(b) On Mars, where gMars = 3.7 m/s2, the length of a pendulum with a 1.9-s period isLMars = gMarsT24𝜋2 = m/s2 s 2 4𝜋2 = m = cm.

Question

(b) On Mars, where gMars = 3.7 m/s2, the length of a pendulum with a 1.9-s period is

LMars = $\frac{g_{Mars} T^2}{4 \pi^2}$
= m/s² s² 4π²
= m = cm.

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Solution

The formula for the length of a pendulum based on its period is given by:

L = g * T^2 / 4π^2

where:

L is the length of the pendulum,
g is the acceleration due to gravity,
T is the period of the pendulum,
π is a mathematical constant whose approximate value is 3.14159.

Given that on Mars gMars = 3.7 m/s^2 and the period T = 1.9 s, we can substitute these values into the formula to find the length of the pendulum on Mars:

LMars = gMars * T^2 / 4π^2 = 3.7 m/s^2 * (1.9 s)^2 / 4π^2 = 3.7 m/s^2 * 3.61 s^2 / 4π^2 = 13.37 m^2/s^2 / 4π^2 = 13.37 m^2/s^2 / 39.4784 = 0.3386 m

So, the length of the pendulum on Mars would be approximately 0.3386 meters or 33.86 centimeters.

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(b) On Mars, where gMars = 3.7 m/s2, the length of a pendulum with a 1.9-s period isLMars = gMarsT24𝜋2 = m/s2 s 2 4𝜋2 = m = cm.

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