Recent content by sungod

In 8 the field points along the +ve x axis. Therefore, this is the direction of force on a positive particle. Since the force is in the direction of decreasing potential, E = -dV/dx, ie V falls, as x increases.

I see what you mean. In that case, perhaps there is only one parabola whose vertex is the top of the log. Any other parabola is either too high or too low. So, (looking at it sideways), you draw the line at x = 2R, you draw the circle (x-R)^2 + y^2 = R^2. The y-axis is one tangent to the...

I don't know if this helps, but this seems to be a trajectory problem with a parabolic trajectory whose vertex is the top of the log. In this case, any trajectory whose highest point is 2R would work. If it takes off at an angle theta with a velocity v and it takes time t for the entire...

Plot sin(theta) against n. This is a linear equation with a slope of lambda/2d. If you know lambda, you can get d.

I'd think you'd need to use Doppler. The frequency changes due to the motion, although it doesn't say which way, and you have the frequency when there's no relative motion.

If the transverse wave is moving assuming it's not a standing wave - wouldn't really matter if it was, all that happens is that the amplitude is now a function of x - the wave obeys: y = A*sin(wt \pm kx) Choosing the origin, it doesn't matter where you look along the wave, the wave equation...

Sorry, I forgot to write down the factor of 1/2 for E(C) and E(D). It's just a typo. The calculation includes the 1/2.

Quick, back-of-an-envelope kind of thing (I'm taking my zero of energy at the bottom of the original path): Energy at A = 79*9.81*6.4 = 4954.88 Energy at B = m*(v_b)^2/2 = 4954.88 -> (v_b)^2 = 125.44 Energy at C = m*(v_b)^2 + mg(0.37) = 5246.68 Energy at D = m*(v_d)^2 + mg(6.4) = 5817.69...

Could it be that by raising himself up, he's given himself some potential energy? Had he never raised himself up, he'd only have the potential energy due to his starting position as the initial energy and his energy at the lowest point would be that of just a particle with that specific...

Let me rewrite this as b = tan(b) Plot y = b and y = tan(b) on the same plot. Where the two plots intersect are the solutions.

You have an equation for the intensity of the diffraction pattern. The maxima and minima of the function will give the maxima and minima of the diffraction pattern. So, simply differentiate and set to zero. You'll get two factors. One of these gives the minima and the other gives the maxima, or...

If it takes 1 second to go 1/4 pi radians, how long does it take to go 2pi radians (the complete circle)? theta = omega*t and theta = 2pi here If it takes t seconds for one revolution, the number of revolutions/min is 60/t

This is the question Einstein asked himself at 16 years old. He concluded that the light would "look like" a stationary, oscillating e-m field - impossible according to Maxwell's equations. Yes, he apparently was familiar with Maxwell's equations at 16! He finally resolved the contradiction by...

It might be a little more complicated. Still water would be a specular reflector, so at midday, all the light in the water would reflect straight up if the water were perfectly clear on a cloudless sky. If the ball were perfectly opaque, the bottom would receive a little blue from the scattered...

It seems to me that they would both have the same magnitude change. If the initial momentum vector has a length n (224 in this case), then the final momentum vector also has a length n. This length n is made up of the length of the vector of the final momentum of the 7kg glider and the length of...