Water flows straight down from an open faucet. The cross-sectional area of the faucet is 1.00 × 10 − 4 m², and the speed of the water is 0.72 m/s as it leaves the faucet. Ignoring air resistance, find the cross-sectional area of the water stream at a point 0.19 m below the faucet.

Answer:[tex]A_2=0.349\times 10^{-4} m^2[/tex]Step-by-step explanation:Givenvelocity [tex]v_1=0.72 m/s[/tex]Area of cross-section [tex]A_1=1\times 10^{-4} m^2[/tex]velocity after Falling 0.19 m[tex]v_2^2-v_1^2=2g\cdot h[/tex][tex]v_2^2=v_1^2+2g\cdot h[/tex][tex]v_2^2=(0.72)^2+2\times 9.8\times 0.19[/tex][tex]v_2^2=0.518+3.724[/tex][tex]v_2=\sqrt{4.24}[/tex][tex]v_2=2.05 m/s[/tex]conserving Flow[tex]A_1v_1=A_2v_2[/tex][tex]1\times 10^{-4}\times 0.72=A_2\times 2.05[/tex][tex]A_2=0.349\times 10^{-4} m^2[/tex]