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Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s. This process takes 10 s. During the time interval what average force is applied to the rocket?

Respuesta :

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

 M v + m v' = 0

 5000 x v - 10 x 4000 = 0

 5000 v = 40000

    v = 8 m/s

speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

[tex]F = \dfrac{m(v_i-v_f)}{\Delta t}[/tex]

[tex]F = \dfrac{5000(8-0)}{10}[/tex]

      F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N

Answer:

4000 N

Explanation:

mass of fuel, m = 10 kg

velocity of ejected fuel, v = 4000 m/s

time, t = 10 s

force, F = velocity x mass of fuel burning per second

F = 4000 x 10 / 10 = 4000 N

Thus, the force is 4000 N.

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