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trapecia [35]
3 years ago
14

In an experiment to measure the acceleration due to gravity, g two values, 9.96 m/s2 and 9.72 m/s2 , are determined. Find (1) th

e percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value: g 5 9.80 m/s2 .)
Physics
1 answer:
alexira [117]3 years ago
8 0

Answer:

(1) Percent Difference = 2.47%

(2) Percent Error (9.96 m/s²) = 1.63 %

    Percent Error (9.72 m/s²) = 0.82 %

(3) Percent Error (Mean) = 0.41 %

Explanation:

(1)

Percent Difference = [(9.96 m/s² - 9.72 m/s²)/(9.72 m/s²)]*100 %

<u>Percent Difference = 2.47%</u>

<u></u>

<u>(2)</u>

Percent Error = (|Measured Value - Original Value|/Original Value)*100%

Therefore,

Percent Error (9.96 m/s²) = (|9.96 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

<u>Percent Error (9.96 m/s²) = 1.63 %</u>

Now,

Percent Error (9.72 m/s²) = (|9.72 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

<u>Percent Error (9.72 m/s²) = 0.82 %</u>

<u></u>

<u>(</u>3<u>)</u>

First we need to find the mean of values:

Mean = (9.96 m/s² + 9.72 m/s²)/2

Mean = 9.84 m/s²

Therefore,

Percent Error (Mean) = (|9.84 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

<u>Percent Error (Mean) = 0.41 %</u>

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