<p>The index of refraction in air is 1, right?</p>
<p>So n2/n1=sin%1/sin%2
Therefore n2=sin%1/sin%2, so n2 is the slope of the graph.</p>
<p>Whatd u guys get for the minimum thickness of the film?</p>
<p>Oh sorry I thought you meant all students press the stopwatch at the same time and end at the same time. So for the tables you put Sin (degree) for each column? I thought part B asked you to label the X axis by degree, not sin degree but I could be wrong.</p>
<p>number 1 was okay</p>
<p>number 2 the experiment was okay</p>
<p>number 3 was electricity and probably my worst 1 i got to part d, and then i started screwing up</p>
<p>then number 4 was okay with the sin stuff, then u graph sin theta r vs sin theta i and take the slope to find the index of refraction between 1.43-1.52 i think then for where it was the interface i didnt know so i just said both, and to find the thickness i had no idea,</p>
<p>thermo question was okay i got negative work and then said heat was added to counteract the work because the internal energy should be 0 in a cyclic process</p>
<p>number 6 was easy and for the experiment part i said shot the electron through a slit, and if u see diffraction pattern then you have wave properties</p>
<p>For the thin film did you have to take into account the change of the wavelength in the oil? I did, but other people were saying you ddint have to.</p>
<p>wait was it between both interfaces or just air-oil?</p>
<p>I believe work is positive because it is counterclockwise, which means that the surroundings do work on the system. For the internal energy to be 0, then, heat must be released from the system to the surroundings. Our teacher absolutely sucks, but for some reason I have a vague feeling of hope ;-).</p>
<p>I think it was air on oil on water.</p>
<p>So does anyone know how to do that one?</p>
<p>I thought all of the FR were fairly easy (easier than our teacher's tests)</p>
<h1>1-easy, SHM/forces problem</h1>
<p>Blanked out on the last part, but I realized later that it was just an energy problem</p>
<h1>2-a joke, I did the 10 m spaced students all starting the stop watch at the same time then stopping them as the runner passed each mark. Then the velocity will be plotted and acceleration can be determined.</h1>
<h1>3-very easy, simple plug-ins, only one i was unsure about was the last part, but I got it right (Work = 0 since the voltage would be 0)</h1>
<h1>4-fairly easy, I considered a few things but realized that it was just sin theta 1 and sin theta 2.</h1>
<p>It's the air/oil interface only since n goes from small to large (which is when you get a 1/2 λ phase shift.
I forgot that it was 2T=1/2 λ and I was off by a factor of 2. Is this a big deal in points? (I put 208 instead of 104 nm)</p>
<h1>5-god this was a joke, work = area of the triangle, and the temperatures were simple plug-ins. I said heat was added, but I dunno.</h1>
<h1>6-moderate, I misread the problem and didn't see that it was talking about an electron for the debroglie so I was thinking photon and used the speed of light...how much do they take off for that?</h1>
<p>Experiment-shoot electrons through a double slit and observe the interference pattern</p>
<p>Our class was very prepared for the FR since our teacher only gave FR type problems on tests (he doesn't believe in MC questions as accurate evaluations of achievement in physics)</p>
<p>Film problem:
600nm/1.43=419nm (converting from wavelength in air to wavelength in oil)
Then,
2T=1/2 λ = 209.79 (Think about it. Oil passes through 2 times the thickness, because it goes in, reflects, then comes back out, and the phase shift is 1/2 λ due to going from a low n to a high n. Going from a high n to a low n is a 0 λ phase shift)
T=104.895
:)</p>
<p>for the experiment you could have just said shoot it through a single slit and oberserve the diffration pattern</p>
<p>so i may not have known a majority of the FRQ's but im positive heat was removed for #5, cyclic process = 0J of internal energy, net work was positive(done on the gas, not by it) so heat must have been removed</p>
<p>does anyone remember any particular answers? if so, care to share? thanks</p>
<p>If I got around a raw score of 40 on the MC and 35 on the FR, then adding the weight factor/curve, what grade is that most nearly on the AP Exam? Does anyone have a grade index or something I can use as an approximation? Thanks in advance ;)</p>
<p>For #2, I'm not sure about the method people are mentioning. Perhaps I'm wrong about this...but there seems to be a serious flaw in the having students stand every 10 meters...What if he reaches the constant speed at 15 meters? How are you supposed to know the exact moment he reaches the constant speed, when you are only taking times at every 10 meters?</p>
<p>I'm not totally sure how to do the problem, but an explanation for that might be useful...I'm pretty sure you need to calculate the time of his constant speed for a full 100 meters, then subtract that from his real time to get the time he speant accelerating. Then plug that into vf=vi+at to find the acceleration. Trouble is, I would think there is an overlap of some sort...so I'm not sure if that works perfectly either.</p>
<p>Anyone have any thoughts on this? Am I just totally wrong about how to approach the problem?</p>
<p>according to my review book, that would be a composite score of 86 which is about a 4 or 3 depending on the curve</p>
<p>Problem number 2 was strange. I think my way of doing it was correct because of the 11 student thing, but it would not yield good results due to the reason confused_student stated.</p>
<p>It reminds me of that calculus problem where you are given the distance vs time chart for a car, you plot it, and then estimate the total distance by looking at the area under the curve. Even though there are gaps in the measurements, you can still estimate the total distance. I think we were supposed to do the same thing here, except take the second derivative of the x vs t graph instead of the integral.</p>
<p>Would vf=vo+at work when the acceleration went from uniform to 0 in the problem?</p>
<p>i think they wanted you to do d=vit+.5<em>a</em>t^2, note that vi=0, and take the time up until when the graph is exponential, plug it in with the distance, and solve for the au</p>
<p>Physicsrocks, thanks. I suspect I'll be getting something on the borderline of a 3 or 4. A 4 is the minimum I need for credit, but even if I do get a 4, I don't plan on skipping into a harder Physics class @ Princeton. I'd rather just take the intro class and do well in it. I have a feeling that if I get a 3 :( that it'll be my lowest AP Exam score this year (everything else, even Calc I'm confident I got AT LEAST a 4 in). So yeah, I guess there's nothing I can do at this point :D but wait until July! Good luck to the rest of you!</p>
<p>-Jon ;)</p>
<p>Regarding that Cyclic Process one, I believe I ended up with a postive Wnet, which meant that work was being done on the system by something else. And because of that, I figured that heat would have to be removed.</p>
<p>The last part of #1 is simple, you just use Vmax=A*Root(k/m)
For #5 I got Work positive +12500J(not sure of exact number) and W=-Q in cyclic, so Heat was removed.</p>
<p>electrostatics...yeah...about that =/</p>