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If the space station is 190 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s² at the rim?

User Smcjones
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Final answer:

The angular velocity that would produce an artificial gravity of 9.80 m/s² at the rim of the space station is approximately 0.465 rad/s.

Step-by-step explanation:

To find the angular velocity that would produce an artificial gravity of 9.80 m/s² at the rim of the space station, we can use the formula for centripetal acceleration:

ac = w²r

Where:

  • ac is the centripetal acceleration
  • w is the angular velocity
  • r is the radius

We can rearrange the formula to solve for angular velocity:

w = sqrt(ac / r)

Substituting ac = 9.80 m/s² and r = 95 m (half the diameter) into the formula, we get:

w = sqrt(9.80 m/s² / 95 m) = 0.465 rad/s

Therefore, the angular velocity that would produce an artificial gravity of 9.80 m/s² at the rim of the space station is approximately 0.465 rad/s.

User ADmad
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