<p>Darcy: WRONG...read the question:</p>
<p>"Derive an expression for the magnitude of the induced current as a function of TIME, T."
"Derive an expression for the magnitude of the magnetic force on the rod as a function of TIME."</p>
<p>WRONG. Why don't YOU, read the question next time?</p>
<p>I know this was mentioned earlier, but I want to reclarify:
Mistakes made in one portion of the problem only deduct points for that portion. Therefore, carried mistakes that are applied correctly receive full credit?</p>
<p>I felt terrible after the test because, for FR 1, I screwed up the normal force, which meant that i screwed up the uk, etc. etc. I thought i was getting a 4 for sure, but if that applies, i still have a chance.</p>
<p>BTW, for #3, was h the unstretched height or the stretched height?
For 2d (slightly elliptical orbit), did it say that the given distances were distances from the SURFACE of mars or from the CENTER or mars? If it was the surface, I got number-screwed...</p>
<p>Thanks</p>
<p>smder99: Sorry, but I'm afraid you're just plain WRONG. You forgot that the vertical wire (not including the conducting rod) has resistance, so there's no way you can cancel the fricking L.</p>
<p>I = (BLv)/(L<em>lambda+2vt</em>lambda)
F = ILB
F = (B^2<em>L^2v)/(L</em>lambda+2vt*lambda)</p>
<p>Well, I probably did this wrong but I had the current as a linear function because I let x = vt so i integrated BLvtdt.... got BLvt^2/2, took the derivative of that (for getting EMF generated) and got BLvt. I then divided that by lambdaL to get Bvt/lambda...I plugged this into F = ILB and I ended up getting a linear function with a y-intercept (this part I know is wrong, so minus one point) and said the speed decreases because now the force is pulling it back in the opposite direction.</p>
<p>Edit: Ahhh ****, I mixed the resistances up, the rod has no resistance...how much is this gonna hurt? 5 points?</p>
<p>Masta P: Don't worry, you're fine. The given distances were from the center, or else it wouldn't have made sense; radius is from the center to the edge.</p>
<p>Thanks dude. Anyone care to estimate the raw score for each grade?</p>
<p>Well, that was the only part I totally messed up and I did state my principles so I'll say 4 pts off for that part as well as for FR 1, so I might end up with a FR score of 35....carried errors FTMFW.</p>
<p>piccolojunior...care to share your answers for Mechanics???</p>
<p>ehhh all I gotta say is I thought the mass of the glider was 20 kg lol, I still have one AP to study for, although I would like to know if:</p>
<p>a. You had to do anything besides sketching the curve of charge with a slope approaching 0 (no asymptote) for the charge v. time graph on 1 d</p>
<p>and</p>
<p>b. if I totally ****ed up 3 (b) by mixing up the resistances, but still applying the formulas I put down correctly, will I get partial credit for that part and full credit for carried errors on the other parts?</p>
<p>Does anyone have the answers to the physics FR? Like for cal BC, theirs a site with all the answers</p>
<p>I am afraid you are wrong, Darcy.</p>
<p>L, the length of the rod, was indeed given. What is varying is the length of the rest of the wire that the current is passing through (READ: NOT THE GIVEN "L"). The resistance of this is lambda times that varying length times two. The rod itself has no resistance. </p>
<p>This length of the wire (not the rod) is a variable, and therefore cannot be in a function that must be in terms of t, time. To convert that to time, you use the kinematic formula x=vt. You know v because it is constant, and now you have t as your variable. The v's cancel out. Peace is restored in the galaxy.</p>
<p>Quod erat demonstrandum.</p>
<p>d4r7h3v1l: Wow, you obviously didn't read my last post. You, like a lot of other people, forgot about the resistance in the vertical wire (not including the conducting rod, which is assumed to have no reisistance). The circuit is a SQUARE. The left vertical wire had resistance L*lambda and was in the denominator, and is independent of v, so V CAN'T CANCEL. I repeat: (quoting from my last post)</p>
<p>I = (BLv)/(L<em>lambda+2vt</em>lambda)
F = ILB
F = (B^2<em>L^2v)/(L</em>lambda+2vt*lambda)</p>
<p>so is it integral of BLvtdt or BLvdt?</p>
<p>Edit: I tried this through the flux approach...so stupid...but anyway do I still get credit (say, a point or two) for the attempt?</p>
<p>Wow, all you guys are wrong here:
The wire is length D, so you can find the acceleration by finding the second time derivative of the wire. Once you have acceleration, multiply it by the mass to get the EMF. Then multiply this EMF by B, the reciprocal of the gravitation constant, and by l, which is 1/2(a)(t)^2. This gives you the field in which you will find the answer. I seriously can't believe you missed that...</p>
<p>FYI: i took mech only.</p>
<p>piccolojunior: I don't think you needed an integral. The flux is increasing into the page, so the induced B field is out of the page. Emf= -dflux/dt. Flux=B<em>A=BxL. Thus, emf= BL</em>dx/dt=BLv. Then, I=emf/R, yada yada yada. I'm not sure where you would get an integral from.</p>
<p>Yeah I was a complete idiot with that **** lol, so any chance I have a shot for a 9 on that problem? (Obviously, I woulda messed up the graph since I dont have the right function...)</p>
<p>I apologize Darcy, I didn't realize that you were referring to the left piece of the wire. Sorry for being a huge dick. I think we were both misunderstanding each other.</p>
<p>I guess my mistake carries through, luckily.</p>
<p>d4r7h3v1l: No problem. I guess my 2nd post was too rude, also. Good luck on the rest of the test!</p>
<p>For EM #1, I created functions for parts b and c. Do you think that I'll get credit for that? Or were we supposed to directly use that graph?</p>
<p>Also, for EM #2, part a. iii, did you get something involving 19?</p>
<p>Well, the way I look at it right now, I got a 10 on the first problem, an 8 on the second, and an 8 on the third...not what I was hoping for but oh well. It's still a comfortable 5.</p>