Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
The velocity at the stagnation point is zero.
The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Thus, the answer is 2.2 psi
We start by finding the angle of inclination with the sine function,
sin θ = 1 m / 4 m
θ = 14.48°
Next, we compute the work done by the movers using the following formula:
W = Fnet * d
We need to first determine Fnet. It is the weight force minus the frictional force.
Fnet = m g sinθ – μ m g cosθ
Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48
Fnet = 84.526 N
The work done is therefore:
W = 84.526 N * 4 m
<span>W = 338.10 J</span>
Answer:
a) 
b) D does not influence the long-term results.
Explanation:
Given that
A = A0 cos(ωt)
This is a linear equation hence the integration factor, I
Now using the characteristics of linear equations
b) At t= 0
Thus, the initial condition
does not affect the long-term outcome.
