almighty3374 almighty3374
  • 17-01-2020
  • Physics
contestada

What is the free-fall acceleration on a planet where the period of a 1.07 m long pendulum is 2.02 s?

Respuesta :

Vespertilio Vespertilio
  • 18-01-2020

Answer:

10.34 m/s^2

Explanation:

length, L = 1.07 m

Time period, T = 2.02 s

The formula of the time period of the pendulum is

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

where, g is the free fall acceleration on that planet.

[tex]g=4\pi ^{2}\frac{L}{T^{2}}[/tex]

[tex]g=4\pi ^{2}\frac{1.07}{2.02^{2}}[/tex]

g = 10.34 m/s^2

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