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Determining the function using graphs

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Shown in picture Attachment
    Question #2.


    2. Relevant equations
    Not sure..That's what I'm supposed to figure out.


    3. The attempt at a solution
    Well I've graphed the points..That's about it hah.
    Seems like a Log graph flipped on the x-axis.
     

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  2. jcsd
  3. Sep 10, 2011 #2

    cepheid

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    What if you assume that the dependence of the weight on distance follows some sort of power law i.e. [itex] w \propto r^a [/itex]? Can you think of something you can do to both quantities before plotting them that would make it easy to measure 'a' just by looking at the plot?
     
    Last edited: Sep 10, 2011
  4. Sep 10, 2011 #3
    I'm still not sure, I haven't done physics in about a year, and my mind is pretty much fresh..
    The answers are ..
    W = Kd^-2

    K = 6.4x10^3 N(MR)^3

    Please help me..I really don't understand anything.. :(
     
  5. Sep 11, 2011 #4

    cepheid

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    What if you plotted both weight and distance on log scales (i.e. take log(w) and log(r)). What happens if you do that (just do it now to both sides of the equation). Can you see how this would make it possible to figure out that a = -2 just from the graph?
     
  6. Sep 11, 2011 #5
    Ohh, yes hahh, I can see a = -2.
    What about K?
     
  7. Sep 11, 2011 #6

    cepheid

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    Well, if w = Kra, where K is some constant of proportionality, then it follows that:

    log(w) = log(Kra) = alog(Kr) by the properties of logs.

    But the log of a product is the sum of the logs of the individual factors so that:

    alog(Kr) = alog(K) + alog(r)

    So, if you plot log(w) vs. log(r), I think you can see that it will be linear and the slope of the line gives you a, while the offset (w-intercept) of the line gives you K.
     
  8. Sep 11, 2011 #7
    Ugh what do you mean it will be linear when I plot log(w) vs. log(r)?

    You mean line of best fit? :S
     
  9. Sep 11, 2011 #8

    cepheid

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    I just showed you that log(w) varies linearly with log(r)! From my previous post:

    log(w) = alog(r) + alog(K).

    This is in the form of the standard equation for a straight line y = mx + b, with:
    y = log(w),
    m = a,
    x = log(r), and
    b = alog(K).

    So, if you plot log(w) vs. log(r), you should get a straight line and you can measure its slope (m) and y-intercept (b). I don't know how to make it any clearer than that.
     
  10. Sep 11, 2011 #9
    oh sorry sorry.
    I got b = 3
    and m = 4
    (assuming these are wrong.. ugh)

    What do I do with these?
    Sorry I'm just really clueless at the moment..
     
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