91Ó°ÊÓ

1. The pressure difference measured by the pitot tube is$180Pa$.

2. The density of air$0.031kg/m^3$.

Using Bernoulli’s equation, find the expression for the airspeed of the plane. Then using this expression, find the airspeed of the plane for the given values. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

The formula is as follows:

$pâˆ\P Ä+\frac{1}{2}\mathrm{ÒÏ}{g}^{2}h+\text{constant}$

Where, p is pressure, v is velocity, h is height, g is the acceleration due to gravity, h is height, and $\mathrm{ÒÏ}$is density.

The airflow obeys Bernoulli’s principle.

So,

$$\begin{gathered} p_A+\frac{1}{2}{\text{ÒÏ}}_{air}^2{\text{ghtÒÏ}}_{air}={\mathrm{ÒÏ}}_B+\frac{1}{2}{\mathrm{ÒÏ}}_{air}^{2}g{h}_{air} \\ {p}_A-p_B=\frac{1}{2}{}_{air}\left({}^2\right) \\ \mathrm{Δ}p=\frac{1}{2}{}_{air}({}^2) \end{gathered}$$

Thus, rearranging,

$$\begin{gathered} v=\sqrt{\frac{2∆p}{{\mathrm{ÒÏ}}_{air}}} \\ =\sqrt{\frac{2×180Pa}{0.031kg/m^3}} \\ =107.76m/s \\ ≈1.1×{10}^{2}m/s \end{gathered}$$

Hence, the plane’s airspeed is $1.1×{10}^{2}m/s$.