During basketball practice shane made a jump shot, releasing a 0.60 kg basketball from his hands at a height of 2.0 m above the floor with a speed of 7.6 m/s. the ball swooshes through the net at a height of 3.0 m above the floor and with a speed of 4.5 m/s. how much energy was dissipated by air drag from the time the ball left shane's hands until it went through the net?

5.373 J

Explanation:

mass of ball (m) = 0.6 kg

initial height (h1) = 2 m

initial speed (u) = 7.6 m/s

final height (h2) = 3 m

final speed (v) = 4.5 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

from the conservation of energy ,

change in kinetic energy + change in potential energy = 0

in this case;

       = 0.5m()

       Uair = -0.5m() - mg(h2-h1)

       Uair = -0.5x0.6() - 0.6x9.8(3-2)

       Uair = -(-11.253)- 0.6x9.8(3-2)

       Uair = 5.373 J

the amount of current in the coil

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.