<p>3 seconds for the… time it takes for the ball to drop? idr
and the Resistor problem [which one adds up to 6 ohms] was the one where 4ohms and 12ohms were in the parallel and that was connected in series to the 3ohm resister
The problem with the pulley and it was asking what the tension on the string connected to the pulley was: I put 2W</p>
<p>@pianofish, the two carts colliding and combining have a velocity of 1.5. I don’t know if we can discuss answers and such yet though so I’ll abstain from all that.</p>
<p>for tension in pulley, i put 1 w.
if you draw a free body diagram, theres weight going down and tension going up. since f net = 0, i assume weight = tension. </p>
<p>for the carts colliding, i think the answer is 6.
2(6) - 2(3) = 2(-3) given + 2 (X)
so x=6. </p>
<p>but i think when you combine it, the answer is 2? </p>
<p>and the kinetic energy for one the problems is 9 right? initially they give you 36 but i think they said they halved the speed or something?</p>
<p>piano fish, i confirm resistor problem that adds up to however many.</p>
<p>envymealways: but the pulley problem was asking for the string pulling the pulley- so wouldn’t it still be the same as the two blocks of W because the pulley doesn’t make a difference here?</p>
<p>anyways, agreed to the kinetic energy of 9 problem</p>
<p>It was “quasar” for the star problem [just looked it up it says a quasar isn’t a star]</p>
<p>quasar: a starlike object that may send out radio waves and other forms of energy; many have large red shifts</p>
<p>grrrr. (10 char)</p>
<p>The charge on the capacitor will go asymptotically to zero. </p>
<p>What was the wave one about the doorway?</p>
<p>No it wouldn’t be 2 m/s. The problem stated that two identical carts (mass = 2kg), one with v= 3 and v =6 moving in opposite directions, collide inelastically. This would mean the sum of momentums before is the same after.
so,
2(3) - 2(6) = (4) (vfinal)
vfinal = -1.5 m/s (the negative means that it is traveling to the left, since the bigger momentum cart hits the smaller momentum cart from the right).</p>
<p>@ banjoman 12348 what is asymptotically to zero?</p>
<p>sorry nvm, i agree with you. i did get 3/2. sorry, brain freeze 
</p>
<p>can anyone respond to the one about tension/weight?</p>
<p>when we draw the free diagram, we have that mg goes down for both masses and T goes up for both masses. However, the question asked for the tension force on the rope connecting the pulley to the CEILING! so, the mg’s become irrelevant, and the answer is simply 2T. But we know that T= mg = W (since neither mass is accelerating), so the final answer is 2W.</p>
<p>For collision of carts, isn’t it
2(6) - 2(3) = 4(x)
x = 1.5?</p>
<p>List:
Force is NOT zero with a nonzero acceleration
Kinetic energy is not a vector
Two carts elastically collide, second one is 6 m/s
Inelastically collide and stick together for 1.5 m/s
Half life is 1 hr 20 min
A is proportional to b^2, straight line when it’s a vs root b.
Constant acceleration graph is straight, displacement is a parabola.
Stars: Quasars
Resistors: 12 and 4 in parallel, 3 ohms in series.
Kinetic energy for half the speed was 9J.
Capacitator: nonlinear, negative decay (doesn’t reach zero, but gets close).</p>
<p>I said pulley tension is 2W, but I’m not sure.</p>
<p>I don’t remember these questions, so I’m not putting them in the list:
minimum speed sq root 2 X v
thick string to thin string: velocity is less in thick string, but frequency same?</p>
<p>Capacitator: it decreases through exponential decay (citation: <a href=“http://www.coilgun.info/theorycapacitors/capacitorcharging.htm[/url]”>http://www.coilgun.info/theorycapacitors/capacitorcharging.htm</a>)</p>
<p>sorry, my bad. i didn’t realize it said tension to ceiling. gosh i wish i had caught that.</p>
<p>stars: quasars (confirm?)</p>
<p>the minimum sq r 2 x v was the one with a downwards slope and then a loop. </p>
<p>the string one had a diagram as well.</p>
<p>the temperature goes to 20 degrees [asymptote]</p>
<p>why is it sqr 2 x v and not 2v or something?</p>
<p>Two loops, one with a current: other loop’s current is opposite, magnetic field to the left</p>
<p>For the downward slope + loop one: wouldn’t you need to know the radius? If it goes too fast it’ll fly off the loop.</p>
<p>And the photoelectric effect one with orange -> blue light or whatever. The choices were photon energy up, electron KE up, and something about the work function. Was it just I and II?</p>