A force of 62 N acts on a 35kg object for 10.0s. What is the object change in velocity?

A force of 62 N acting on a 35 kg object applies an acceleration a of

62 N = (35 kg) a

a = (62 N) / (35 kg) ≈ 1.8 m/s²

If v₀ is the object's initial velocity, then after 10.0 s it attains a final velocity v of

v = v₀ + a t

v = v₀ + (1.8 m/s²) (10.0 s)

v = v₀ + 18 m/s

so that the change in velocity is

∆v = v - v₀ = 18 m/s

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm