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A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:________

Respuesta :

Answer:

[tex]v_2 =2414\ m/s[/tex]

Explanation:

given,

exhaust velocity of fuel(v_e) = 1500 m/s

initial speed of rocket,v₁ = 0 m/s

final speed of rocket, v₂ = ?

fuel weigh = 80 % of total weight

using  Tsiolkovsky rocket equation

[tex]\Delta v = v_e ln(\dfrac{m_1}{m_2})[/tex]

Δ v = v₂ - v₁

v_e is the exhaust speed

m₁ is the  initial total mass.

m₂ is the is the final total mass without propellant.

m₂  = m₁ - 0.8 m₁

m₂  = 0.2 m₁

[tex]v_2-v_1 = 1500\times ln(\dfrac{m_1}{0.2 m_1})[/tex]

[tex]v_2 = 1500\times ln(\dfrac{m_1}{0.2 m_1})[/tex]

[tex]v_2 =2414\ m/s[/tex]

When all the fuel is exhausted speed of the fuel is equal to [tex]v_2 =2414\ m/s[/tex]

samueladesida43

The speed when all the fuel has been exhausted is 2415m/s

  • According to this question, the following information was given:

  1. Exhaust velocity of fuel, V(e)= 1500 m/s
  2. Initial speed of rocket, V₁ = 0 m/s
  3. Final speed of rocket, V₂ = ?
  4. Fuel weight = 80% of total weight

  • Using Tsiolkovsky rocket equation as follows:

∆V = V(e) ln(m1/m2)

  1. m1 = initial mass
  2. m2 = final mass without repellant

  • m2 = m1 - 80%m1

  • m2 = m1 - 0.8m1

  • m2 = 0.2m1

  • ∆V = V2 - V1

Hence;

  • V2 - 0 = 1500 × ln (m1/0.2m1)

  • V2 = 1500 ln(1/0.2)

  • V2 = 1500 × 1.609

  • V2 = 2415m/s.

  • Therefore, the speed when all the fuel has been exhausted is 2415m/s.

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