michoyak
Answered

A hose directs a horizontal jet of water, moving with a velocity of 20m/s, on to a vertical wall. The

cross-sectional area of the jet is 5 x 10-4m2

. If the density of water is 1000kgm-3

, calculate the force on a

wall assuming the water is brought to rest there



Answer :

Demiurgos
Force is defined as the rate of change of momentum.
The initial amount of momentum is [tex]mv[/tex] because water stops when it hit the wall total change of momentum must be [tex] \Delta p=mv [/tex].
Now let's calculate the force.
[tex]F= \frac{dp}{dt}=\frac{d(mv)}{dt}=\frac{dm}{dt}v[/tex]
We need to find [tex]\frac{dm}{dt}[/tex]. This is the amount of water hiting the wall per second.
[tex]\frac{dm}{dt}=\rho Av[/tex]
Our final formula would be:
[tex]F=\rho Avv=\rho Av^2[/tex]
And now we can calculate the answer:
[tex]F=1000\cdot5\cdot 10^{-4}\cdot(20)^2=200 N[/tex]


snehashish65

The magnitude of force  exerted by the jet of water on the wall is 200 N.

Given data:

The velocity of jet is, v = 20 m/s.

The cross-sectional area of jet is, [tex]A = 5 \times 10^{-4} \;\rm m^{2}[/tex].

The density of water is, [tex]\rho = 1000 \;\rm kg/m^{3}[/tex].

The force exerted by the jet of water on the wall is given as,

[tex]F = \rho \times A \times v^{2}\\F = 1000 \times (5 \times 10^{-4}) \times 20^{2}\\F = 200 \;\rm N[/tex]

Thus, the magnitude of force  exerted by the jet of water on the wall is 200 N.

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