A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a speed of 5.0 ✕ 103 m/s relative to the rocket. The mass of the rocket at this time is 6.5 ✕ 104 kg, and its acceleration is 4.0 m/s2.
What is the velocity of the exhaust relative to the solar system? (B) At what rate was the exhaust ejected during the firing?

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aristocles aristocles · Answer 1 · 2019-11-24T12:05:20+01:00

Answer:

Part a)

[tex]v_E = 2.5 \times 10^3 m/s[/tex]

Part b)

[tex]\frac{dm}{dt} = 52 kg/s[/tex]

Explanation:

Part a)

As we know that the velocity of exhaust with respect to the rocket is given as

[tex]v_{ER} = v_E - v_R[/tex]

so we will have

[tex]-5 \times 10^3 = v_E - 7.5 \times 10^3[/tex]

[tex]v_E = 2.5 \times 10^3 m/s[/tex]

Part b)

As we know that

[tex]F = ma[/tex]

here F = thrust force on the rocket

So we have

[tex]v_{ER} \frac{dm}{dt} = ma[/tex]

[tex](5 \times 10^3) \frac{dm}{dt} = (6.5 \times 10^4)(4)[/tex]

[tex]\frac{dm}{dt} = 52 kg/s[/tex]