_2=5k$ (for some $k$), the distance will be minimum when $n$ is minimum, so $k=1$ and $n_1=4$ and $n_2=5$.

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> ...

0

Hint. ( I give absolutely no guarantee about not making calculational mistakes!) \begin{eqnarray*} L(\tau ) &=&\frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dy\left( 1-\frac{1}{ \sqrt{\pi }\sigma }\int_{-\tau /2}^{+\tau /2}dx\exp [-\frac{(y-x)^{2}}{ \sigma ^{2}}]\right) e^{-iPy}\rho e^{+iPy}=L_{1}(\tau )-L_{2}(\tau ) \\ L_{1}(\tau ) ...

0

I'll give you a hint, since this is a homework question. There are many types of resistors. One is the ohmic resistor (it corresponds to one of your graphs, I won't tell you which one) and it has constant resistance but most resistors have increasing resistance as temperature increases. That means the current decreases as temperature increases. (the ...

0

Expanding on Kavan's answer, I'd say that since the two ends of the cell are facing each other (as in they cancel out) that the dots could also be interpreted to do the same. Usually dots correspond to a series that follows the same pattern given by the first term (in this case the cells that cancel out). So I'd say 0 is your best shot. If the cells aren't ...

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