Recent content by asrodan

The frequency you found from the equation I gave you was angular frequency.

I dug out some of my textbooks to check the equation, and it is correct. However, the v should actually have been w (angular frequency) and the question seems to be asking for normal frequency in which case you need to divide by 2*Pi

The answer I'm getting is 1337.71 Hz If your using a program to calculate v, the way you have your equation (v=) written will put the (33*0.778)^(.5) in the numerator instead of the denominator.

Almost forgot, some general advice always check to make sure units work out, if K is the wave number in the equation your prof gave you then the equation gives units (1/s*1/m*m^2)=m/s, assuming there is no mistake in the equation (I think the frequency should be squared) then K would have to...

You could try this equation Average power per unit length = .5*A^2*v^2*(Tl)^.5 where A is amplitude, v is frequency, T is tension, and l is linear mass density

The only reason I can think of right now is that Psi has to go to zero faster than any power of x goes to infinity is so that expectation values of x. Since the wave function has no physical meaning by itself, there is the obvious requirement that it be a function such that it is possible to...

The gravitational field is calculated from g = -GM/r^2 where the negative sign indicates that the field is toward mass M, which in this case would be the mass of the Earth. G is the gravitational constant and r is the distance from the center of the object of mass M and the point you are...

Actually it's been awhile since I've done any problems like this, but it seems to me that it would, instead of re-writing B, be better to make a change of variables, from dA to dx where x is the distance from the wire then integrate "toward" the wire.

You need to write B in terms of d and a because it's value depends on the distance from the wire, when you evaluate the resulting integral you don't get any b*a

F = \lambda \frac {Ax}{L} = T F = force, \lambda = modulus of elasticity, A = cross sectional area x = displacement L = natural length T = tension E = \int F dx = \lambda \frac {Ax^2}{2L} E = elastic potential energy

Calculate the gravitational field due to a homogeneous cylinder at an exterior point on the axis of the cylinder. Perform the calculation by computing the force directly. I'm not sure if I did this right, here's what I did z = distance from point to center of cylinder r = distance from...

The initial angular momentum is mvR assuming the the particle hits right at the top edge of the cylinder. L = I\omega = mvR

Question 1: Correct Question 2: Knowing the specific heat capacity of water and the heat of vaporization you can find the amount of energy required to vaporize the water. You will also need to find the power dissapated by the elements in the water. Question 3: Assuming the 6 ohms is...

r= 2sin(\theta) is an ellipse so r= 6-2sin(\theta) is just shifting and stretching it. Therefore the bounds on \theta are 0 \leq \theta \leq 2 \pi

s is often used as displacement, which is what Benny is using it to represent. s = ut + \frac {1} {2} at^2 = v_o t + \frac {1} {2} at^2 = x