Water flows steadily through a horizontal pipe of varying cross section. at one place the pressure 15 1.5 atm and the speed is 1.0 m/sec. determine the pressure at another place where the speed is 10 m/sec. a. 0.5 atm b. 0.6 atm c. 0.7 atm d. 0.8 atm e, 1.0 atm ap = z (v₂²-42) - 1000 ( 10212) marr is removed

 P₂ = 1.011 atm       The correct answer is E

Explanation:

For this problem we must use Bernoulli's equation, which is the conservation of energy for fluids

           P₁ + ½ ρ v₁² +ρ g y₁ = P₂ + ½ρv₂² + ρ g y₂

Let's write the data and reduce to the SI system

The initial pressure P₁ = 1.5 atm (1,013 10⁵ Pa / 1 atm) = 1.52 10⁵ Pa

The initial velocity v₁ = 1.0 m / s

The final speed v₂ = 10 m / s

The pipe is horizontal y₁ = y₂

We apply this to the equation

          P₁ -P₂ = ½ ρ (v₂² - v₁²)

          P₂ = P₁ - ½ ρ (v₂² - v₁²)

Let's calculate

          P₂ = 1.52 10⁵ - ½ 1000 (10²- 1²)

          P₂ = (1.52- 0.495) 10⁵

          P₂ = 1.0245 10⁵ Pa

Let's reduce to atm

         P₂ = 1.0245 10⁵ (1 atm / 1.013 10⁵ Pa)

         P₂ = 1.011 atm

The correct answer is E

i guess it’s a and b