# Tag Info

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 ...
An angular momentum eigenstate can be rotated using, $$\left| J , m \right\rangle \rightarrow e ^{ i {\vec S} \cdot {\vec \theta} } \left| J , m \right\rangle$$ where ${\vec S}$ is the $2J+1$ dimensional Pauli matrices. For spin $1/2$ for example, ${\vec S}$ are just the ordinary Pauli matrices, $\frac{1}{2} ... 4 The equation comes from Newton's second law: $$F = ma$$ Galileo didn't know calculus (because Newton and Leibniz hadn't discovered it yet) so he couldn't derive the equation mathematically. Since we do know calculus we know that acceleration is the variation of velocity with time: $$a = \frac{dv}{dt}$$ And also the gravitational force$F$is equal to ... 2 A general strategy for these questions is to start at the battery and trace the current through the circuit. So, starting from the batter, we can see that the entire current passes through$R_1$. After that, the current hits a split (at the top of the circuit in your drawing), where part of it goes to the left through$R_2$and the other part of it goes to ... 2 First: The Schrödinger equation is a equation using functionals. Solutions to this equation are such$\Psi(r,t)$, that fulfill this equation. The finite square well 1.0 fm wide means you have a potential$V(r)$which is zero for r<0 and r>1fm and -d in between. d is your depth. Now you have to determine d such, that only two of the resulting$\Psi(r,t)$... 2 I believe in the last line, the plane-wave functions$u_k(x)$should carry different coordinates and momenta, e.g $$[a(k)^\dagger,a(k')]u_k(x)u_{k'}(x')$$ You may note that the commutator$[\phi(x),\pi(x')]=i\hbar\delta(x-x')$holds if one choses$[a_k,a_{k'}^\dagger]=\delta_{kk'}$. However, this indirect reasoning is no proof that this choice is unique. ... 1 Is it correct that the zero atmospheric pressure occurs at A? Yes, approximately. In a real system the pressure at A would equal the vapor pressure of the liquid. Your "other resources" are referring to the pressure of air in the atmosphere. In the problem, the diagram represents a situation like mercury in a barometer, where the weight of the mercury ... 1$x=x(y,z)$,$y=y(x,z)$,$z=(x,y)$$$dx= (\frac{\partial x}{\partial y})_z dy + (\frac{\partial x}{\partial z})_y dz$$ $$dy= (\frac{\partial y}{\partial x})_z dx + (\frac{\partial y}{\partial z})_x dz$$ $$\therefore dx= (\frac{\partial x}{\partial y})_z [(\frac{\partial y}{\partial x})_z dx + (\frac{\partial y}{\partial z})_x dz] + (\frac{\partial ... 1 In a 1D geometry the problem is mostly analytically solvable. Start with the total energy of the system,$$ \frac12\mu\dot x^2-\frac kx=E, $$where E is the total energy, k is the interaction constant, \mu is the reduced mass, x is the relative separation, and the usual variable separation into centre-of-mass and relative coordinates has already been ... 1 In Mathematica, using the formula \sqrt{u_0^2-v^2}=\begin{cases}\\v\tan(v) \\-v\cot(v)\end{cases} where u_0=mL^2V_0/(2\hbar^2), we can first compute u_0: << PhysicalConstants` u = Convert[ NeutronMass (1 Femto Meter)^2 V0 ElectronVolt/(2 \ PlanckConstantReduced^2), 1] Out: 1.20649*10^-8 V0 Since u_0 is typically on the order of an ... 1 When you solve a problem like this, you are using a system of reference (actually you use one in all problems, but here it is very explicit). In this case, the easiest one is y in the vertical and x in the horizontal. Almost all the forces are already in one of these 2 directions. Namely, you have all the weights pointing downwards, so in the -y direction ... 1 It will in general depend on the shape of the object. If it has a large concentration of mass at the edge you are lifting then the force will be close to its weight; if its mass is concentrated near the other edge then it will be very small. The general case is solved using the law of levers: If d is the distance from the fulcrum to the centre of mass ... 1 If you are lifting on one edge and it is resting on the other edge, and the edges are an equal distance from the center of mass, then the answer is$$\boxed{F = \frac{1}{2} W}$$If you are lifting with a distance of \ell_1 from the center, and the pivot is \ell_2 from the center then$$\boxed{ F = \frac{\ell_2}{\ell_1+\ell_2} W } $$This is commonly ... 1 Open or not open, touching or not touching, ice cube above is better than ice cube below where cooling of the pot is concerned. Your choice is correct but it is more accurate to say that air cooled by the ice cube sinks onto the teapot thereby cooling it (note that this mechanism cannot occur if the cube was under the teapot). This isn't a question about ... 