<p>wrong thread^?</p>
<p>what is their about electric dipole that is tested on physics c?</p>
<p>^^Yeah, (for the 1998 exam) it was 55/90 for Mechanics and 49/90 for E&M</p>
<p>i don’t think dipoles can appear on the c test… isn’t that more like chem? also thanks salzahrah!</p>
<p>@salzahrah: Yes the answer for 6 is D, but could you explain more? I know it seems obvious to you, but I’m getting all confuzzled by the other batteries in the circuit…</p>
<p>Also for 7, why is there an initial bump in the graph? I do love your trick though :D</p>
<p>haha no problem</p>
<p>well for 6, you see the middle branch has a battery and a resistor. Now the potential difference between points X and Y is equal to the total potential differences created by the battery and the resistor. The resistor will cause a potential difference of V=IR=2. However, this potential drop is negative because whenever current flows across a resistor, the potential drop is always negative. Now, the battery creates a potential difference that is equal to the Voltage of the battery, which is 10V. However, if you look closely you see that the top is + and the bottom is - . This means that the potential at the lower side is smaller than the potential at the top side. Also the current flows FROM the TOP to the bottom. So the final potential-initial potential = negative number because the final potential is less. Therefore, the potential difference is -10V.
sooooooooooooo
-2V + (-10V) = -12V</p>
<p>7 is kinda hard to explain, but i hope you understand B can be the ONLY right answer because E=0 at x=0 and you should just put it down instead of evaluating the rest of the graph. BUTTT if you do look at the rest then here’s the explanation for the bump in the graph…
Basically, the electric field starts off at zero and increases because the ring has an net electric field component in the x-direction after x=0. However, the electric field is equal to KQ/r^2 and as distance increases the electric field decreases. The decreasing starts to counter the increasing due to the component of the E-field getting larger. The decreasing starts to happen at a certain spot which is the maximum value of the electric field, aka the bump on the graph.</p>
<p>can you guys help me with a few problems? </p>
<p>1) An object released from rest at time t=0 slides down a frictionless incline a distance of 1 meter during the first second. the distance traveled by the object during the time interval t=1 to t=2 is a) 1 m b) 2 m c) 3 m d) 4 m e) 5 m<br>
Why is the answer c?</p>
<p>2) Two artificial satiellites, 1 and 2, orbit the Earth in circular orbits having radii R(1) and R(2). If R(2)=2R(1), the accelerations a(2) and a(1) of the two satellites are related by which of the following?
a) a(2)=4a(1)<br>
b) a(2)=2a(1)
c) a(2)=a(1)
d) a(2)=a(1)/2
e) a(2)=a(1)/4</p>
<p>I got d, but the answer is e, can someone explain why?</p>
<ol>
<li><p>use kinematics for t=0 to t=1.
Find the final velocity of the object in that time interval^
Then use that final velocity as the initial velocity from t=1 to t=2 and see what the distance is.
remember to use the acceleration that you get from t=0 to t=1 when you evaluate for the particle at t=1 to t=2.</p></li>
<li><p>GMM/R^2 = MA
therefore acceleration is GM/R^2 where M is the mass of the Earth. Therefore if the distance from Earth doubles to 2R then the acceleration decreases by a factor of 4 relative to the acceleration at a distance of R.</p></li>
</ol>
<p>^thank you :)</p>
<p>salzahrah you are so ready for this exam lol</p>
<p>THANK YOU</p>
<p>Does anyone have any past EM tests?</p>
<p>^check out this thread:</p>
<p><a href=“http://talk.collegeconfidential.com/sciences/1141111-need-ap-physics-c-mechanics-2004-2005-mc.html[/url]”>http://talk.collegeconfidential.com/sciences/1141111-need-ap-physics-c-mechanics-2004-2005-mc.html</a></p>
<p>is the 1998 test similar to tests now?</p>
<p>A 100 newton weight is suspended by two cords. The tension in the slanted cord is? (angle with horizontal is 30 degrees). Theres a figure but i dont think its necessary. Its from 1984 #32</p>
<p>^Decompose it into vectors. The y component of the slanted cord is equal to the weight of the mass (100N). And the x component of the slanted cord is equal to the total tension of the horizontal cord (because there’s no y component for this) </p>
<p>I hope this helps, if my explanation doesn’t make sense, I’d be willing to draw a picture and post it</p>
<p>A spring has a force constant of 100 N/m and an unstretched length of 0.07 m. One end is attached to a post that is free to rotate in the center of a smooth table, as shown in the top view above. The other end is attached to a 1 kg disc moving in uniform circular motion on the table, which stretches the spring by 0.03 m. Friction is negligible.</p>
<p>What is the centripetal force on the disc? </p>
<p>Ahh, I can’t get it. I know the force uses mv^2/r.
For v I tried doing this: 1/2kx^2 = 1/2mv^2, and then I got v^2 = 9, which would make the answer 90 (m=1 and r=0.03+0.07=0.1) but the answer is 3.</p>
<p>Centripetal Force = the spring force, which can be found using Hooke’s Law (F= -kx). Plugging in the values for k and x, we get (100N/m)(.03m) = 3 Newtons.</p>
<p>Wow, that was a lot easier than I had thought then.</p>
<p>How about this one:
A target T lies flat on the ground 3 m from the side of a building that is 10 m tall.A student rolls a ball off the horizontal roof of the building in the direction of the target. Air resistance is negligible. The horizontal speed with which the ball must leave the roof if it is to strike the target is most nearly:
There’s a diagram but it’s unnecessary.</p>
<p>Does it involve conservation of energy? I thought no, because it would only apply to the vertical motion, right? Like mgh=1/2mv^2 would only solve for vertical velocity?
I thought I was supposed to somehow use d=st.</p>
<p>Find the time it will take for the ball to hit the ground using conservation of energy or kinematics - either works, but I would suggest kinematics - then plug it into the equation for horizontal distance, which is just d=rt. Solving for rate, we will have r=d/t.</p>
<p>Not sure how to use conservation of energy- would you mind explaining?</p>
<p>This is what I tried for kinematics, but my final answer is wrong:
vf^2 = v0^2 + 2ad, vf = 0.
v0^2 = 2(10)(3)
vo = sqrt60.</p>
<p>I used d= 1/2at^2 though, and got the right answer.
Is my approach above incorrect because I set vf=0?</p>
<p>First, find the time it takes the object to fall from the top of the building to the ground, which is 10m=1/2(10 m/s^2)(t^2) so t= sqrt2 seconds. Then, the ball has to travel 3m horizontally in sqrt2 seconds so the horizontal speed it must have is 3/sqrt2 m/s</p>
<p>I think that’s right…</p>
<p>Yep, that’s right!</p>