June 2010: Physics

<p>only a few days before the test, from what ive seen its always helpful if we post random questions and discuss the answers or post any topics you want explained.</p>

<p>yeah---what is the curve like to get an 800? Princeton Review says at least 60/75......but that seems suspiciously low. So I don't trust it at all.</p>

<p>The official College Board book says 59/75 for an 800.</p>

<p>I have 3 questions from that book that I need explained, please!</p>

<ol>
<li>When coal burns, it produces heat in the amount of 2.5 x 10^4 joules per gram. About 4000 joules of heat is required to raise the temperature of one kilogram of water by one degree. The amount of coal required to heat 5 kilograms of water from 10 degrees Celcius to 60 degrees Celcius is most nearly:</li>
</ol>

<p>The answer is 40 grams.</p>

<ol>
<li>Two positive charges of magnitudes q and 4q are 6 centimeters apart. If the electric field is zero at point P (not shown) located on the line segment joining the charges, what is the distance of point P from the charge of magnitude q?</li>
</ol>

<p>q - - - - - 6cm - - - - - - 4q</p>

<p>The answer is 2cm.</p>

<ol>
<li>The density of a certain material is 3 grams per cubic centimeter. What is the density of the material expressed in kilograms per cubic meter?</li>
</ol>

<p>The answer is 3000kg/m^3</p>

<p>Thanks!</p>

<p>As for number one, there might be a more efficient way to solve it, but here's how I did it.</p>

<p>You need 4000 J to raise 1 kg of water 1 degree. So, to raise 5 kg of water 1 degree, you need 4000 x 5 J, which is 20000 J. To raise these 5 kgs 50 degrees (60-10), you multiply that number by 50, to get 1000000 J. Divide this number by 2.5 x 10^4, and you'll get 40, which is the answer.</p>

<p>I would like someone to explain number two for me, too.</p>

<p>kind of laborious but here's number two</p>

<p>q----p-------4q with 6 cm separation, break up the distances from p into d and 6-d</p>

<p>E=kq/d^2=k(4q)/(6-d)^2
cancel k and q and rearranging yields
100d^2+4d-0.12=0
solve the quadratic yields 2cm</p>

<p>alternatively i guess you can observe that since distance squared is related to charge that the p must be twice as far from 4q as from q, thus 2 cm</p>

<p>for three, do a dimensional analysis, so:
(3g/1cm^3)(1kg/1000g)(100cm/1m)^3
this gives the answer</p>

<p>Hey hahagooman, I think you made a mistake in number 2. This is what I did, I did kq/x^2 = 4kq/(6-x)^2. Therefore, 4x^2 = (6-x)^2, then take the square root of both sides, you get 2x = 6 - x, x = 2.</p>

<p>A satellite moves in a circular orbit of radius r around a planet of mass M and radius R. The speed of the satellite would be greater if M and r were changed in which of the following ways?</p>

<p><strong>M</strong><strong><em>r</em></strong>__
A decreased no change
B decreased increased
C no change no change
D no change increased
E increased no change</p>

<p>I thought it was D, but apparently it's something else.. Can somebody please explain this to me?</p>

<p>^
is the answer e?
F=GMm/r^2=mv^2/r
canceling you get v=sqrt(GM/r)
so either increase M or decrease r</p>

<p>^^
yeah, i used 0.06 instead of 6 cuz i converted it into meters first but that doesn't change the answer. method is still right</p>

<p>Oh God I can't believe how stupid I am that was very simple.. The answer is indeed E. Thanks for your help.</p>

<p>now i have a question, for someone who doesn't really know waves/optics/modern physics and is just reading the pr for them, is it enough? i think i'm fine for em and mech and the other thermo stuff (took chem, bio, and physics c both parts already)</p>

<p>I'm happy.
Did half a sparknotes test and got 13 out of 30 and didn't even do the other 45 and got a 500.... is it really that easy to get a 650+?</p>

<p>@ Malfunction
You setup a system of equations.</p>

<p>G (Mm)/r^2 = mv^2/r
simplify both sides and you get
GM/r = v^2
if r gets smaller v gets bigger
if M gets smaller v gets smaller.</p>

<p>can anybody explain the difference between all the electric potential stuff. this is killing me: electric potential energy, electric potential, voltage/potential difference, change in potential energy??</p>

<p>Electric Potential=Potential Difference=Voltage=EMF (at least for SAT II and AP Physics B)</p>

<p>Change in Potential Energy=Work</p>

<p>A bar magnet is approaching a loop of wire from above, south pole first. Which statement describes the current induced before the magnet reaches the coil?</p>

<pre><code> (A) It flows clockwise as viewed from above.
(B) It flows counterclockwise as viewed from above.
(C) It first flows clockwise and then flows counterclockwise.
(D) It first flows counterclockwise and then flows clockwise.
(E) No current is induced before the magnet reaches the coil.
</code></pre>

<p>Explain? I have the answer.</p>

<p>I know that Magnetic Flux is proportional to the Magnetic Field and the Cross Sectional Area. But it doesn't seem like either of those two variables are changed.</p>

<p>Is there a picture? The question seems kind of ambiguous, because I need to know whether part of the magnet field is near the coil.
If the magnet field is not covering the wire its E.
If it is covering the wire it is B. I think Not sure. Whats the answer.</p>

<p>Edit: I'll go with B using the Right hand rule.</p>

<p>Is 59/75 for an 800 true for most of the physics tests? Or is it just an average?</p>

<p>There is no picture.
Answer is A.
Here is the book's explanation: As the magnet moves down, flux is decreasing in the downward direction. By the right-hand rule, the induced current must flow clockwise to create flux downward.</p>

<p>Care to explain what that means? I don't see why flux is decreasing in the downward direction.</p>

<p>
[quote]
Is 59/75 for an 800 true for most of the physics tests? Or is it just an average?

[/quote]
</p>

<p>I'd say it's an average. Like the AP exam curves, rarely will 2 different years have the EXACT same curves (that we'll know about, of course). So, the SAT II's probably have curves that vary, but 59 or 60 out of 75 is probably the lowest you want for an 800.</p>