<p>^oh man...i didnt think about the radius of mars either..</p>
<p>I believe Smders answers are right. I screwed up the problem, but I believe that was what my teacher had. </p>
<p>I can't believe I forgot about angular momentum!!</p>
<p>smder99, would you please explain to me in detail how you solved 2c?</p>
<p>Certainly! There are two ways and both yield the same answer.</p>
<p>E = K + U</p>
<p>K = 1/2 * m * v^2 where v is velocity and m is mass of GS
U = - GMm/r where M is mass of Mars and m is mass of GS</p>
<p>Please note that U is negative. From that K = 5.38 * 10^9 J and U = - 1.07 * 10^10...therefore K = -5.37*10^9 J</p>
<p>Also, more simply, E = - GMm/(2r)....yields same answer (with correct sigfigs)</p>
<p>Here are answers to #3 Mech...</p>
<p>(a) 1/2<em>m</em>v^2 = 1/2<em>k</em>x^2</p>
<p>(b) Plot the points from the table above and label axes (should form straight line with positive association)</p>
<p>(c) (i) Draw line of best fit (should pass through at least 2 points with equal number of points above and below line...its only an estimation)</p>
<p>(c) (ii) m = 0.20 kg...found using the formula from part A and points given in table</p>
<p>(d) (i) 1/2<em>m</em>v^2 = 1/2<em>k</em>x^2 + mgh
(d) (ii) It will NOT be a straight line because h can be expressed as x*tan(theta)...therefore the conservation of mech energy expression is no longer a linear expression because it has a term that is not of power x^2.</p>
<p>Any comments/questions? I will put up #1 later...
Can anyone corroborate or confirm these answers? admanrich maybe??</p>
<p>thank you so much smder99 that was very helpful!</p>
<p>Yes those are right. I don't remember the explanation in part d exactly, but your's is probably right.</p>
<p>come on someone's gotta know the answers to E&M #s 1 and 2...... i can't wait til they post answers on collegeboard. will that take what 10 years...</p>
<p>im going insane with wonder...... which is actually.... really sad..</p>
<p>Okay I have EM #1...can anyone confirm?</p>
<p>EM1.
(a) emf = 1237.5 V
(b) voltage at 4s = 1017.5 V
(c) q at 4s = 4.1 C
(d) graph should have exponential shape (increasing but concave down) with asymptote labeled at emf*C (should start from 0 though)</p>
<p>(e) Power is 73.4 Watts</p>
<p>(f) Would be the same (I got this wrong) because dielectric causes C to increase but causes V to decrease..therefore they cancel out.</p>
<p>Anyone get the same stuff?</p>
<p>so what did you write on the optics question
P.S. Darn the questions i answered kinda ok where all 10 points not 15 ergggg!</p>
<p>too lazy to redo my answers, smder99, but part D looks correct.</p>
<p>yeah everything looks good but.. how'd u get power... and finally. what was the answer to paart f. on the test i had put greater.... but i realize now i hadn't read the question that well</p>
<p>btw i had gotten 88 for power. ughh</p>
<p>opps just realized this is physics C HAHA</p>
<p>P = I^2*R...right?</p>
<p>Use the table to get I at 4s...or the solved differential equation..I used the differential equation which yields 0.365 A</p>
<p>Were you even supposed to solve the differential equation??? I realize now that u could've done the WHOLE problem without ever deriving the equation.</p>
<p>yeah if u used 0.4 A for I...you get 88 Watts</p>
<p>how about the part whole dialectric thing. you said you messed it up, but do you know for certain the right answer?</p>
<p>come on guys we only need E& M #2!!!!!!. alright ill try to recreate my methods and answers... but ill need everyone's input. esp the genius ppl that um. whats the phrase.... "know the right answers"</p>
<p>E&M #2 a. i. E= (kQr)/ (a^3)
ii. E= (kQ)/ (r^2)
iii. E=(kQ)/(27a^3)-- i got this one wrong but redid it now and came that that answer as a conclusion....
iv. E=0
b. didn't know this one. put some equations down. and showed that when adding the Vs you keep the negatives in the Q's..... umm yeah
c. i wrote down some calc equations in case i was wrong i.e. V= integral Edr. ... ok but my (i think too simplistic to be correct) answer was= V= KQ/2a</p>
<p>come on people. i see there are viewings... someone have teh cahones to defend, reject or qualify my answers...</p>