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an airplane has a mass of 1.9E6 kg and the air flow past the
lowersurface of the wings at 90 m/s. If the wings have surfacearea
of 1.6E3 m^3, how fast must the air over the upper surface ofthe
wing if the plane is to stay in the air. Consider onlythe Bernoulli
effect.

Respuesta :

whitneytr12

Answer:

[tex]v_{1}=164.4 m/s[/tex]  

Explanation:

The Bernoulli equation is:

[tex]P+\frac{1}{2}\rho v^{2}+\rho gh=constant[/tex] (1)

  • P is pressure related to the fluid
  • ρ is the density of the fluid (ρ(air)=1.23 kg/m³)
  • v is the speed of the fluid
  • h is the displacement from one position to the other

Now let's apply the equation (1), for our case:

[tex]P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2}[/tex] (2)

here we assume that h is the same in both cases so it canceled out.

  • subscript 1 is related to the upper surface of the wings
  • subscript 2 is related to the low surface of the wings

Solving the equation for v₁ we have:

[tex]v_{1}=\sqrt{\frac{2(P_{2}-P_{1})}{\rho}+v_{2}^{2}}[/tex] (3)

Now, we know that pressure P=F/A (force over area)

[tex]\Dela P=\frac{W}{A}=\frac{mg}{A}=\frac{1.9\cdot 10^{6}\cdot 9.81}{1.6\cdot 10^{3}} =11649.4 N/m^{2}[/tex] (4)

Combining (3) and (4), we can find v1.

[tex]v_{1}=\sqrt{\frac{2(11649.4)}{1.23}+90^{2}}[/tex]  

[tex]v_{1}=164.4 m/s[/tex]  

I hope it helps you!  

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