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Please check my solutions on the following scans.

  1. Dec 15, 2008 #1
    I'm sure that I have all the answers right except for number 5 and number 10.
    http://i68.photobucket.com/albums/i15/monkey13330gg/scan0001.jpg
    http://i68.photobucket.com/albums/i15/monkey13330gg/scan0002.jpg

    On number 5, I think the answer is either B or E.

    I think it is B because if the cart is already in motion when the timer starts, then the graph would start on the y-axis instead of all the way over to the right.

    I also think that the answer is E because the distance between the graph and the y-axis indicate the friction that the force have to overcome to be able to make the cart go.

    For number 10.

    Somehow my answer does not seem right. I have E as an answer because the only thing I could get out of that graph is acceleration.

    Please check on these two problems (number 5 and 10) and do any problem that you are interested in. Thank you very much.
     
  2. jcsd
  3. Dec 15, 2008 #2

    cepheid

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    I'm only gonna look at 5 and 10.

    For number 5, it means that there is zero acceleration for a non-zero force. I'll leave it up to you to determine what answer choice that leads to.

    Hint: In the formula F = ma, F is the NET force acting on an object.
     
  4. Dec 15, 2008 #3

    cepheid

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    10 looks good. Understand why:

    It's not A because distance covered would be the area underneath each curve (clearly not the same for I and II)

    It's not B because the inertial mass is the 'm' in F = ma, and although the two objects appear to have roughly the same acceleration, suggesting that there is a constant force acting on each one, that does not necessitate them having the same mass. It could be that they have different masses and are being acted upon by two *different* forces (eliminating D as well). These two different forces could differ by just the right amount in order to have the two objects of differing mass accelerate at the same rate.

    It's not C because gravity is not relevant to the problem.
     
  5. Dec 15, 2008 #4
    Thank you!
     
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