When the atmospheric pressure is 1 atm a water fountain ejects a stream of water that rises to a height of 5 m. there is a 1.9-cm-radius pipe that leads from a pressurized tank to the opening that ejects the water. 1. what would happen if the fountain were operating when the eye of a hurricane passes through? 2. assume that the atmospheric pressure in the eye is 0.877 atm and the tank's pressure remains the same. what will be the height?

1) When we are in the eye of a hurricane the atmospheric pressure decreases so the water must rise more         2) y = 25.3 m

Explanation:

In this exercise we must use the concepts and mechanics of fluids and kinematics to solve them. Let's start by finding the speed (v₁) so the atmosphere left the pipe, for this we use the vertical launch equations

           v² = v₀² - 2 g y

At the highest point the speed is zero

          v₀ = v₁ = √ 2 g y

          v₁ = √ 2 9.8 5

          v₁1 = 9.90 m / s

Now let's use the continuity equation to find the velocity (v₂) in the wide part of the tank

         A₁ v₁ = A₂ v₂

They indicate that the radius of the pipe r₂/r₁ = 1.9, in area of a circle is

        A = π R²

       A₂ / A₁ = π r₂² /π r₁²

        A₂ / A₁ = (r₂ / r₁)²

       v₂ = A₁ /A₂   v₁

       v₂ = (1 / 1.9)² 9.9

       v₂ = 2.74 m / s

Now let's use Bernoulli's equation, to find the pressure, point 1 is at the exit and point 2 is in the tank,

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

In the problem it indicates that P₁ = 1 atm, they do not give explicit indications about the height of the pipe, suppose that it is horizontal (the most common) y₁ = y₂

      P₂ - = ½ ρ (v₁² - v₂²)

      P₂ = + ½ ρ (v₁² -v₂²)

Let's calculate

      P₂ = 1.013 10⁵ + ½ 1000 (9.90² - 2.74²)

      P₂ = 1.013 10⁵ + 4.525 10⁴

      P₂ = 1.47 10⁵ Pa

1) When we are in the eye of a hurricane the atmospheric pressure decreases so the water must rise more

2) Let's use Bernoulli's equation to find the output velocity (v₁)

       ½ ρ v₁² = P₂- P₁ + ½ ρ v₂²)

Let's calculate

       P₂ = 0.877 1.013 10⁵ Pa

       P₂ = 0.888 10⁵ Pa

       ½ 1000 v₁² = 1.47 10⁵ - 0.888 10⁵ + ½ 1000 2.74²

      500 v₁² = 0.582 10⁵ + 3753.8

      v₁ = √ (61953.8 / 500)

      v₁ = 11.13 m / s

now we use kinematics

     v₀ = 11.13 m/s

     v² = v₀² - 2 g y

    y = v₀² / 2g

    y = 11.13 2/2 9.8

    y = 25.3 m

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)