Question

I need this now

On the planet Xenos, an astronaut observes that a 2.0 m long pendulum has a period of 2.2 s.
What is the free-fall acceleration on Xenos? *

Answer 1

`g=16.31\ m/s^2`

Step-by-step explanation:

Given that,

The length of the pendulum, l = 2 m

The period of the pendulum, T = 2.2 s

The formula for the time period of a pendulum is given by :

`T=2\pi \sqrt{(l)/(g)}`

or

`T^2=4\pi ^2(l)/(g)\\\\g=(4\pi ^2l)/(T^2)\\\\g=(4\pi ^2* 2)/((2.2)^2)\\\\g=16.31\ m/s^2`

So, the free fall acceleration is
`16.31\ m/s^2`.