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A person trying to lose weight by burning fat filts a mass of \({10}{k}{g}\) upto a being of \({1}{m}{1000}\) time . Assume that the potential energy lost each time be lower the mass is dissipated . How much far will be use up considering the work done only when the weight is lifted up ? Far supplies \({3.8}\times{10}^{{{7}}}{J}\) of energy per kg wich is canverted to mechanical energy with \({x}{20}\%\) efficiency rate Take \(={9.8}{m}{s}^{{-{2}}}\)
(A) \({9.89}\times{10}^{{-{3}}}{k}{g}\)
(B) \({12.89}\times{10}^{{-{3}}}{k}{g}\)
(C) \({2.45}\times{10}^{{-{3}}}{k}{g}\)
(D) \({6.45}\times{10}^{{-{3}}}{k}{g}\)


Answer: B
\({F}{a}{t}{u}{s}{e}{d}=\frac{{{49}\times{10}^{{{4}}}}}{{{3.8}\times{10}^{{{7}}}}}={12.89}\times{10}^{{-{3}}}{k}{g}\) .
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