<p>
[quote]
i hope you are right.
[/quote]
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<p>What did you put for that? Anyway, you don't have to hope I'm right, the right answer is, it's greater than .6c, but less than c. As conker said, if you have learned special theory of relativity and understood the concept well, the answer is pretty obvious -though the mathematics part can be quite complicated-. One piece of advice when learning it: ignore what common sense.</p>
<p>thats what i put... greater than .6c but less than c, no way i could have calculated the number, but i knew that nothing can travel as fast as, or faster than light.</p>
<p>any other questions you people can think of?</p>
<p>I am still sticking with D on the blob question.</p>
<p>F = Kq1q2/r^2 where r is the distance between the centers of mass. D was the closest one to the center of mass.</p>
<p>yea i think so too</p>
<p>I could have sworn I have seen the "blob" question before...not sure if it was on an AP test or in a review book...maybe AP Physics B/C PR review book?</p>
<p>Anyone else seen something similar to it before?</p>
<p>And, I think it's D as well.</p>
<p>For the blob question, here's proof that the answer was C, which was the point at which the curvature of the blob was greatest.</p>
<p>Scroll down about one third of the way to the portion of the page where they discuss surface curvature.</p>
<p>well crap.. haha o well i can miss 9.. thats 1... hopefully no more than 8 others hahaha... i wish</p>
<p>My take on the relativity question:</p>
<p>The formula you guys are using is the ADDITION of velocities formula:</p>
<p>v = (x + y)/(1 + (xy/c^2))</p>
<p>This formula "corresponds to the theorem of addition for velocities in one direction according to the theory of relativity". The problem on the test did NOT deal with velocities in ONE direction, but instead two: east and west. The problems found within most of the review books are with a ship traveling at some fraction of c, and then additionally, in the same direction, they fire some particles at another fraction of c; therefore, in that case, you can't just add the two values and must use that specific equation. But, that was not the type of question on the test.</p>
<p>Personally, I think that the answer is just c.</p>
<p>youre right...dont listen to the other people</p>
<p>u sure hes right</p>
<p>well if anybody else would like to say what the answer is go ahead, but i just don't know which one's right, i just know that i put c...</p>
<p>i'll ask my physics professor for my UVA class.. he has a phd from yale he might know lol!</p>
<p>yea i'll ask my physics teacher too he's awesome hehe</p>
<p>can you rephrase the whole question though i just wanna make sure i got it right?</p>
<p>two spaceships going towards each other, each going .6c. what's the velocity observed by someone on one of the ships??</p>
<p>thanks</p>
<p>yep... my physics professor might know more than yours... lol :)</p>
<p>or vice-versa...lol
we'll see if they agree it would be funny if they don't</p>
<p>haha if they dont.. i think my professor is right... Richard Lindgren, PHD from Yale, Research Professor, University of Virginia... lol too bad he has NO common sense..</p>
<p>
[quote]
The problem on the test did NOT deal with velocities in ONE direction, but instead two: east and west. The problems found within most of the review books are with a ship traveling at some fraction of c, and then additionally, in the same direction, they fire some particles at another fraction of c; therefore, in that case, you can't just add the two values and must use that specific equation. But, that was not the type of question on the test.
[/quote]
</p>
<p>It doesn't matter if the ships are going to different directions, you will still have to use the same formula. Allright then, let me try to explain it one more time and see if it will be clear this time (unlikely though as my writing skill sucks). I'm actually not sure if they are going toward each other, or moving away, but either way, the answer will still be the same.</p>
<p>Suppose they're moving apart from each other. And suppose we have a static frame of reference A, from which the speed of .6c for both ships is measured. Let's say ship 1 moves to the left, and ship 2 to the right. Since we are to assume them to move in the constant speed, it is ok to pick either ship 1 or ship 2 as the new static frame of reference. </p>
<p>Let's say, we're trying to measure the speed of ship 2 with respect to ship 1. Then we pick ship 1 as the static frame of reference. For the pilot of ship, there are two objects moving to the right direction, and they're ship 2, and frame A (which is the original frame of reference). We know the velocity of ship 2 as measured from the frame A (.6 c), and we know the velocity of frame A according to ship 1(which is also .6c). These .6c are the values of x & y. </p>
<p>That said, all you have to do is plug in .6c as x, and y values into that equation, the end result, v, will be the speed of ship 2 with respect to ship 1. As I have calculated on my previous post, the answer wouldn't be c. It would still be greater than .6c though. I don't know if anybody would even bother reading my post (I don't know if it even makes sense at all to begin with).</p>
<p>yep, my high school physics teacher said the same thing .6 as x and y..... im waiting on my uva professor now.</p>
<p>Maybe I just read the question wrong, the reference was from inside one of the ships and was looking at the other one...correct?</p>