Question

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 1

Answer:$A_2=0.349\times 10^{-4} m^2$

Step-by-step explanation:

Given

velocity $v_1=0.72 m/s$

Area of cross-section $A_1=1\times 10^{-4} m^2$

velocity after Falling 0.19 m

$v_2^2-v_1^2=2g\cdot h$

$v_2^2=v_1^2+2g\cdot h$

$v_2^2=(0.72)^2+2\times 9.8\times 0.19$

$v_2^2=0.518+3.724$

$v_2=\sqrt{4.24}$

$v_2=2.05 m/s$

conserving Flow

$A_1v_1=A_2v_2$

$1\times 10^{-4}\times 0.72=A_2\times 2.05$

$A_2=0.349\times 10^{-4} m^2$