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But isn't the intensity given by \Psi \Psi^*? This gives an intensity equal to 1/r^2 for the wave described by a complex exponential function but an intensity equal to 1/r^2 cos^2(k r - w t) for the other one.

Okay, so if they both had time dependence -i \omega t so that \Psi = \frac{1}{r} e^{i ( k r - \omega t)} and \Psi = \frac{1}{r} \cos( k r - \omega t)}, but they still don't have the same intensity, since the intensity of the second one is an oscillating function of r and t...

Hey! Im quite confused about spherical waves. I mean, I understand that a spherical wave can be described by \Psi = \frac{1}{r} e^{i r}, because the intensity of such a wave decreases as 1/r^2. The intensity of such a wave is given by I = 1/r^2 which makes sense to me. But a...

Find the energy carried by one photon and multiply by Avogadro's number. In terms of frequency the energy carried by a photon is: E=h \nu Where h is Planck's constant and \nu is the frequency. So you have to express the frequency through the wavelength.

Hey... I think you should subtract the kinetic energy gained by M1 as well. It does indeed have a kinetic energy when M3 has fallen a distance of 2.5m... Sorry, I forgot about this :-)

The equation you wrote comes from: \frac{1}{2}m v^2 = m g h This equation represents the fact that potential energy (right hand side) is converted to kinetic energy (left hand side) on falling a distance h through the gravitational field. Cancelling m on both sides and multiplying by 2...

The concept would be the same even if the horizontal surface was not frictionless. Then you would just subtract the energy delivered to the surface due to friction as well, and this energy would be the the frictional force times the distance traveled (2.5m).

Just write the complete equation: \frac{1}{2}M_3 v^2 = M_3 g \Delta h - \frac{1}{2}M_2 v^2 - M_1 g \Delta h Where the term om the left is the kinetic energy of M3 (the one you want to calculate), the first term on the right is the total energy available (with \Delta h=2.5m, the second...

First of all you should realize that all the masses have the same speed because the wire is unstretchable. Next, try to imagine what happens when the system is released. Because M3 is heavier than M1, and because the horizontal surface is frictionless, M3 will fall. Now, when I solve problems...

What is the electric field inside a sphere of radius R when the charge density is given by: \rho=\frac{A}{r} Where A is a constant, and r is the radius at where the charge density is to be evaluated. By Gauss' law I have calculated the field to be equal to: E=\frac{A}{2\epsilon_0} But that...