<p>No problem. Heh, my class has to cover sound, light, and modern. I did modern over spring break though. It still doesn’t look like a fun next two weeks. :-/</p>
<p>Oh, it makes sense now Keasbey Nights. Thanks!</p>
<p>OK I have one.</p>
<p>Which of the following is an expression for the acceleration of gravity with uniform density ρ and radius R?</p>
<p>A) G(4piρ/3R^2
B) G(4piR^2/3)
C) G(4piρ/3R)
D) G(4piRρ/3)
E) None of these is correct.</p>
<p>This question sucks. A lot. Anyway, acceleration is measured in m/s^2.</p>
<p>(A) comes out to 1/m^2<em>s^2
(B) comes out to m^5/kg</em>s^2; did you forget to include the ρ? If so, it would come out to m^2/s^2
(C) comes out to 1/m*s^2
(D) comes out to m/s^2</p>
<p>The answer is (D), judging by units.</p>
<p>My, that certainly was some good thinking! Actually, that question is pretty simple given you know the volume of a sphere.</p>
<p>Given that
ρ=M/V
M=ρV
Substitute in the volume of a sphere
M=ρ((4piR^3)/3)</p>
<p>The formula for g is Gm/R^2
If you substitute the mass
g= G(ρ((4piR^3)/3))/R^2
This gives us
g= G(ρ((4piR)/3))
g= G(4piRρ/3) (Choice D)</p>
<p>And yes, it seems I forgot to include ρ in choice B.</p>
<p>^That problem was cruel. I just realized that G was the gravitational constant. Where did you get this problem from? Barrons?</p>
<p>Well, you didn’t mention that it was a sphere. (: I kind of figured it was but didn’t want to take any chances.</p>
<p>I like how you realized that it was a question from Barron’s. Actually, it’s a question from the SAT book. It’s marked as a “challenge question”, though.</p>
<p>The question never said whether it was a sphere or not. But I’m assuming you can just assume it’s a sphere when it’s a question about gravity.</p>
<p>OK guys, I have another one.</p>
<p>A 5 meter uniform plank of mass 100 kilograms rests on the top of a building with 2 meters extended over the edge. How far can a 50 kilogram person venture past the edge of the building on the plank before the plank just begins to tip?</p>
<p>(A) 0.5 m<br>
(B) 1 m<br>
(C) 1.5 m<br>
(D) 2 m
(E) It is impossible to make the plank tip (the person would have to be more than 2 meters from the edge of the building)</p>
<p>I believe this is a torque problem. </p>
<p>torque on the side of the 100 kg: t(torque)=100(9.8)(5)=49000 torque</p>
<p>49000/(9.8x50)=10m from the axis of rotation.</p>
<p>E is the answer.</p>
<p>But Jerry, you’ve neglected the fact that the plank extends two meters off the roof. If we choose the axis of rotation to be the point where the plank goes over the roof, our torque equation is:</p>
<p>(100)(9.8)(.5) = (50)(9.8)(x), where x is the distance from the edge of the roof. x comes out to 1, so the answer should be B. The reason that we multiply by .5 is because the center of mass of the object is .5 meters away (at 2.5 meters) from the axis of rotation (at 3 meters).</p>
<p>Actually, we’re both wrong. the answer is B. It’s a center of a mass problem, something that is not tested in the AP Exam anymore. We can’t solve using the application of torque.</p>
<p>Well, I got B, and I just did solve it using torque, haha. My teacher has always had us assume for AP Physics B that density and mass are uniform and that the center of mass/gravity is at the center of the object, so using those assumptions, I got it.</p>
<p>From a Princeton Review book (not going to use the MC from a recent exam yet but this should do)</p>
<p>A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is perpendicular to the 4 m displacement, the work it does is equal to</p>
<p>A) 0 J
B) 60 J
C) 80 J<br>
D) 600 J
E) 2400 J</p>
<p>^^Yeah, i saw the explanation on the internet.</p>
<p>Jerry - the answer is A. No work is done when the force is perpendicular (as well as when there is no displacement).</p>
<p>Yeah, the answer is obvious, so i wouldn’t be surprised if you got that right.</p>
<p>A container holds a mixture of two gases CO2 and H2, in thermal equilibrium. Let Kc and Kh denote the average kinetic energy of a CO2 molecule and an H2 molecule, respectively. Given that a molecule of CO2 has 22 times the mass of a molecule of H2, the ratio Kc/Kh is equal to</p>
<p>A) 1/22
B) 1/(sq 22)
C) 1
D) sq 22
E) 22</p>
<p>(hmm, after looking through the PR book, it seems that it’s a good book after all)</p>
<p>I’m not really sure on this problem. Are we to assume that rms velocities are equal because the mixture is in thermal equlibrium? If so, then the ratio is 22, or E. If not, I have no idea what to do at all for this problem.</p>
<p>(Formula used: K = (1/2)mv^2, where v is the root-mean-square velocity)</p>
<p>This is a kinetic theory problem. One equation that would help you is KE=(3/2)kT. If I remember from previous threads, you said you own a Cutnell book? If you do, Chapter 14.</p>
<p>So, you can solve the problem from there.</p>
<p>were you guys kidding with the explanations to the first question on this thread 
mg is the force required to lift the object at constant velocity… if its accelerating at g then if we assume the force required to be F(=normal reaction by the hand on the block).
F-mg=ma(a=g). F=2mg
are questions on physics B really that easy?</p>
<p>Ah alright, so the answer is C then, since obviously the temperature and Boltzman’s constant are equal. I did remember that formula but stupidly disregarded it. Excellent question!</p>
<p>Which of the following best describes a perfectly inelastic collision free of external forces?</p>
<p>(A) Total linear momentum is never conserved.
(B) Total linear momentum is sometimes conserved.
(C) Kinetic energy is never conserved.
(D) Kinetic energy is sometimes conserved.
(E) Kinetic energy is always conserved.</p>