A cylindrical opaque drinking glass has a diameter of 4.7 cm and a hieght h. An observer's eye is placed to where they are just barely looking over the rim of the glass. When the glass is empty, the observer can just barely see the edge of the bottom of the glass. When the glass is filled to...
Thanks for your help. I came up with something close to Gellman's solution using Galileo's hint. What I came up with is:
U = Vq (potential energy)
1/2 mv^2 (kinetic energy)
Vq = 1/2 mv^2 (then solve for v)
v = SQRT(2qv/m)
F = qB(SQRT(2qv/m))sin0 (0 = theta)
Here is the problem word for word:
An electron in a vacuum is first accelerated by a voltage of 61500 V and then enters a region in which there is a uniform magnetic field of 0.477 T at right angles to the direction of the electron's motion. The mass of the electron is 9.11e-31 Kg and its...
My physics class is not a calculus based class; therefore, the formulas you see here will not be calculus. My problem states:
A parallel-plate capacitor has 3.56 cm^2 plates that are separated by 6.08 mm with air between them. The permittivity of a vacuum is 8.85419e-12 C^2/Nm^2. If a 10.4 V...
I believe this may be a problem with the website where I do my homework. Please check my solution and see if there is a mistake or if I am missing something. The following is the arrangement of point charges in the problem:
8.26 micro C ---2.93 cm---> P1 ---1.59 cm ---> 4.94 micro C ---2.73...
Here is the problem word for word:
A charged cork ball of mass 1.53g is suspended on a light string in the presense of a uniform electric field. When the electric field has an x-component of 346000 N/C and a y-component of 383000 N/C, the ball is in equilibrium at 37.6151 degrees. The...
I have been working on this problem off-and-on for a couple of days now and cannot seem to get the correct answer. The problem is an Atwood machine problem. The following are the variables:
Mass of pulley = .20 kg
Radius of pulley = .15 m
Clockwise frictional torque = .35 m*N
Mass 1 on...
I finally got the correct answer. I hate it when you are on the right track, but can't find the missing part of the equation. My missing part of the equation was the moment of inertia of the runner. After using the formula mvr = (Imgr + Ir)w (where Imgr is inertia of merry-go-round and Ir is...
If L = r * p then I have already tried this as it would be the same as L = mvr. I have tried mvr = Iw and solved for w and got the formula w = mvr/I. When I plug the numbers into this formula I do not get the correct answer. Am I missing something?
The problem reads as follows:
A runner of mass m=36 kg and running at 2.9 m/s runs and jumps on the rim of a playground merry-go-round which has a moment of inertia of 404 kgm^2 and a radius of 2 m. Assuming the merry-go-round is initially at rest, what is its final angular velocity to three...