Question

Make "l" the subject of T = 2pi to the sqrt of l/g

(yes, that is 2 pi to the square root of "L" OVER "G") Confused? Hit me up.

Answer 1

period of a pendulum.

Answer 2

\[T= 2 \pi \sqrt{ \frac{l}{g}}\]

Answer 3

How are you making those nice fractions?

Answer 4

\[\sqrt { \frac{l}{g} } = \frac{T}{2\pi}\]

Answer 5

\[\frac{l}{g} = \frac{ T^2}{ 4 \pi^2} \]

Answer 6

\[l = \frac{ g T^2}{4 \pi^2} \]

Answer 7

\[\color{red}{\text{the pit and the pendulum}}\]