Question

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) the 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 .)

Answer 1

(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 %

Step-by-step explanation:

(1)

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

Percent Difference = 2.47%

(2)

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%

Percent Error (9.96 m/s²) = 1.63 %

Now,

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

Percent Error (9.72 m/s²) = 0.82 %

(3)

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%

Percent Error (Mean) = 0.41 %