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Calculate the force exerted on a rocket when the propelling gases are being expelled at a rate of 1300 kg/s with a speed of 4.5 × 104 m/s.

Respuesta :

skyluke89
Considering the rocket+gases as an isolated system, the variation of momentum of the rocket should be equal to the variation of momentum of the gases, which is
[tex]\Delta p = \Delta m \cdot v[/tex]
since the speed v at which the gas is expelled is constant.
If we divide both terms per the time interval, [tex]\Delta t[/tex], we get
[tex] \frac{\Delta p}{\Delta t}= \frac{\Delta m}{\Delta t} v [/tex]
But [tex] \frac{\Delta p}{\Delta t}[/tex] is equal to the force F exerted on the gas by the rocket (and for Newton's third law, this is equal to the force exerted by the gas on the rocket), while [tex] \frac{\Delta m}{\Delta t} [/tex] is the rate at which the gas is expelled, 1300 kg/s. Therefore, the force exerted on the rocket is
[tex]F= \frac{\Delta m}{\Delta t} v = (1300 kg/s)(4.5 \cdot 10^4 m/s)=5.85 \cdot 10^7 N[/tex]
hamzaahmeds

This question involves the concept of Newton's Second Law of Motion.

The force exerted on the rocket is "5.85 x 10⁷ N".

NEWTON'S SECOND LAW OF MOTION

According to Newton's Second Law of Motion, whenever an unbalanced force is applied to an object, it produces an acceleration in the direction of the force. Mathematically,

[tex]F=ma[/tex]

where,

  • F = force = ?
  • a = acceleration = [tex]\frac{\Delta v}{t}[/tex]

Therefore,

[tex]F = \frac{m\Delta v}{t}[/tex]

where,

  • [tex]\frac{m}{t}[/tex] = mass flow rate of gases = 1300 kg/s
  • Δv = change in velocity = 4.5 x 10⁴ m/s

Therefore,

F = (1300 kg/s)(4.5 x 10⁴ m/s)

F = 5.85 x 10⁷ N

Learn more about Newton's Second Law of Motion here:

https://brainly.com/question/23845187

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