What is the free-fall acceleration on a planet where the period of a 1.07 m long pendulum is 2.02 s?

10.34 m/s^2

Explanation:

length, L = 1.07 m

Time period, T = 2.02 s

The formula of the time period of the pendulum is

where, g is the free fall acceleration on that planet.

g = 10.34 m/s^2

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y