<p>The density ratio raft : water was equal to the percentage of the raft submerged, M17.</p>
<p>i got 8 people too..idont remember the other stuff</p>
<p>the questions are on collegeboard.com now.. if you all need a reference</p>
<p>Lets compile a complete list of answers and solutions</p>
<p>yeah either complile a list of answers, or help find a site that has answers posted.</p>
<p>i know usually after like a couple days, some college professor will post answers online, just like for calc.</p>
<p>Ack, my explanation for the raft problem was wrong, sorry.
Number of people * m of person *g + m of raft * g = Bouyant force.
Solve for n, get 8.4</p>
<p>And BTW, <a href="http://apcentral.collegeboard.com/repository/_ap05_frq_physicsb_45645.pdf%5B/url%5D">http://apcentral.collegeboard.com/repository/_ap05_frq_physicsb_45645.pdf</a></p>
<p>I'll do questions one and two... and let someone else verify my answers.. i think i nailed these 2 questions</p>
<p>Question 1:
A. Graph is at +1.5 from 0 to 9, then it slopes down to 0 at 10, then stays at zero across to 18, then goes down to -2.4 from 18 to 19, then it stays at -2.4 from 19 to 25
B. I. Speed goes from 1.5 to 0, so 1.50-0 = 1.5 is difference, 1.5/2 = .75 m/s^2
II. Acceleration Vector is down.
C. No acceleration, so the passenger feels his own weight. W=MG=70*9.8=686Newtons</p>
<p>Question 2:
A. 3 Forces, MG downward, Tension to the left, and Tension up at 60 degrees from the horizontal.
B. Weight offsets vertical component of angled string. so W=MG=1.8<em>9.8=17.64 Newtons. Draw the same triangle, but with 17.64 as vertical leg of triangle. Find the horizontal leg. Tan(30)=Horizontal/17.64, H=17.64</em>Tan(30) so H=10.2Newtons, and since forces are equal, 10.2N = tension in horizontal string
C. Conservation of Energy, MGH=1/2MV^2, GH=1/2V^2, H is found by finding the leg of the triangle down from 30degrees. Cos30 = X/2.3, 1.99 = length, so 2.3-1.99 = Change in Height = .31 meters. so 9.8<em>.31=1/2</em>V^2, V=2.5 m/s</p>
<p>for question 2c, i thought the h would just be the length of the string (2.3) since lengths don't change when swung down.</p>
<p>i found the change in height from the beginning to end... that is .31 meters.. 2.3-1.99</p>
<p>The ball wasn't completely horizontal, however, so you had to find the actual height displacement.</p>
<p>Sorry to digress, but I've got a question on the raft problem. Goshamoossy wrote above "The density ratio raft : water was equal to the percentage of the raft submerged," and that's what I did: 650/1000 = .65 , and then the problem said that the volume of the raft was 1.8 m^3, so .65 x 1.8 = 1.17. Therefore, 0.63 m^3 of the raft was NOT submerged (1.8 - 1.17). Then, it said the surface area was 8.2 m^2, so .63/8.2 is about .08m.... why isn't that the height of the raft submerged? Sounds right to me... at least, it DID...</p>
<p>i dont know, i guess its just the other way, id have to work through it.</p>
<p>anyone wanna do question 3 and 4 for me.. i think i lost most of my points on those two.</p>
<p>M17...65% of the raft submerged, that means 35% above water, right? The height of the complete raft is the volume divided by the surface area, so 1.8/8.2=0.219 meters. 35% of 0.219 is .077 meters, or 7.7 cm, which is what the answer was. So about 8 cm if rounded. Yep. </p>
<p>bsbllallstr8: Didn't you have to label the reaction force of the string that was holding the ball to the wall?</p>
<p>oops, i meant tension to the right, i typed to the left.</p>
<p>3:
a) Electric field = k<em>q/r^2. The field towards the top charge is k</em>q/a^2, the field towards the bottom is k<em>2q/a^2. Subtracting, we get magnitude k</em>q/a^2, downwards/
b) Since V = Ed, we substitute E from problem above, and d=a, and get V=k<em>q/a.
c)
i) F=k</em>q1<em>q2/r^2=k</em>-q<em>q/(x^2+a^2)=-kq^2/(x^2+a^2)
ii)F=k</em>q1<em>q2/r^2=k</em>-q*2q/(x^2+a^2)=-2kq/(x^2+a^2)
NOTE: the x^2+a^2 comes from the pythagorean theorem, just in case you don't see that.
d) at A and C, slighty downwards, point to somewhere between the origin and the 2q charge, and B, directly downwards.</p>
<p>50/ 70 mc and -25 points on FR enough for a 5? 4?</p>
<p>Thanks, looks like i did OK... perhaps 9-12/15</p>
<p>but.. on part A the force is upwards, i think... because the 2q charge repels a positive charge more than a q charge...</p>
<p>Also, for part B, I dont think V=ED works b/c it is not in a uniform electric field, the field varies with the distance from the fixed charges</p>
<p>crichessill... looks like you would have a shot at a 5.. i think it depends on the curve, but most likely a 4</p>