# Question #a4647

Mar 4, 2016

(a) $1.0 \times {10}^{13} e r g$

#### Explanation:

Work done in SI system of units$= 1 \textrm{m e g a j o \underline{e}}$
Since mega$= {10}^{6}$, hence Work done $1.0 \times {10}^{6} J$

We need to convert Joule to, CGS system of units where erg is the unit of work, dyne is the unit of Force and centimetre is the unit of distance.
Now $1 J = 1 N \times 1 m = {10}^{5} \mathrm{dy} n e \textcolor{red}{\diamond} \times 100 c m = {10}^{7} e r g$
$\textcolor{red}{\diamond}$ Force$= m a$
$1 N = 1 k g \times 1 m {s}^{-} 2 = 1000 g m \times 100 c m {s}^{-} 2 = {10}^{5} \mathrm{dy} n e$

In the given problem
$1.0 \times {10}^{6} J = 1.0 \times {10}^{6} \times {10}^{7}$
$= 1.0 \times {10}^{13} e r g$