Average Force on a Rocket During Gas Exhaustion

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The discussion focuses on calculating the average force exerted on a rocket during gas exhaustion, specifically when 1150 kg of gas is expelled at a velocity of -4.50×10^4 m/s over one second. The formula used is F = m(Δv) / (Δt), leading to a calculated force of -5.175 x 10^7 N. However, it is noted that the negative sign may not be necessary as the problem seeks the magnitude of the force. Additionally, the importance of conservation of momentum in understanding the relationship between the rocket and expelled gas is emphasized. The calculation and underlying principles are crucial for accurately determining the force involved in rocket propulsion.
subopolois
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Homework Statement


Rocket engine expends 1150 kg of gas in 1 second, with a velocity of -4.50×10^4 m/s. Calculate the average force exerted on the rocket by the gas during that 1.0 second interval

Homework Equations


F= m(delat)v / (delta)t

The Attempt at a Solution


F= (1150kg)(-45000m/s) / 1 s
-5.175 x 10^7 N
 
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I would use a different sign for deltam, but apart from that it is right.
 
subopolois said:

Homework Statement


Rocket engine expends 1150 kg of gas in 1 second, with a velocity of -4.50×10^4 m/s. Calculate the average force exerted on the rocket by the gas during that 1.0 second interval

Homework Equations


F= m(delat)v / (delta)t

The Attempt at a Solution


F= (1150kg)(-45000m/s) / 1 s
-5.175 x 10^7 N
Your answer is OK except no - sign. The problem implicitly asks for a magnitude.
But, you should give a bit more detail on why that formula works, starting with conservation of momentum of the rocket/gas system:
force x 1 sec. = change per second in momentum of rocket = - change per second in momentum of the gas
etc.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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