Question

Ch11 problem # 62

Answer 1

Answer:

a) v₂ = 32.83 m / s, b)   y = 54.99 m

Explanation:

a) For this problem we must use Bernoulli's equation, we define the point as the surface of the water in the tank and point 2 is the point of the nozzle

          P₁ + ½ ρ v₁² +ρ g y₁ = P₂ + ½ρ v₂² + ρ g y₂

As the nozzle is open to the outside it has the atmospheric pressure (P₂ = $P_{atm}$) and the height   y₂ = -4.00 m, since the tank is very large the water velocity inside the tank is very small, we will assume zero (v₁ = 0), replace

         P₁ + ρ g (y₁-y₂) =$P_{atm}$ + ½ ρ v₂²

        v₂² = 2 /ρ [(P₁ -$P_{atm}$) + ρ g (0 - y₂)]

Let's calculate

         v₂² = 2/1000 [(6.01 -1.013) 10⁵ + 1000 9.8 4]

         v₂ = √ (2 10⁻³ [4,997 10⁵ + 3.92 10⁴]) = √ 1077.8

        v₂ = 32.83 m / s

b) the height at which the water reaches can be found with kinematics at this point the speed is zero

          Vy² = v₂² - 2 g y

           vy = 0

           y = v₂² / 2g

Let's calculate

           y = 32.83²/2 9.8

           y = 54.99 m