A gardener holds a pipe of inside cross-sectional area `3.60 cm^(2)` at a height of `1.50 m` from ground. The opening at the nozzle of pipe has cross-sectional area `0.250 cm^(2)`. Water velocity in the segment of the hose that lies on the ground is `50.0 cm//s`. a. What is the velocity of water leaving the nozzle? b. What is the thrust force experienced by the gardener due to flow of water through nozzle. c. What is the water pressure in the hose on the ground?

Explanation

Figure illustrates the situation. From continuity equation we can find the velocity of the fluid at the nozzle.

`v_(2)-(A_(1))/(A_(2))v_(1)=(3.60/0.250)(50.0)=720cm//s=7.20m//s`
b. Thrust force experienced by gardener at hand due to flow of water
`F_("thrust") =rhoA_(2)v_(2)^(2)=1000x0.25xx10^(-4)xx(7.2)^(2)=1.3N`
c. we will apply bernoull's equation to determine pressure `P_(1)`
`h_(1)=0.00m, h_(2)=1.50m, P_(2)=1.01xx10^(5)Pa`
`v_(1)=0.50m//s, v_(2)=7.20m//s,rho=1.00xx10^(3)kg//m^(3)`
`P_(1)+1/2 rho (v_(1)^(2)+rhogh_(1))=P_(2)+1/2rho_(2)^(2)+rhogh_(2)`
`P_(1)=P_(2)+1/2rho(v_(2)^(2)-v_(1)^(2))+rhog(h_(2)-h_(1))`
`=(1.01xx10^(5))+1/2(1.00xx10^(3))[(7.20)^(2)-(0.50)^(2)]`
`+(1.00+10^(3))(9.80)(1.50-0.00)`
`=1.41xx10^(5)Pa`