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nuttyg3705
21.06.2021 •
Physics
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 centripetal acceleration was equal to the gravitational acceleration at the equator, what would be the tangential speed of a person standing at the equator?
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Ответ:
v = 6295.55 m/s
Explanation:
Given that,
The radius of a planet,![r=5.32\times 10^6\ m](/tpl/images/1380/4433/638ee.png)
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,
So, the tangential speed of the person is equal to 6295.55 m/s.
Ответ: