Q9P - Fundamentals Of Physics

Question

Question: Antarctica is roughly semicircular, with a radius of 2000 km (Fig. 1-5). The average thickness of its ice cover is 3000 m. How many cubic centimeters of ice does Antarctica contain? (Ignore the curvature of Earth.)

Figure 1-5 Problem 9

Step-by-Step Solution

Answer

Answer

 

The volume of ice is $1.9\times {10}^{22}c{m}^3$

Step 1: Given data

The radius of a semicircular Antarctica is 2000km .

The average thickness of the ice cover is 3000m.

Step 2: Understanding the conversion factors

Unit conversion is the conversion between different units of measure for the same quantity, usually through multiplier conversion factors. Use the given conversion factors to convert the quantities from one unit to other.

 

The expression for the volume of semicircle is given as follows: 

   $V=\frac{\pi R^2}{2}t$                                                                       … (i)

 

Here, r is the radius and t is the thickness. 

Step 3: Conversion of radius and thickness into cm

Convert the radius and thickness into cm.

 $$\begin{gathered} {\text{ 1 km=10}}^3{\text{cm and 1 m = 10}}^2\text{m} \\ \text{ Radius of the semicircle is, } \\ R= 2000km\left(\frac{{10}^{3}m}{1 km}\right)\left(\frac{{10}^{2}cm}{1 km}\right) \\ =2000 \times {10}^{5}cm \\ \text{ Thickness of semicircle is, } \\ t=3000m\left(\frac{{10}^{2}cm}{1 m}\right) \\ =3000\times {10}^{2}cm \\ \text{ Thus, the radius is }2000 \times {10}^{5}cm\text{ and thickness is }3000\times {10}^{2}cm\text{.} \end{gathered}$$

 

Step 4: Determination of the volume

Using equation (i), the volume of the ice is calculated as follows: 

$$\begin{gathered} V=\frac{\pi R^2}{2}t \\ =\frac{\pi {\left(2000\times {10}^{5}cm\right)}^2}{2}\times 3000\times {10}^{2 }cm \\ =1.9\times {10}^{22}c{m}^3 \end{gathered}$$

 

Thus, the volume of ice is$1.9\times {10}^{22}c{m}^3$ .