Question

What is the period, in seconds, of a simple pendulum of length 5 meters? Use

the gravitational constant g = 9.8 m/s2 and round your answer to two decimal
places.

Answer 1

`T=2\pi \sqrt{((l)/(g) )} \\T=2\pi \sqrt{(5m)/((9.8m/s^(2)) ) } \\`

`T` ≈
`4,49s`