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

the results is a decrease in air temperature with an increase of altitude.

explanation:

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards