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9

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 ...


8

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

1. Since $x\gg p$, we see that $\sin(px)$ is highly oscillatory. In fact, the integral becomes $$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}\sim \int_{-\infty} ^\infty \mathrm{d}p\ p\ e^{ipx-it\sqrt{p^2 +m^2}}$$ modulo some factor of $\pm2/i$. Observe now this integral resembles $\int f(p)\exp(g(p))\,\mathrm{d}p$. We find the point ...


5

First, a note about the Hamiltonian and its time derivatives. I think that it is misleading to write that the Hamiltonian $$ H = i\hbar\frac{d}{dt}, $$ although the time-dependent Schrodinger equation is of course $$ H\psi = i\hbar\frac{d}{dt} \psi. $$ To evaluate e.g. $\frac{d}{dt}H$ you should consider $H=H(p, q, t)$, rather than $H = i\hbar\frac{d}{dt}$. ...


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

In Mathematica, assuming a spherical ball of ice with a rough surface (so that drag coefficient is 1), the position as a function of time in meters is: m = 1;(*Mass in kg*) \[Rho] = 1.2;(*Density of air in kg/m^3*) A = 0.012;(*Projected area of ice sphere in m^2*) Cd = \ 1;(*Conservative estimate of 1, might be lower*) g = 9.81;(*Gravity \ acceleration in ...


3

What you need for circular motion is Centripetal Force. Definition: Centripetal force is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion. If the road is ...


3

We don't normally answer homework question in full, but I'll answer because I'm very eager to make the point that the best, by far, method of answering questions like this is to choose relevant spacetime points in $S$, apply the Lorentz transformations and find where the points are in $S'$. This normally makes the answer obvious. In this case we have a pair ...


3

1.Uncertainty in position and in momentum If you've taken a class of statistics, you should recall that for some variable $y$, \begin{align} \sigma_y^2 = \sum (y_i-\bar y_i)^2 = \sum y_i^2 - \left(\sum y_i\right)^2 = \langle y^2\rangle - \langle y\rangle^2. \end{align} In other words, the expectation value of the position of the harmonic oscillator ...


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

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

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 ...


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

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

Use Lagrangian mechanics method to answer this problem because it is easier than Newtonian mechanics (IMHO). Let $T$ be the kinetic energy, $V$ be the potential energy then the Lagrangian $L$ is given by $$ L=T-V $$ and the Lagrangian equation is $$ \frac{d}{dt}\left(\frac{dL}{d\dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=0, $$ where it is assumed that ...


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

The simplest way to approach this is to note that the molar volume of an ideal gas (helium and air are close to ideal at STP) is $22.4$ litres. This means that $22.4$ litres of helium weighs $4$g and similarly $22.4$ litres of air (average $M_W = 28.8$) weighs $28.8$g. Archimedes' principle tells us that the upthrust is equal to the weight of fluid ...


2

You need to draw a Gaussian surface. In this case, you can use a rectangular box and look at the electric flux out the top and bottom of the surface of that Gaussian box. The usefulness of Gauss's is that it exploits symmetry in the problem allowing you to treat the electric field as a constant when you take the surface integral of E dot da. Which equals Q ...


2

Lets do a Free Body Diagram of the lift (NOTE: Always do A FBD first). What are the forces acting on the lift? $$\sum \vec{F} = \begin{pmatrix} -T \sin\psi \\ T \cos\psi - W \end{pmatrix} $$ What is the acceleration on the lift? $$\vec{a} = \begin{pmatrix} -a \cos \theta, a \sin \theta \end{pmatrix} $$ Combine them with $\sum \vec{F} = m \vec{a}$ and ...


2

I believe spontaneous means it happens on its own. You don't need any outside influence to get the isotope to decay. This term is sometimes used in contrast to stimulated. Random means one cannot know precisely when the next decay will happen, though one can predict the probability of such events occurring in some time interval. A decay process can be both ...


2

In theory, yes: If you assume a point-like positive nucleus and use Feynman's QED theory, you can predict the electron configuration of any element. Feynman even gives some rough examples in his Lectures on Physics, which now are online (yahoo!... as in yippee!, versus a certain web site). Since this is a homework question, I heartily recommend looking up ...


2

I depends on the pressure ratio between the chamber and ambient . As long as the ratio is critical, the ambient air will flow through the orifice at the speed of sound (with respect to the state in the chamber). For undercritical pressure ratio the flow can basically be found by the Bernoulli Equation.


2

The state from $1\text{ atm}$ to $2\text{ atm}$ is normally called decompression or contraction. An equation you can use going from one state to the next is: $$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$ Where $P$ is pressure, $V$ is volume and $T$ is temperature. Now if you want to calculate the force you have to know the surface area of what you are ...


2

There are basically two kinds of addition here (ordinary addition and inverse-sum-inverse), representing series and parallel arrangements. You can represent the thing as a tree with alternating nodes of addition and ISI layers. The thing resolves pretty much down to a tree with N leaves. The magic is dealt with here http://oeis.org/A000669 . It talks ...


2

Got it, I found a much better way to solve this problem, which eliminates my wish to confirm my previous reasoning and it partially answers the question the moderators' policy forced upon me, which was only a side-question to the main thing I wanted to ask, namely to help me solve this problem... That's why this answer might look like missing the point, but ...


2

We use the notation $[X] = L^{a} T^{b} M^{c}$ to denote the dimension of the quantity $X$, which is of length to the power $a$, time to the power $b$ and mass to the power $c$. In your problem, you are supplied with Mass, $[m] = M$ Radius of the sphere, $[r]=L$ Planck's constant $[h] = L^2 M T^{-1}$ We seek a quantity, potential energy, with dimensions ...


2

The Torricelli Formula you used is certainly a good way to start if the level of the water source is stationary. In addition there will be losses which depend on geometry of your flow restriction and on Reynolds Number. Loss coefficients $\zeta_{loss}$ for free jet discharge, valves, etc. can be found in literatute, for example here: ...


2

$|\Psi|^2=\Psi^*\Psi=(c_1^*\Psi_1^* + c_2^*\Psi_2^*)(c_1\Psi_1 + c_2\Psi_2)=c_1^*\Psi_2^*c_1\Psi_1+c_2^*\Psi^*c_2\Psi_2 + c_1^*\Psi_1^*c_2\Psi_2 + c_1\Psi_1c_2^*\Psi_2^* = |c_1\Psi_1|^2+|c_2\Psi_2|^2 + c_1\Psi_1c_2^*\Psi_2^* + (c_1\Psi_1c_2^*\Psi_2^*)^* = |c_1\Psi_1|^2+|c_2\Psi_2|^2 + 2Re(c_1\Psi_1c_2^*\Psi_2^*)$ Than substituting 2.9, you get 2.10.



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