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,
[tex]F_D = \frac{1}{2} c \rho A V^2[/tex]
Where,
c is the drag coefficient
[tex]\rho[/tex] is the density
A cross sectional area
V the velocity.
Our values for this problem are divided for Passenger and Driver.
For Jet are:
[tex]V_1 = 1000km/h \\h_1 = 10*10^3m\\\rho_1 =0.38kg/m^3[/tex]
For prop-driven are:
[tex]V_2 = 500km/h\\h_2 = 5km\\\rho_2 = 0.67kg/m^3[/tex]
From the problem we need to assume that [tex]A_1 = A_2[/tex] and [tex]c_1 = c_2[/tex], THEN
Applying to each case the Drag force equation we have,
[tex]F_{D1} = \frac{1}{2} \rho_1 V^2_1\\F_{D2} = \frac{1}{2} \rho_2 V^2_2[/tex]
The ratio between the two force is,
[tex]\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[/tex]
Therefore the force experienced by a Jet pilot under these conditions is 2,268 times greater than that of a prop-driven
