The free-fall acceleration on Mars is 3.7 m/s2.
(b)What length of pendulum would have a 0.9 s period on Mars?
T^2 = 4*(pi)^2(L/g) = 90.9)^2
4*(3.14)^2(L/g) = 0.81
39.48(L/g) = 0.81
L/g = 0.02052
L = 0.02052*3.7 = 0.076 m.
The freefall acceleration is 9.8m/s square.what period has pendulum of 2.5m length?
To find the length of the pendulum that would have a 0.9 s period on Mars, we can use the formula for the period of a pendulum:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, the period given is 0.9 s and the acceleration due to gravity on Mars is 3.7 m/s^2. We need to solve for L.
Rearranging the formula, we have:
L = (T/2π)^2 * g
Substituting the values we have:
L = (0.9/2π)^2 * 3.7
Now, we can calculate the length:
L ≈ (0.9/2π)^2 * 3.7 = 0.0407 m (rounded to four decimal places)
Therefore, the length of the pendulum that would have a 0.9 s period on Mars is approximately 0.0407 meters.