- #1
jakerue
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This is an issue I am running into at the beginning of my physics course.
Given distance and time in minutes, calculate the time in hours (part of a larger average velocity question) and graph over 170minutes. Include error bars in the graph
So if tm = 10.0 +/- 0.1min and I use min -> hr conversion as 1hr/60min = 0.0167
th = 10.0min * 0.0167 hrs/min = 0.167 min
The error in min is +/- 0.1 min. Now if I am using 0.0167hrs/min as an exact constant factor I should use z=k*x and [itex]\Delta[/itex]z= k [itex]\Delta[/itex]x
So I will use hours = 0.0167 * minutes and my [itex]\Delta[/itex]hours = 0.0167 * 0.1min making the [itex]\Delta[/itex]hours = 0.00167hours.
By sig fig this value is 0.002 hours correct?
At 10 minutes th = 0.167 +/- 0.002hrs
at 120 minutes th = 2.00 hrs +/- 0.002 hrs
I think I am right but I want to make sure my answer is correct and that this error holds true for all values of minutes from 0-170min.
Thanks for any help, most appreciated.
JakeRue
Homework Statement
Given distance and time in minutes, calculate the time in hours (part of a larger average velocity question) and graph over 170minutes. Include error bars in the graph
Homework Equations
So if tm = 10.0 +/- 0.1min and I use min -> hr conversion as 1hr/60min = 0.0167
th = 10.0min * 0.0167 hrs/min = 0.167 min
The Attempt at a Solution
The error in min is +/- 0.1 min. Now if I am using 0.0167hrs/min as an exact constant factor I should use z=k*x and [itex]\Delta[/itex]z= k [itex]\Delta[/itex]x
So I will use hours = 0.0167 * minutes and my [itex]\Delta[/itex]hours = 0.0167 * 0.1min making the [itex]\Delta[/itex]hours = 0.00167hours.
By sig fig this value is 0.002 hours correct?
At 10 minutes th = 0.167 +/- 0.002hrs
at 120 minutes th = 2.00 hrs +/- 0.002 hrs
I think I am right but I want to make sure my answer is correct and that this error holds true for all values of minutes from 0-170min.
Thanks for any help, most appreciated.
JakeRue