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

Answer:

$8\times10^5m^2$

Step-by-step 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

$A_1= \frac{A_2v_2}{\sqrt{v_2^2+2ay} }$

now putting the values we get

$A_1= \frac{1.9\times10^{-4}\times0.75}{\sqrt{0.75^2+2\times9.80\times0.10} }$

A_1= 0.00008= $8\times10^5$