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8

In addition to Ali's answer, here are some pictures which may be helpful in convincing people that the origin is not the only point inside the polygon where $\mathbf{E}=\mathbf{0}$. Letting the charges be located at $(\cos(2\pi k/N),\sin(2\pi k/N))$ for $k\in\{1,2,...,N\}$, we can generate plots of $|\mathbf{E}|^{-1}$ for various $N$. The zeros of ...


6

One can do the calculation(expand the potential to the second order around the center) and show that the center of the polygon is a minimum of potential. We are free to choose $V(\infty)=0$, if we do so, then it would be easy to show that the potential at the center of the polygon is positive. Combining the results above with the fact that the potential is ...


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

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

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

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 ...


3

That $C$ is the specific heat for the given cycle, i.e. $$dQ=nCdT$$ This is for $n$ moles of gas.(not the $n$ you stated in question) I will assume $$PV^z=\text{constant}$$ $$nCdT=dU+PdV$$ $$\int nCdT=\int nC_vdT+\int PdV$$ We will integrate it using Pranjal's method : $$nC\Delta T=nC_v \Delta T+\int \frac{PV^z}{V^z}dV$$ As numerator is a constant, ...


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 rate of formation is much higher in the presence of dust. There needs to be a mechanism for the energy of formation of the hydrogen molecule to be dissipated. Dissipating energy via a photon involves a forbidden transition. Instead, the energy can be transferred to the vibrational lattice of a dust particle. See The Interstellar Abundance of the ...


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

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 ...


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

Yes, you're right ! The uncertainty principle tells us that the thickness of the energy state $\Delta E$ is linked to the typical decay time of this energy level. If $\Delta E$ is large then $\Delta t \sim \tau$ (decay constant of the energy level) is small and then this energy state is very unstable.


2

As the formula you wrote suggests, it depends on $r$, the distance between the satellite and the centre of mass. $3.1\:\mathrm{km/s}$ is for geostationary orbit, and $8000\:\mathrm{m/s} = 8\:\mathrm{km/s}$ is for low Earth orbit. Please see Wikipedia for more details.


2

The first equation is for a hollow cylinder that has a thickness, hence the $r_i$ and $r_o$ (inner and outer radii respectively). The second equation is for a thin hollow cylinder, for which the thickness is negligible, so $r_i$ can be assumed to equal $r_o$. As expected, setting $r = r_i = r_o$ and substituting into the first equation yields the second ...


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

Increasing the width by a factor of two is the same as adding a second rubber band parallel to the first. For each, $\Delta F=-k\Delta x$. To stretch the combined system a distance $\Delta x$, you have to apply a force $F$ to the first, and $F$ to the second, doubling the needed force. Thus, for the combined system you have $\Delta F_\mathrm{combined} = ...


1

Take a rubber band. Stretch it by a distance $x$ with your hands. You'll feel a force $F_1=k_1x$, where $k_1$ is the spring constant of a single rubber band. Now take two rubber bands, and hold them side by side. Stretch it by a distance $x$ with your hands. Since you're stretching two of them, you'll feel twice the force, so $$F_2=2F_1=2k_1x=k_2x$$ where ...


1

The velocity after a time t1 of accelerating is the starting velocity of the deceleration phase. Thus $$a t_1 = b t_2$$ (not worrying about the sign here. I suppose you could) Further you have $$\frac12at_1^2+\frac12bt_2^2=s$$ Now you have two equations with two unknowns. Solving: rearrange first equation $$ \frac{a}{b} = \frac{t_2}{t_1}\\ t_2=\frac ab ...


1

I thought a cross-section would be $cm^2$, not $cm^{-2}$. The information you really lack is probably the macroscopic cross section. That is, $\Sigma = n \sigma$, where n is the number density of the target objects, in number per cubic centimeter. Thus, this gives a value of inverse length units. Using this, you can readily speak in terms of the ...


1

I'll answer my own question. Since $\boldsymbol{\omega }=\omega _{z}\hat{\boldsymbol{k}}$ the angular momentum reduces to $ \boldsymbol{L}_{O}=-I_{xz}\omega _{z}\hat{\mathbf{i}}-I_{yz}\omega _{z}\boldsymbol{\hat{j}}+I_{zz}\omega_{z}\boldsymbol{\hat{k}}$. We can split the rod in three pieces, calculate moment(product of inertia for each body and sum up. ...


1

As an alternate answer using parallel axis theorem, note that the inertia tensor of a rod pointing in the $y$-direction rotating about its center is $$\mathbf{I}_\text{rod,y}=\left( \begin{array}{ccc} \frac{b^3 \rho }{12} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \frac{b^3 \rho }{12} \\ \end{array} \right)$$ and similar for the $x$ and ...


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: ...



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