Answer:
v₂ = 63.62 m / s
Explanation:
For this exercise in fluid mechanics we will use Bernoulli's equation
P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂
where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.
We will assume that the distance between the two parts is small, so y₁ = y₂
P₁-P₂ = ρ g (v₂² - v₁²)
pressure is defined by
P = F / A
we substitute
ΔF / A = ρ g (v₂² - v₁²)
v₂² = 
suppose that the area of the wing is A = 1 m²
we substitute
v₂² =
v₂² = 79.10 + 3969
v₂ = √4048.1
v₂ = 63.62 m / s
Answer:
50m/s²
Explanation:
cross muliplyy write down the values acceleration and force
Answer:
1.991 × 10^(8) N/m²
Explanation:
We are told that its volume increases by 9.05%.
Thus; (ΔV/V_o) = 9.05% = 0.0905
To find the force per unit area which is also pressure, we will use bulk modulus formula;
B = Δp(V_o/ΔV)
Making Δp the subject gives;
Δp = B(ΔV/V_o)
Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²
Thus;
Δp = 2.2 × 10^(9)[0.0905]
Δp = 1.991 × 10^(8) N/m²