A space station in the form of a large wheel, 222 m in diameter, rotates to provide an "artificial gravity" of 9.9 m/s 2 for people located on the outer rim. Find the rotational frequency of the wheel that will produce this effect. Answer in units of rpm.

Answer:0.0078 rpmStep-by-step explanation:Diameter of the wheel = 222mRadius of the wheel = 222/2= 111mArtificial gravity = 9.9m/s^2a= V^2/r (V= velocity) V^2 = a*rV = √a*rV = √ 9.9*111V= √1098.9V = 33.15m/s T= πd/VT = (π * 222) / 33.15 = 21.04Frequency(f) = 1/Tf = 1/21.04f = 0.046Rotational frequency (w) = 2πf = 2π * 0.046 = 0.2986 rad/s1 rev = 2π radians0.2986 rad/s = 0.2986/2π rev/s= 0.4690 rev/sRecall that 60 secs = 1 minute = 0.4690/60= 0.0078rev/min= 0.0078rpm