Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10^4 pa.
if the pipe widens to twice its original radius, what is the pressure in the wider section?

9.8 × 10⁴Pa

Explanation:

Given:

Velocity V₁ = 12m/s

Pressure P₁ = 3 × 10⁴ Pa

From continuity equation we have

                              ρA₁V₁ = ρA₂V₂

                                 A₁V₁ = A₂V₂

making V₂ the subject of the equation;

the pipe is widened to twice its original radius,

                                r₂ = 2r₁          

then the cross-sectional area A₂ = 4A₁

                           ⇒  

This implies that the water speed will drop by a factor of   because of the increase the pipe cross-sectional area.  

 The Bernoulli Equation;

     Energy per unit volume before = Energy per unit volume after    

        p₁ + ρV₁²  + ρgh₁ = p₂ + ρV₂²  + ρgh₂  

Total pressure is constant and = P = ρV₂²ρV²  

        p₁ + ρV₁²  = p₂ + ρV₂²

Making p₂ the subject of the equation above;

        p₂ = p₁ + ρV₁² - ρV₂²

But so,

        p₂ = p₁ + ρV₁² - ρ      

       p₂ = 3.0 x 10⁴ + ( × 1000 × 12²) - ( × 1000 × 12²/4² )

      P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³    

       P₂ = 9.79 × 10⁴Pa      

      P₂ = 9.8 × 10⁴Pa                      

he made 11 calls in one day

neither accurate nor precise.