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HACTEHA [7]
3 years ago
12

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 Xenophou

s?
Physics
1 answer:
myrzilka [38]3 years ago
8 0
The period of the pendulum is given by the following equation

T = 2<span>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</span>π)<span>^</span><span>2 = L/g

g = L/(</span>T /2π)^2...........> g = 22.657 m/s^2
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