Answer:
v₂ = 97.4 m / s
Explanation:
Let's write the Bernoulli equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Index 1 is for tank and index 2 for exit
We can calculate the pressure in the tank with the equation
P = F / A
Where the area of a circle is
A = π r²
E radius is half the diameter
r = d / 2
A = π d² / 4
We replace
P = F 4 / π d²2
P₁ = 397 4 /π 0.058²
P₁ = 1.50 10⁵ Pa
The water velocity in the tank is zero because it is at rest (v1 = 0)
The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa
Since the pipe is horizontal y₁ = y₂
We replace on the first occasion
P₁ = P₂ + ½ ρ v₂²
v₂ = √ (P1-P2) 2 / ρ
v₂ = √ [(1.50-1.013) 10⁵ 2/1000]
v₂ = 97.4 m / s
