Now consider a finite-span wing with the same profile, chord and attack angle flying at 5 km altitude with a speed of 50 m/s. The wing aspect ratio is 5, area is 5 m2 , and span efficiency factor is 0.9. Determine the wing lift.

Wing lift, L = 211.57kg

Explanation:

We are given:

•Attitude= 5Km

•V = 50m/s => 164.042 ft/s

• Area= 5m² = 53.8196ft²

•Aspect ratio = 5

•span efficiency factor = 0.9

Table is attached

For density at 5km, we have: d = 16404ft

To find wing lift, we use the formula:

Substituting figures, we have:

(1/2)*(0.0014314)*(164.042)²*(53.8196)*0.5*0.9

= 466.4382 lb

= 211.57

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.