Question

Calculate the ratio of the drag force on a passenger jet flying at an altitude of 10 km to the drag force on a prop-driven plane flying at two-fifths the speed and half the altitude of the jet. At 10 km the density of air is 0.380 kg/m3 and at 5.0 km it is 0.670 kg/m3. The prop-driven plane has one-third the cross-sectional area of the passenger jet. The passenger jet has two-fifths the drag coefficient of the prop-driven plane.

Answer 1

Answer:

2.268

Explanation:

The concepts that we need to use here for give a solution are Drag coefficient and drag force.

Drag force is given by,

$F_D = \frac{1}{2} c \rho A V^2$

Where,

c is the drag coefficient

$\rho$ is the density

A cross sectional area

V the velocity.

Our values for this problem are divided for Passenger and Driver.

For Jet are:

$V_1 = 1000km/h \\h_1 = 10*10^3m\\\rho_1 =0.38kg/m^3$

For prop-driven are:

$V_2 = 500km/h\\h_2 = 5km\\\rho_2 = 0.67kg/m^3$

From the problem we need to assume that $A_1 = A_2$ and $c_1 = c_2$, THEN

Applying to each case the Drag force equation we have,

$F_{D1} = \frac{1}{2} \rho_1 V^2_1\\F_{D2} = \frac{1}{2} \rho_2 V^2_2$

The ratio between the two force is,

$\frac{F_{D1}}{F_{D2}}=\frac{\rho_1 V^2_1}{\rho_2 V^2_2}\\\frac{F_{D1}}{F_{D2}}=\frac{0.38*1000^2}{0.67*500^2}\\\frac{F_{D1}}{F_{D2}}= 2.268$

Therefore the force experienced by a Jet pilot under these conditions is 2,268 times greater than that of a prop-driven