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
Answer:
An object is called a horizontal projectile if it is launched from a certain height with some initial horizontal velocity only. The initial vertical velocity of such an object is zero. But as the object falls through the atmosphere the horizontal component of velocity remains constant but vertical component increases due to gravitational acceleration.
Explanation:
Answer:
d = 76.5 m
Explanation:
To find the distance at which the boats will be detected as two objects, we need to use the following equation:

<u>Where:</u>
θ: is the angle of resolution of a circular aperture
λ: is the wavelength
D: is the diameter of the antenna = 2.10 m
d: is the separation of the two boats = ?
L: is the distance of the two boats from the ship = 7.00 km = 7000 m
To find λ we can use the following equation:
<u>Where:</u>
c: is the speed of light = 3.00x10⁸ m/s
f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz
Hence, the distance is:

Therefore, the boats could be at 76.5 m close together to be detected as two objects.
I hope it helps you!
The distance the spring stretches is the answer.
I hope this helps.
Have a nice day.