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DaniilM [7]
3 years ago
8

A motorcycle traveling at 15 m/s accelerates at a constant rate of 4.0 m/s over a distance of 50 meters. What is the final veloc

ity of the motorcycle
Physics
1 answer:
ololo11 [35]3 years ago
5 0
From equations of motion,

v^2 = u^2 + 2as ----- v = final velocity, u = initial velocity, a = acceleration, s = distance covered.

Using the vales given;

v^2 = 15^2 + 2*4*50 = 625
v = Sqrt (625) = 25 m/s
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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
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Answer:

 v_{1f} = +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 v_{1f} + m₂ v_{2f}

     p₀ = [texpv_{f}[/tex]

     (m₁ + m₂) v₀ = m₁ v_{1f} + m₂ v_{2f}

Let's calculate the final speed (v1f) of the first stage

     v_{1f} = ((m₁ + m₂) v₀ - m₂ v_{2f}) / m₁

     

     v_{1f} = ((2100 +1160) 4300 - 1160 5940) / 2100

     v_{1f} = (14,018 10 6 - 6,890 106) / 2100

     v_{1f} = 7,128 106/2100

     v_{1f} = +3,394 103 m / s

come the same direction of the final stage, but more slowly

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