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HACTEHA [7]
2 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]2 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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Answer:

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Explanation:

First we should define the variables

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By plugging in our variables we can get 100=4(v+v0)/2

Which is 50=v+v0

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5 0
3 years ago
Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.
lapo4ka [179]

Answer:

This is how I do it:

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Hope this helps you.

Explanation:

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