ANSWER:
224.32 m/s
STEP-BY-STEP EXPLANATION:
We can use Bernoilli's theorem:
[tex]P_1+\frac{1}{2}\rho(V_1)^2=P_2+\frac{1}{2}\rho(V_2)^2[/tex]We replace and solve for V2, like this:
[tex]\begin{gathered} P_1-P_2=\frac{1}{2}\rho(V_2)^2-\frac{1}{2}\rho(V_1)^2 \\ 9460=\frac{1}{2}\cdot0.376\cdot(V_2)^2-\frac{1}{2}\cdot0.376\cdot(0)^2 \\ 9460=\frac{1}{2}\cdot0.376\cdot(V_2)^2 \\ (V_2)^2=50319 \\ V_2=\sqrt[]{50319}=224.32\text{ m/s} \end{gathered}[/tex]The airplane’s speed relative to the air is 224.32 m/s