Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.00 10-4 m2, and the speed of the water is 0.94 m/s as it leaves the faucet. Ignoring air resistance, find the cross-sectional area of the water stream at a point 0.11 m below the faucet.

10.79×10^-5 m^2

Explanation:

v2 = 0.94m/s

Cross sectional area(A) = 2 × 10^-4m^2

v1A1 = v2A2 (continuity equation used to compare any two points of the flow of an incompressible fluid. )

A1 = (v2A2) ÷ v1

v1^2 = v2^2 + 2as

v1 = sqrt(v2^2 + 2as)

distance below faucet(s) = 0.11m

Acceleration during to gravity (a) = 9.8m/s^2

Substituting v1 into the continuity equation :

A1 = (v2A2) ÷ sqrt(v2^2 + 2as)

v2A2 = 0.94 × 2 × 10^-4m^2 = 1.88×10^-4

sqrt(0.94^2m/s + (2×9.8m/s^2×0.11m)) =

sqrt(0.8836 + 2.156) = 1.743

A1 = (1.88×10^-4)÷1.743

A1 = 0.000188 ÷ 1.743

A1 = 10.79×10^-5 m^2

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