Question

Water sprays out of a faucet at a rate of 1.5 m/s1.5 \mathrm{~m} / \mathrm{s} . After a brief distance of falling, the gravitational acceleration causes its speed to climb to  4.5 m/s4.5 \mathrm{~m} / \mathrm{s}. How would you calculate the area at the lower point by multiplying the stream's initial cross-sectional area by that number?

Solution

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On page 182 of the book, we find the expression for the rate of flow of a fluid with a velocity vv in a pipe with cross-sectional area AA as,

Q=Av(1)Q = Av \hspace{0.5 cm} (1)

For a streamlined flow, the rate of flow remains constant, i.e.

Q=constantAv=constantA1v1=A2v2A2=v1v2A1(2)\begin{aligned} Q & = \text{constant}\\ Av & = \text{constant}\\ A_1 v_1 & = A_2 v_2\\ A_2 & = \frac{v_1}{v_2} A_1 \hspace{0.5 cm} (2) \end{aligned}

Given: v1=1.5v_1 = 1.5 m/s, v2=4.5v_2 = 4.5 m/s

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