Recent content by Jared944

A coin is located 20.0 cm to the left of a converging lens (f=16.0 cm). A second identical lens is placed to the right of the first lens, such that the image formed by the combination has the same size and orientation as the original coin. Find the separation between the lenses. So, f=...

Homework Statement A piston (with an area of 3.14 x 10^-2 m/s^2) is pushing down on a vertical cylinder (with a pressure of 1.01 x 10^5 Pa) which contains an monotomic ideal gas. 2093 J of heat is removed from the gas. Ignoring the mass of the piston or friction, find how far the piston...

Homework Statement .005 m of Ice is in contact with .0015 m of Aluminum. The temp of the Ice is -10.0 degrees C, and the temp of the Aluminum is -25.0 degrees C. I need to calculate the heat per second per square meter that is conducted through the Ice - Aluminum combination Homework...

of course... I was thinking w was the work done by the tension of the system.

So, KE = mgh + w KE = 2*-9.8*5+14 KE = -84J? How can I reach a negative KE?

The velocities should be the same. However, I've calculated the acceleration and the Tension of the blocks in motion. The acceleration and T of block 2 can be illustrated as T-w2=m2(-a) And Block 1 can be illustrated as 0+T=m1*a The tension turnes out to be 14 N and the acceleration is 2.8...

Oh, ok. Block 1 is sitting on a (frictionless) table attached to block 2 which is hanging off of the table. Both blocks are attached by a rope and a (frictionless) pully. Before the system is released, block 2 is 5m from the ground (h = 5m) Thanks for the obervation about the pic...

Homework Statement Two boxes are set up in a system as shown in the attached picture. Box 1 - mass = 5 kg Box 2 - mass = 2 kg h = 5m The system is released, and box 2 hits the ground. With a frictionless system, find the total KE before box 2 hits the ground. Homework Equations...

Ahh, I just figured it out. I was using the wrong set of equations, I am sorry! Part a) can be figured by using the equation - .5 m(vf^2)+m(g)hf = .5m(v0^2)+m(g)h0 Or, in other words, .5(1.5)(vf^2)+(1.5)(9.8)(0)=.5(1.5)(5^2)+(1.5)(.3) vf=5.56! Thanks for the nudge in the right direction!

Ok, so this problem is kind of complicated, so ask me if you need clarification. This problem deals with Impulse and Momentum. There are two balls hanging vertically from a horizontal plane. Ball A is pulled back and is set up to strike ball B. Given for ball A - mass = 1.5 kg. Height...