Recent content by ch3570r

does the internal energy equal the potential energy?

ok, I would multiply. If I multiply 4186J/kg by 505kg, I get 2113930.

I would think that in order to find the Joules per Kg, I would have to divide the given ratio by the given mass. The problem itself states that 4186 J/kg is needed to raise the temp. 1 degree

well let's see. It takes 4186 J/kg to increase the temperature by 1 degree. So if I take that 4186 and divide it by 505, that gives me about 8.3.

does that mean that first i divide the 505 kg by 4186, then use (505/4186) x 50 x 9.8? That gives me 59, added to the original 10, 69?

1. Water at the top of Niagara Falls has a temperature of 10.0 degrees celsius. Assume that all of the potential energy goes into increasing the internal energy of the water and that it takes 4186 J/kg to increase the water's temperature 1 degree celsius. If 505 kg of water falls a distance of...

Ive been messing around with the problem for a while, but have had no luck thus far. I'll try to figure it out tomorrow, but thanks for the help guys.

ok, I am not sure how to set up the equation if its wrong. I can change the equation to CpH20 * H20M * (Tf - Ti) = CpAu * AuM * (Tf - Ti). That would mean that energy is conserved. Is that enough to solve the problem. I could try to get Tf by itself, and get the answer that way. Well, I...

1. What is the final temperature when a 3.0 kg gold bar at 99 degrees celsius is dropped into 0.22 kg of water at 25 degrees celsius. H20 Heat Capacity (CpH20)= 4186 H20M (mass) = .22 kg H20Ti (initial temperature) = 25 degrees celsius Au (CpAu)= 129 AuM (mass) = 3.0 kg AuTi (initial...

hey, you're right Phanthom, it is 0.4. Sorry for the mistake guys! As for your question, I don't know why I doubt my work, just trying to make sure I am going in the right direction. Thanks again guys

one of the things I was trying is that by calculating the product of the component of the force (58.8) and the 15m displacement, which gives me 882. Its close to 890, but I still doubt its the answer.

"What is the work doen by friction on a 15 kg object pulled horizontally in a straigth line for 15 meters, if the coefficient of friction between the object and the surface is given by uk = .04?" a) -59 J b) -91 J c) -145 J d) -590 J e) -890 J Im not sure how I would solve this...

by using F=ma, 5*12, I get 60. For the uk force, I multiplied 49*.07. then uk force + force = 63.4, answer D

"For a 5.0 kg object on a horizontal surface that has a coefiicient of static friciton where us = 0.15 and a coefficient of kinetic friction where uk = 0.07, what is the parallel force necessary to accelerate the object at 12 m/s^2." a) 53.6 N b) 56.6 N c) 60.0 N d) 63.4 N e) 67.4 N I...

hold on, I was mistaken when I said 71.8 was the normal force. 7.5 * g = 73.5...which is the weight = FN. then, FN * 1 = FN, which is 73.5. Thanks for the help moose