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Phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I've been reviewing for Test 1. First off, I'm very nervous about it and I have a few questions and may have even more before the night's out. I'm hoping to take the test tomorrow (Friday) depending on how comfortable I feel by the end of tonight's review.
Below is part of a question from a test:
We know that for an air cart gliding along an incline, the force that accelerates it is equal to W sin(`theta), where W is the weight of the cart and `theta the angle of the incline with horizontal.
If the force on the glider was W, what would be its acceleration?
****My answer/question: Wouldn't it be Wsin(`theta)? is that one of those obvious answers?
acceleration = net force / mass
W sin(theta) is a force, not an acceleration
Since W = m g, though, the net force is
F_net = W sin(theta) = m g sin(theta).
What do you get if you divide this by m?
Additionally, there is a question to list the 4 equations of uniform motion, which I know. Here's the question:
Specify which two are considered the most basic, and explain their meaning in commonsense terms. Explain how the third fourth equations demonstrate the truth or falseness the argument 'if an automobile roles down a constant incline, since it experiences the same acceleration no matter how fast it is going when it starts down the incline, then as long as a resistance and other extraneous frictional forces are not significant, its velocity will increase by a lesser amount when the initial velocity is higher'.
****I'm assuming the most basic are equations 1 and 2, but I don't understand the last portion about the 3rd and 4th demonstrating the truth or falseness of the argument??
Presumably the length of the incline doesn't change, so that vf^2 = v0^2 + 2 a `ds and vf = sqrt(v0^2 + 2 a `ds). Note that for simplicity we have assumed that v0 and vf are both positive.
So the change in velocity is vf - v0 = sqrt( v0^2 + 2 a `ds ) - v0.
v0^2 + 2 a `ds = v0^2 ( 1 + 2 a `ds / v0^2 ), so that sqrt( v0^2 + 2 a `ds ) = v0 sqrt( 1 + 2 a `ds / v0^2 ).
It follows that the change in velocity is
vf - v0 = v0 sqrt ( 1 + 2 a `ds / v0^2 ) - v0 = v0 ( sqrt( 1 + 2 a `ds / v0^2 ) - 1 ).
Then the algebra starts to get complicated, and we have to introduce some confusing notation.
At this point it's OK to default to the much simpler argument that greater v0 implies greater vAve and hence less time on the incline. Less time for the acceleration to change the velocity implies less change in velocity.
I also have answer the question about the 80 megaJ of energy being released in the gallons of gasoline. I think I've approached that problem correct, but just to verify, I should first convert the megaJ to J?
You are correct.
80 megaJoules is 80 * 10^6 Joules = 8 * 10^7 Joules, or 80 000 000 Joules.
I'm sorry for all the questions, but I want to do well and I'm having a harder time grasping this section. You may get more questions before I'm done with this reviewing!
Thanks!
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Questions are always welcome.
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