<p>Hey!
Guys… any physics genius there?
2010 AP Physics E and C — question 1 (part E) …
Have been trying for an hour or so. Saw the solution,din’t get it!!
can any one explain that in a simple lang.
ANd…
is it enough if I solve 2008-2012 FRQ. I usually get around 10-12 every question.
Thanks for your time.</p>
<p>Awesome, thanks for the info! That pulls my FRQ scores a lot higher.</p>
<p>any one know the general curves for E and M, for a 4 or 5?</p>
<p>^ Just over 50% correct usually gets you a 5 on E&M.</p>
<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board; </p>
<p>ques1,Part e
Please help!!! how do i do this? </p>
<p>any idea?</p>
<p>You need to calculate the electric field ‘dE’ due to an infinitesimally small charge element of the arc ‘dQ’. Then, realize that because of symmetry, the net E-field is along the +x axis (the vertical elements cancel).</p>
<p>You then set up the integral that E = ∫dE*cosø, including cosø because we only want the horizontal components.</p>
<p>Then, you replace dE in the integral with k*dQ/r^2. You need to replace dQ in terms of dø, because you are integrating over ø as you move along the arc. dQ = 2Q/pi * dø</p>
<p>Just substitute that into the integral, pull out your constants, and you integrate from -pi/4 to pi/4 because you are integrating over a quarter circle. The rest is math/simplification.</p>
<p>You need to replace dQ in terms of dø, because you are integrating over ø as you move along the arc. dQ = 2Q/pi * dø</p>
<p>What’s this? Totally went over my head… ??
The rest is crystal clear :)</p>
<p>When you do an integral (in general), you integrate over something that varies. For example, the integral of 2x from 0 to 3 means that x is varying from 0 to 3. In other words, you are integrating over x.</p>
<p>In our example, dQ isn’t changing. It’s a constant. BUT, theta is changing. So, we need to integrate with respect to theta (instead of Q). To change from dQ to dø, we know that the arc (a quarter circle) has length (pi/2)<em>R. We also know that arc length in general is theta</em>R. So, Q/((pi/2)<em>R) [which is the linear charge density] = Q/(ø</em>R). You can rearrange that to find that Q = 2<em>ø</em>Q/pi [R cancels from both sides of the equation]. Take the derivative of both sides with respect to ø. You get that dQ/dø = 2<em>Q/pi. Solve for dQ = (2</em>Q/pi) * dø.</p>
<p>Now, you can effectively substitute dQ in the integral with what we just found above. Everything except cos(ø) is a constant here [and you pull it out of the integral], so the only thing left to integrate is ∫cos(ø)dø. You evaluate that integral from -pi/4 to pi/4 which is the interval over which theta varies [or from 0 to pi/4 and then multiple it by 2].</p>
<p>Hey, does anyone understand parts e and f of #1 on the 2012 E&M FRQ?</p>
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<p>For some reason, the scoring guidelines have the electric field from 0.1<r<0.2 as negative, which I don’t quite understand.</p>
<p>The electric field is radially inward between the two shells because the outer shell is at a higher potential than the inner shell. You know that a proton goes from high to low potential, and also in the direction of the electric field. So, the E-field must point inward between the shells. Above the axes it says that radially outward is positive, so you know that inward must be negative when you make the sketch.</p>
<p>Aha! Thank you WiseGuy, I understand now. Also, quick question on the voltage/radius graph. I never understood why the electric potential inside a conductor has to be constant. Is that because if there is a difference in electric potential, a current will flow, and everything gets messed up? Also, on the Voltage/radius graph, why is the line from 0.10<r<0.20 meters concave down as opposed to concave up?</p>
<p>@FastNeutrino Yes, that’s exactly what we did. We had to pick the direction of the current ourselves and apply all of Kirchhoff’s laws. After taking a few practice released exams, I notice a fun quirky problem from one of them. If anyone has time, one should take a look at the “wheatstone bridge” in Google.</p>
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<h1>3, part b:</h1>
<p>Why is work only the change in KE?
Should work also account for the change in PE (which is integral of v from 0 to T times sin(theta))?</p>
<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<h1>1 part e)</h1>
<p>Isn’t some of the energy lost due to the collision (sound, shape deformation) and therefore not completely due to retarding forces?
Also, doesn’t the frictional force act on both the projectile and the block (because the projectile is embedded in the block)?</p>
<p>@Wise Guy… you are my new bestest friend…!! I understand that so much better now!
Thanks a ton dude. :)</p>
<p>If you could just pick in a last query…
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>Ques 3 ,Part b) How did we calculate the value of resistance?</p>
<p>I have solved the whole question but this small value bothered the whole question 
Thanks for your time Bro!</p>
<p>
</p>
<p>They are telling you that resistance per unit length is λ. So, to get the whole resistance you need to multiple λ by the length. But here, the length is changing in time since the sides are getting longer as the rod moves to the right. The rod itself has negligible resistance, so we need to add the left side (side PQ), which is constant length L to the changing length of the rails. Remember that v = d/t, so d = vt. Then, you multiply by 2 because you have resistance on both the top and bottom rails. So, your final R = λ*(L + 2vt).</p>
<p>For the Mechanics questions, I’m a little rusty on Mech so I can’t pick up those answers :(</p>
<p>Anyone have any good practice for Kirchoff’s rules?</p>
<p>Could someone check if this logic is correct? For 2d on 2006 (<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board), the reason why I2 starts at a value and I1 starts at 0 is because at t=0, the capacitor blocks the circuit, but within its little loop which contains switch 2, R2, and the capacitor, it begins discharging as switch 2 is closed. This causes the current in R2 to be E/R2, right? At t=0, the current of R1 is zero, but rises as the capacitor discharges and the opposition to the battery’s EMF decreases. Both I2 and I1 approach E/R1+R2 as t=infinity. Is my reasoning correct?</p>
<p>For mechanics, is it necessary to know the equations for satellites?</p>
<p>The FRQs seem so difficult. Are there any tips for the various FRQs? Specific tips for work, energy, when to use those vs momentum, etc. and is there a lot of gravity/oscillations on the exam?</p>