6) Joe throws a 0.15-kg rubber ball directly downward to the floor. The ball’s speed just before impact is 6.5 m/s and just after the impact is 3.5 m/s. If the ball is in contact with the floor for 0.025 s, what is the magnitude of the average force that the floor exerts on the ball during the time of contact? A. 133 N B. 60 N C. 30 N D. 18 N E. 3.0 N

B. 60 N

Explanation:

Change in momentum after ball bounces back = m ( v - u )

v is velocity after the bounce and u is before the bounce

= .15 ( 6.5 + 3.5 )

1.5 kg m /s

If F be the average force acting on the ball which changed its velocity

impulse of force = force x time

= F x .025

Impulse of force = change in momentum

F x .025  = 1.5

F = 1.5 / .025

= 60 N .

hèllo, sorry i really need some more points so if you need this qustion answered than retype it

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.