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The period of rotation of Mercury in hours is 1407.6 hr. Given that Mereury has a radius of 2.4 X 10^6 m at its equator, what is the linear velocity (in m/s) at Venus's surface? (Enter the magnitude of the linear velocity at the equator.)

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Hania12

Answer:

v=2.9578 m/s

Explanation:

t= 1407.6 hr

t = 3600 * 1407.6 = 5067360 s

w = 2π/T

    =2π/5067360

w = 1.24 x 10^-6 rad/s

v = r*w

v = 2.9758 m/s

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