Recent content by matthew1991

I made this way more complicated than it needed to be. Charge is always conserved, so Q=CV at 180 deg is the same Q still at 0 deg. I had pF as E-9 instead of E-12

Homework Statement The capacitance of the variable capacitor of a radio can be changed from 100 to 350 pF by turning the dial from 0° to 180°. With the dial set at 180°, the capacitor is connected to a 130 V battery. After charging, the capacitor is disconnected from the battery and the dial...

sorry, that was isothermal. adiabatic expansion work can be found by W = (3/2)nR(deltaT)

If you replace the P from the integration formula a few posts earlier with (nRT/V) and integrate the formula, you will get W=nRT*ln[V(init)/v(fin)]

Okay, i knew total energy is conserved. for anyone who needs it in the future, the closest distance of approach is when all kinetic energy is converted to potential energy. the problem can be solved using this equation: U = K*[q*q/r] where U = potential energy, K = coulomb's constant...

I don't understand. Initially total energy and ending with total potential correct? How is this solved?

the object moves with SHM, but energy is conserved. The object will reach the other side of the planet at the heigt it was dropped from at the exact moment its speed reaches zero. for more information look up Hooke's theory of gravity trains.

Q2: Start by turning the square into a diamond by rotating it so that the middle ball lies at the origin and te others lay on axis in an xy plane. Separate the forces into components and allow them to cancel where able. Then subtract the forces from each other until you are left with net...

Problem solved. 1.65*10E3 m/s^s. For all who may come along seeking help: initial velocity is the velocity when the ball first contacts the floor and final velocity is the velocity at which the ball leaves. Delta T is the time in contact with the floor. With the distanc equation you can...

a= (v final - v initial) / (t final - t initial) a = dv/dt a d^2(s)/d(t)^2

[b]1. A golf ball released from a height of 1.50 m above a concrete floor, bounces back to a height of 0.62 m. If the ball is in contact with the floor for 5.39 ms, what is the magnitude of the average acceleration a of the ball while it is in contact with the floor? [b]2...