Recent content by joseg707

For the second container, can I assume that pressure will be held constant as the volume of the container increases? The heat put into the system will then be converted into work as well and so \DeltaU will equal q+w. The temperature will increase, but not as dramatically as the first container...

Homework Statement Two containers each contain 0.500 moles of N2 initially at 300. K. One container has a fixed volume, and the other container has a movable frictionless piston. If both containers are heated by 1000. J energy, calculate the final temperature of N2 for each container...

The 1/2 magically disappeared. :) You just forgot to multiply by 2, but you got essentially.

If you can find the gravity on Neptune, you can use conservation of energy to determine it's velocity.

Homework Statement A 4 meter long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end is allowed to fall. What is the angular acceleration as it is released? Homework Equations Ang. Accel. = (Tan. Accel.)/(radius) The...

Ah! Thank you so much! I don't know why I was thinking are was 2*pi*r^2. Thanks a lot!

Oh, yeah. Well how do I calculate that? I know that the change is .2% if I use .002 or .2 as a value for \DeltaL/L I don't get the right answer. What is wrong with my calculation?

Homework Statement A wire of diameter 0.2mm stretches by 0.20% when a 6.28-N force is applied. What is the elastic modulus of the wire? Homework Equations Elastic Modulus=F/A/L/\DeltaL The Attempt at a Solution r=.1mm=1x10-4m 6.28N/(2\pi1x10-8/.02=5x109N/m2 The right answer...

So does this hold true? m(9.8)(520)=4.32e11 m=84772370.49 kg*1m3/1000kg=84772 m3

Oh! Haha yeah, I completely jumped right over that didn't I. (120e6W/s)(3600s)=work work=4.32e11J

Are you saying that the kinetic energy is equal to the output energy of the turbine, 120MW, and the the potential energy is equal to the kinetic energy? KE=120 MW PE(2)=KE(2) mgh=120W m(9.8)(520)=120MW? I don't feel like I understood that right.

Normal force and the force of gravity. N-mg=ma N-mg=mv2/r N-mg=m5gr/r N=5mg+mg N=6mg I see now! Thank you very much! =)

Ok so v22=gr. I substitute that into my energy equation and I get: .5mv12=2mgr+.5mgr After all the algebra I get v=\sqrt{5gr} I don't see how this is a step in proving that a person's weight is 6 times their normal weight at the top of the loop. Could you explain what is happening please?

Homework Statement Some electric-power companies use water to store energy. Water is pumped by reversible turbine pumps from a low to high reservoir. To store the energy produced in 1.0 hour by a 120*106 electric-power plant, how many cubic meters of water will have to be pumped from the lower...

Homework Statement Show that on a roller coaster with a circular vertical loop, the difference in your apparent weight at the top of the circular loop and the bottom of the loop is 6 g's--that is six times your weight. Ignore friction. Show also that as long as your speed is above the minimum...