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12

You basically just need to be careful about the distinction between velocity and speed. In particular, you say that Won't the particles change velocity when exposed to the magnetic field, and therefore change KE? A change in velocity is not necessarily accompanied by a change in speed, and it's the speed that determines the kinetic energy. The ...


11

If the cage is completely closed, it doesn't make a difference if the bird is hovering inside it or if it sits on the ground. When flying, the bird pushes air to the ground which will exert a downward force on the cage exactly equal to the weight of the bird. This is a direct consequence of the conservation of momentum and Newton's second & third law. ...


11

It seems that the question (v1) is caused by the fact that there are two different notions of the commutator: One for group theory: $$\tag{1} [A,B] ~:=~ ABA^{-1}B^{-1}$$ (or sometimes $[A,B] := A^{-1}B^{-1}AB$, depending on convention), which is relatively seldom used in physics. One for rings/associative algebras: $$\tag{2} [A,B]:=AB-BA,$$ which is ...


10

Batteries do not behave in such an ideal way across all conditions. The simplest model of a battery as a circuit element is the one you describe - a pure voltage source. A slightly-more sophisticated model is as a voltage source connected to a fixed resistor, called the battery's internal resistance. A typical battery has an internal resistance of between 1 ...


9

Hints: Prove that the angular momentum $L^{ij}:=x^ip^j-x^jp^i$ is conserved for a central force law in $d$ spatial dimensions, $i,j\in\{1,2,\ldots ,d\}.$ Choose a 2D plane $\pi$ through the origin that is parallel to the initial position and momentum vectors. Deduce (from the equations of motion $\dot{\bf x} \parallel {\bf p}$ and $\dot{\bf p} \parallel ...


9

This problem is generally called propagation of error / uncertainty. You can google it and find a lot of info (I'd also recommend Taylor's "Introduction to Error Analysis"). Here's the gist of it, though. If you have independent measured quantities $x, y, z, \ldots$ with errors $ \sigma_x, \sigma_y, \sigma_z, \ldots$, then the error on a function ...


9

The superscript $^2$ in $1750\text{ mm}^2$ refers to a squaring of the units, not the number $1750$. A more transparent way to write this is $1750\text{ mm}\cdot\text{mm}$. The idea is now to multiply by $1$, but $1$ written in a clever way: $$1=\frac{1\text{ m}}{1000\text{ mm}}$$. Can you see how that number is conceptually equal to $1$? The top and ...


8

When quoting results, there are a few good rules to follow: Avoid rounding errors in intermediate calculations. Write your error to 1 significant figure if your data set is smaller than $10^2$, 2 if it's smaller than $10^4$ etc. Write your estimate and its error with the same number of decimal places. Rules 1. and 3. are simple to understand. Rule 2. ...


8

You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...


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 ...


7

Actually the conducting disk problem is solved very easily in the so-called oblate spheroidal coordinates. First, alter the coordinates so that your disc is centered at the origin and is orthogonal to the $z$-direction. I will follow the notation of the Wiki article: $$ x=a\cosh\mu\cos\nu\cos\phi\\ y=a\cosh\mu\cos\nu\sin\phi\\ z=a\sinh\mu\sin\nu $$ where ...


7

This is a newton's third law problem, I'm having a hard time thinking of a way to explain this but because the forces of you pulling up will also be equal to the force of your feet pushing down the net force is equal to zero and there is no net external force, there will be no change in acceleration. I find it best to draw out free body diagrams for problems ...


7

Yes, with a pulley: (I'm aware this is cheating. This is my lawyerly interpretation of "pulling the handles of the basket.") It's instructive to consider why this works while pulling directly on the handles doesn't... or prove that this doesn't work :D


7

The left-hand integration is to be interpreted as over a domain $R$ in the set $\{(z,\bar z)\,|\, \bar z = z^*\}$ which defines a copy of $\mathbb R^2$ in $\mathbb C^2$. Let $\sigma^1$ and $\sigma^2$ be real coordinates on this surface. Using the results on page 33, we find that \begin{align} \partial_zv^z &= \frac{1}{2}(\partial_1 ...


7

Freely-moving charges placed on a line will tend to fly away from each other - with no equilibrium position possible - unless there is some potential that confines them to a specific region. Enforcing the charges to lie within an interval $[0,L]$ will always mean one charge is at either end, so you might as well consider $n-2$ charges confined by the ...


7

This problem has been solved by Griffiths in Charge density of a conducting needle. David J. Griffiths and Ye Li. Am. J. Phys. 64 no. 6 (1996), p. 706. PDF from colorado.edu. The problem is nontrivial.


7

This is a more down-to-earth answer as opposed to the fancy mathematics in the other one. This problem is easily solved numerically. The equations are easily stated: inverse-square forces to the right from the particles to the left and to the left from the particles to the right. Thus, for a system of $n+2$ charges where the first and last are fixed at $x=0$ ...


