Calculate Percent Error: Measuring 75cm with Meter Stick

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Homework Help Overview

The discussion revolves around estimating the percent error in measuring a distance of approximately 75 cm using a meter stick. Participants are exploring the implications of measurement precision and the calculation of percent error based on different assumptions about uncertainty.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Some participants discuss the accuracy of the measurement, suggesting that the uncertainty could be around ±0.2 cm or ±1 mm. Others question how the book arrived at a specific percent error of 0.13% and explore different methods of calculating percent error based on varying assumptions of uncertainty.

Discussion Status

The discussion is active, with participants providing different perspectives on how to approach the problem. Some guidance has been offered regarding the calculation of percent error, but there is no explicit consensus on the best method or the correct interpretation of the problem.

Contextual Notes

Participants note that the question refers to a distance of "about" 75 cm, suggesting that a rough estimate of the error may be sufficient. There is also mention of differing conventions in expressing fractional and percentage errors, which may contribute to confusion.

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Homework Statement


Estimate the percent error in measuring a distance of about 75cm with a meter stick.



Homework Equations





The Attempt at a Solution


I'm assuming one could accurately measure this value to a couple mm. so it would be 75+-0.2cm. Now the book gives an answer of 0.13% i just have no clue how they get there. Thanks in advance.
 
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If the meter stick has mm's as the smallest division, you would be rounding at most .5 mm on each end. Divide by 75 and multiply by 100 to convert to per cent.
 
sergey90 said:

Homework Statement


Estimate the percent error in measuring a distance of about 75cm with a meter stick.



Homework Equations





The Attempt at a Solution


I'm assuming one could accurately measure this value to a couple mm. so it would be 75+-0.2cm. Now the book gives an answer of 0.13% i just have no clue how they get there. Thanks in advance.

Your uncertainty of 0.2 cm means 0.2 in 75, which is 2 in 750 which is 1 in 375

1/375 * 100 = 0.26666 %

If you had thought you could get it to within 1 mm, the uncertainly would have been 0.13 %
 
I would say the error on any measuring instrument is +/- 1 scale division or +/1 1 in the last digit of a digital meter.
If the metre stick has mm divisions then I would say the error is +/- 1mm which is +/-1 in 750 (similar to PeterO with his +/-)
The question refers to a distance of ABOUT 75cm so it seems to me they want a rough estimate of the error rather than an answer to several significant figures.
 
thanks for the help guys. One last clarification on the matter: in previous problems of fractional errors if for example we have a value of 9.69+-0.07cm^2 the book just divides 0.07 by 9.69 and says that's my fractional error % without multiplying by 100. Why so?
 
sergey90 said:
thanks for the help guys. One last clarification on the matter: in previous problems of fractional errors if for example we have a value of 9.69+-0.07cm^2 the book just divides 0.07 by 9.69 and says that's my fractional error % without multiplying by 100. Why so?

Perhaps it was given as a fractional error, to convert a fraction to a percentage, you multiply by 100.

It is unusual to fractional error % - usually either fractional error or percentage error.
 

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