Aspace 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.

0.0078 rpm

Step-by-step explanation:

Diameter of the wheel = 222m

Radius of the wheel = 222/2

= 111m

Artificial gravity = 9.9m/s^2

a= V^2/r (V= velocity)

V^2 = a*r

V = √a*r

V = √ 9.9*111

V= √1098.9

V = 33.15m/s

T= πd/V

T = (π * 222) / 33.15

= 21.04

Frequency(f) = 1/T

f = 1/21.04

f = 0.046

Rotational frequency (w) = 2πf

= 2π * 0.046

= 0.2986 rad/s

1 rev = 2π radians

0.2986 rad/s = 0.2986/2π rev/s

= 0.4690 rev/s

Recall that 60 secs = 1 minute

= 0.4690/60

= 0.0078rev/min

= 0.0078rpm

29 is the correct answer

step-by-step explanation:

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752