After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm . The explorer finds that the pendulum completes 108 full swing cycles in a time of 132 s . Part A What is the magnitude of the gravitational acceleration on this planet

The magnitude of the gravitational acceleration on this planet is 14.54 m/s².

Explanation:

given information:

pendulum length, L = 55 cm = 0.55 m

total swing, n = 108

time, t = 132 s

to calculate the gravitational acceleration of the planet, we can use the simple pendulum formula

T = 2π√(L/g)

where

T = period (s)

L = the length of pendulum (m)

g = gravitational force (m/s²)

thus,

T² = (2π)²(L/g)

g = (2π)²(L/T²)

we know that

T = t/n

so,

g = (2π)²(L/T²)

  = (2π)²[L/(t/n)²]

  = (2π)²[L (n/t)²]

  = (2π)² [0.55 (108/132)²]

  = 14.54 m/s²

answer; i believe that the correct answer is (d) a very small amount of mass

more than, less

less than, more