A 0.53 kg arrow leaves a bowstring at a velocity of 63 m/s. If the arrow was initially at rest and then the string applied a force on it for 7 x 10-3 seconds, what was the approximate average force in Newtons that acted on the arrow during this time?

4770 N

Explanation:

Momentum is the product of mass and velocity and force is the change in momentum divided by change in time.

Given from the question;

Mass of arrow= 0.53 kg

Velocity of arrow = 63 m/s

Initial velocity of arrow = 0 m/s

Change in time = 0.007 s

Finding momentum after the arrow is released as;

p=m*v

p= 0.53 * 63

p= 33.39 kg*m/s

Force is the change in momentum divided by change in time;

F= 33.39 / 0.007

F= 4770 N

option number c is the right answer.

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.