<p>Oh wow, I had no idea what I was doing for the density (I did use two given points though because they were on the line, and that’s what I got)… [901…]</p>
<p>wow i feel stupid. i forgot to do the inverse of the slope…</p>
<p>For the density, it’s delta x/delta y. I got 875, but I used points on the line.</p>
<ul>
<li>I meant 909, it feels odd to estimate point on the line when two points actually seemed like they fit. Idk, I hope I don’t lose points.</li>
</ul>
<p>Also, for the area of ABCDA, I put work done ON the gas. Will I lose the point since I probably forgot to put “negative of work done on gas?”</p>
<p>“For the density, it’s delta x/delta y. I got 875, but I used points on the line.” - same here. </p>
<p>For 7, I used the kinetic energy of the electron to find it’s momentum.</p>
<p>Anyone think I still have a chance at a 5? I’m assuming that I got about 40-45 correct on MC with 9 skipped. Here’s what I did on Free Response. </p>
<p>1a.) .4 s
1b.) 3 m/s
1c.) .235 m
1d.) .6 m
1e.) E2<E1 because kinetic energy was lost in the inelastic collision.</p>
<p>2a.) Buoyant force up, weight down
2b.) (mass of sample + mass of cup)/density of oil
2c.) It’s a line…
2d.) I screwed up and forgot to take the inverse so I got .0011. Any chance I get partial credit for still taking the slope?
2e.) I put that the y-intercept represented the volume of the cup that was submerged in the oil.</p>
<p>3a.) Q1 is negative, Q2 is positive.
3b.) F2 was up and to the left, F1 was down and to the left.
3c.) .00282 N
3d.) 2820 N/C (Over here, I did something really stupid and instead of just dividing 3c by the charge on the particle, I recalculated the forces if particle 3 was replaced by a proton and then divided the new force by the charge of a proton. I got the same answer and indicated that E=f/q but I did a lot of unnecessary work. Will I get points taken off?)
3e.) The top left box</p>
<p>4a.) 3.75 x 10^7 W
4b.) Skipped 
4c1.) I think I put net work done on the system, will I get points taken off since I didn’t say that it was net work done by the gas?
4c2.) 320,000 W
4d.) I put AB and DA. Will I get partial credit for at least getting the AB part correct?</p>
<p>5a.) 22.9 degrees
5b.) I screwed up and put that angle 1 should be increased. Will I get partial credit for at least finding the critical angle?
5c1.) 4.81 x 10^-7 m
5c2.) 2.41 x 10^-7 m. I know that this is incorrect because this would be the thickness needed for constructive interference (the question wants destructive interference) but is there any chance I get partial credit?</p>
<p>6a.) Clockwise (apparently wrong). My explanation included magnetic flux increasing and right hand rule.
6b1.) .18 A (I know that this is the wrong answer. I did everything correctly but I incorrectly used .12 as the length in the formulas.)
6b2.) .0432 N (again, I used .12 as the length so my calculations are slightly off).
6c.) Net force = 0 because there is no change in flux and therefore no induced current.</p>
<p>7a.) 7.5 x 10^-14 Hz
7b.) 3.873 x 10^-19 J
7c.) I had no idea what it was asking for so I put the minimum frequency/x-intercept.
7d.) -9.243 x 10^-28 kg x m/s (This answer is different than what someone else on this forum got. I took the given Kmax and plugged it into the E=pc formula to solve for p. Is this correct?)</p>
<p>I’m guessing I got 15/15 on question 1, about 12/15 on question 2, 10/10 on question 3, 6/10 on question 4, 5/10 on question 5, 5/10 on question 6, and 5/10 on question 7. So if I got a 55/70 on Free Response and a Raw score of 35 on MC (this is my worst-case scenario IMO) is there still a chance I’d get a 5?</p>
<p>Ahh, darn. I didn’t know that if stuff stuck together, it would be deemed inelastic afterwards.</p>
<p>For 2e, would it be alright just to say that the intercept represented the submerged volume of the cup alone, or the fluid displaced solely by the cup?</p>
<p>For 7c, I thought of gravitational potential…where E= mgh so E/m = gh = gravitational potential. So when it asked for stopping potential, I set E = Kmax and divided both sides by m to get E/m=(1/2)v^2. Would that be appropriate at all(because the voltage way is for sure right)</p>
<p>I thought it was going clockwise.</p>
<p>Here’s another disclaimer - I’m not sure on a few of my answers, including on 5b, saying that angle 1 must be decreased. So I wouldn’t agree with people who say they “screwed” it up, unless I can get affirmation from others that decreasing theta is correct.</p>
<p>I said angle 1 should be decreased. Decreasing Angle 1 = Decreasing Angle 2 = Increasing Angle 3 </p>
<p>right… I hope?</p>
<p>Right. The formula is as follows:
theta 3(angle of incidence for glass->air) = 60 - theta 2(angle of refraction for air->glass)</p>
<p>theta 3 is the critical angle, and u wanna increase it. that means u wanna make sure theta 2 is smaller.</p>
<p>Using snell’s law:
n1sin(theta1)=n2sin(theta2)</p>
<p>sin(theta1)=(n2/n1)sin(theta2)</p>
<p>as sin(theta) increases, theta increases for the range of 0 to 90, so to decrease sin(theta2)(and thus theta2), decrease sin(theta1)(and thus theta 1). This can be done by shining the incident light more perpendicularly to the surface of the prism.</p>
<p>Yeah, my teacher went over most of the problems today. Theta 1 should be decreased for exactly the reason you mentioned.</p>
<p>BTW, I’m an idiot, FRQ is out of 80 lol. So would about a 63/80 and 40ish raw score on MC get me a 5?</p>
<p>What do you think the cutoff for a five is this year?</p>
<p>@Helljumper: That score would have a good chance of being a 5. Good luck.</p>
<p>Shouldn’t E1=E2 since mechanical energy (spring, kinetic, potential) is conserved because none is lost to friction or air resistance?</p>
<p>No, 314159265. The collision is inelastic, so energy is not conserved. Also, calculate the energies before and after yourself. E1 is much greater than E2.</p>
<p>Yeah, E1 is greater than E2. But where does the energy go?</p>
<p>The kinetic energy that is lost goes towards vibrating the atoms of the collided bodies, producing heat and (unlike the example in the free response) deforming them like the crunching in a car accident.</p>
<p>Oh, I see. I guess I didn’t think too much about this part, since I had used conservation of energy to do parts c and d, so I thought energy was conserved in this situation.</p>
<p>For 7d, shouldn’t you equate KEmax=1/2mv^2?</p>
<p>so,
1.1x10^-19J=.5(9.11x10^-31)(v^2)
solve for v, then find momentum with p=mv and you get p=4.5x10^-25 kgm/s</p>
<p>You get different answers if you use E=pc…which ends up with 3.7x10^-28.</p>
<p>What is the correct way to solve this?</p>