Since you put it that way. Δv=v. However, the problem states that the speed doubled plus its going in the opposite direction, that's where I got the -2.
3. I'm going to try. Going back to pre-calculus-solving exponential equations w/ same base. Let original velocity = u.
v=\frac{-2u}{.45}
a=\frac{-3u}{.45}
Then \frac{-2}{-3} = .66666667 or 6.7s^{-1} or 6.7*10^{-1}...
I have to admit this one was tricky because normally (well the way I learned in chemistry) the given units come first then the desired conversion factor comes next. In this case the desired conversion factor came first (or as you say take the reciprocal) then the given. Well We knocked #2 out...
Oh SHISHKABOB Your the best! I found my mistake I shouldn't have been separating 6.7 * 10^{7}. thanks again SHISKABOB I was making it too complicated! I'm trying not to I always make the easy stuff complicated, but the complicated stuff comes easy... why me :rolleyes:
2. Sorry, if you haven't figured it out already I'm a Visual Learner. But Thank that you God, a Light bulb just went off in my head. \frac{1 cal}{10^{-9} sec} \frac{1 sec}{6.7 *10^{7}} When you add the exponents you get \frac{1}{6.7*^10{-2}sec }. YAY!
2. Here's how I wrote it down on paper: 1 cal , 1 ns / 10^-9 sec, 10 ^7 sec/ 6.7 cal = .15 cal / 10^-2. Everything cancels out expect for nanosecond. Then after subtracting the exponents I get 10^-2. Instead of dividing by .15 cal / 10 ^-2 I flipped the numerator and denominator. 10^-2/ .15...