Chapter 4.2, Problem 41IP

a

To determine

To find : how many times as long is a virus than an atom

Answer to Problem 41IP

A virus is ${10}^3$ times longer than an atom

Explanation of Solution

Given information :

The table shows the average lengths of different objects

  

Calculation :

The length of virus is ${10}^{-7}$ and the length of atom is ${10}^{-10}$

  $\begin{array}{l}\frac{\text{the length of virus }}{\text{the length of atom}}\\ \frac{{10}^{-7}}{{10}^{-10}}\\ ={10}^{-7-\left(-10\right)}\\ ={10}^{-7+10}\\ ={10}^3\end{array}$

Hence, a virus is ${10}^3$ times longer than an atom.

b

To determine

To find : how many viruses would fit across a pinhead

Answer to Problem 41IP

  ${10}^4$ Viruses would fit across a pinhead

Explanation of Solution

Given information :

The table shows the average lengths of different objects

  

Calculation :

The length of a pinhead is ${10}^{-3}$ and the length of avirus is ${10}^{-7}$

The viruses would fit across a pinhead are

  $\begin{array}{l}\frac{\text{the length of pinhead }}{\text{the length of virus }}\\ \frac{{10}^{-3}}{{10}^{-7}}\\ ={10}^{-3-\left(-7\right)}\\ ={10}^{-3+7}\\ ={10}^4\end{array}$

Hence, a pinhead ${10}^4$ times is greater than a virus, so ${10}^4$ viruses would fit across a pinhead.

c

To determine

To find : how many time as long is a football field than a cell

Answer to Problem 41IP

A football field is ${10}^8$ times longer than a cell

Explanation of Solution

Given information :

The table shows the average lengths of different objects

  

Calculation :

  $\text{the length of a football feild }$ is ${10}^2$ meter that is ${10}^4$ cm.

  $\begin{array}{l}\frac{\text{the length of a football feild }}{\text{the length a of cell}}\\ \frac{{10}^4}{{10}^{-4}}\\ ={10}^{4-\left(-4\right)}\\ ={10}^{4+4}\\ ={10}^8\end{array}$

Hence, a football field is ${10}^8$ times longer than a cell.