Answer:
A). 2.3 s
Explanation:
Given that,
Length of the string = 130cm or 1.3 m
g near Earth's surface = 9.8 [tex]ms^{-2}[/tex]
To determine the period of a pendulum near Earth’s surface;
T = 2 π [tex]\sqrt{\frac{L}{g} }[/tex]
T = 2π[tex]\sqrt{\frac{1.3}{9.8} }[/tex]
T = 2.3 sec
Therefore, the period of the pendulum near the earth's surface is 2.3 sec.
