Answered

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A pendulum is placed on a distant planet. The length is one meter, and the measured period is 1.4 seconds, what is the acceleration of gravity on that planet?

Sagot :

Answer:

[tex]a=20.14\ m/s^2[/tex]

Explanation:

The time period of the simple pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is the length of the pendulum

g is the acceleration due to gravity

We have,

T = 1.4 s, l = 1 m

So,

[tex]T^2=\dfrac{4\pi^2 l}{g}\\\\g= \dfrac{4\pi^2 l}{T^2}\\\\g= \dfrac{4\pi^2 \times 1}{(1.4)^2}\\\\g=20.14\ m/s^2[/tex]

So, the acceleration due to gravity of that planet is [tex]20.14\ m/s^2[/tex].

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