Abasketball rolls onto the court with a speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right, and slows down with a constant acceleration of 0.50\,\dfrac{\text m}{\text s^2}0.50 s 2 m 0, point, 50, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction over 14\,\text m14m14, start text, m, end text. what is the velocity of the basketball after rolling for 14\,\text m14m14, start text, m, end text?

1.4 m/s

Explanation:

We can solve the problem by using the following suvat equation

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the basketball in this problem, we have

u = 4.0 m/s

s = 14 m

Solving for v, we find the final velocity:

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j