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Interperating a graph with logarithmic scales

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    see attachment


    2. Relevant equations
    log10x=y
    10y=x


    3. The attempt at a solution
    Since I'm rather confused on reading a logarithmic scale, I thought I'd post this to see if I'm doing this right and/or if there's a better way to do the problem.

    a) T=500 C
    using the # of intervals we can say that if d=.01, x=.5 intervals so...
    log(t)=.5
    t=3.16 min

    similarly for d=.1
    log(t)=3.5 intervals
    t=3162

    change in time = 3162-3.16 = 3158.8min

    b) same process but use x=-.5 intervals and x=2.1

    If this method does work, is there any way to interpolate values so I can be more exact? or is there a better way to do the problem in general?
    thanks!
     

    Attached Files:

  2. jcsd
  3. Mar 11, 2010 #2

    Mark44

    Staff: Mentor

    I think you're on the right track, but maybe just a little off. For the 500 degree track (part a), the starting time is about 3.2 min, but the ending time looks to me like about 3.6, which corresponds to about 4000 min. Subtracting gives a duration that's again close to 4000 min. Unless you can really measure the position of the points very precisely, it doesn't make sense to end up with answers with more than one or two places of precision.
     
  4. Mar 11, 2010 #3
    yah, I was worried about being able to get a precise enough answer to matter. I think later I'll find a ruler and figure out more exact data points
    thanks!
     
  5. Mar 11, 2010 #4

    Mark44

    Staff: Mentor

    Yes, a ruler would help you get more precise answers.
     
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