Dimensionally, Energy`=mass xx("velocity")^(2)`
`=massxx(("length")/"time")^(2)=ML^(2)T^(-2)`
Thus , `1 joule`=(1kg)(1m)^(2)(1s)^(-2)`
and `1 erg =(1g)(1cm)^(2)(1s)^(-2)`
`(1"joule")/("1 erg")=((1kg)/(1g))((1m)/(1cm))^(2)((1s)/(1s))^(-2)`
` =((1000g)/(1g))((1000cm)/(1cm))^(2)=1000xx10000=10^(7)`.
So `1 "joule"=10^(7)` erg.