Answer:
(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)
(b). P₂ = 236500 Pa
Explanation:
This is quite straight-forward so let us begin by defining the terms given.
Given that;
The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.
Let us make the velocity of water inside the house be V₁
such that the Volume of water entering the per second is = A₁V₁
Therefore, in 90sec:
45 L = 90 A₁V₁
V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴
V₁ = 10m/s (velocity of water inside the house)
From the continuity equation we have that;
A₁V₁ = A₂V₂
0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂
V₂ = 5m/s (velocity at ground level)
(b). We are told to calculate the water pressure in the pipeline at the ground level.
Using Bernoulli's equation;
P₁ + pgh₁ + 1/2PV₁² (inside) = P₂ + pgh₂ + 1/2PV₂² (ground level)
1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²
P₂ (pressure) = 1.01*10⁵Pa
Therefore we have;
101000 + 98000 + 50000 = P₂ + 12500
P₂ = 236500 Pa
cheers I hope this helped !!
