Respuesta :
Answer:
ΔP = 97.93 Pa , T₂-T₁ = 71.5° C
Explanation:
For this problem let's use Bernoulli's relationship, as point 1 we will take the plane and as point 2 the air
P₁ + ½ ρ g v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂
As the whole process occurs at the same height y₁ = y₂ = 11 km. We will consider that the air goes in the opposite direction to the plane
P₂ –P₁ = ½ ρ (v₁² - v₂²)
Let's reduce the magnitudes to the SI system
v₁ = 935 km / h (1000 m / 1 km) (1h / 3600s) = 259.72 m / s
v₂ = 280 m / s
The density of air at 11000 m is
Rho = 0.3629 kg / m
P₂-P₁ = ½ 0.3629 (259.72 + 280)
ΔP = 97.93 Pa
The variation of the temperature with the altitude is 0.65 per 100 m
T₂ –T₁ = (0.65 / 100) 1000
T₂-T₁ = 71.5° C
The temperature has decreased this value
