Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven transport flying at half that speed and altitude. The density of air is 0.53 kg/m3 at 7.5 km and 0.74 kg/m3 at 3.8 km. Assume that the airplanes have the same effective cross-sectional area and drag coefficient C.

Respuesta :

Answer:

[tex]\frac{D_{jet}}{D_{prop}}=2.865[/tex]

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

[tex]D=\frac{1}{2}CpAv^2\\[/tex]

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

[tex]\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\ \frac{D_{jet}}{D_{prop}}=2.865[/tex]

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