Water flowing through a garden hose of diameter 2.77 cm fills a 25.0 L bucket in 1.20 min.
(a) What is the speed of the water leaving the end of the hose?
(b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?
a) 0.58 m/s
b) 5.19 m/s
To determine the speed of the water leaving the end of the hose, we first need to calculate the flow rate of the water through the hose.
Flow rate is given by the equation:
Flow rate = Volume / Time
(a) To find the speed, we need to convert the flow rate to speed. The flow rate is the volume of water passing through a certain point per unit time.
Given:
Diameter of the hose (d) = 2.77 cm
Radius of the hose (r) = d/2 = 2.77 cm / 2 = 1.385 cm = 0.01385 m (converting cm to m)
Volume of water (V) = 25.0 L = 25.0 * 10^-3 m³ (converting L to m³)
Time taken (t) = 1.20 min = 1.20 * 60 s (converting min to s)
Flow rate = Volume / Time = (25.0 * 10^-3 m³) / (1.20 * 60 s)
Simplifying the expression, we have:
Flow rate = 0.34722 * 10^-3 m³/s
To find the speed, we divide the flow rate by the cross-sectional area of the hose.
Cross-sectional area of the hose (A) = π * r²
Substituting the values, we have:
Cross-sectional area = π * (0.01385 m)²
Simplifying the expression, we have:
Cross-sectional area ≈ 0.000603 m²
Now we can find the speed:
Speed = Flow rate / Cross-sectional area
Speed ≈ (0.34722 * 10^-3 m³/s) / 0.000603 m²
Speed ≈ 0.574 m/s
Therefore, the speed of the water leaving the end of the hose is approximately 0.574 m/s.
(b) Now let's calculate the speed of the water leaving the nozzle, where the diameter of the nozzle is one-third the diameter of the hose.
Diameter of the nozzle = 2.77 cm / 3 = 0.9233 cm
Radius of the nozzle = 0.9233 cm / 2 = 0.46165 cm = 0.0046165 m (converting cm to m)
Cross-sectional area of the nozzle = π * (0.0046165 m)²
Simplifying the expression, we have:
Cross-sectional area ≈ 0.000067 m²
Using the same flow rate obtained earlier, we can now calculate the speed:
Speed = Flow rate / Cross-sectional area
Speed ≈ (0.34722 * 10^-3 m³/s) / 0.000067 m²
Speed ≈ 5.18 m/s
Therefore, the speed of the water leaving the nozzle is approximately 5.18 m/s.