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
D. Extreme temprature and cloudy waters
Answer:
(a). The spring compressed is 
.
(b). The acceleration is 1.5 g.
Explanation:
Given that,
Acceleration = a
mass = m
spring constant = k
(a). We need to calculate the spring compressed
Using balance equation
....(I)
The spring compressed is .
(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,
The compression is given by
Here, acceleration is zero
So, 
We need to calculate the acceleration
Put the value of x in equation (I)
Hence, (a). The spring compressed is .
(b). The acceleration is 1.5 g.
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows:
