## Related questions with answers

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

Solution

VerifiedOn page 182 of the book, we find the expression for the rate of flow of a fluid with a velocity $v$ in a pipe with cross-sectional area $A$ as,

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

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

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

Given: $v_1 = 1.5$ m/s, $v_2 = 4.5$ m/s

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