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elixir [45]
1 year ago
13

The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in

Physics
1 answer:
pochemuha1 year ago
7 0

Since g is constant,  the force the escaping gas exerts on the rocket will be 10.4 N

<h3>What is Escape Velocity ?</h3>

This is the minimum velocity required for an object to just escape the gravitational influence of an astronomical body.

Given that the velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in 0.60s. The gravitational field intensity is 9.8N/kg.

To calculate the force the escaping gas exerts of the rocket, let first highlight all the given parameters

  • Mass (m) of the rocket 0.25 Kg
  • Initial velocity u = 15 m/s
  • Final Velocity v = 40 m/s
  • Time t = 0.6s
  • Gravitational field intensity g = 9.8N/kg

The force the gas exerts of the rocket = The force on the rocket

The rate change in momentum of the rocket = force applied

F = ma

F = m(v - u)/t

F = 0.25 x (40 - 15)/0.6

F = 0.25 x 41.667

F = 10.42 N

Since g is constant,  the force the escaping gas exerts on the rocket is therefore 10.4 N approximately.

Learn more about Escape Velocity here: brainly.com/question/13726115

#SPJ1

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Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

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E₁: Mechanical energy at the end (bottom of the incline)

then

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If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

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b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

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d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

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f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

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