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5

Since the gravitational force only pulls the ball down, but not back or forth, it will not experience any acceleration changing its forward velocity but only downward acceleration. Thus, the ball will return to the thrower. You can also imagine the train to have no windows and be moving extremely smoothly. The thrower won't know if the train is moving or ...


5

$\int^t_0 A x^2 dt = x_0 + A x^2 t$ is incorrect. You are assuming $x$ as a constant. $x$ is a function of time x(t). Try $\dfrac{dx}{dt}=Ax^2 \implies \dfrac{dx}{x^2}=Adt$. Now integrate both the sides in appropriate limits. $$\int_{x_0}^{x(t)}\dfrac{dx}{x^2}=\int_0^t Adt$$ $$\int_{x_0}^{x(t)}x^{-2}dx=\int_0^t Adt$$ ...


3

First problem: you say $v(t) = A x^2$, but that is a function of position, not time. Putting the definition right: $$ v = \frac{dx}{dt} = A x^2 $$ You can regroup terms on the same variable: $$ \frac{dx}{x^2} = A dt$$ And then do the integration: $$ \int \frac{dx}{x^2} = \int A dt$$ This is homework, so I will leave the integral limits and the ...


3

Think of it in terms of current, V, W, and Z are in series, so each is equally bright. X and Y are in parallel, so each gets half the current of the others. If you assume each bulb is a constant resistance R (not true for incandescent bulbs, by the way), then V,W and Z will each dissipate $i^2R$. For X and Y, since each has a current i/2, the power will be ...


3

You're running into a tricky property of statistical variables: what is true for an individual particle is not necessarily true when averaged across a distribution. In particular, you can say that the distance one particle travels between collisions, its free path length $\ell$, is equal to that one particle's speed times its free path time $t$: $$\ell = ...


3

The easiest thing for this exercise is to use Levi-Civita symbol for the vector product: $$\vec{a} \times \vec{b} = a_i b_j e_k \varepsilon_{ijk},$$ where I denote by $e_i$ the canonical basis of $\mathbb{R}^3$. Using this notation, we have: $$[L_j,p_i]=[r_k p_l \varepsilon_{klj},p_i]= i \hbar p_l \varepsilon_{ilj}.$$ and $$[L^2,\vec{p}]=e_i[L_j ...


2

The particle continues moving until it reaches the maximum height. At maximum height its velocity vanishes. In return, it again takes $T1$ to travel from $B$ to the ground. So it takes $\tfrac{T2 -T1}{2}$ for it to travel from B to the maximum height. $V_0$ denotes for initial velocity of particle on the ground and $H(B)$ the height of B, ...


2

The point is: you can apply Newton's law $dp/dt= F = mg - T$, but you can't assume $dp/dt = m dv/dt$ as $dm/dt$ is nonzero. The fundamental law in Newtonian dynamics is the object's rate of change in momentum equals the net force exerted on it. Only in the special case of a constant mass object can this be translated into a statement about mass times ...


2

Simply consider Maxwell equation : $$\vec{\nabla}\wedge\vec{E}=-\frac{\partial\vec{B}}{\partial t}$$ If you interger this on a given closed surface $\Sigma$, it follows : $$\oint_\Sigma \left(\vec{\nabla}\wedge\vec{E}\right) \cdot d\vec{S} =-\frac{\partial}{\partial t}\oint_\Sigma \vec{B}\cdot d\vec{S}$$ where $d\vec{S}=dS\,.\vec{n}$ with $dS$ the ...


2

Your expression for the velocity looks right; but we have to get a few other things taken care of. First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression. On the other hand, you need to take account of the energy of the sphere ...


2

The vertically moving object is an Atwood machine and the two masses have their own accelerations that are in different directions. The acceleration of $m_2$ and $m_3$ (separate from the total system) is given by $$ a=\frac{m_3-m_2}{m_2+m_3}g\tag{1} $$ Mass $m_2$ is accelerating upwards, hence the acceleration in your case of $a_0-a$; likewise mass $m_3$ is ...


2

Yes , that's easy because mAh is presented for a specific voltage that battery can output. Because the battery must output at a specified voltage. Your 5600mAh is for 5 volts. If it's not 5 volts. You must recalculate it. But the voltage shouldn't be change if the battery is not adjustable. Hope you understand


2

I'm assuming the proton at x=0 and the alpha particle at x=2 are held in place. I'll call these the "held" particles. There will be two forces which act on the "free" proton (at x=-1). I'm also assuming that this is just a classical mechanics problem, and will solve it that way. Before we go into the calculations, let's think about what is going to happen. ...


2

You didn't really simplify the expressions in the same manner, so they're hard to compare. Even if they're not the same, it's often useful to have a clear idea of what differs in the final result. Simplifying the first expression to share the denominator $(b^2-c^2)^2$: $$ \begin{align} W&=\frac{1}{2}\int\rho V\,d\tau\\ &=\frac{\pi ...


1

We will work with the metric supplied, $$\mathrm{d}s^2 = A(r)^2\mathrm{d}t^2 -B(r)^2 \mathrm{d}r^2 -r^2 \mathrm{d}\theta^2 - r^2 \sin^2\theta \, \mathrm{d}\phi^2$$ I will assume that omitting the fourth spatial coordinate (which I have added) is simply a typo in the original post. I have also redefined the arbitrary functions for convenience. We choose a ...


