# A small hole in the wing of a space shuttle requires a 12.8 cm^2 patch. What is the patch's area...

## Question:

A small hole in the wing of a space shuttle requires a {eq}12.8\ cm^2 {/eq} patch. What is the patch's area in square kilometers {eq}(km^2) {/eq}?

## Unit Conversion:

There are many ways to express a single quantity in physics. For example, let us choose length. Length can be measured in inches, yards, centimeters, kilometers, among many others. Now if we wish to express our lengths in order units, we convert the units. We use conversion factors like having 1000 meters in one kilometer or having 2.54 centimeters in one inch.

Given:

• {eq}\displaystyle A = 12.8\ cm^2 {/eq} is the area of the patch

In an area we have squared units. So when we do conversions, we convert both units in the squared term which essentially means two conversions per unit. So let us recall the following equalities:

{eq}\displaystyle 1\ m = 100\ cm {/eq}

{eq}\displaystyle 1\ km = 1000\ m {/eq}

So here when we do the conversions,

{eq}\displaystyle A = 12.8\ cm^2 \left(\frac{1\ m}{100\ cm} \right)^2 \left(\frac{1\ km}{1000\ m} \right)^2 {/eq}

We cancel like units here:

{eq}\displaystyle A = 12.8\ \require{cancel}\cancel{cm^2} \left(\frac{1\ \require{cancel}\cancel{m}}{100\ \require{cancel}\cancel{cm}} \right)^2 \left(\frac{1\ \require{cancel}\cancel{km}}{1000\ \require{cancel}\cancel{m}} \right)^2 {/eq}

We will thus get:

{eq}\displaystyle \boxed{A = 1.28\ \times\ 10^{-9}\ km^2} {/eq} 