Question

Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it exits from the faucet is Blank
1. Calculate the answer by read surrounding text. cm3/s. As the water falls from the faucet with the given speed, it accelerates due to gravity and reaches a speed of _______
2. Calculate the answer by read surrounding text. after it has moved 0.2 m downward. With this change in speed of the water, the diameter of the stream 0.2 m below the faucet is _______
3. Calculate the answer by read surrounding text. _________ cm.

Answer 1

Answer:

Q = 165.95 cm³ / s,  1)    v = $\sqrt{0.55^2 + 19.6 y}$,  2)  v = 2.05 m / s,

3)  d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

         Q = v A

where Q is the flow rate, A is area and v is the velocity

         

the area of ​​a circle is

        A = π r²

radius and diameter are related

        r = d / 2

substituting

       A = π d²/4

       Q = π/4   v d²

let's reduce the magnitudes

       v = 0.55 m / s = 55 cm / s

let's calculate

       Q = π/4   55   1.96²

       Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

        v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

        v = $\sqrt{0.55^2 + 2 \ 9.8\ y}$

        v = $\sqrt{0.55^2 + 19.6 y}$

2) ask to calculate the velocity for y = 0.2 m

        v = $\sqrt{0.55^2 + 19.6 \ 0.2}$

        v = 2.05 m / s

3) We write the continuous equation for this point 2

        Q = v₂ A₂

        A₂ = Q / v₂

let us reduce to the same units of the SI system

        Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

        A₂ = 165.95 10⁻⁶ / 2.05

        A₂ = 80,759 10⁻⁶ m²

area is

        A₂ = π/4   d₂²

        d₂ = $\sqrt{4 A_2 / \pi }$

        d₂ = $\sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }$

        d₂ = 10.14 10⁻³ m

        d₂ = 1.014 cm