African solar systems part 2: A problem on available energy

There are 1000 x 1000 = 1000000 $m^{2}$ per $km^{2}$ so 0.16 $km^{2}$ = 0.16 x 1000000 = 160000 $m^{2}.$

There are 1370 watts per $m^{2}$ so there are 1370 x 160000 = 219200000 watts in total.

Watts are a measurement of how many Joules hit a certain area per second so the final answer is 219200000 Joules.

The question wasn't that bad, it's typing in the right amount of 0s that's the tricky bit.

(1,000m)^2 = 1,000,000m^2

0.16 * 1,000,000m^2 * 1,370J/s/m^2 = 219,200,000J/s

$\Rightarrow 1370\frac { w }{ { m }^{ 2 } } =1370\frac { J }{ s } \frac { 1 }{ { m }^{ 2 } }$

$\Rightarrow \left( 1370\frac { w }{ { m }^{ 2 } } \right) \left( 0.16km^{2} \right) =\left( 1370\frac { w }{ { m }^{ 2 } } \right) (160000{ m }^{ 2 })=219200000\frac { J }{ s }=219200000\frac { J }{ s }$

I dont know ..i just simply use unit cancelling method

Energy per second will be = 0.16 * 10^6 * 1370.... = 219200000 J .... for converting sq kilometres to sq metres we need the term 10^6...