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In an experiment to measure the acceleration of g due to gravity, two values, 9.96m/s (s is squared) and 9.72m/s (s is squared), 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=9.80m/s (s is squared))

Respuesta :

Answer:

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Explanation:

Given that

g₁ = 9.96 m/s²

g₂ = 9.72 m/s²

The actual value of  g = 9.8 m/s²

a)

The difference Δ g =  9.96 -9.72 =0.24  m/s²

[tex]The\ percentage\ difference=\dfrac{0.24}{9.72}\times 100=2.46\ percentage\\[/tex]

b)

For first one :

[tex]Error\ in\ the\ percentage =\dfrac{9.96}{9.81}\times 100 =101.52\ perncetage[/tex]

For second  :

[tex]Error\ in\ the\ percentage =\dfrac{9.72}{9.81}\times 100 =99.08\ perncetage[/tex]

c)

The mean g(mean )

[tex]g(mean )=\dfrac{9.96+9.72}{2}\ m/s^2\\g(mean)=9.84\ m/s^2[/tex]

[tex]The\ percentage=\dfrac{9.84}{9.8}\times 100=100.40\ percentage[/tex]

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

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