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) \({6.45}\times{10}^{{-{3}}}{k}{g}\)

(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

(b)\({n}=\frac{{{W}}}{{\in{p}{u}{t}}}=\frac{{{m}{g}{h}\times{1000}}}{{\in{p}{u}{t}}}=\frac{{{10}\times{9.8}\times{1}\times{1000}}}{{\in{p}{u}{t}}}\)

\(\in{p}{u}{t}=\frac{{{98000}}}{{{0.2}}}={49}\times{10}^{{{4}}}{J}\)

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

(b)\({n}=\frac{{{W}}}{{\in{p}{u}{t}}}=\frac{{{m}{g}{h}\times{1000}}}{{\in{p}{u}{t}}}=\frac{{{10}\times{9.8}\times{1}\times{1000}}}{{\in{p}{u}{t}}}\)

\(\in{p}{u}{t}=\frac{{{98000}}}{{{0.2}}}={49}\times{10}^{{{4}}}{J}\)

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

To Keep Reading This Answer, Download the App

4.6

Review from Google Play

To Keep Reading This Answer, Download the App

4.6

Review from Google Play

Correct60

Incorrect0

Still Have Question?

Load More

More Solution Recommended For You