# Tag Info

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If you're trying to simulate a 2D solution of the Laplace equation (which is the only unambiguous reading of your post as currently stated; if that's not what you're doing then you should clarify your question with exactly what it is you're doing and how), then your code is wrong. The reason is that your results don't obey the maximum principle: a harmonic ...

8

A human can produce about 100 W power continuously. That is equivalent to a couple of lightbulbs. So, while pedalling continuously you can keep one room reasonably lit. In one hour, that means you can generate a total energy of 0.1 kWh. A top athlete will be able to do better, so let's say it is possible to generate 0.2 kWh in 1 hour. That is still nowhere ...

4

$1$ litre of water will remain almost $1$ litre as long as it is in the liquid state, no matter what the temperature is. The following formula gives you an order of magnitude estimate of the expansion: $$\Delta V=V_0\ \Delta T \ \beta$$ where $\beta$ is the coefficient of thermal expansion and $V_0$ is the initial volume. For water, $\beta \approx 10^{-... 3 First you should understand why gravity does not change as you move around on Earth's surface. The easiest way to explain this is using symmetry. The Earth is roughly a sphere. On a sphere there are no special points so gravity must act the same everywhere. A flat planet can actually be made to have constant gravity. If you had a flat planet that extended ... 3 No you cannot assume that. The initial rotation is about the major axis, and it will continue to be so (in the absence of torque, and since you were already rotating about the major axis). Instead, since$\omega_2=\omega_3=0$, your equations for the evolution of the angular momentum don't require the moments of inertia to be the same. 3 All you need to do is calculate the perigee distance$r_p$that is the distance of closest approach. Then if$r_p < R_A$your satellite will crash and burn. Once again we start from the vis-viva equation: $$v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right) \tag{1}$$ The parameter$a$is the semi-major axis of the ellipse, and it is related to the ... 3 As indicated in the answer to your previous question, since as$E(=30\:\mathrm{eV})$is higher than$V_2(=20\:\mathrm{eV})$, that particle is not bound, it's not an eigenstate of the system's Schrödinger equation (it's a scattered state). Its wave function for$x \to +\infty$would something like: $$\psi=c_1e^{-ikx}+c_2e^{+ikx}$$ ... where both complex parts ... 2 I have just now finished an article, "Geometry of the 3-sphere", in which at the end of the paper I give a simple derivation of the Riemann curvature bivector for the unit 3-sphere, using (Clifford) geometric algebra. I also discuss the Lie group$SU(2)$and Lie algebra$SU(2)$on the unit 3-sphere, using the powerful, but still rather unknown geometric ... 2 $$A\sin(k_iL)=De^{-qL}$$ $$Ak_i\cos(k_iL)=-Dqe^{-qL}$$ $$k_i\cot(k_iL)=-q$$ Insert the values for$k_i$and$q$: $$[2m(V_1-E_i)/\hbar^2]^{1/2}\cot[2m(V_1-E_i)/\hbar]^{1/2}L=-[2mE_i/\hbar^2]^{1/2}$$ The allowed energy levels ($E_i<V_2$) for bound states can be determined by numerical solution of that equation. But particles with energy$E>V_2$can ... 2 You are combinig two question, I am combining two answers. The system you describe consists of two points with masses. We know that: Every two points$A$and$B$lay on a single line; When line is rotated around axis intersecting it, all points of the line have same angular velocity$\omega$, except for the intersection with$\omega_0=0$; Centre of mass$...

2

Spin operator of the total 2 electron system is tricky: the statistical requirement reduces the Hilbert space to a 3-d rather than 4-d version. Like a spin-1 system, the $S_z$, as represents in basis $|S_z=1,0,-1\rangle$ is (see http://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html, for example) $$S_z=\hbar\left(\begin{matrix}1&0&0\\0&0&... 2 I think you made an mistake in the algebra. Let's focus on the PDE$$p \frac{\partial \rho }{\partial q} - q \frac{\partial \rho }{\partial p} = 0.$$You claim that \rho(p,q) = \exp^{\, f(p+q)} is a solution to this equation for any function f, but that's simply not true. If you plug that Ansatz in the PDE, you get$$ (p-q)\; f'(p+q) = 0 $$which is ... 2 As drawn the disc's angular speed \omega is too fast as related to the velocity of the centre of mass of the disc v for the no slipping condition (v = R \omega, with R the radius of the disc) to be satisfied. You can think if the frictional force as trying to accelerate the centre of mass whilst at the same time the frictional force applies a torque ... 1 What you want is \rho(\textbf{r})=Cr^2, where C is a constant. Find C by integrating this function over cylinder volume, and equating the result to total mass m. 1 For the horizontal portion, there is no acceleration, initial velocity is 12 m/s and final, I have no idea. See Newton's first law an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Your rock moving at 12 m/s horizontally has momentum in that direction. That momentum is ... 1 Assuming no air resistance, the only force acting on the satellite is the force of gravity, \bf{F_g}. The stated circular orbit also means that centripetal force, \bf{F_c} is involved, which causes the satellite to follow the circular path. Due to this, the centripetal acceleration can be directly equated to g, which allows you to immediately solve for ... 1 Thanks to Ján Lalinský for getting me to double check my faulty memory. My assumption of \langle\chi_n|[H_0,z]|\chi_m\rangle = 0 was wrong. Here is the solution after double checking my work.$$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\chi_n|H_0\,\,z|\chi_m\rangle-\langle\chi_n|z\,\,H_0|\chi_m\rangle\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\...

