Water flows through a fire hose of diameter 6.35 cm at a rate of 0.015 m3/s. The fire hose ends in a nozzle of inner diameter 1.750 cm. At what speed does the water exit the nozzle?
water speed*area=volume speed
waterspeed*(1.750/2)^2*Pi=.015m^3/sec
solve for water speed
To find the speed at which the water exits the nozzle, we can apply the principle of conservation of volume. Since the water is incompressible, the volume of water flowing into the hose must be equal to the volume flowing out of the nozzle.
First, let's find the cross-sectional area of the hose and the nozzle using their respective diameters:
Area of hose = π * (diameter of hose/2)^2
Area of hose = π * (6.35 cm/2)^2
Area of nozzle = π * (diameter of nozzle/2)^2
Area of nozzle = π * (1.750 cm/2)^2
Next, let's calculate the velocity of the water as it flows through the hose. We can use the equation:
Velocity of water = volume flow rate / cross-sectional area
Velocity of water = 0.015 m^3/s / Area of hose
Similarly, the velocity of water as it exits the nozzle can be calculated using:
Velocity of water = 0.015 m^3/s / Area of nozzle
Now that we have the velocities of water in the hose and the nozzle, we can compare them. Since the volume of water entering and exiting remains the same, the velocities must be inversely proportional to the cross-sectional areas.
Velocity of water in hose / Velocity of water in nozzle = Area of nozzle / Area of hose
Substituting the values we calculated earlier, we can solve for the velocity of water in the nozzle.
Velocity of water in nozzle = (Velocity of water in hose) * ( Area of nozzle / Area of hose )
Hence, by plugging in the values, we can find the speed at which the water exits the nozzle.