B. Wagon B will accelerate at twice the rate as Wagon A. hope this helps!
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:x(t)= Acos(wt)
Explanation:
According to Newton's 2nd law,a particle of mass m acted on by a force is given by:Fs=-kx
Where x is displacement from equilibrium
K = spring constant
Therefore X(t) = Acos(2pit/T)
X(t)= Acos(wt)
Explanation:
Let's say right is positive and left is negative.
F₁ = -150 N
F₂ = 220 N
Fnet = F₁ + F₂
Fnet = -150 N + 220 N
Fnet = 70 N
The magnitude of Fnet is 70 N, and since it's positive, the direction is to the right.
And since Fnet isn't 0, the force is unbalanced and the motion is changing.