<p>Time for "Q of the Day" ...</p>
<p>A 600 kg car goes from rest to 20 m/s in 30 sec, at constant acceleration. What power is being generated by the engine at the end of the 30 secs?</p>
<p>A) 1000 W
B) 2000 W
C) 4000 W
D) 8000 W
E) 10000 W</p>
<p>(someone other than ZeD, please ;)</p>
<p>Ugh, I just failed the Sparknotes online test...
Am I mistaken, or does the first practice test contain several mistakes?(such as exchanging lambda and l, an image not showing up even after refreshing, ...?)
Guess I wont be able to take the test this saturday:(</p>
<p>fignewton, I believe the answer is D.</p>
<p>You would first find acceleration by doing-</p>
<p>a = (final velocity - initial velocity)/time</p>
<p>which is-</p>
<p>a = 20 m/s - 0 m/s / 30 sec</p>
<p>a = .6666667 m/s^2</p>
<p>Then you would do F = ma</p>
<p>F = 600 kg * .6666667 m/s^2</p>
<p>F = 400 N</p>
<p>Then you would do-</p>
<p>P = Fv</p>
<p>P = 400 N * 20 m/s</p>
<p>P = 8000 W</p>
<p>I hope I'm right, otherwise that's how I did it >__<</p>
<p>When I first saw what they were looking for, I thought of the two power formulas that could relate -> Power = Work * displacement or P = Force * velocity.</p>
<p>I thought F*v was the one to use..</p>
<p>Hope I helped ^^</p>
<p>i think its c... but whos right, me or ya-alim??
i just said that there is 0 kinetic energy to start.
and at the end there is 1/2mv^2 so, .5(600)(20^2)= 120,000
and power is work over time. since it did 120,000 J of work in 30 secs
we divide 120,000J/30s = 4000 W</p>
<p>Ohh, (sorry, I messed up in my other post, saying P = F*d, that's actually work)</p>
<p>Sorry if I'm wrong >__<</p>
<p>anyone else have time difficulty on the test?
I feel like I understand everything now that I self-reviewed, but it's difficult to answer all the questions in the time limit.</p>
<p>The answer to the car problem was D, 8000 watts.</p>
<p>Note that 4000 W is the <em>average</em> power generated, not the power at t=30.</p>
<p>Sorry if I made this a bit hard, P=Force*velocity probably isn't on the test.</p>
<p>BTW, actual cars tend to accelerate at constant P, not constant a!</p>
<p>I'm taking physics this saturday, and when I take sparknotes/barron's practice tests I get around 700. Would I get similar score on the real test?</p>
<p>*A 600 kg car goes from rest to 20 m/s in 30 sec, at constant acceleration. What power is being generated by the engine at the end of the 30 secs?</p>
<p>A) 1000 W
B) 2000 W
C) 4000 W
D) 8000 W
E) 10000 W</p>
<p>(someone other than ZeD, please *</p>
<p>F = the change in p / time
F = 600(20) / 30
F = 400N</p>
<p>P = Fv
P = 400N(20)
P = 8000W</p>
<p>Now if the questions were only that easy on the real thing =[</p>
<p>Can you not also use F = ma when a = v/t?</p>
<p>Yes, you could. </p>
<p>You would do:</p>
<p>F = (mv) / t to find force, then do P = Fv.</p>
<p>Putting it together, you would have:</p>
<p>P = (mv^2) / t = [(600 kg)(20 m/s)^2]/30 s = 20^3 = 8000 W.</p>
<p>I took Physics B (5) and C (5/5) but I'm only taking the SAT II this year (yeah, i know...) and I don't really remember much of the Physics B topics; ie. sound, thermo, waves, interference, etc. How much emphasis is there on those topics?</p>
<p>^ Rupac, how are you so good at Physics?</p>
<p>XD, I understand what's going on, I just can't formulate my thoughts into formula's and math for some reason..=</p>
<p>Good luck to all on the SAT II Physics this Saturday!</p>
<p>Haha, its physics. There are thousand ways to get to the same destination =)</p>
<p>And, anyone else have problems with the sparknotes online test?(As I mentioned, I failed at it, mainly because images werent shown and some variables were exchanged)</p>
<p>A person is pushing a 2 kg crate at constant velocity along the floor of an elevator, which is accelerating up at 30 m/s^2 (ouch!). If the person is applying a horizontal force of 20 N, the coefficient of kinetic friction is:</p>
<p>A) 0.10
B) 0.20
C) 0.25
D) 0.40
E) 0.50</p>
<p>Is the answer E?</p>
<p>I was a little stumped on the question, I'm not sure if it's correct though..</p>
<p>coefficient of kinetic friction is (frictional force/ normal force).</p>
<p>frictional force = 20 N
normal force = (2*30) + (9.8 * 2)</p>
<p>so it would be 20/79.8</p>
<p>and the answer would be C.</p>
<p>I'm not sure if I'm right or not... I will be taking the SAT II Physics tomorrow and it will be my first time! I haven't started studying for it yet.. or even know how to..</p>
<p>im not taking the sat 2 physics, but ya youre right promemorex. normal force here is not equal to weight of block - rather its weight + force needed to generate upward acceleration.</p>
<p>Okay, I was taking a Barrons Practice Test for physics and came across this annoying question about ammeters.</p>
<p>Essentially, the question said that an ammeter is composed of a galvanometer and a resistor. What magnitude should the resistor have (high/low) and how should it be placed (series/parallel) so that it will function properly?</p>
<p>I got the answer right (low/parallel), but will questions like this come up on the actual exam?</p>
<p>Q of the night? (answer to last one: 0.25)</p>
<p>Pigeons (2 kg total) are roosting in a large, 20 kg closed-in coop. The whole thing is on a scale. Suddenly, a loud noise causes all the pigeons to fly, but they remain inside the coop. The scale reads:</p>
<p>A) 190 N
B) 200 N
C) 210 N
D) 220 N
E) Not enough info. Depends on whether birds are hovering, or accelerating up.</p>
<p>just curious fignewton, where do you get all these questions from?? do you make them up?</p>