Given the formula:
[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]
Where:
• Period, T = 4.9 seconds
,
• Length, L = 1 meter
Let's find the acceleration due to gravity, g, on the moon.
To solve for g, let's first rewrite the formula for g.
Take the following steps:
• Square both sides:
[tex]\begin{gathered} T^2=(2\pi\sqrt{\frac{L}{g}})^2 \\ \\ T^2=4\pi^2*\frac{L}{g} \\ \\ T^2=\frac{4\pi^2L}{g} \end{gathered}[/tex]
• Cross multiply:
[tex]g=\frac{4\pi^2L}{T^2}[/tex]
To solve for g, substitute 4.9 for T and 1 for L:
[tex]\begin{gathered} g=\frac{4\pi^2*1}{4.9^2} \\ \\ g=\frac{4\pi^2}{24.01} \\ \\ g=1.64\approx1.6\text{ m/s}^2 \end{gathered}[/tex]
Therefore, the acceleration due to gravity on the moon is 1.6 m/s².
ANSWER:
C. 1.6 m/s²
