In an experiment to measure the acceleration g due to gravity,two values, 9.96 m

Question

In an experiment to measure the acceleration g due to gravity,two values, 9.96 m/s2 and 9.72 m/s2 aredetermined. [Note that the accepted value of g is 9.80m/s2 ] A. Find the percent error of the first measurement B. Find the percent error of the second measurement C. Find the percent difference of their mean In an experiment to measure the acceleration g due to gravity,two values, 9.96 m/s2 and 9.72 m/s2 aredetermined. [Note that the accepted value of g is 9.80m/s2 ] A. Find the percent error of the first measurement B. Find the percent error of the second measurement C. Find the percent difference of their mean

Explanation / Answer

(A). the percent error of the first measurement = [( 9.96 -9.8 ) / 9.8 ] * 100                                                                      = 1.632 % (B). the percent error = [( 9.8-9.72 ) / 9.8 ] *100                                  = 0.816 % (C). mean of 9.96 , 9.72 is = (9.96+9.72 ) /2                                           = 9.84 So, percent error = ( 9.84 - 9.8 ) / 9.8 *100                            = 0.408 %                                  = 0.816 % (C). mean of 9.96 , 9.72 is = (9.96+9.72 ) /2                                           = 9.84 So, percent error = ( 9.84 - 9.8 ) / 9.8 *100                            = 0.408 %

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