Answer:
102000 kg
Explanation:
Given:
A total Δν = 15 km/s
first stage mass = 1000 tonnes
specific impulse of liquid rocket = 300 s
Mass flow rate of liquid fuel = 1500 kg/s
specific impulse of solid fuel = 250 s
Mass flow of solid fuel = 200 kg/s
First stage burn time = 1 minute = 1 × 60 seconds = 60 seconds
Now,
Mass flow of liquid fuel in 1 minute = Mass flow rate × Burn time
or
Mass flow of liquid fuel in 1 minute = 1500 × 60 = 90000 kg
Also,
Mass flow of solid fuel in 1 minute = Mass flow rate × Burn time
or
Mass flow of solid fuel in 1 minute = 200 × 60 = 12000 kg
Therefore,
The total jettisoned mass flow of the fuel in first stage
= 90000 kg + 12000 kg
= 102000 kg
Answer:
the answer is b because it does not show evidence
They do. mostly because they want validation from others
Answer:
a) v = 2.4125 m / s , b) Em_{f} / Em₀ = 0.89
Explanation:
a) This is an inelastic crash problem, the system is made up of the four carriages, so the forces during the crash are internal and the moment is conserved
Initial
p₀ = m v₁ + 3 m v₂
Final
= (4 m) v
p₀ =p_{f}
m (v₁ + 3 v₂) = 4 m v
v = (v₁ +3 v₂) / 4
Let's calculate
v = (3.86 + 3 1.93) / 4
v = 2.4125 m / s
b) the initial mechanical energy is
Em₀ = K₁ + 3 K₂
Em₀ = ½ m v₁² + ½ 3m v₂²
The final mechanical energy
= K
Em_{f} = ½ 4 m v²
The fraction of energy lost is
Em_{f} / Em₀ = ½ 4m v² / ½ m (v₁² +3 v₂²)
Em_{f} / Em₀ = 4 v₂ / (v₁² + 3 v₂²)
Em_{f} / Em₀ = 4 2.4125² / (3.86² + 3 1.93²)
Em_{f} / em₀ = 23.28 / 26.07
Em_{f} / Em₀ = 0.89