<p>Ok. Thanks! </p>
<p>One last one, which is number 6 c.</p>
<p>set K from part b equal to U
U = k(q)(Q)/D
where q is charge of the proton (which is e)
and Q is the charge of the nucleus (which is 92e)
solve for D</p>
<p>There were two ways to do number 3 part (a)
firstly, since the pressures are equal on both sides, you can set the absolute pressures equal:</p>
<p>D and H: Density and height for oil
d and h: Density and height for water</p>
<p>Patm + DgH = Patm + dgh
DgH = dgh
DH = dh
D = (dh)/H or (h/H)d</p>
<p>the second way to answer the problem was to see that the force due to gravity both water and oil exerted on the Mercury was equal. Therefore the mass of oil was equal to the mass of water. And so:
density = mass/volume
mass = (density * volume)
you can set the product of the density and volumes of both equal and you’d still get that the density of oil, D, is equal to the density of water, d, times the ratio h/H (height of water/height of oil)</p>
<p>The correct answer for this part was 900 kg/m^3</p>
<p>the next parts for number three (c and d) were “below A” and “increases”, respectively.</p>
<p>The second part of that question but it was quite easy.
P(absolute) = P(atmospheric) + (density)(g)(h)
you just had to remember to convert the centimeters to meters for h.
I got 1.0245 x 10^5 N/m^2
that should be right unless i made a careless calculation error</p>
<p>if anyone else wants the answers for any of the questions (excluding the lab question) just hit me up.</p>
<p>What were some people on the other page saying? That number 1 part b (the acceleration) was -3m/s^2???
How can you possibly get that number?
the correct answer should be 5m/s^2.</p>
<p>For number 1 part f, I wrote that the string’s mass is not negligible and, therefore, the string-pulley system impedes the motion of the 1.5kg block.</p>
<p>well depending on how you defined the direction in your problem it will beeither postive or negative</p>
<p>ok im freaking out about this. I realized today all of the dumb mistakes I made on the frqs. My teacher worked them all out. I kind of guessed the point values and graded myself. I ironically did bad on the first one ( sooo stupid) and I bombed the thermo part. I am estimating about 54/80 frq points. does that fallmwithin the five range with a decent mc?</p>
<p>Did anyone else notice there wasn’t even one question on springs–the formulas F = kx and U = (1/2)kx^2 were never used not even once.
I found that to be weird.</p>
<p>Aside from this seclusion, this year’s test was balanced out: there were many more magnetism questions in the multiple choice with no electromagnetism question in the FRQs.</p>
<p>I also found that the lab question was way more difficult than any other but that doesn’t help or hurt anyone since the test is relative.</p>
<p>Furthermore, the two 15 point questions were probably the easiest questions, the momentum problem took time to figure out but was doable, the oil-water-mercury system question was different then previous hydro-static questions but was easy, and three out of four parts of number 7, including the last part which incorporated Einsteins equation, should have been easy points for everyone.</p>
<p>@buildthewall99, #1 part b is 5 m/s^2. We were discussing #1 part c</p>
<p>Check this score calculator: [AP</a> Pass - AP Physicsb Calculator](<a href=“http://appass.com/calculators/physicsb]AP”>http://appass.com/calculators/physicsb)
it’s pretty reliable.</p>
<p>Yeah but part c should have gave the same acceleration as part b.</p>
<p>Because the 15 Newton force is now replaced by a block that still gives the string a tension of 15 Newtons (mg = (1.5kg)(10m/s^2) = 15N)</p>
<p>I know now that I’m starting to think about it…you might be right…argh</p>
<p>c is a different acceleration because you have to incorporate the weight of the 1.5kg block into F = ma whereas in b you did not. The mass for c would be 3.5kg, but the net force is still the same, 10N.</p>
<p>EDIT: That is, using g = 10m/s^2</p>
<p>Never mind. to solve that part (part c), you would set the forces equal.</p>
<p>For the 2.0kg block set the horizontal forces equal:
Ft - Ff = ma ; where Ft is the tension in the string and Ff is the force of friction. m=2.0kg
Ft = Ff + ma</p>
<p>For the 1.5kg block:
Mg - Ft = Ma
Ft = Mg - Ma</p>
<p>Now since the Ft’s for both are equal, you can set both sides equal to each other.
Ff + ma = Mg - Ma (the accelerations are equal and Ff = 5N)
5 N + 2a = (1.5)(10) - (1.5)(a)
-10 N = -3.5a
a = 2.9 m/s^2</p>
<p>I just did this right now and got this answer but on the test I assumed since the force was the same (15N) then the acceleration of the 2.0kg block was the same as before (as in part b). Therefore it would have had the same acceleration as before but I guess not.</p>
<p>So is my answer correct or not?? I am so confused haha</p>
<p>it should be correct
or around correct
i guess i was wrong for that part
but i got 2.9 not 3.something</p>