Q46P - Fundamentals Of Physics

Question

A unit of area often used in measuring land areas is the hectare, defined as${\mathbf{10}}^{\mathbf{4}}{\mathbf{m}}^{\mathbf{2}}$ . An open-pit coal mine consumes $\mathbf{75}$ hectares of land, down to a depth of $\mathbf{26}\mathbf{ }\mathbf{m}$ , each year. What volume of earth, in cubic kilometers, is removed in this time?

Step-by-Step Solution

Answer

The volume of earth removed is $0.020 {\mathrm{km}}^3$ .

Step 1: Given data

Area of land,$A=75 hectares$ 

Depth, $\mathrm{d}=26 \mathrm{m}$ 

Step 2: Understanding the concept of volume

The volume of the earth removed is equal to the product of area and depth. Use the conversion of hectares into square meters and square meters into square kilometers, to find the volume of earth removed.

The expression for volume is given as: 

$\mathbf{V}\mathbf{=}\mathbf{A}\mathbf{\times }\mathbf{d}$                                                             … (i)

Here, A is the area and d is the depth.

Step 3: Determination of the volume

The conversion factors are as follows:

$$\begin{gathered} 1 \mathrm{hectare} ={10}^4{\mathrm{m}}^2 \\ 1 {\mathrm{m}}^2={10}^{-6} {\mathrm{km}}^2 \end{gathered}$$

 

Using equation (i), the volume of earth removed is calculated as: 

 

$$\begin{gathered} \mathrm{V}=\mathrm{A}\times \mathrm{d} \\ =(75\times {10}^4 {\mathrm{m}}^2)\times \left(26\mathrm{m}\right) \\ =1.95\times {10}^7 {\mathrm{m}}^3 \end{gathered}$$

 

Convert the volume from  to .

 

$$\begin{gathered} \mathrm{V}=(1.95\times {10}^7 {\mathrm{m}}^3){\left(\frac{1 \mathrm{km}}{1000\mathrm{m}}\right)}^3 \\ \approx 0.020 {\mathrm{km}}^3 \end{gathered}$$

 

Thus, the volume of earth removed is $0.020 {\mathrm{km}}^3$.