1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Significant Figures

  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data
    I am starting to confuse myself with the proper use of SF. I am to convert time from minutes to hours, keeping in mind proper SF

    2. Relevant equations
    conversion factor: 1 min = 1/60 hr

    3. The attempt at a solution

    The timing error is +/- 0.2 (1SF) = 0.003hr (1SF)

    20.0 min (3 SF) = 0.333 hr (3 SF), but the conversion factor has 1 SF??? would my answer be 0.3 hr? I'm confused because now I have varying decimal places and when I say 0.3 +/- 0.003 hr doesn't seem corect at all because when adding or subtracting I would drop the 0.003 therefore never being relevant???

    Please help.
     
  2. jcsd
  3. Jan 30, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    Conversion factors that are defined rather than measured are taken to have infinite precision. Thus the constant "60" in the conversion of minutes to hours or vice versa is exact.
     
  4. Jan 30, 2016 #3
    Ok, so if I am converting 20.0 min using the "infinite" 60, the answer would reduce to the 3SF from the minute value?
     
  5. Jan 30, 2016 #4
    As a general rule, you need to quote your answer to the same number of decimal places as your uncertainty.

    If you have an uncertainty of 0.55 on 10, I would expect to see ##x = (\mathrm{10.00} \pm \mathrm{0.55}) \mathrm{units}##
     
  6. Jan 30, 2016 #5
    Right. Just so I am understanding that correctly.

    In minutes my uncertainty is +/- 0.2, therefore on 20.0 minutes, it would remain as 20.0 +/- 0.2 min (both have the same amount of decimal places)
    In hours the uncertainty would be +/- 0.003, where 20.0 min = 0.333 hr, would be 0.333 +/- 0.003 hr.

    Now what if distance comes to play and it was only one decimal place? Am I still ok to leave the above as 3 decimal places if i intend to plot d v t on a graph?
     
  7. Jan 30, 2016 #6

    gneill

    User Avatar

    Staff: Mentor

    As long as the different data is kept separate they keep their own significant figures and accuracy. So in a plot where the data are on separate axes they have their separate uncertainties and significant figures. If you use the data to calculate the slope of the resulting curve or the area under the curve, then you are "mixing" the data and you need to heed the significant figure rules for any results.
     
  8. Jan 30, 2016 #7
    Thank you :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Significant Figures
  1. Significant Figures (Replies: 5)

  2. Significant figures (Replies: 1)

  3. Significant figures (Replies: 13)

  4. Significant figures. (Replies: 5)

  5. Significant figures (Replies: 6)

Loading...