Chapter 2.1, Problem 53PS

Solution

a.

To find: Volume of a penny.

Volume of a penny is equal to $4.64\times {10}^{-7}\text{ cubic meters}$ .

Given:

A penny is a cylinder with a radius of about $9.53$ millimeters and a height of about $1.55$ millimeters

Calculation:

Let $V$ denote the volume of the penny.

Then,

  $\begin{array}{c}V=\pi \left(9.53\times {10}^{-3}\right)\left(1.55\times {10}^{-3}\right)\\ \approx (3.14\times 9.53\times 1.55)\times \left({10}^{-3}\times {10}^{-3}\right)\\ \approx 46.4\times {10}^{-3-3}\text{ (Using product of powers property) }\\ =46.4\times {10}^{-6}\\ =4.64\times {10}^{-7}\text{ cubic meters}\end{array}$

b.

To find: The volume of your classroom in cubic meters.

The volume of your classroom in cubic meters is equal to $60\text{cubic}\text{meters}$

Given:

A penny is a cylinder with a radius of about $9.53$ millimeters and a height of about $1.55$ millimeters

Calculation:

The dimension of my classroom is $5\times 3\times 4$

So, the volume of my classroom is

  $\begin{array}{c}{V}_C=5\times 3\times 4\\ =60\text{cubic}\text{meters}\end{array}$

c.

To find: Number of pennies required to fill your classroom.

Number of pennies required to fill your classroom is equal to $129310344$

Given:

A penny is a cylinder with a radius of about $9.53$ millimeters and a height of about $1.55$ millimeters

Calculation:

Number of pennies required is $=\frac{60}{4.64\times {10}^{-7}}$

  $\approx 129310344$

The above approximation is an under estimated of the actual number.