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So there is not an easy way - I had to grind out all the algebra. Starting with $$\nabla^2 E +k^2E = 0$$ The Electric field forms are $$E_r = P_n^0(cos\theta)[\frac{A_n j_n(k r)}{k r} + \frac {B_n y_n(k r)}{k r}]$$ $$E_\phi = P_n^1(cos\theta)[C_n j_n(k r) + D_n y_n(k r)]$$ Using $\nabla \cdot E = 0$ and the symmetry over $\phi$ I find $$E_\theta = -\frac ...


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The continuous stream of air that you are blowing in, it doesn't enter the pipe continuously. When the stream of air hits the hard edge in an organ pipe, it flaps in and out of it due to the difference in the density of the air outside and inside the pipe. This oscillation of the air in and out, it will be a periodic energy supply for the standing wave in ...


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No. When matter and antimatter annihilate, they do so particle by particle. Each electron annihilates with an anti-electron, protons with anti-protons, etc. And in each annihilation the total rest mass of the two particles is converted to energy in photons. Even the lightest annihilation, that of an electron with an anti-electron, must put over $1\ ...


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I think you mean: for a given loudness, which frequencies involve greater physical movement, high frequencies or low? And that is simple - the lower the frequency the greater the amplitude of the movement. Here's a simple demonstration. Take the grille cloth off a speaker with a woofer. Play music through the speaker, something with sustained notes like ...



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