The water leaves the sprinkler holes at a speed of approximately 0.53 meters per second.
At what speed does it leave the sprinkler holes?
Radius of the hose: r hose = 1.2 cm / 2 = 0.6 cm
Area of the hose: A hose = π * r hose² ≈ 1.13 cm²
Radius of a hole: r hole = 0.29 cm / 2 = 0.145 cm
Area of one hole: A hole = π * r hole² ≈ 0.066 cm²
Total area of all holes: A holes = 27 * A hole ≈ 1.78 cm²
This principle states that the mass flow rate of an incompressible fluid remains constant throughout a closed system.
In this case, the system is the hose and the sprinkler container.
Mass flow rate = density * velocity * area
Assuming the water density remains constant (around 1000 kg/m³), we can equate the mass flow rate in the hose to the total mass flow rate through all the sprinkler holes:
ρ * v hose * A hose = ρ * v holes * A holes
v holes = (v hose * A hose) / A holes
v holes = (0.83 m/s * 1.13 cm²) / 1.78 cm²
v holes ≈ 0.53 m/s
Therefore, the water leaves the sprinkler holes at a speed of approximately 0.53 meters per second.