<p>I just compared our frq’s to Physics B’s
lol it’s like middle school vs high school</p>
<p>I was looking at that too. I almost cried. Anyone have any insight on 2 for Mech? For the differential equation I did dv/dt = a
And then a is (Fa-kv)/m?</p>
<p>@Samson that’s what i got! I hope you’re right!</p>
<p>@nutmeg22:</p>
<p>b. I didn’t use the gravitational force in the torque equation, I just substituted a/r for alpha. c. w=w0+a/r<em>t
r=.1 so a/.1 is the same as a</em>10</p>
<p>Anyone want to work out the E&M FRQ’s?</p>
<p>I’ll give it a shot
1.
a. So first draw a cylindrical gaussian surface in the diagram with a smaller radius than the actual cylinder. For epsilon, ill use a lower case e
E<em>dA=Qenc/e
E(2r</em>l<em>pi)=Qenc/e
Qenc=pdV
Qenc=p</em>2pi<em>r</em>l<em>dr from 0 to r
Qenc=p</em>pi<em>r^2</em>l
Answer:
E=p*r/(2e)</p>
<p>b. same basic concept except Qenc=p <em>2pi * l * rdr from 0 to R
so E(2pi</em>r<em>l)=p</em>pi<em>l</em>R^2/e
Answer:
E=p<em>R^2/(r</em>2*e)</p>
<p>c. a graph: from O to R, should be linear and in the first quadrant and past R should be an inverse curve with asymptote at E=0, max value at R is equal to p*R/(2e)</p>
<p>d. (i) V= - integral from R to 0 (the answer to part a) dr
Answer:
V=p/(2e) * R^2/2</p>
<p>(ii) The answer is at r=0</p>
<p>e. Another graph. From 0 to R, should be a line at E=0 and then from R on, should be an inverse curve</p>
<ol>
<li>
a(i) B
(ii) Voltmeter in parallel with the capacitor</li>
</ol>
<p>b. (i)ln(v) vs. time
(ii) Just take the ln(v) for each voltage value and put into the table</p>
<p>c. Graph of ln(v) vs. time
d. i got something along the lines of RC=12
e. (i) R=8 million ohms or something like that
(ii) The graph has a less steep slope than the original because RC is greater and the slope =-1/RC, so RC increases, which means the slope decreases.</p>
<p>3.
a. (i) emf= derivative of magnetic field with respect to time multiplied by area
=.09e^(-.05t)*.25
=.0225e^(-.05t)</p>
<p>(ii) I=.0225e^(-.05*4)/12
I=.0015 Amps</p>
<p>b. (i) another graph. I’ll explain this in intervals
from (0,8) should be the equation in a(ii) with 4 replaced by t
from (8,12) and (16,18), should be a line at y=0
from (12,16) should be a straight line
all these should be in the fourth quadrant because current is clockwise</p>
<p>(ii) Current is clockwise because magnetic field is decreasing into the page. By Lenz Law, the induced magnetic field opposes this change in flux, so the induced current must create a magnetic field into the page which occurs when current is clockwise.</p>
<p>c. E=Pdt from 0 to 8
P=I^2<em>R
i integrated this using my calculator and came up with 2.32</em>10^-4 Joules</p>
<p>let me expand on 3b(i) from (12,16) should be a horizontal line</p>
<p>Woohoo, apart from the Gaussian disaster on number 1 for me, looks like I worked out everything else fine! Do you think I would get marked off a lot if I accidentally solved for power in 3c? I did everything correctly, just forgot to integrate it to get joules.</p>
<p>Also, for #1, for comfort’s sake - since I completely blanked on how to do Gauss for a cylinder, I didn’t know how to do 1a or 1b. However, I still attempted them and wrote down equations and tried deriving them from Gauss’ law - ended up using 4pi<em>r^2 instead of 2pi</em>r*l. If I ended up getting all the wrong answers for those, but integrated the wrong answer correctly in 1d to get the potential difference, does anyone know how much partial credit would I get?</p>
<p>Guys if regular is
0-22
23-30
31-38
39-49
50-90</p>
<p>I expect for curve
0-18
19-26
27-35
36-43
44-90</p>
<p>@SAT200 why is it at r = 0 not r = R</p>
<p>I think 55-90 is usually a good estimate for a 5. APPass says 50-90 though so idk</p>
<p>@SAT200: Ahh, I see now. About part b, I didn’t see that you used sum of the torques as well as forces. I didn’t use sum of the forces… I still don’t understand how to do part b, but my friends are telling me they got a little over 1 (you got 4/3) so I think you did it correctly.</p>
<p>Anyone want to go over the other to mechanics free response?</p>
<p>Are they going to release the other form for Mechanics free response?</p>
<p>@guyco: I’ll try #1</p>
<p>a) graph is concave down approaching asymptote at y = 0.5 m/s
b)
i. take the integral of the recorded data set with the initial condition at t = 0, v = 0
ii. graph is concave up until t = 0.79, then it is linearly increasing
c) 0.79 + (1.75/0.5) = 4.29 seconds
d) 1/2mv^2 = 1/2kx^2
(0.4)(0.5)^2 = k(0.25)^2
k = 1.6 N/m
e)
i. 0.25 m
ii. T = (2pi)sqrt(m/k)
T = pi sec or T = 3.1 sec</p>
<p>Somebody else want to do #2?</p>
<p>@Nutmeg22
Awesome. I got everything right except for part C, which I blanked on.</p>
<p>On the actual test I don’t remember what i got for my answer but I think I forgot the 0.79 seconds part, but as long as you get a little more than half the points, you’re on pace for a 5!</p>
<h1>2.</h1>
<p>a) Fa going right and kv going left. I also forgot to put in on the test the mg going down and Fn going up, which is probably required.</p>
<p>b)netF = ma = Fa-kv
a = (Fa-kv)/m
a = dv/dt
dv/dt = (Fa - kv)/m</p>
<p>c)a = 0
Fa = kv
v = Fa/k</p>
<p>d) (Hello Calc AB…)
dv/(Fa - kv) = dt/m
-ln(Fa - kv)/k = t/m + C
ln(Fa - kv) = -kt/m + C
Fa - kv = Ce^(-kt/m)
Set v and t equal to 0 to find C
Fa = C
Fa(1 - e^(-kt/m)) = kv
I left it in that form, but you can divide by k if you so wish.</p>
<p>e) Begins at zero, increases while concave down with an asymptote at Fa/k.
(If you know your RC/RL circuits, you should know what this function looks like.)</p>
<p>it’s really weird that some of my classmates have totally different form as mine.
And theirs aren’t released.
got form o.hate it.</p>
<h1>2 is awful…</h1>