A0.25 kg ball bearing is released from rest at the surface of the water in the tank of water.

after falling 2 m in the water, its speed is 5 m/s.

what is the amount of work done by the water on the ball bearing?

amount of work done = -1.77 J

Explanation:

given data

mass = 0.25 kg

speed = 5 m/s

height = 2m

to find out

amount of work done by the water on the ball bearing

solution

we get here potential energy when ball fall

potential energy = mgh

potential energy = 0.25 × 9.8 × 2

potential energy = 4.9 J

and that lose potential energy gain by kinetic energy that is express as

kinetic energy = 0.5 × m × v²

kinetic energy = 0.5 × 0.25 × 5²

kinetic energy = 3.13 J

so we get now work done that is

work done = kinetic energy - potential energy

work done = 3.13 J - 4.9 J

amount of work done = -1.77 J

continuous conducting path for current, power supply

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min