Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cmcm . The explorer finds that the pendulum completes 100 full swing cycles in a time of 134 ss . Part A What is the magnitude of the gravitational acceleration on this planet

Answer 1

Answer:

≅12m/s²

Explanation:

T = $2\pi \sqrt{\frac{L}{g} }$

where T = time taken for no of oscillation/no of oscillation

T = 134/100

= 1.34

g = (2π)²L/T²

g = (2π)² 0.55/1.34²

 = 39.5 ×0.306

 = 12.099

 ≅ 12m/s²