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

the answers c

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.