22.07.2022 - 05:16

# Water flows straight down from an open faucet. The cross-sectional area of the faucet is 1.70 times 10^{-4} m , and the speed of the water is 0.82 m / s as it leaves the faucet. Ignoring air resistanc

Question:

Water flows straight down from an open faucet. The cross-sectional area of the faucet is {eq}1.70 times 10^{-4} m {/eq}, and the speed of the water is {eq}0.82 m / s {/eq} as it leaves the faucet. Ignoring air resistance, find the cross-sectional area of the water stream at a point {eq}0.16 m {/eq} below the faucet.

The continuity equation states that the mass flow rate of a fluid is constant throughout a pipe or stream, assuming no sources or sinks of mass. This can be expressed as: $A_1 v_1 = A_2 v_2$ Where $A_1$ and $A_2$ are the cross-sectional areas of the pipe or stream at two points, and $v_1$ and $v_2$ are the velocities at those points. In this problem, we know $A_1 = 1.70 times 10^{-4} m^2$ and $v_1 = 0.82 m/s$. We want to find $A_2$ at a point $0.16 m$ below the faucet, so we can write: $A_1 v_1 = A_2 v_2$ $A_2 = \frac{A_1 v_1}{v_2} = \frac{(1.70 times 10^{-4})(0.82)}{v_2}$ To find $v_2$, we can use the fact that the water is flowing under the influence of gravity, so its speed will increase at a rate of $g = 9.81 m/s^2$. Using kinematic equations, we can find $v_2$: $v_2^2 = v_1^2 + 2gh$ $v_2^2 = (0.82)^2 + 2(9.81)(0.16)$ $v_2 = 1.50 m/s$ Substituting this into the equation for $A_2$, we get: $A_2 = \frac{(1.70 times 10^{-4})(0.82)}{1.50} = 9.29 times 10^{-5} m^2$ Therefore, the cross-sectional area of the water stream 0.16 m below the faucet is $9.29 times 10^{-5} m^2$.