Recent content by variation

Thank you all.

Some examples in textbook make me confused when these two works are discussed at the same time. One of the works is the (mechanical) work in work-energy theorem: \Delta K = \sum_iW_i, where K is the kinetic energy and W_i was the work done by the i-th force. The other is the...

Hello, The problem you meet seems "how to do the integral \int_{-\infty}^{+\infty}H_n(x)e^{-Ax^2+Bx}dx" I try your problem as far as i can ( with brute force >"< ): In the begining, use the generating function of Hermite polynomials e^{-t^2+2tx}=\sum_{n=0}^{\infty}\frac{1}{n!}t^nH_n(x)...

The force \vec{F} is along the direction of the string. \vec{r} is the positin vector from the pivot A to B. They are vectors. Therefore, \frac{6}{5}\times7=0.6\cdot J\cdot\sin\theta where \theta is the angle of the two vectors. \frac{6}{5}\times7=0.6\cdot...

Acceleration direction is constant ?

Or simply, \vec{N}=\frac{\Delta\vec{L}}{\Delta t}=\vec{r}\times\vec{F}\quad\Rightarrow\quad\Delta\vec{L}=\vec{r}\times(\vec{F}\Delta t)=\vec{r}\times\text{the impulse}

The impulse is defined by \vec{F}\delta t. You have calculated the change in angular momentum, of course use the correct one. If you like, the change in angular momentum is "the angular impulse". Therefore \text{the impulse}\times\text{arm length}=\text{the angular momentum change}

The height that the center of mass drop is 0.3(m). Why not 0.6(m)?

Hello, How do you obtain the angular speed \sqrt{24.5} ? I have a different result about the angular speed. Please check the quantity again. Best regards

Hello, There are different current for different point on the spining solid sphere. The magnitude of current density J you have is correct. But the total magnetic moment should be \vec{m}=\int d\vec{m} ,where d\vec{m} is a thin ring with radius r\sin\theta rotating along z-axis. The...

Hello, The amplification I/O also equls to \frac{|D_i|}{|D_o|} when we talk about near-axis light. Be careful and recall that the image is a virtual one and therefore D_i is a negative value in Gauss' formula. Best regards

Hello, I have tried to prove some relation that you want by some simple assumptions: (1)two identical lenses (2)the curvatures of two surface of one lens are the same (3) lenses are thin. But no such a general relation is satisfied. More intuitively, you can consider some special examples...

The four momentum of the decay chanel is: \mathbf{P}_\Lambda=\mathbf{P}_{n}+\mathbf{P}_{\pi} that is \left(\begin{array}{c}m_{\Lambda}c^2/c\\0\end{array}\right)=\left(\begin{array}{c}\gamma_nm_{n}c^2/c\\\gamma_nm_n\vec{v}_n\end{array}\right)+\left(\begin{array}{c}\gamma_\pi...

Hello, For the second question, you can use the invariance of four-momentum and obtain simple relations. For example, \mathbf{P}_\Lambda=\mathbf{P}_{n}+\mathbf{P}_{\pi} here, it can be \mathbf{P}_\Lambda-\mathbf{P}_{n}=\mathbf{P}_{\pi} and one can take inner pruoduct of each side...

Sorry, i did not expain very detail. I mean that: \vec{F}=-\hat{r}\frac{\partial}{\partial r}U(r)=\frac{4A}{r^5}(-\hat{r}) The acceleration in 2-D polar coordinate \vec{a}=(\ddot{r}-\dot{\theta}^2r)\hat{r}+(r\ddot{\theta}+2\dot{r}\omega)\hat{\theta} The central acceleration in the case...