Question

to water the yard, you use a hose with a diameter of 3.4 cm. water flows from the hose with a speed of 1.1 m/s. if you partially block the end of the hose so the effective diameter is now 0.57 cm, with what speed does water spray from the hose?

a. 24 m/s
b. 56 m/s
c. 72 m/s
d. 39 m/s
e. 11 m/s

Explanation:

Step1: Recall the continuity equation

The continuity equation for fluid flow is $A_1v_1 = A_2v_2$, where $A$ is the cross - sectional area and $v$ is the speed of the fluid. The cross - sectional area of a circular hose is $A=\pi(\frac{d}{2})^2$, with $d$ being the diameter.

Step2: Express the areas in terms of diameters

$A_1=\pi(\frac{d_1}{2})^2$ and $A_2=\pi(\frac{d_2}{2})^2$. Substituting into the continuity equation gives $\pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2$. The $\pi$ and the $(\frac{1}{2})^2$ terms cancel out, so we have $d_1^{2}v_1 = d_2^{2}v_2$.

Step3: Solve for $v_2$

We are given $d_1 = 3.4\ cm$, $v_1=1.1\ m/s$ and $d_2 = 0.57\ cm$. Rearranging the equation $d_1^{2}v_1 = d_2^{2}v_2$ for $v_2$ gives $v_2=\frac{d_1^{2}v_1}{d_2^{2}}$.
Substitute the values: $v_2=\frac{(3.4)^{2}\times1.1}{(0.57)^{2}}$.
$(3.4)^{2}=11.56$, $(0.57)^{2}=0.3249$.
$v_2=\frac{11.56\times1.1}{0.3249}=\frac{12.716}{0.3249}\approx39\ m/s$.

Answer:

D. $39\ m/s$