Recent content by houseguest

Wow! I'm sorry, that was definitely one of my stupid things. Thanks for pointing that out! The container is filled with a monatomic gas. So I should be using C_p for that. So, since C_p = R + C_v and C_v = 3/2 * R ( for a monatomic gas) then C_p = 5/2 * R So that would give ( with the R's...

Sorry, I forgot to mention that the container is filled with water.

Homework Statement A container has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction and is placed below a heater. Suppose the heater has 25 W of power and are turned on for 15 s. What is the final volume of the container? Initial Volume: 800 cm^{3}...

Ooooh yeeeeeaaah. THANKS!

Your problem is not in calculating the mass of O2, rather your problem is that it's O2 and not O. The equipartition theorem states that a system has an average energy of kT/2 in each quadratic degree of freedom (mode in which the gas can store the energy). A monatomic gas can store energy as...

Homework Statement At 100 celsius the rms speed of nitrogen molecules is 576 m/s. Nitrogen at 100 celsius and a pressure of 2.0 atm is held in a container with a 10cm x 10cm square wall. Homework Equations rate of collisions: N_coll/ \Deltat = (N*A*v_x)/(2V) (this is from my textbook) ideal...

oh yeah, duh. Thank you so much for your help! Seriously.

Although, it retrospect this can't be right since deltaL = ( (3 * 10^4 * 9.8) / .00576 * .8) / (.00144 * 10^10) meters = 2.84 m and that is way to large to be realistic

Hang on, I see the mistake I made earlier: 3.8 cm x 3.8 cm = .038m x .038m = .00144 m^2 (thanks btw), but wouldn't then total area then be 4 * .00144 = .00576 m^2 ? Then the pressure would be 3 * 10^4 * 9.8 / .00576. Then plug in that pressure for F in deltaL = F*L_0/(Y*A) Is that right?

4*.144 = .576 m^2 but, isn't it asking the deltaL for an individual post? I thought that I might have to divide the force by 4, but that answer was also incorrect. Thanks

So I took the mass of water to be 1Kg per liter. So the m = 3 * 10^4 Then the equation comes out as: deltaL = ( 3 * 10^4 * 9.8 * .8)/(.144 * 10^10) meters = 1.63 * 10^-4 m But I am told this is the incorrect answer. Any help? Thanks!

Homework Statement A large 3.00×10^4 L aquarium is supported by four wood posts (Douglas fir) at the corners. Each post has a square 3.80 cm x 3.80 cm cross section and is 80.0 cm tall. By how much is each post compressed by the weight of the aquarium? Homework Equations deltaL =...

Hello, I am attaching what was an extra credit question in my physics class which I didn't understand at all. The topic isn't in the book and all the internet searchs I read confuse me. I was hoping someone might give me a walk through. Thanks!

Ooooh yeah. Thanks!

image It's been pending approval for longer than my patience. I got imageshack to host. URL: http://img88.imageshack.us/my.php?image=prb1gr7.jpg Thanks