Au-tube open at both ends is partially filled withwater. oil
(density = 750kg/m3) is then pouredinto the right arm
and forms a column l = 5.00 cm high. (a)determine the difference
in, h, in the heights of the twoliquid surfaces. assume
that the density of air is1.29kg/m3, but be sure to
include differences in theatmospheric pressure due to changes in
altitude. (b) the right armis then shielded from any air motion
while air is blown across thetop of the left arm until the surfaces
of the two liquids are atthe same height. determine the speed of
the air being blownacross the left arm.

a)  h_w = 0.02139 m , b)      v₁ = 9.74 m / s

Explanation:

For this exercise we use the pascal principle that states that the pressure at one point is the same regardless of body shape.

At the initial moment (before emptying the oil), we fix the point on the surface of the liquid, in this case the left and right sides are in balance.

      P₁ = P₂ = P₀

Now we add the 5 cm (h₂ = 0.05 m) of oil, in this case the weight of the oil creates an extra pressure that pushes the water, let's look for how much the water moved (h_w). The weight of oil added is equal to the weight of displaced water

      W_w = W_oil

      m_w g= m_oil g

Density is defined

      ρ = m / V

we replace

        ρ_w V_w =  ρ_oil W_oil

       V = A h

       ρ_w A h_w =  ρ_oil A h_oil

      h_w = h_oil  ρ_oil /  ρ_w

Now let's analyze the pressure at the initial reference height for both sides two had in U

Right side

We have the atmospheric pressure, with its decrease due to the lower height, plus the oil pressure above the reference level

       h’= 0.05 cm - h_w

      P = (P₀ -  ρ_air g (0.05-h_w)) +  ρ_oil g (0.05-h_w)

Left side

We have at the same point, the atmospheric pressure with its reduction due to the height change plus the water pressure

        P = (P₀ -  ρ_air h h_w) +  ρ_w g h_w

As we have the same point we can equalize the pressure

(P₀ -ρ_air g (0.05-h_w)) +ρ_oil g (0.05-h_w) = (Po -ρ_air h h_w) +ρ_w g h_w

        ρ_air g (h_w - (0.05-h_w)) =  ρ_w g h_w - ρ_oil g (0.05-h_w)

       - ρ_air g 0.05 = h_w g ( ρ_w +  ρ _oil) - rho_oil g 0.05

       h_w g ( ρ_w +  ρ_oil) = g 0.05 ( ρ_air -  ρ_oil)

calculate

       h_w = 0.05 (ρ_ oil- ρ_air)  / ( ρ_w +  ρ_oil)

       h_w = 0.05 (750 - 1.29) / (1000-750)

      h_w = 0.02139 m

The amount that decreases the height on one side is equal to the amount that increases the other

b) cover the right side and blow the air on the left side, let's use Bernoulli's equation, where index 1 will be for the left side and index 2 for the right side

      P₁ + 1/2 ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since they indicate that both sides are at the same height y₁ = y₂, the right side is protected from the wind speed therefore v₂ = 0, let's write the equation

      P₁ + ½ ρ_air v₁² = P₂

    v₁² = (P₂-P₁) 2 / ρ_air

Let's analyze the pressure on each side of the tube, since the new equilibrium height is the height that was added of oil distributed between the two tubes, bone

       h’= 2.5 cm = 0.025 m

       P₁ = P₀ - ρ_w g h’

      P₂ = P₀ - ρ_oil g h ’

      P₂-P₁ = g h’ (rho_w - Rho_oil)

We replace

     v₁² = 2g h’ (ρ_w –ρ_oil) / ρ_air

calculate

    v₁² = 2 9.8 0.025 (1000 - 750) /1.29

    v₁ = √ 94.96

    v₁ = 9.74 m / s

-4 on

dang that balls went far