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Determine the period of a 1.7-m-long pendulum on the Moon, where the free-fall acceleration is 1.624 m/s2. Express your answer with the appropriate units.

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asuwafoh

Answer:

6.42 s

Explanation:

Period: This can be defined as the time taken for a wave or an oscillating body to complete on oscillation. The S.I unit of period is Second (s).

The period of a simple pendulum is represented as,

T = 2π√(L/g)....................................... Equation 1.

Where T = period, L = length of the pendulum, g = acceleration due to gravity. π = pie.

Given: L = 1.7 m, g = 1.624 m/s² and π = 3.14

T = 2(3.14)√(1.7/1.624)

T = 6.28√(1.047)

T = 6.28×1.023

T = 6.42 s.

Thus, the period of the pendulum = 6.42 s

abidemiokin

The period of a 1.7-m-long pendulum on the Moon, where the free-fall acceleration is 1.624 m/s² is 6.425 secs

The formula for calculating the period of a simple pendulum is expressed using the formula:

[tex]T=2 \pi\sqrt{\frac{l}{g} }[/tex]

l is the length of the pendulum

g is the acceleration due to gravity

Given the following parameters

g =  1.624 m/s²

l = 1.7m

Substitute the given values into the formula:

[tex]T=2(3.14)\sqrt{\frac{1.7}{1.624} } \\T=6.28\sqrt{1.0468}\\T=6.28(1.0231)\\T = 6.425secs[/tex]

Hence the period of a 1.7-m-long pendulum on the Moon, where the free-fall acceleration is 1.624 m/s² is 6.425 secs.

Learn more here: https://brainly.com/question/2936338

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