A heavy box of mass 20 kg is placed on a horizontal surface . If coefficient of kinetic friction between the box and the horizontal surface . Is 0.25 calculate the force of kinetic friction Also calculate acceleration produced under a force of 98 N applied horizontally ?
A heavy box of mass 20 kg is placed on a horizontal surface . If coefficient of kinetic friction between the box and the horizontal surface . Is 0.25 calculate the force of kinetic friction Also calculate acceleration produced under a force of 98 N applied horizontally ?
Answer
(a) The force on 7 th coin is due to weight of the three coins lying above it Therefore \({F}={\left({3}{m}\right)}{k}{g}{f}={\left({3}{m}{g}\right)}{N}\) where g is acceleration due to gravity . This force acts vertically downwards . (b) The eighth coin is alredy under th
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What volume of \({3}\) molar \({H}{N}{O}_{{{3}}}\) is needed to oxidise \({8}{g}\) of \({F}{e}^{{{3}+}}\), \({H}{N}{O}_{{{3}}}\) gets converted to \({N}{O}\) ? (A) \({8}{m}{L}\) (B) \({16}{m}{L}\) (C) \({32}{m}{L}\) (D) \({64}{m}{L}\)
One mole of \({N}_{{{2}}}{H}_{{{4}}}\) loses ten moles of electrons to form a new compound \({A}\). Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in \({A}\)? (There is no change in the oxidation state of hydrogen.)
A man throws the bricks to a height of 12 m where they reach with a speed of \({12}{m}/{s}\). If he throws the bricks such that they just reach that height, what percentage of energy will be saved? \({\left({g}={9.8}{m}/{s}^{{{2}}}\right)}\) (A) \({9.5}\%\) (B) \({19}\%\) (C) \({38}\%\) (D) \({76}\%\)