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finlep [7]
4 years ago
9

Astronauts on a distant planet set up a simple pendulum of length 1.20 m. The pendulum executes simple harmonic motion and makes

100 complete oscillations in 450 s. What is the magnitude of the acceleration due to gravity on this planet?
Physics
1 answer:
Ray Of Light [21]4 years ago
4 0

Answer:

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

Explanation:

Time period of simple pendulum is given by

         T=2\pi\sqrt{\frac{l}{g}}, l is the length of pendulum, g is acceleration due to gravity value.

We can solve acceleration due to gravity as

            g=\frac{4\pi^2l}{T^2}

Here

  Length of pendulum = 1.20 m

  Pendulum executes simple harmonic motion and makes 100 complete oscillations in 450 s.

  Period, T=\frac{450}{100}=4.5s

Substituting

         g=\frac{4\pi^2\times 1.2}{4.5^2}=2.34m/s^2

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

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The velocity of a particle moving along the x axis is given by vx = a t − b t3 for t > 0 , where a = 27 m/s 2 , b = 3.1 m/s 4
Elanso [62]

Answer:

-54 m/s²

Explanation:

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0 = at-bt³

t(a-bt²) = 0

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a = bt²

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Acceleration = a - 3bt²... (2)

Substituting t² = a/b into equation 2 will give;

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4 0
4 years ago
PLZ HELP WILL GIVE BRAINLIEST
xxTIMURxx [149]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of the diver = 77kg

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Unknown:

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Solution:

To solve this problem, we use one of the motion equations.

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u is the initial velocity

g is the acceleration due to gravity

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             v² = 160.3

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