Answer:
Step-by-step explanation:
Given:
Angle, θ = 35°
Vertical distance, Δx = 6 m
Diameter, d = 1.9 cm
= 0.019 m
A.
When the water leaves the sprinkler, it does so at a projectile motion.
Therefore,
Using equation of motion,
(t × Vox) = 2Vo²(sin θ × cos θ)/g
= Δx = 2Vo²(sin 35 × cos 35)/g
Vo² = (6 × 9.8)/(2 × sin 35 × cos 35)
= 62.57
Vo = 7.91 m/s
B.
Area of sprinkler, As = πD²/4
Diameter, D = 3 × 10^-3 m
As = π × (3 × 10^-3)²/4
= 7.069 × 10^-6 m²
V_ = volume rate of the sprinkler
= area, As × velocity, Vo
= (7.069 × 10^-6) × 7.91
= 5.59 × 10^-5 m³/s
Remember,
1 m³ = 1000 liters
= 5.59 × 10^-5 m³/s × 1000 liters/1 m³
= 5.59 × 10^-2 liters/s
= 0.0559 liters/s.
For the 4 sprinklers,
The rate at which volume is flowing in the 4 sprinklers = 4 × 0.0559
= 0.224 liters/s
C.
Area of 1.9 cm pipe, Ap = πD²/4
= π × (0.019)²/4
= 2.84 × 10^-4 m²
Volumetric flowrate of the four sprinklers = 4 × 5.59 × 10^-5 m³/s
= 2.24 × 10^-4 m³/s
Velocity of the water, Vw = volumetric flowrate/area
= 2.24 × 10^-4/2.84 × 10^-4
= 0.787 m/s
