Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at one-fourth 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. (drag on jet / drag on transport)

Explanation:

Drag force is given by

C = Drag coefficient is constant

A = Area is constant

= Velocity of the passenger jet = 1200 km/h =

= Velocity of the prop plane =

= Density of the air where the jet was flying =

= Density of the air where the prop plane was flying =

The ratio of the drag forces is .

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

answer(20) psi;