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In an experiment to measure acceleration g due to gravity, two values, 9.95m/s^2 and 9.82 m/s^2, are determined. Find: (a) the value of g for this experiment (b) percent error of their mean

Answer :

Answer:

Part a)

g = 9.885 m/s/s

Part b)

error = 0.66 %

Explanation:

Part a)

The experimental values of acceleration due to gravity is given as

[tex]g_1 = 9.95 m/s^2[/tex]

[tex]g_2 = 9.82 m/s^2[/tex]

now we know that the value of g is the mean value of all the readings

so we have

[tex]g = \frac{g_1 + g_2}{2}[/tex]

[tex]g = \frac{9.95 + 9.82}{2}[/tex]

[tex]g = 9.885 m/s^2[/tex]

Part b)

error in both readings of gravity is given as

[tex]\Delta g_1 = 9.95 - 9.885[/tex]

[tex]\Delta g_1 = 0.065[/tex]

[tex]\Delta g_2 = 9.885 - 9.82[/tex]

[tex]\Delta g_2 = 0.065[/tex]

now mean error in both the readings is given as

[tex]\Delta g_m = 0.065[/tex]

now percentage error is given as

[tex]error = \frac{\Delta g}{g}\times 100[/tex]

[tex]error = \frac{0.065}{9.885} \times 100[/tex]

error = 0.66 %

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