Question

what is the period (in s) of a simple pendulum of length 0.14 m and mass 0.49 kg near the surface of a planet where the gravitational acceleration is 11.9 m/s2?

Answer 1

The period of a simple pendulum with a length of 0.14 m and mass of 0.49 kg near the surface of a planet with gravitational acceleration of 11.9 m/s² is approximately 0.6826 seconds.

Step-by-step explanation:

The period of a simple pendulum can be calculated using the formula:

T = 2π√(L/g)

Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

Using the given values, we can calculate the period:

T = 2π√(0.14 / 11.9)

T ≈ 2π√(0.0118)

T ≈ 2π * 0.1086

T ≈ 0.6826 s

Therefore, the period of the simple pendulum is approximately 0.6826 seconds.