# Calculated Water Flow Rates and Speeds

## Question:

If it takes 1.25 min to fill a 22.0 L bucket with water flowing from a garden hose of diameter 3.50 cm, determine the speed at which water is traveling through the hose. What happens if a nozzle with a diameter three-fifths the diameter of the hose is attached to the hose?

## Answer:

First, let's calculate the speed at which water is traveling through the hose. Then, we will determine the speed of the water leaving the nozzle after the attachment of a smaller diameter nozzle.

## Calculation of Water Flow Rate and Speed:

To begin, we need to calculate the volume flow rate of the water through the hose. This can be done by dividing the volume of water (22.0 L) by the time taken to fill the bucket (1.25 min or 75.0 s).

The volume flow rate is calculated as follows:

(22.0 L) / (1.25 min) = 17.6 L/min

Converting the volume flow rate to liters per second, we get:

17.6 L/min * (1 min / 60 s) = 0.293 L/s

Now, we can determine the speed of the water traveling through the hose using the equation Q = v * A, where Q is the volume flow rate, v is the speed, and A is the cross-sectional area of the hose.

The diameter of the hose is 3.50 cm, which equals a radius of 0.0175 meters. Therefore, the area A is calculated as:

A = πr^2 = 3.14 * (0.0175)^2 = 0.000962 m^2

Substitute the values into the equation Q = v * A to find the speed v:

0.293 L/s = v * 0.000962 m^2

v = 304.35 m/s

**The speed of the water traveling through the hose is 304.35 m/s.**

## Calculation of Water Speed through Nozzle:

When a nozzle with a diameter three-fifths of the hose diameter is attached, the area of the nozzle becomes (3/5)^2 times smaller than the area of the hose.

This means the area of the nozzle is A_n = (3/5)^2 * 0.000962 m^2 = 0.000347 m^2

Since the volume flow rate is conserved, we can set up the equation Q = v * A for both the hose and the nozzle:

0.293 L/s = v_n * 0.000347 m^2

v_n = 843 m/s

**Therefore, the speed of the water leaving the nozzle is 843 m/s.**