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Calculate the ratio of the drag force on a passenger jet flying with a speed of 950 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at one-fifth the speed and half the altitude of the jet. At 10 km the density of air is 0.38 kg/m3 and at 5.0 km it is 0.67 kg/m3. Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient C.

Respuesta :

Answer:

14.18

Explanation:

given,

Speed of passenger jet,v₁ = 950 Km/h

Altitude of flight, h₁ = 10 km

Prop-driven speed,v₂ = 950/5 = 190 Km/h

altitude,h₂ = 5 Km

density of air at 10 Km = 0.38 Kg/m³

density of air at 5 Km = 0.67 Kg/m³

Drag force formula

[tex]F = \dfrac{1}{2}C\rho Av^2[/tex]

now, ratio of drag Force

[tex]\dfrac{F_{jet}}{F_{prop}}=\dfrac{\dfrac{1}{2}C\rho_{10} Av_{1}^2}{\dfrac{1}{2}C\rho_{5} Av_2^2}[/tex]

[tex]\dfrac{F_{jet}}{F_{prop}}=\dfrac{\rho_{10} v_{1}^2}{\rho_{5}v_2^2}[/tex]

[tex]\dfrac{F_{jet}}{F_{prop}}=\dfrac{0.38\times 950^2}{0.67\times 190^2}[/tex]

[tex]\dfrac{F_{jet}}{F_{prop}}=14.18[/tex]

Hence, the ration of the drag force is equal to 14.18.

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