Question

Water flows straight down from an open faucet. The cross-sectional area of the faucet is 1.9 × 10-4m2 and the speed of the water is 0.75 m/s as it leaves the faucet. Ignoring air resistance, find the cross-sectional area of the water stream at a point 0.10 m below the faucet.

Answer 1

`8*10^5m^2`

Explanation:

To find the cross-section area at a point below the faucet

we can use following equations

`v_1^2=v_2^2+2ay`

and equation of continuity

`A_1v_1=A_2v_2`

v_1= velocity at the out let

v_2= velocity at the inlet (faucet)= 0.75 m/s

y = distance below the faucet = 0.10 m

A_1= cross-sectional area of the water stream at a point 0.10 m below the faucet.

A_2= area of faucet= 1.9 × 10-4m2

from above two equation we can write

`A_1= (A_2v_2)/(√(v_2^2+2ay) )`

now putting the values we get

`A_1= (1.9*10^(-4)*0.75)/(√(0.75^2+2*9.80*0.10) )`

A_1= 0.00008=
`8*10^5`