The shape of a four-spoked wheel with a diameter of 185 m. If the space station rotates at a rate of 2.35 revolutions per minute, what is the magnitude of the artificial gravitational acceleration provided to a space tourist walking on the inner wall of the station

The magnitude of the artificial gravitational acceleration provided to a space tourist walking on the inner wall of the station is 5.6 m/s²

Explanation:

Given-

Diameter, d = 185 m

radius, r = d/2 = 185/2 = 92.5 m

Frequency, f = 2.35 rev/min = 2.35/60 rev/sec

Magnitude of gravitational acceleration, a = ?

Centripetal acceleration, a = v²/r       -1

and v = 2πrf

Substituting the value of v in equation 1

a = (2πrf)²/r

a = 2πrf X 2πf

a = 4π²rf²

a = 4 X 3.14 X 3.14 X 92.5 X 2.35 X 2.35 / 60 X 60

a = 5.6 m/s²

Therefore, the magnitude of the artificial gravitational acceleration provided to a space tourist walking on the inner wall of the station is 5.6 m/s²

they should be similar or the same

i think its d if wrong im sorry