Question

Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.401 × 10 − 3 m 3 / s and the diameter of the nozzle you hold is 5.27 × 10-3 m. At what speed does the water exit the nozzle?

Answer 1

To solve this problem it is necessary to apply the concepts related to the Flow Rate. It is understood as the volumetric amount transported during a certain time and can be calculated as

$Q = AV$

Where

A = Area

V = Velocity

Our values are given as

$r = 5.27*10^{-3}m \rightarrow A = \pi r^2 = \pi ( 5.27*10^{-3})^2 = 8.72*10^{-5}m^2$

$Q = 0.401*10^{-3}m^3/s$

Replacing and rearranging to find the velocity

$V = \frac{Q}{A}$

$V = \frac{0.401*10^{-3}}{8.72*10^{-5}}$

$V= 4.59m/s$

Therefore the speed of the water at the end of the Nozzle is 4.59m/s