The mean pressure is `(6K)/5`, which rain renders to vertical windshield of automobile, moving with constant velocity of magnitude `v=12m//s`. Consider that raindros fall vertically with speed `u=5m//s`. The intensity of rainfall deposits `h=2cm` of sediments in time `tau=1` minute: [ `rho=10^(3)kg//m^(3))` is the density of liquid] (Assume collisions are inelastic). Calculate `K`.

Answer

8

Explanation

Let in time `t, m` mass of liquid collides with windshield. So
`F=|(p_(f)-p_(i))/t|=|(0-mv)/t|=(mv)/t`
`lamdaimplies` density of liquid molecules in air so
`m=Svtxxlamda=Svlamdat`
Where `Simplies` surface area of the windshield
`F=Sv^(2)lamda=` Mean pressure `=F/S=lamdav^(2)`.........(1)
Now as we assumed that `lamda` is the density of liquid molecules of drops in air so in `tau` time `utauS_(0)` volume of liquid will strikes on the surface of earth so mass of liquid strikes on the earth in time internval `tau` will be
`m_(0)=S_(0)utau lamda`
`impliesS_(0)rho h=S_(0) u tau lamda`
`implieslamda=(rhoh)/(u tau)`
Putting this value in equation (1), we have
Mean pressure `=(rho h v^(2))/(u tau)=48/5impliesK=8`