What is the period, in seconds, of a simple pendulum of length 2 meters? Use the gravitational constant g = 9.8 m/s^2 and round your answer to two decimal places.

Answer: The period, in seconds to two decimal places is, 2.84 secStep-by-step explanation:Using formula:[tex]T = 2 \pi \sqrt{\frac{l}{g}}[/tex]             ......[1]where T represents the Time period in secondl represents the length of the pendulum/As per the statement:Use the gravitational constant g = 9.8 m/s^2 , length of the simple pendulum(l) = 2 meters and [tex]\pi = 3.14[/tex]Substitute the given values we have;[tex]T = 2\cdot 3.14 \cdot \sqrt{\frac{2}{9.8}}[/tex]⇒[tex]T = 6.28 \cdot \sqrt{\frac{1}{4.9}}=6.28 \cdot \frac{1}{2.21359436212}= \frac{6.28}{2.21359436212} = 2.83701481512[/tex]Therefore, the period, in seconds to two decimal places is, 2.84 sec