<p>oh okay then ... sorry!</p>
<p>so was my method correct for 2a? I have no clue what a kepler law is.</p>
<p>T^2 = 4 pi ^2 / (GM) * r^3</p>
<p>u had to use that...although ur answer may be correct if u derived it correctly from the given information</p>
<p>What was the mass of the glider in the 3rd problem?</p>
<p>20 kg, or any number close to that</p>
<p>I got 0.2 kg</p>
<p>Because you use the table to find x^2/v^2 and multiple that by k (40)...which equals like 0.19.. or 0.2 kg</p>
<p>For mechanics 2a, couldn't you have you use the equation
T= (2pi / ω) = 2pi*R/v
so R= Tv/2pi?
That's what I 've used to find radius.. which then was used to find the mass of Mars in b...
N for 2c.. did u guys use Energy of circular orbit = GMm/2R?
I memorized the equation right b4 the test.. not sure it is the right one to use but anyway..
Yeah... I typed in something wrong for 2b and it affected b and c, but since I put the right formulas for both I think I should be ok.. I calculated the values now using those formulas I wrote in my exam, and it seems very close to the actual mass of the mars I searched in Internet..
Really hope I did get 5 on this exam.</p>
<p>wow, i felt so good about the test after i finished, and the more i talk on this discussion, the more i realize how many i got wrong. I thought you were supposed to multiply k by the inverse slope of the line (something like .5).</p>
<p>for 2c i used 1/2mv^2 - GMm/r</p>
<p>the inverse slope of the line was 0.005....times 40 equals 0.2 kg....20 kg is illogical...the k of the spring is only 40 N/m......</p>
<p>ahhhh, i just remembered that the x^2 axis had units of m^-2. well, at least i'll get partial credit.</p>
<p>Hmm I didn't get that. I got 5.18...</p>
<p>I took the slope of the line and divided by k. Was there another way we were supposed to do it?</p>
<p>Bts7390-- I don't think you'll get points off for that. It technically was m^2 because they squared the actual meters value--it was just a really small number to begin with. I think the data table even said m^2.</p>
<p>I got .2kg as well. Did anyone here take E&M. That test destroyed me.</p>
<p>o god i was demolished by E&M</p>
<p>anyway for E&M #3, did u get something along the lines of EMF=BLV for the first part? it was a pure guess ;)</p>
<p>yes but u have to change it into terms of t....</p>
<p>I got EMF = BL / (llamda * t)</p>
<p>For that question, I got:</p>
<p>EMF=BLv</p>
<p>Ohm's law: V=IR or I=V/R</p>
<p>I=(BLv)/R</p>
<p>since the wire has resistance per length of lambda, R=2(length)(lamba) and length = vt at constant velocity. </p>
<p>Therefore, I=(BLv)/(2(lambda)vt)=(BL)/(2(lambda)t)</p>
<p>I forgot to cancel out the velocities though on the actual test. I hope I don't get points off for not having the answer in simplest form.</p>
<p>poster above me: there's no way you could have canceled the v's without making the equation more complex. v doesn't factor into the L*lambda in the denominator. You're fine; v was a given constant.</p>
<p>I don't get what you mean, Darcy. L in the denominator is variable and must be converted to velocity * time. From that, the v's cancel out.</p>
<p>nooo u could cancel v...cuz u make a substitution of x/t for v...and x cancels with the length dimension of R. </p>
<p>what do we get for the magnetic force as a result???</p>
<p>F = ILB
F = (BL)/(lambda<em>t)</em>L<em>B
F = B^2</em>L^2/(lambda*t)</p>
<p>is what I got...anyone concur???</p>
<p>what does the graph and (e) look like for #3 EM??? I bombed that part.</p>
<p>also? is the current counterclockwise or clockwise??</p>
<p>d4r7h3v1l: WRONG. L is not a variable, it's a given quantity. Read the question.</p>
<p>smder99: WRONG. t is not a given quantity.</p>