91Ó°ÊÓ

First, convert the speed of the airplane from km/hr into m/s. This is done by multiplying the given speed \(200 \mathrm{km/hr}\) by \(\frac{5}{18}\) to obtain \(55.56 \mathrm{m/s}\).

Next, apply Bernoulli’s principle to solve for the speed of air relative to the wing. Bernoulli's Principle \( P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \), where \(P_1\) and \(P_2\) are the pressure at the two points, \(\rho\) is the fluid density, \(v_1\) and \(v_2\) are the velocities at the two points, \(g\) is the acceleration due to gravity, and \(h_1\) and \(h_2\) are the heights at the two points. Ignoring the heights (since height change is negligible for flow over airplane wing) and the initial velocity (which is zero), simplifying the equation gives \(v_2 = \sqrt { \frac{2 (P_1 - P_2)}{\rho}} \). The air density \(\rho\) can be found using the Ideal Gas Law \( \rho= \frac {P_{air}}{RT}\), where \(P_{air}\) is the pressure of the air, \(R\) is the specific gas constant, and \(T\) is the temperature in Kelvin. Convert -10°C to Kelvin to get 263.15 K. The specific gas constant for air is 287 J/(kg K). Using this information, calculate \(\rho\) to get roughly 0.816 kg/m\(^3\).

Use the simplified Bernoulli's equation and the found density value to calculate \(v_2\), thus determining the relative speed of the air to the wing. Use \(P_1=65 KPa\) and \(P_2=60 KPa\). Be sure to convert these pressures into Pascal (Pa) to maintain consistent units. This results in a relative speed of approximately 289.15 m/s.

Subtract the speed of the airplane from the calculated relative speed to obtain the actual speed of the air or absolute speed. This results in an absolute speed of approximately 233.59 m/s.