1

The problem in yours is that you are taking the net force acting downward to be $(m_2+m_3)g$ is incorrect and that led you to take the total mass to be $m_1+m_2+m_3$ which is again incorrect because $m_2\neq m_3$. If $m_2=m_3$ then the center of mass of $m_2$ and $m_3$ will lie on the straight vertical line through the center of the pulley B and the force ...


1

As you have written, $$ \epsilon = {\Delta \phi \over \Delta t}$$ This equation says that $ \epsilon $ is greatest when the change in flux with respect to time is greatest, not when the flux itself is greatest. In order to find when the change in flux is the greatest, you need to come up with an equation for the flux, then take the derivative with respect to ...


1

Spin1/2 particle Ususally, in this kind of Hamiltonian, people uses $s=s_z$, where $$s=s_z=\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right].$$ Then, your unperturbed hamiltonian $H_0$ is: $$H_0=-\mu s\cdot B_0 = -\mu \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right]B_{0,z}. $$ Then the eigen vectors of energy are: ...


1

It sounds like you're trying to find the shifts in the energy levels caused by $H_V$. To find the energy shift that goes as the first order in your small parameter you compute $$E_n^{(1)} = \langle \psi_n^{(0)}|H_V|\psi_n^{(0)} \rangle \quad (1)$$ as you noted. In this expression $|\psi_n^{(0)}\rangle$ means "the $n^{\textrm{th}}$ eigenstate of the ...


1

Maybe you have seen the experiment where a permanent magnet is dropped through a pipe made of conducting metal. The magnetic flux through a cross-section of the pipe will be changing, so a current is induced. The induced magnetic field is such as to oppose the change in magnetic flux, so it will slow down the falling magnet. Eventually an equilibrium is ...


1

Let's assume the light from the Sun is parallel, then the shadow of Earth looks like this: The dotted line is the orbit of the satellite at a height $h$ (I've exaggerated the height a bit to make the diagram clearer). All we have to do is calculate the angle $\theta$, because the time the satellite is in the Earth's shadow is simply: $$ t = \tau ...


1

Well, I guess the answer is it becomes both! You're almost there! But then the third and fourth terms in your equation will vanish. So, since $\hat{n}$ is the number operator, a number state, such as $\vert m \rangle$ say will be an 'eigenfunction' or the $\hat{n}$ operator with 'eigenvalue' m. It seems that you probably already know this judging by how ...


1

Unfortunately I cannot comment due to insufficient reputation, so here a comment on the question. There are three cases: $\frac{1}{2}mv_A^2>2mgR$ In this case the pearl has a velocity $v>0$ in the top point and will continue its movement. $\frac{1}{2}mv_A^2<2mgR$ In this case the pearl won't reach the top and will oscillate around point $A$. ...


1

Firstly, since "The projectile starts on the ground and gets launched at the specified angle over a flat surface", $y_0 = 0$ and the equation reduces to: $$ d = \frac{v^2sin(2\theta )}{g} $$ Now you can see that $d$ will be the same for $\theta $ and $90 - \theta$, such as 15 and 75 degrees; however, this is not true for the more general equation in the ...


1

Actually, this is a very general issue that affects all continuous measurable quantities. Whenever you make a measurement of a continuous quantity, you are choosing from all possible real numbers within the allowed range for that quantity. There are an infinite number of possible values, and thus the probability of getting a specific one of them - like $\xi ...


1

Regarding your "at least" vs "greater than": I think your intuition about "at least" vs "greater than" works for discreet outcomes, but not continuous ones like what you are considering. Well, at least I think your parameter squiggle is continuous? Read on if so. Think about a six-sided die. This is a discreet case. The probability to roll at least 3 is ...


1

With $$u(n)=u_0e^{-i(\omega t+k n a)}$$ we have $$p=\sum_{n=1}^Nm\frac{d}{dt}u(n)=i\omega m u_0e^{-i\omega t}\left(\frac{1-e^{-iakN}}{1-e^{iak}}\right).$$ With cyclic boundary conditions we have $$u(N+1)=u(1)\Rightarrow k=\frac{2\pi j}{Na}\text{ for }j\in\mathbb{Z}$$ and inserting $k$ into $p$ gives $$p=i\omega m u_0e^{-i\omega t}\left(\frac{1-e^{-2 i \pi ...


1

Consider two masses M and m in circular motion with same velocity,v. Both has acceleration v^2/R. The forces acting on the two masses are different. Force will become more on the greater mass. But acceleration of both are same. Because, if you put M and m in the following relation, you get same v^2/R. $$(mv^2/R)/m=v^2/R$$ since we know $$F/m=a$$ where ...


1

You can think of acceleration from a purely mathematical context - it's the rate of change in velocity. (If you're familiar with calculus, you can say that the acceleration is the derivative of velocity.) Because of this, you don't need any mechanics to determine acceleration, so the mass is irrelevant. More concretely, a mathematician could calculate the ...


1

Mass and acceleration are two independent variables. You need to consider them together to arrive at a force with the direct relationship F=ma. In other words, if it's mass doubles, so does the force and the acceleration is the same. Gravity works exactly this way. If you are considering a fixed force from something other than gravity (a force of constant ...



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