<p>Haha yeah, I thought everything else was fair except that one... lol it was the one topic I really didn't review... and it was on the FR and all over the MC L:(L</p>
<p>number 5)
part a - no problem, that's just pv=nRT for each of those
part b - the net work done on the gas.... state 1 to 2 was positive area under the curve, which is work done BY the gas... that amounts to 25,000 J. state 2 to 3 was work done ON the gas... that amounted to 31,250 J. state 2 to 3 was isochoric, meaning no area under the curve, and no work. so net work done ON the gas in the system = (-25,000) + 31,250 = 6,250 J
part c - you use delta U = Q + W (W being the work that is done ON the system). delta U = 3/2 x n x R x delta T, and since it's a cyclic process, delta T is 0, meaning the change in internal energy is 0. that means 0 = Q + 6,250. you do the algebra, and you end up getting -6,250 J for Q, meaning heat was removed.</p>
<p>the tricky thing about this problem is the language, that's all. just remember you can look at the law of thermodynamics in two ways = U = Q - W, but that means work is the work done BY the gas, or U = Q + W, but that means work is the work done ON the gas</p>
<p>for electrostatics, i got 4.8 x 10^-8 for the charge of q2, and it was positive. i used kq1q2/r^2 to calculate the force on q2, and the direction of the force was to the left on the x-as, since it had an opposite charge and opposites attract. i didn't get to finish d, but i was on the right track to figure it out :(</p>
<p>sunkist u coundnlt use PV=nRT for number 5, it would be P1V1/T1=P2V2/T2 cause no moles were given</p>
<p>ehh, the combined gas law is just another version of pv = nRT, just that it freezes the number of moles. more specifically, i did V1/T1 = V2/T2 for the temperature of state 2, and i did P1/T1 = P2/T2 for the temperature of state 3</p>
<p>sunkist -- i got everything you did for 5 and for the charge on q2... sweet</p>
<p>for number 5, i think indeed you could use the ideal gas law, as the question gives P, V, and T for the original state (if i remember correctly) which would allow a calculation for n.</p>
<p>and i got a positive work--heat removed, as well</p>
<p>Does anyone remember the answers for #1 or the temperatures for #5?</p>
<p>tension was equal to weight of the block, so it was (9.8)(4), which was 39.2</p>
<p>force constant was 784</p>
<p>to find the time it took for the block to reach the ground, you use x = vot + 1/2 x a x t², so when you do the math, it was the square root of (1.4/9.8)</p>
<p>frequency was 1/2pi x (square root)(784/8)</p>
<p>to find the maximum speed, you find the potential energy of the block in its position, which is PE = 1/2 x k x x². x at that point is 0.05 m, because unstretched length is 0.2 m and stretched length is 0.25 m. PE at that length would all become kinetic energy at its maximum speed, so you set that energy equal to 1/2 x m x v², and you'll get your answer.</p>
<p>yikes, is it bad if i approximated gravity to be 10. m/s/s?? I guess I was used to doing that because the practice questions I got in class told me to</p>
<p>no, the free response scoring guidelines say that approximating g to be 10 m/s² is fine</p>
<p>For number 5 heat was definitely removed. P1v1/t1. For the experiment i did what everyone else did however i found my friends experiment pretty funny. He colored the floor with chalk and as the runner finished running he measured his stride and found the length of it and made a proportion to frequency tehn he determined how fast he was running based on the stride and the stop watches the 11 studnets had... it was too funny but nonetheless it all made sense.</p>
<p>c? finding the spot where potential energy feels z?</p>
<p>it was something like that. That was the question shich many students had hard time solving it.</p>
<p>What i did is i used many algebra tricks</p>
<p>R2-R1 = .3m</p>
<p>V= k*(q/r^2)
V1+V2=0 </p>
<p>and then solve for r which is the distance between the charges where it would feel 0 potential energy</p>
<p>and then adding R1+r </p>
<p>does this look familiar to anyone here?</p>
<p>i was really confident when i was solving this
:) i figured #3 out when the proctor said "5min left"</p>