<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>They are posted, but not linked yet.</p>
<p>Do they have the answers up? And do they have the ones for Physics C up (I’m taking the make up so I want to see what the test on monday was like)</p>
<p>They release the scoring guidelines after we get our scores, at least that’s how it was done last year.</p>
<p>Before I start, I’d like to say that you shouldn’t freak out if your answer differed from mine by an infinitesimal amount. The College Board generally gives leniency on how many sig figs are used (and consequently, how exact the answer is). In fact, they specifically state so in their scoring rubrics.</p>
<p>Also, keep in mind that I obviously can be wrong with my answers, haha. These are just what I got.</p>
<p>1a. t = 0.404 seconds using y = .5at^2
1b. v = 2.97 m/s using more kinematics
1c. x = 0.223 m using conservation of energy
1d. d = 0.6 m using conservation of momentum and then kinematics
1e. E2 < E1 since the collision is inelastic, so energy is not conserved</p>
<p>2a. Buoyant force going up, weight going down
2b. Vo = (ms + mc)/po; started with buoyant force = weight, so pgV = (ms + mc)g
2c. It was a line hahahaha, and it intercepted slightly above the origin
2d. p = 625 kg/m^3; since V is on the y-axis and m is on the x-axis, and p = m/V, 1/p = slope, then I used the table to find the slope and took the inverse of that answer to find density.
2e. The experimenter didn’t account for the mass of the cup on his data table</p>
<p>3a. Q1 is negative, Q2 is positive
3b. Too lazy to describe the arrow directions
3c. F = 5.5 * 10^-4 N; since we know that the y-component is 0, I just calculated the x-component
3d. 2820 N/C (found the x and y components of both fields, added, used Pythagorean’s theorem)
3e. Upper-left hand corner</p>
<p>4a. 3.75 * 10^7 J/s; e = W/Qh and P = W/t so W = Pt, solve for Qh/t
4b. F = 6.43 * 10^5 N; P = Fv
4c. The area represents the work done by the system, and the power output is 3.2 * 10^5 W; one cycle is in .25 seconds, so you just find the area of the box and divide by .25
4d. AB and BC</p>
<p>5a. 22.9 degrees using Snell’s Law
5b. The critical angle is slightly greater than angle 3, so I said that you need to decrease angle 1 to increase angle 3
5c. 4.82 * 10^-7; wavelength(flim) = wavelength(vacuum)/n; and then I skipped the second part haha</p>
<p>6a. Counterclockwise by Lenz’s law and the right hand rule
6bi. I = 0.15 A; V = Blv and then I = V/R
6bii. F = 0.036 N; F = ILB
6c. F = 0 since there is no induced currend (flux is constant) and F = ILB</p>
<p>7a. 7.5 * 10^14 m; c = fw (where w is wavelength), so f = c/w
7b. work = 3.87 * 10^-19 J; K = hf - work (was I supposed to set K = 0 for this? if so, then I did it wrong)
7c. V = 0.688 V; K = qV
7d. p = 1.66 * 10^-27; de Broglie (not sure if I was supposed to do that, either)</p>
<p>The current in 6a was clockwise. To oppose the increased flux into the page, the loop had to create flux out of the page on its right side. This could only be done with the current going downward, and thus clockwise through the loop.</p>
<p>Most of our answers are the same. So far, check your density on the 2nd question, I got 909.091. (84e-6 - 29e-6)/(0.070 - 0.020) = 0.0011</p>
<p>Then take the inverse of that and you get 909.091</p>
<p>Also the physical significance of the y-intercept was the amount of water that overflowed due to the mass of the cup alone. (I’m not sure if this is what you meant) I got the same answers for question 1. Also your calculation for Force on question 3 looks incorrect. The force should be 0.0028N. F = qE. I had the same answer for the electric field you did, and the charge is 1 x 10^-6. The electric field at that location multiplied by the charge must equal the force. To calculate the force without the field, you had to just add the vectors F = sqrt(F1^2 + F2^2). Notice how both arrows geometrically add to give you the same vector. Your thermo looks like what I got. #5 looks good. I didn’t answer the minimum thickness because I didn’t know the equation. I had the same answers you had for #6, that looks good as well. The last question on modern physics looks ok except for part 7d. I used the kinetic energy to solve for the velocity of the electron K = 1/2mv^2. Then I simply multiplied the velocity by the mass of an electron p=mv</p>
