For those who think electric cars are sissy, Keio University in Japan has tested a 22-ft long prototype whose eight electric motors generate a total of 590 horsepower. The "Kaz" cruises at 175 mi/h. If the drag coefficient is 0.35 and the frontal area is 23.5 ft2, what percent of this power is expended against sea-level air drag? Round the final answer to three decimal places.

The percent of power extended against sea-level air drag is 50.15%

Explanation:

Length L is 22 ft, Power P is 590 hp,

Speed of air V = 175 mi/h. To convert to ft, we will multiply by 1.46

V = 175 × 1.46 = 255.5 ft

Frontal Area A is given as 23.5 ft

Coefficient of drag, Cd is 0.35

From text, Density of Air at sea level ρ = 0.00237 slug/ft³

The formula for calculating Drag Force F is

F = Cd × × ρAV²

Substitute the values above into the formula

F = 0.35 × × 0.00237 × 23.5 × 255.5²

= 0.35 × × 3635.78

= 636.26 lbf

But Power P = FV

Therefore we substitute for F and V to get P

P = 636.26 × 255.5

= 162564.43 ft.lbf/s

To convert to horsepower, we multiply by 0.00182 hp

= 162564.43 × 0.00182

= 295.87 hp

The percent of power extended against sea-level air drag is

= (295.87/590) × 100

= 50.15%

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

1/0.14nm = 10^10/1.4 a lot