1 Whenever it seems like two water levels should be equal but aren't, either there is a physical restriction preventing flow (like a dam keeping upstream waters higher than downstream, or surface tension causing meniscuses or capillary action), or there is energy being expended to put water back upstream as fast as gravity is pulling water downstream. In a ... 1 I didn't read your answer, but let's think about just computing the operator \partial_x^2 f. First we need to compute the operator \partial_x f. Now I am saying "the operator" because we are viewing \partial_x f as a composition of first multiplying by f and then taking the derivative. By the product rule, we know \partial_x f = (\partial_x f) + f ... 1 I am not sure what you meant by: "I figured I could simply calculate the magnitude of the components since that will give me the distance" But the idea is use the kinematics equations for x and y: x(t)=x_{0}+v_{x0}t+1/2at^2 and y(t)=y_{0}+v_{y0}t+1/2at^2 These equations are derived from integrating the acceleration function ... 1 You can't talk about "correct" without a definition of what you are trying to achieve. However, you can rule out some obviously bad arrangements if the placement of the meters keeps the circuit from working, even though we have to guess at what the intended "working" is. This is a poorly specified problem, or there is context surrounding it that you ... 1 I might be able to get you started a direction. Not necessarily the right one or a good one, but a direction. First, I chose a different place for \theta=0. It appears you chose it at the top of the loop, while I chose it on the right side. Oh well. Also, it's likely I have a typo somewhere... Newton's second law for this problem is ... 1 You can solve for A and B by using r(0) = a, \theta(0) = 0 and evaluating \dot{r} and \dot{\theta} in your solution and using the initial conditions for velocity. You will need an equation for velocity as a vector in polar coordinates. Furthermore, while you don't have \theta(t) explicitly, it is a useful exercise to consider what happens to ... 1 Define the potential energy$$U(r) = -\int_{\infty}^r F(\bar r)\cdot(-d \bar r) = -\int_r^\infty F(\bar r)\cdot d \bar r$$with the condition that it is zero at infinity. If the initial overall energy of the particle E_0 = U(a) + \frac m2 V^2 is larger than zero the particle will escape. (The only exception is the case that k<0 and the path is ... 1 The canoe has a relative velocity to the river which we simply call \vec{v}_{c/r} =( \dot{x}, \dot{y} ). The motion of the river relative to the earth is \vec{v}_{r/e} = ( 0.54, 0 ). The canoe relative to the earth is \vec{v}_{c/e} = ( 0.55\cos(-45^\circ), 0.55\sin(-45^\circ) ) . All together you have$$ \vec{v}_{c/e} = \vec{v}_{c/r} + \vec{v}_{r/e} ... 1 your initial energy of$m_1$is all potential$m_1gh$, where$h$is the elevation difference from where$m_1$at the start to where it will go at the bottom of its travel. at the bottom the spring will contract by$x$along the ramp. now you need an angle between the ramp and the vertical, let's say$\theta$. so the initial energy of$m_2$is also all ... 1 The derivation of the Fermi-Dirac distribution using the density matrix formalism proceeds as follows: The setup. We assume that the single-particle hamiltonian has a discrete spectrum, so the single-particle energy eigenstates are labeled by an index$i$which runs over some finite or countably infinite index set$I$. A basis for the Hilbert space of the ... 1 To give you a hint: Without losing generality you can use the given standard-basis. What is the dimension of the density matrix? How many complex numbers do you need to determine this matrix? Has the density matrix some special properties, you can use to get more information on these complex number and reduce them to three real numbers? What are the ... 1 The simplest setup is for small displacements. Suppose the spring rest lengths are$L_1,L_2,L_3$, the mass has mass$m$, the springs have constants$k_1,k_2,k_3$, the angle is 120 degrees between attachments, and the attachment points are set up so that at rest, the springs are all unstretched. The potential becomes ... 1 Well, without the delta potential the wave function is $$\tag{1} \psi_0(x,t)~=~\exp\left[ -\frac{iE_1 t}{\hbar}\right] \phi(x) ,$$ where $$\tag{2} \phi(x)~:=~\sqrt{\frac{2}{L}}\sin\frac{\pi x}{L}, \qquad E_1~:=~ \frac{\hbar^2}{2m}\frac{\pi^2}{L^2}.$$ Next we are supposed to incorporate the "full" effect of the delta function$\delta(t)\$ (as opposed to ...