During take-off a 8kg model rocket is burning fuel causing its speed to increase
1 answer:
The strength of the thrust is 122 newtons.
The motion of the rocket is described by the second Newton's law, whose model is shown below:
(1)
Where:
- Thrust, in newtons
- Drag, in newtons
- Mass of the rocket, in kilograms
- Net acceleration of the rocket, in meters per square second
If we know that ,
and
, then the strength of the thrust is:
The strength of the thrust is 122 newtons.
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Answer:
dR/dt = 10.2 ft / s
Explanation:
Let's work this problem by finding the distance between the balloon and the motorcycle and then drift for the speed change of the distance
Balloon
y = y₀ + t
Motorcycle
x = v₀ₓ t
Distance, let's use Pythagoras' theorem
R² = x² + y²
R² = (v₀ₓ t)² + (y₀ + t)²
v₀ₓ = 88 ft / s
= 8 ft / s
y₀ = 150 ft
R² = (8 t)² + (150 + 8 t )²
R² = 64 t² + (150 + 8t )²
This is the expression for the distance between the two bodies, the rate of change is the derivative with respect to time (d / dt)
2RdR / dt = 64 2 t + 2 (150 + 8t) 8
dR / dt = [64 t + (1200 + 64t )] / R
dR/dt = (1200 +128 t)/R
Let's calculate for the time of 10 s
dR / dt = (1200 + 128 10) / R = 2480 /R
R = √ [64 10² + (150 + 8 10)²
R = √ [6400 + 52900]
R = 243.5 ft
dR / dt = (2480) / 243.5
dR / dt = 10.2 ft / s