<p>Israel - oh yeah you’re right, my bad on that one. I can’t remember what I put on the test, no worries. I had a right had rule freakout on Sunday night, as evidenced by a few posts, so I think I just confused myself even more for the exam, hahaha.</p>
<p>eg2333 - for density, I did the same thing, but using different values on the table. However, I think it still made a difference. And yes, that is what I meant for the intercept. As for force, who knows?</p>
<p>EDIT: Yeah, I used the points where m = 0.03 and m = 0.04.</p>
<p>I got counterclockwise current. Check your right hand rule again lol. Current flowing downward would increase the flux going into the page. Wrap your fingers around the wire and point your thumb in the direction of the current.</p>
<p>Here are the forces </p>
<p><a href=“http://img31.imageshack.us/img31/3637/charges.jpg[/url]”>http://img31.imageshack.us/img31/3637/charges.jpg</a></p>
![http://img31.imageshack.us/img31/3637/charges.jpg[/url]](http://img31.imageshack.us/img31/3637/charges.jpg%5B/url%5D){kind=link}
<p>Notice how the addition of both vectors perfectly add up to the horizontal resultant vector in the -x direction. All you had to do was find the magnitude of the vector.</p>
<p>I did just that, and it looks like it creates flux into the page on the left side of the wire, and flux out of the page on the right side of the wire. Right side is what we’re looking for, I think? I don’t know. I had a freak-out the night before the test with some questions in PR where my Lenz’ Law logic didn’t work, and no one gave me a satisfactory answer.</p>
<p>Israel, Current going downward on the right side would produce flux into the page, check it again. Look at this</p>
<p><a href=“http://scienceblogs.com/startswithabang/upload/2009/04/the_left-hand_rule/Right_hand_rule.png[/url]”>http://scienceblogs.com/startswithabang/upload/2009/04/the_left-hand_rule/Right_hand_rule.png</a></p>
![http://scienceblogs.com/startswithabang/upload/2009/04/the_left-hand_rule/Right_hand_rule.png[/url]](http://scienceblogs.com/startswithabang/upload/2009/04/the_left-hand_rule/Right_hand_rule.png%5B/url%5D){kind=link}
<p>eg2333 - yeah, that makes sense. I just redid your math to confirm it. <em>shrugs</em> Oh well, I anticipated a few mistakes. Hopefully I will glean some partial credit.</p>
<p>Yeah with the answers you have, as long as you did not fail the multiple choice you should have a 5</p>
<p>Well, that’s assuming that I’m right on the rest. 
I don’t want to jinx myself. Well, that’s a lie. I don’t want to feel like I got a 5 and then not get one. That’s a bad feeling. I’m sure I passed though, and I’m really glad for that since physics is a hard subject.</p>
<p>5) c. ii. 2t + (m+0.5)wavelength(flim) = wavelength(vacuum). t = (wavelength(vacuum) - (m+0.5)wavelength(flim))/2 = any integer multiple of 89 nm.</p>
<p>how do you remember exactly what you put?..seriously, some of these remind me what I put, but still I couldn’t just state all of these off the top of my head…</p>
<p>Was it just me, or did they give us very little room to work out some problems (especially parts of number 7 i think)</p>
<p>I reworked the problems quickly, hahaha.</p>
<p>I thought that for the most part, we had enough room, and I wrote my math large.</p>
<p>oh, haha, that explains it…</p>
<p>yeah I tried to put down all the units and formulas I used in every step. I know in the last problem I had an issue with the space where it asked for the wave speed or something in the film. Instead of putting the fomulas together I did it step by step. I probably should have simplified the problems beforehand, but I hope they don’t deduct for that.</p>
<p>@eg2333
“Most of our answers are the same. So far, check your density on the 2nd question, I got 909.091. (84e-6 - 29e-6)/(0.070 - 0.020) = 0.0011”</p>
<p>You used two of the points they gave you to do this? Didn’t you have to use the slope of the line of best fit, unless you had these two points on it?</p>
<p>definitely counterclockwise induced current. as to the density of the liquid, I got around 909kg/m^3 as well, but now that i think of it, i used two of the points they gave. hope i don’t lose credit on that…</p>