Torque [Nm] = Force [N] x Force arm [m]
C=F*b=20*0.20=4 Nm
The correct answer is C
Answer:
The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.
Explanation:
It is given that,
A high jumper jumps over a bar that is 2 m above the mat, h = 2 m
We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

g is acceleration due to gravity

v = 6.26 m/s
So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.
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.
Newton's 2nd law of motion: Force = (mass) x (acceleration)
Divide each side by (mass) : Acceleration = (force) / (mass)
Acceleration of the truck = (4,200 N) / (2,000 kg) = 2.1 m/s²