Relating to Newton’s Third Law, action and reaction, what is the reaction when a rocket expels gas, smoke and flames from the no
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Answer:
θ = 28°
Explanation:
For this exercise We will use the second law and Newton, let's set a System of horizontal and vertical.
X axis
Fₓ = m a
Nₓ = m a
Where the acceleration is centripetal
a = v² / r
The only force that we must decompose is normal, let's use trigonometry
sin θ = Nₓ / N
cos θ = / N
Nₓ = N sin θ
= N cos θ
Let's replace
N sin θ = m v² / r
Y Axis
- W = 0
N cos θ = mg
Let's divide the two equations of Newton's second law
Sin θ / cos θ = v² / g r
tan θ = v² / g r
θ = tan⁻¹ (v² / g r)
We reduce the speed to the SI system
v = 61 km / h (1000 m / 1 km) (1h / 3600 s) = 16.94 m / s
Let's calculate
θ = tan⁻¹ (16.94 2 / (9.8 55.1)
θ = tan⁻¹ (0.5317)
θ = 28°