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Help with graphs

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to know the units of the slope. What the slope should be. And the percent error.
    Pictures:
    http://imageups.com/files/101/456.PNG

    Graph 1 is function of period(1/s^2) v weight of washers(N)
    r=.77m
    mass of rubber stopper= .02kg

    2. Relevant equations

    f(t)=1/t^2
    w=mg
    1/4pie^2(mass of rubber stopper)r

    3. The attempt at a solution

    would the slope be m/s*kg? or (1/s^2)/N?

    I did 1/4pie^2(.02)(.77)= 1.644
    (1.644-1.892/1.644)x100=15.02%

    I think that is right
     
    Last edited: Oct 21, 2009
  2. jcsd
  3. Oct 21, 2009 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Are you sure the abscissa has units of 1/s^2? You said period, which implies units of 1/time, not 1/time2.


    Dimensional analysis to the rescue:
    You have a line, y=mx+b. However, these are not just numbers. They are things with dimensions. So, apply the units operator to the equation of a line:

    units(y) = units(m)*units(x) + units(b)

    Each term on the right must agree with the left-hand side:

    units(b) = units(y)
    units(m)*units(x) = units(y)

    The first simply says that the intercept has the same units as does y. The second says that the slope has units equal to units(y)/units(x).

    Example: Suppose the graph is of position in meters (abscissa) versus time in seconds (ordinate). Thus in this example, the y-intercept has units of meters and the slope has units of meters/second.
     
  4. Oct 21, 2009 #3
    Yeah its suppose to be 1/t^2. I got that from Ac=(4pie^2r)/(t^2)
    (4pie^2r) is constant in the lab so it would be one. I had to do this since the graph that was just period v weight wasn't a straight line. And we had to find a way to get a straight line.

    1/t^2=(1/t^2)*Nx+1/t^2

    so it would be (1/t^2)*N?
     
    Last edited: Oct 21, 2009
  5. Oct 21, 2009 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Correct.
     
  6. Oct 21, 2009 #5
    Thanks :)
     
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