The terminal speed of a sky diver is 122 km/h in the spread-eagle position and 308 km/h in the nosedive position. Assuming that the diver's drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position.

The ratio of the cross-sectional area inthe slower position to that in the faster position is

Explanation:

The terminal velocity of any object is given by

where '' is the mass of the object, '' is the acceleration due to gravity, '' is the drag coefficient, '' is the density and '' is the cross-sectional area of the object.

For spread-eagle position if be the cross-sectional area and for nosedive position if be the cross-sectional area, then from the above expression

where are the terminal velocities for spread-eagle position and nosedive position respectively.

Taking the ratio of both the velocities,

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.