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

2.268

Step-by-step 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 = (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) = (1)/(2) \rho_1 V^2_1\\F_(D2) = (1)/(2) \rho_2 V^2_2`

The ratio between the two force is,

`(F_(D1))/(F_(D2))=(\rho_1 V^2_1)/(\rho_2 V^2_2)\\(F_(D1))/(F_(D2))=(0.38*1000^2)/(0.67*500^2)\\(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