<p>U = Vq = (kQ/r)q. Here Q = 92 protons, which have a charge of +92e. q is a proton, which has a charge of +1e. r is what we’re trying to solve, which is just D. So, U = (92ke^2)/D</p>
<p>I never learned that. Got that wrong.</p>
<p>Can someone please do the Lab problem? Thanks.</p>
<p>Does anyone else think FRQ Question #3 was unfair (the one about the cylinder in the bathtub)? It doesn’t tell you if pressure or volume or whatever remains the same (all it says is that the piston moves slowly or quickly); how are we supposed to know?</p>
<p>yeah man that one was tricky. and it said equilibrium was reached as it wanted to know what happened to the piston???</p>
<p>Equilibrium implies that thermal equilibrium must have been reached with the tube and its surroundings. Because the volume of the tube couldn’t change, and the temperature had returned to its initial value- (PV/T=PV/T) Soooooo pressure would have to equal the original value as well, which was atmospheric… so the position of the piston remains the same. </p>
<p>Lab problem:
- push piston all the way down
- set known frequency
- move piston until standing wave appears on display(first resonance)
- measure L(length of tube-well distance to end of piston) with meter stick
- With a closed end pipe… L=wavelength/4 for first resonance and velocity=wavelength*frequency
- plug yo the calculated wavelength and known frequency…then BAM</p>
<p>part c) vary frequency(independent) and measure corresponding lengths(dep) of the tube for the first resonance… L=(v/4)f sooooo take slope of yo line and multiply it by four and you have the velocity of sound</p>
<p>welp, that was my interpretation anywho… correct me if I’m wrong</p>
<p>@woahgoah I wasn’t sure about what happened when it was pushed slowly, so I just guessed that the pressure stayed the same, but I think it may actually be temperature… so perhaps it was an isotherm? I drew an isobar though 
Second part was adiabatic fa sho though, cause it happened quickly, and no heat was lost… the graph of an adiabatic has same shaped curve as isotherm only steeper</p>
<p>oops on my first post I meant (v/4)*(1/f)</p>
<p>do you guys remember the answer for the acceleration of the block on the right in #1??</p>
<p>I didnt have this form, but i just solved it and i got 2.857m/s^2. Can anyone confirm this?</p>
<p>tdk… I got 5</p>
<p>I think i got -3.something…maybe 9.8 instead of 10?</p>
<p>oh wait you mean c…no dont remember</p>
<p>i got that for the 2.0 kg block, i’m asking about the 1.5 kg one blonde</p>
<p>Wait… I think I got 3 tooooooo</p>
<p>is it not negative?</p>
<p>ok I got 2.8 cause I got tension to equal 10.5 N in the next part… and my friend asked my physics teacher about that part today(I wasn’t there) and he said that tension was correct… so (1.5*9.8)-10.5=4.2 then 4.2/1.5=2.8</p>
<p>Yes, it is negative. eh hope I remembered to put that, lol</p>
<p>I think I got 10 for the tension…and i used 10 instead of 9.8…so my answer came out to be -3.3</p>
<p>They give you a pretty good range I think, so we are probably both within it. Yippeee :3</p>