Answer:
3.73 m/s^2
Explanation:
The period of a simple pendulum is given by
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex] (1)
where
L is the length of the pendulum
g is the gravitational acceleration on the planet
The pendulum in the problem has length
L = 0.640 m
and makes 10 oscillations in 26.0 s; it means that its frequency is
[tex]f=\frac{10}{26 s}=0.385 Hz[/tex]
And so its period is
[tex]T=\frac{1}{f}=\frac{1}{0.385 Hz}=2.6 s[/tex]
So now we can solve equation (1) using L=0.640 m and T=2.6 s, so we can find the value of g on the planet:
[tex]g=(\frac{2\pi}{T})^2L=(\frac{2\pi}{2.6 s})^2 (0.640 m)=3.73 m/s^2[/tex]
