25 August, 02:02
On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophous?
25 August, 03:47
The period of the pendulum is given by the following equation
T = 2 π * sqrt (L/g)
Where g is the gravity (free fall acceleration)
L is the longitude of the pendulum
T is the period.
We find g ... > (T / 2 π) ^ 2 = L/g
g = L / (T / 2π) ^2 ... > g = 22.657 m/s^2
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» On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophous?