The period of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.

On Earth, we know that T = 2 seconds and g = 9.8 m/s². We can rearrange the formula to solve for L:

L = g(T/2π)² = 9.8*(2/2π)² = 0.993 m

Now, we can use this length to find the period of the pendulum on the Moon, where g = 1.6 m/s²:

T = 2π√(L/g) = 2π√(0.993/1.6) = 2.44 seconds

So, the correct answer is c. 2.44 s.

Question

The period of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.

On Earth, we know that T = 2 seconds and g = 9.8 m/s². We can rearrange the formula to solve for L:

L = g(T/2π)² = 9.8*(2/2π)² = 0.993 m

Now, we can use this length to find the period of the pendulum on the Moon, where g = 1.6 m/s²:

T = 2π√(L/g) = 2π√(0.993/1.6) = 2.44 seconds

So, the correct answer is c. 2.44 s.

Knowee AI · Accepted Answer

The period of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.

On Earth, we know that T = 2 seconds and g = 9.8 m/s². We can rearrange the formula to solve for L:

L = g(T/2π)² = 9.8*(2/2π)² = 0.993 m

Now, we can use this length to find the period of the pendulum on the Moon, where g = 1.6 m/s²:

T = 2π√(L/g) = 2π√(0.993/1.6) = 2.44 seconds

So, the correct answer is c. 2.44 s.

A simple pendulum has a period of 2 seconds on Earth (g=9.8m/s²). What would be its period on the Moon where g is approximately 1.6m/s²?Select one:a.12.25 sb.0.65 sc.2.44 sd.7.75 s