A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive ch
1 answer:
Answer:
= +3,394 103 m / s
Explanation:
We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.
The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s
The moment before the explosion
p₀ = (m₁ + m₂) v₀
After the explosion
pf = m₁ + m₂
p₀ = [texpv_{f}[/tex]
(m₁ + m₂) v₀ = m₁ + m₂
Let's calculate the final speed (v1f) of the first stage
= ((m₁ + m₂) v₀ - m₂
) / m₁
= ((2100 +1160) 4300 - 1160 5940) / 2100
= (14,018 10 6 - 6,890 106) / 2100
= 7,128 106/2100
= +3,394 103 m / s
come the same direction of the final stage, but more slowly
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Answer:
y = 77.74 10⁻⁵ m
Explanation:
For this exercise we can use Newton's second law
F = m a
a = F / m
a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹
a = 0.538 10¹⁵ m / s
This is the vertical acceleration of the electron.
Now let's use kinematics to find the time it takes to move the
x= 29 mm = 29 10⁻³ m
On the x axis
v = x / t
t = x / v
t = 29 10⁻³ / 1.7 10⁷
t = 17 10⁻¹⁰ s
Now we can look for vertical distance at this time.
y = t + ½ a t²
y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²
y = 77.74 10⁻⁵ m