Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 52.0 cm. The explorer finds that the pendulum completes 91.0 full swing cycles in a time of 140 s. What is the value of the acceleration of gravity on this planet?

Answer 1

8.67340834769 m/s²

Step-by-step explanation:

Time period is given by

`T=(140)/(91)\\\Rightarrow T=1.53846153846`

L = Length of pendulum = 52 cm

g = Acceleration of gravity on the planet

Time period also is given by

`T=2\pi\sqrt{(L)/(g)}\\\Rightarrow g=4\pi^2(L)/(T^2)\\\Rightarrow g=4\pi^2(52* 10^(-2))/(1.53846153846^2)\\\Rightarrow g=8.67340834769\ m/s^2`

The value of the acceleration of gravity on this planet is 8.67340834769 m/s²