Answer:
A) θ = 13.1º , B) E
Explanation:
A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.
In this reference system the only force that we must decompose is the Normal one, let's use trigonometry
sin θ = Nₓ / N
cos θ = / N
Nₓ = N sin θ
Ny = N cos θ
x-axis (radial)
Nₓ = m a
where the acceleration is centripetal
a = v² / R
we substitute
-N sin θ = -m v² / R (1)
the negative sign indicates that the force and acceleration towards the center of the circle
y-axis (Vertical)
Ny - W = 0
N cos θ = mg
N = mg / cos θ
we substitute in 1
mg / cos θ sin θ = m v² / R
g tan θ = v² / R
θ = tan⁻¹ (v² / gR)
we calculate
θ = tan⁻¹ (25² / 9.8 274)
θ = 13.1º
B) when comparing the equations the correct one is E
The amplitude is equivalent to the coefficient of the sine function, so:
Amplitude = 1
The time period of a Sin(x) = 2π
The time period of Sin(4x) = 2π/4 = π/2
The period is π/2
Explanation:
I assume the acceleration calculated in part (b) is the 3.33 m/s² from your other question.
Use Newton's second law to find the total force:
F = ma
F = (60,000 kg) (3.33 m/s²)
F = 200,000 N
Since there are 2 engines, the thrust from each is half of this:
F = 100,000 N
In reality, there are forces other than thrust. There are also drag forces (rolling friction and air resistance).
From Newton's second law, if we increase the mass and keep the force the same, the acceleration decreases. So it would take longer to reach the take-off speed.
Answer:
The flow area at the location where the Mach number is 0.9 is 25.24 cm²
Explanation:
Here we have for isentropic flow;
Where:
A = Area of flow = 36 cm²
M = Mach number at section of = 1.8
k = Specific heat ratio = 1.4
A* = Area at the throat
Therefore, plugging the values we get
Therefore, A* = 36/1.439 = 25.01769 cm²
Where the Mach number is 0.9, we have
Therefore A = 25.020× 1.009 = 25.24 cm²
The flow area at the location where the Mach number is 0.9 = 25.24 cm².