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morpeh [17]
3 years ago
7

If the rotation of a planet of radius 5.32 × 106 m and free-fall acceleration 7.45 m/s 2 increased to the point that the centrip

etal acceleration was equal to the gravitational acceleration at the equator, what would be the tangential speed of a person standing at the equator?
Physics
1 answer:
leonid [27]3 years ago
3 0

Answer:

v = 6295.55 m/s

Explanation:

Given that,

The radius of a planet, r=5.32\times 10^6\ m

The free fall acceleration of the planet, a = 7.45 m/s²

We need to find the tangential speed of a person standing at the equator.

Also, the centripetal acceleration was equal to the gravitational acceleration at the equator.

We know that,

Centri[etal acceleration,

a=\dfrac{v^2}{r}\\\\v=\sqrt{ar}\\\\v=\sqrt{5.32\times10^6\times 7.45}\\\\v=6295.55\ m/s

So, the tangential speed of the person is equal to 6295.55 m/s.

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