A force of 200N was required to pull the arrow back 0.25m in the bow. If the archer were to release the bowstring how much energy would the arrow gain?

50 Joules

Explanation:

Work (J) = Force (N) x Distance (m)

Work (J) = 200 N x 0.25 m

Work = 50 Joules

Work is converted to kinetic energy when the arrow is released.

frequency or wavelenth of em radiation in a vacuum

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s