After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 . The expl?

We can use the fact that:

P = 2π√(L/g).

Solving this for g:

g = (4π^2*L)/P^2.

I am assuming that the pendulum as a length of 49.0 m and that the pendulum completes 104 full swings in 133 s.

Since the pendulum completes 104 full swings in 133 s:

P = 133/104 s.

Then, given that L = 49.0 m:

g = (4π^2*L)/P^2

= [4π^2*(49.0 m)]/(133/104 s)^2

= 1.18 x 10^3 m/s^2.

This value of g seems rather large. Did you happen to have 49.0 cm = 0.49 m? If so, then:

g = (4π^2*L)/P^2

= [4π^2*(0.49 m)]/(133/104 s)^2

= 11.8 m/s^2.

I hope this helps!

period of one oscillation = 133/104 = 1.28 sec

Recall that the formula for the period for simple pendulum with small angle oscillation = 2(pi)(sqrt(L/g))

Assuming the length you gave is in cm, L = 0.49

Hence, rearranging, g = (4 * pi * L)/period.

= (4 * 3.14 * 0.49)/1.28

= 4.8 ms^-2