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3 years ago

How many grams of butter, which has a usable energy content

Physics

1 answer:

VARVARA [1.3K]3 years ago

4 0

Answer:

$m_{butter} = 156.93 grams$

Explanation:

Given data;

energy content = 6.0 Cal/g = 6000 cal/g

mass of man 60 kg

climb height = 6.70 km

g = 9.80 m/s^2

we know that

potential energy = mgh

$E_t = mU$

wehre U - usable energy constant

equating both energy

$60 \times 9.8 \times 6.70\times 10^3 = m_{butter} 6000\times \frac{4.184 j}{cal}$

$m_{butter} = 156.93 grams$

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A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci

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Answer:

Approximately $\rm 2.5\; m \cdot s^{-1}$ .

Explanation:

Let the increase in the rocket's velocity be $\Delta v$ . Let $v_0$ represent the initial velocity of the rocket. Note that for this question, the exact value of $v_0$ doesn't really matter.

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Mass of the rocket with the 5 kg of fuel: $1000$ .
Initial velocity of the rocket and the fuel: $v_0$ .
Hence the initial momentum of the rocket: $1000\,v_0$ .

Mass of the rocket without that 5 kg of fuel: $1000 - 5 = 995$ .
Final velocity of the rocket: $v_0 + \Delta v$ .
Hence the final momentum of the rocket: $995\,(v_0 + \Delta v)$ .

Mass of the 5 kg of fuel: $5$ .
Final velocity of the fuel: $v_0 - 500$ (assuming that the the 500 m/s in the question takes the rocket as its reference.)
Hence the final momentum of the fuel: $5\,(v_0 - 500)$ .

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

$1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500)$ .

Note that $1000\, v_0$ appears on both sides of the equation. These two terms could hence be eliminated.

$0 = 995\, \Delta v - 5\times 500$ .

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Hence, the velocity of the rocket increased by around 2.5 m/s.

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AlladinOne [14]

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espero que esto te ayude! buena suerte!

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