A stream of water flowing horizontally with a speed of `15 ms^(-1)` pushes out of a tube of cross sectional area `10^(-2)m^(2)` and hits a vertical wall near by what is the force exerted on the wall by the impact of water assuming.that it does not rebound? (Density of water `=1000 kg m^(3)`)

Explanation

In one second , the distance travelled is equal to the velocity v. Volume of wate hitting the wall per second , V = av where a is the cross-sectional area of the tube and v is the speed of water coming out of the tube.
V = `10^(-2)m^(2) xx 15 ms^(-1)` = `15 xx 10^(2) m^(3)s(-1)`
Mass of water hitting the wall per second = `15 xx 10^(-2) xx 10^(3) kg s^(-1)` = 150`kg s^(-1)` [v density of wate = 1000 kg `m^(-3)`]initial momentum of water hitting the wall per second
= `150 kg s^(-1) xx 15 ms^(-1)` = 2250 `kg ms^(-2)` or 2250 N Final momentum per second = 0 . Force excerted by the wall = 0 - 2250 N = 0 2250 N Force excerted on the wall = -(-2250 ) N = 2250 N