Question 1.30 - Chemistry: Molecular Nature Of Matter And Change

Question

A small hole in the wing of a space shuttle requires a $\mathbf{20}\mathbf{.}\mathbf{7}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{2}}$ patch.

(a) What is the patch’s area in square kilometres ( $\mathit{k}{\mathit{m}}^{\mathbf{2}}$)?

(b) If the patching material costs NASA $\mathbf{\$}{\raisebox{1ex}{$\mathbf{3}\mathbf{.}\mathbf{25}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{i}\mathbf{n}\mathbf{c}\mathbf{h}$}\right.}^{\mathbf{2}}$ , what is the cost of the patch?

Step-by-Step Solution

Answer

Answer:

                

The cost of patching $3.208 i{n}^2$will be $\$ 10.426$ .

Conversion unit involved for converting cm 2 to km 2

Conversion units are used to convert the measurement to the same unit. In the given question, there is a need for conversion of cm² can be converted to km^2, which can be formulated as

$$\begin{gathered} 1 c{m}^2 = 1\times {10}^{-10} k{m}^2 \\ Conversion factor =\frac{1\times {10}^{-10} k{m}^2}{1 c{m}^2} \\ Length in k{m}^2 = Length in c{m}^2 \times Conversion Factor \\ Length in k{m}^2 = 20.7 \cancel{c{m}^2} \times \left(\frac{1\times {10}^{-10} k{m}^2}{1\cancel{ c{m}^2}}\right) \\ On further simplification, \\ length = 20.7 \times {10}^{-10} k{m}^2 \\ = 2.07 \times {10}^{-9} k{m}^2 \\ Thus the patch area in k{m}^2 will be 2.07\times {10}^{-9} k{m}^2 \end{gathered}$$

Conversion unit involved for converting cm 2 to inch 2 .

Since the given length is in $i{n}^2$ which can be converted to cm² as:

$$\begin{gathered} 1 c{m}^2 = 1.555 i{n}^2 \\ Conversion factor=\frac{1.555 i{n}^2}{1 c{m}^2} \\ Length in i{n}^2 =length in c{m}^2 \times conversion factor \\ Length in i{n}^2 =20.7 \cancel{c{m}^2} \times \left(\frac{1.555 i{n}^2}{1\cancel{ c{m}^2}}\right) \\ On further simplification, \\ Length =3.208 i{n}^2 \end{gathered}$$

Calculate the cost of patching material

The given cost of patching material is . Thus, the cost of patching material for  can be calculated as

$$\begin{gathered} Cost of patching per i{n}^2 =\frac{\$3.25}{i{n}^2} \\ Cost of patching 3.208 i{n}^2= 3.208 i\cancel{n^2}\times \frac{\$3.25}{\cancel{i{n}^2}} \\ =\$10.426 \\ \mathbf{Thus}\mathbf{,}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{cost}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{patching}\mathbf{ }\mathbf{3}\mathbf{.}\mathbf{208}\mathbf{ }{\mathbf{in}}^{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{be}\mathbf{ }\mathbf{\$}\mathbf{10}\mathbf{.}\mathbf{426}\mathbf{ }\mathbf{.} \end{gathered}$$