7

When $v\ll c$, the ratio $\beta = v/c$ is small, so we perform a Taylor expansion about $\beta = 0$; \begin{align} \frac{1}{\sqrt{1-\beta^2}} = (1-\beta^2)^{-1/2} = 1+\frac{1}{2}\beta^2+\frac{3}{8}\beta^4+\cdots \end{align} Now plug this into your expression and simplify.


7

You're on the right track. Complete the square on $x$ and you'll have some newly defined Harmonic oscillator whose position operator you have found. The additional constant that comes from completing the square will add to your ground state energy. (Once you have completed the square, you should have something of the form $H = constant + \frac{P^2}{2M} + ...


7

To show that this measure is Lorentz invariant you first need to explicitly write your integral as an integral over mass shell in 4D k-space. This could be done by inserting Dirac delta function $\delta[k^\mu k_\mu-m^2]$ and integrating over the whole 4D space. Then you could apply the following transformations: \begin{align} \theta(k_0)\cdot\delta[k^\mu ...


7

This vector potential can be written in every point on the plane except the origin as: $$ A_x = -\frac{\partial \psi}{\partial y}$$ $$ A_y = \frac{\partial \psi}{\partial x} $$ with $$\psi = \frac{1}{2}\mathrm{log}(x^2+y^2)$$ $A$ is not exact, because $\psi$ is singular at the origin. But this means that the magnetic field is zero at every point except ...


6

think about this with an example: the sine and cosine functions. They both average individually to zero over an interval. You can multiply those averages and still obtain zero. But if you multiply sin by itself and then average, you get a very distinct non-zero result. When the functions are arbitrary, the average of the product quantifies statistical ...


6

As you have noticed yourself, your system is simply underdetermined. In order to find a unique solution you need to add some extra constraints in addition to Newton's equations. Imagine a table with more than four legs: the more legs you add, the more unknown forces you have. But the number of equations does not change. If we instead remove a leg we find a ...


6

For reversible processes, we may derive that $$ d S = \frac{\delta Q}{T} $$ (note that it's $dS$, not $\delta S$, on the left hand side because $S$ itself is actually a well-defined property of the state, namely the amount of disorder in the system, and not the "rate of chaos", as you misleadingly wrote; on the other hand, there is no unique $Q$ "overall ...


6

This is your circuit: The current that comes from the source, when reaches the point that must choose it's way, sees no difference between the two paths (symmetry) , so half of it flows through one way and the other part flows in the second way. It means that, $I_1=I_2$ , So the potential difference across yellow resistors is the same. It means that the ...


6

You have to use the eigenstates $|n\rangle $ of the operator $\hat{n} = a^\dagger a$. You have, then, that $a \sqrt{\hat{n}} ~|n\rangle = a \sqrt{n} ~|n\rangle = \sqrt{n} ~ a |n\rangle = \sqrt{\hat{n}+1} ~ a |n\rangle ,$ where the last equality is because $a |n\rangle \sim |n-1\rangle$. So, $\left[a, \sqrt{\hat{n}}\right]~ |n\rangle = ...


6

I'll risk answering a homework problem since I think there's a principle that we experimental physicists take for granted, but which may not be obvious to budding physicists. When doing experiments it's not uncommon for the experimenter to have no way of knowing if a result is correct, but you can usually work out whether it's reasonable. If you look at the ...


6

The voltage across either horizontal resistor is zero so they can be removed from the circuit without changing the solution. This is most easily seen by simply removing the two horizontal resistors and then it's clear that the nodes the horizontal resistors connect to each have the same voltage. Thus, by Ohm's law, there is no current through either ...


6

With a potential $V(x) = - \frac{\alpha}{|x|}$, with the notation $a = \large \frac{\hbar^2}{m \alpha}$, solutions are : $$u^+_n(x,t) \sim x e^{ - \large \frac{x}{na}} ~L_{n -1}^1(\frac{2x }{na}) e^{ -\frac{1}{\hbar} \large E_nt}~~for~~ x>0$$ $$u^+_n(x,t) = 0~for~~ x\le0$$ and : $$u^-_n(x,t) \sim x e^{ + \large \frac{x}{na}} ~L_{n -1}^1(\frac{2x ...


6

Key is to notice that your steps provide you with a unit length as well as a unit time. So, let's measure distance in $steps$ and time in $ticks$, with your speed being $1 \ step/tick$. The length of the train is $x$ steps, and its speed is $v \ steps/tick$ ($v<1$). It follows that $$x \ + \ 18 \ v \ = \ 18 $$ $$x \ - \ 11 \ v \ = \ 11 $$ Adding 11x ...



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