Answer:
ω = 630.2663 = 630[rad/s]
Explanation:
Solution:
- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.
1rev(20 hole) -> 20(cycle)/rev
2006.2(cycle) -> f ?
f = 2006.2/20 = 100.31rev at second
- The relation between angular frequency and angular speed is given by:
ω = 2πf
ω = 2*3.14*100.31
ω = 630.2663 = 630[rad/s]
125 because f=ma so you would use 100=mx0.75
Work = (force) x (distance)
The worker does (40N) x (4m) = 160 joules of work.
Friction eats up (27N) x (4m) = 108 joules of that energy,
generating 108 joules of heat.
The remaining (160J - 108J) = 52 joules of energy moves the box.
It shows the ray passing through the boundary.