CHEGG You have two pipes, having the same elevation, connected to each other that transport water. The given density of water is rho = 997 kg/m . The flow rate of water through the pipes is R = .009463522m3/s, but they have different diameters. The first pipe has a measured pressure P 1=13.785Pa and a velocity of v = 41.221 m/s. The second pipe has a measured velocity of v = 55.654m/s. A) What is the pressure in the second pipe? Provide your answer in MPa.

P = 0.6814 MPa

Explanation:

For this exercise we must use the Betuliano relation

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

where we use the subscript 1 for the first pipe and the subscript 2 for the second pipe

indicate that the two pipes are at the same level therefore

        y₁ = y₂

       P₂ = P₁ + ½ ro (v₁² - v₂²)

let's calculate

       P₂ = 13,785 10⁵ + ½ 997 (41,221² - 55,654²)

       P₂ = (13.785 - 6.97) 105

       P₂ = 6.814 10⁵ Pa

let's reduce to Mpa

       P₂ = 6.814 10⁵ Pa (1 MPa / 10⁶ Pa)

       P₂ = 6.814 10-1 MPa

       P = 0.6814 MPa

area of focus

there are no contact forces mentioned on the list of choices that you've provided.