Recent content by dgl7

Ok nope. Nevermind, I'm still confused. This is what I've been doing and attempting: E=Eleft+Eright E=kQleft/r^2+kQright/r^2 (r=r and Qleft=Qright) E=2kQ/r^2 E=2*8.98755e9*2.2e-6/(1.8^2+3^2) E=3230.818627 V/m Not sure what I'm doing wrong.

AHHH that makes sense. Thanks very much!

Two charges create an electric field--electric field strenght at a point above h.fiel Homework Statement Two charges are located on a horizontal axis. The Coulomb constant is 8.98755x10^9 Nm^2/C^2. a) Determine the electric field at p on a vertical axis as shown in the attachment. Up is...

It is a pendulum, thus O is at the opposite end of the rod.

Homework Statement Consider a light rod of negligible mass and length "L" pivoted on a frictionless horizontal bearing at a point "O." Attached to the end of the rod is a mass "m." Also, a second mass "M" of equal size (i.e., m=M) is attached to the rod (0.2L from the lower end). What is the...

I have a question about physics in real life. Other than the fact that the area under a force versus time curve is a more accurate way to measure impulse because it's not an average like calculating impulse from a change in velocity is, is there any other reason that the integrated impulse would...

Yes that would make a lot more sense to just use Uspring+Ugravitational+Wnc=K. I didn't/don't know how to solve for Wnc, so I just used Ffriction*r*costheta, which brought me to the cos20...

1. A spring with a spring-constant 3.6 N/cm is compressed 39 cm and released. The 9 kg mass skids down the frictional incline of height 36 cm and inclined at a 20◦ angle. The acceleration of gravity is 9.8 m/s^2. The path is frictionless except for a distance of 0.6 m along the incline which...