# A horizontal jet of water coming out of a pipe of area of cross - section 20cm^2 hits a vertical wall with a velocity of 10m/s and rebounds with the same speed. The force exerted by water on the wall is

1
by crmail

2015-10-16T04:28:27+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Volume of water coming out of the pipe = A * v
= 20 cm² * 10 m/sec
= 0.02  m³/sec

mass of water hitting the wall  = m = 0.02 m³/sec * 1000 kg/m³
= 20 kg/sec

As the water hits the wall normal to the wall and gets rebounded at the same speed, the change of momentum of water hitting the wall is:
= m (v  - (-v))  =  2 m v
= 2 * 20 kg /sec *  10 m/sec
=  400 kg-m/sec²

Since, the change of momentum obtained is per unit time,  it is same as the force.  Thus the force acting on the wall by water = force experienced by water jet.

= 400 Newtons.

can you solve same problem by using 'F=density of water*area of cross section*velocity^2
force will be twice that product.. i explained in steps..
ok.i understod