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3 years ago

9

A space station of diameter 20.0 meters is turning about its axis to simulate gravity at its center rim. How fast must it rotate

to produce an outer rim acceleration of 9.80 m/s^2 ?

1 answer:

bezimeni [28]3 years ago

4 0

Answer:

9.89 m/s

Explanation:

d = diameter of the space station = 20.0 m

r = radius of the space station

radius of the space station is given as

r = (0.5) d

r = (0.5) (20.0)

r = 10 m

a = acceleration produced at outer rim = 9.80 m/s²

v = speed at which it rotates

acceleration is given as

$a = \frac{v^{2}}{r}$

$9.80 = \frac{v^{2}}{10}$

v = 9.89 m/s

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Answer:

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