Answer:
P2 = 5.18 10⁴ Pa
Step-by-step explanation:
This is a fluid mechanics exercise where underneath use Bernoulli's equation
Let's use subscript 1 for the thick part of the hose and subscript 2 for the thin part
P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂
Since the system is at the same height y₁ = y₂
Now we must use the continuity equation to find the muzzle velocity (v₂)
v₁ A₁ = v₂ A₂
v₂ = v₁ A₁ / A₂
The nozzles are circular, so the area of a circle is
A₁ = π r₁² = π d₁² / 4
A₂ = π d₂² / 4
v₂ = v₁ d₁² / d₂²
We substitute in the first equation
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁ d₁² / d₂²²
We clear the pressure P₂
P₂ = P₁ + ½ ρ v₁² (1 - d₁² / d₂²)
Let's reduce the magnitudes to the SI system
d₁ = 3 cm = 0.03 m
d₂ = 0.3 cm = 0.003 m
P₁ = 1 atm = 1,013 10⁵ Pa
Let's calculate
P2 = 1,013 10⁵ + ½ 1 10³ 0.65 (1 - (3 / 0.3)²)
P2 = 1,013 10⁵ - 4.95 10⁴
P2 = 5.18 10⁴ Pa