1

Volume is not a meaningful measure of quantity, for the reason you hint at in your question. You can say how many moles (or grams) of water you drank - more useful if you want to know about the impact on your body chemistry. This is related to my answer about scales measuring in grams rather than Newtons. Can you see how?

1

Write $$U = \frac{1}{2}(I_1 + Z_1) \otimes U'_{00} + \frac{1}{2}(I_1 - Z_1) \otimes U'_{11} + (X_1 + iY_1) \otimes U'_{01} + (X_1 - iY_1) \otimes U'_{10} = \left(\begin{array}{cc} U'_{00} &U'_{01} \\U'_{10}& U'_{11}\end{array}\right)$$ where $U'_{ij} \in G_n$. In terms of the latter,  U' |\Psi\rangle = \sqrt{2} \;\langle 0 | U | 0\otimes \Psi \...

1

Here is how to answer this question (rather than the specific answer, which I don't think you're after). Firstly they tell you that the two bits of the path are the same length ('the first half of the path') and are straight. Call the length $l$. They also give you the angles, $\alpha_1$ and $\alpha_2$ of the two bits of the path to some reference line. ...

1

For inelastic scattering, the initial momentum is $m_b v_{b_i}$. After collision, both $m_b$ and $m_c$ move together, with a velocity $v_{b_f}=v_{c_f}=v_{cm}$. By conservation of momentum $m_b v_{b_i}=m_b v_{b_f}+m_c v_{c_f}=(m_b +m_c)v_{cm}$, whichyield the equation that you are looking for

1

The systematic way to set up the equations is to draw a free body diagram for each mass. The FBD shows all the forces acting on that particular mass. You can then use Newton's second law to get the acceleration from the resultant force. The weight of mass 2 acts on mass 2, not on mass 1. Of course the weight of mass 2 will affect the motion of the system, ...

1

Advantages Very accurate ( In fact its accuracy allows it to be utilised to calibrate other thermometers) Wide Range Independent of gas used Disadvantages Large and bulky(inconvenient to carry and handke) Slow to Respond (due high heat capacity) Expensive to manufacture and keep I'm not sure about its use in any specific industries.

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You must look at all forms of energy. Just before the explosion, the projectile has gravitational potential energy GPE, kinetic energy KE, and also chemical potential energy CPE stored in the dynamite. Just after the explosion the 3 fragments all have the same GPE as before. The CPE has disappeared in the explosion. As Jim says, we must assume that it ...

1

I believe Jordans lemma, https://en.wikipedia.org/wiki/Jordan%27s_lemma, provides that the integral II goes to 0 as $R\rightarrow \infty$.

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Lagranges equations are: ${\partial L \over \partial q_i} = {d \over dt }{ \partial L \over \partial q_i'}$ where $q_i'={dq_i \over dt}$ You can find constants of motion using lagranges equations and Hamiltons equations. You already know that the Hamiltonian is conserved-time is not explicit. (The energy is not always equal to the Hamiltonian) You can ...

1

There is extensive (100s of articles) lit. on the web regarding atmospheric drag and drag in general. A spherical bob and a very thin rod makes the problem very simple. One may use this graph: http://eis.bris.ac.uk/~memag/Teaching/Multi/dragcurve.pdf , and then calculated the Reynolds number to find the drag force using the formula given by the first ...

1

The above isn't drawn correctly. And it is far too complicated in my opinion. I think I was able to solve this using high school physics: This bugged me a long time. I am surprised no one ever posted an answer on the internet. The length of his rope was 60.8 meters, assuming a drop above the target building of about 12 meters. The difference in height ...

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