Answer:
The speed of the aircraft is [tex]V_1 = 92.0 \ \frac{m}{s}[/tex]
Explanation:
Assumptions.
1. flow of air is steady & incompressible.
2. Frictional effects are neglected.
From the Bernoulli's equation
The velocity of the jet is given by
[tex]V_{1} = \sqrt{2(\frac{P_2 - P_1}{\rho}) }[/tex]
Here [tex]P_2 -P_1 = 3500 \ Pa[/tex]
[tex]\rho = 0.82 \frac{kg}{m^{3} }[/tex]
Thus velocity
[tex]V_{1} = \sqrt{2(\frac{3500}{\ 0.82}) }[/tex]
[tex]V_1 = 92.0 \ \frac{m}{s}[/tex]
Therefore the speed of the aircraft is [tex]V_1 = 92.0 \ \frac{m}{s}[/tex]
