A pendulum clock was moved from a location where g = 9.8135 m/s 2 to another location where g = 9.7943 m/s 2. During the move, the length of the clock's pendulum did not change; nevertheless, the clock lost accuracy. Assuming the clock was perfectly accurate at its previous location, how many seconds a day does it lose at the new location?

Question

Monica Jennings

Physics

4 July, 14:28

A pendulum clock was moved from a location where g = 9.8135 m/s 2 to another location where g = 9.7943 m/s 2. During the move, the length of the clock's pendulum did not change; nevertheless, the clock lost accuracy. Assuming the clock was perfectly accurate at its previous location, how many seconds a day does it lose at the new location?

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Answers (1)

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Sadie Wagner · Answer 1

The pendulum clock does it loses 83.0304 seconds a day at the new location.

Explanation:

T1 = 1 s

T2=?

g1 = 9.8135 m/s

g2 = 9.7943 m/s

L=?

L = (T1/2π) ² * g1=

L = 0.24857m

T2 = 2π * √ (L/g2)

T2 = 1.000961 s

ΔT = T2 - T1

ΔT = 0.000961 s

seconds lost a day = 24 * 3600 * ΔT

seconds lost a day = 24 * 3600 * 0.000961 s

seconds lost a day = 83.0304 s