<p>not sure if it was on this test or the chem, but wasn’t one answer the curve with the inverse relationship with T and R</p>
<p>@alar03 I think the heavier object goes down but also continues to the right as well…</p>
<p>In the q vs time graph, it should have a changing curve, as well as the current vs time one. The first is increasing, the second decreasing. (sorry for answering my own questions, but I thought it might help someone)
The relativistic problem, was the speed <0.1c or between 0.1c and 0.6c?</p>
<p>PE was an upside down V.</p>
<p>So capacitence should change with time?</p>
<p>And for acceleration on the vertically rotating object, I put the arrow that pointed NW.
The centripetal force was pointing inward (left) and the velocity was was pointing up. So the resultant would be pointing NW.</p>
<p>I think for the relativistic problem c<v was the answer.</p>
<p>@eevvaann22 yes, I checked now. since Vx’=2*(MV)/(m+M)-MV=v(M-m)/(m+M) and M>m Vx’>0, so the object will not only go down, but also to the right.</p>
<p>For the question with the block moving up and down the incline without friction,</p>
<p>the magnitude of the net force is constant(straight line),F=mgsintheta
the velocity is a V (because the acceleration is constant if the net force is constant F=ma), and the slope of the velocity graph gives the acceleration. If the velocity graph was a parabola, the acceleration would change because the slope is always changing</p>
<p>total mechanical energy is a straight line too</p>
<p>potential energy is a upside down U, because time does not have a direct relationship with height because it is ACCELERATING, </p>
<p>then for the pendulum on an elevator, T=2pi rad(l/g), when the elevator accelerates up, g increases, so T must decrease</p>
<p>the object goes down and to the right after the collision, and I also got 3/2 ohms for resistance</p>
<p>for the relativistic problem, 0.7c<v<c was the answer. V cannot be greater than C because no object can move faster than the speed of light</p>
<p>@abatis no object can have speed higher than c. That is what relativity is (partially) about. I got confused between v’=(v-u)/(1-uv/c^2) and v=(v’+u)/(1+uv’/c^2) I said the first, but now I think the 2nd is right.</p>
<p>@esplin not capacitance, but charge on the capacitor
the acceleration doesn’t have to point in the same direction with the speed, but actually it points in the direction speed is changing. If you imagine a frictional force on an object with initial speed to the right, its speed is to the right and the acceleration to the left.</p>
<p>Then for the first question, which of these shows when velocity is perpendicular to acceleration, the answer was none of the above</p>
<p>perhaps I misread the relativity question since i was in a rush near the end, but if i remember correctly the problem said that a probe moving at 0.6c PASSED a spaceship moving at 0.7c? How does that work?</p>
<p>Well, both were moving in the same direction. the spacecraft had a velocity of ~0.6 c and then the person in the space craft observed the probe moving at around 0.6c again. So, wouldn’t that make the actual speed of the probe (0.6+0.6)c and thus the answer would become 1.2c.
Thus, velocity of the probe be>c.</p>
<p>no, an observer on earth observed the spaceship moving at 0.7 c, but an observer on the probe observed the spaceship moving at 0.6 c</p>
<p>so what speed would an observer on earth see the probe moving?</p>
<p>NOTHING is faster than speed of light…</p>
<p>so… i originally put v<c or something like that</p>
<p>but if the space ship saw the probe passing at .6c… then it could go either way, we just know that ONE of the 2 moving objects is .6 faster than the other, right?</p>
<p>and since spaceship was .7c, then the probe is .1c… so i put 0<v<.1c</p>
<p>not sure I was right, it made sense (and still does) hm.</p>
<p>when things are moving at relativistic speeds, you can’t just add the velocities. </p>
<p>you use: V(observer)=UV/(1+(UV)^2), where U is the velocity the other object moves at
and nothing can ever move faster than the speed of light</p>
<p>So, i guess i messed up that question as well.</p>
<p>it’s okkk physics has a nice curve</p>
<p>I said speed is .6c < v < .7c (nothing is faster than c) but why is it greater than .7c?? damnitttt i thought speeds close to c appear to slow down…</p>
<p>i missed so many…</p>
<p>what was the answer to the one with resistor and capacitor?</p>
<p>If I omitted 3 and missed ~10-15 what would be my approx. score?</p>
<p>you can get 15 wrong and get like a 770~</p>