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How many grams of butter, which has a usable energy content
of6.0 Cal/g (= 6000 cal/g), would be equivalent to the change
ingravitational potential energy of a 60 kg man who climbs 6.70
km?Assume that the average gfor the climb is 9.80 m/s2.

Respuesta :

Answer:

[tex]m_{butter} = 156.93 grams[/tex]

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

[tex]E_t = mU[/tex]

wehre U - usable energy constant

equating both energy

[tex]60 \times 9.8 \times 6.70\times 10^3 = m_{butter} 6000\times \frac{4.184 j}{cal}[/tex]

[tex]m_{butter} = 156.93 grams[/tex]

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