Chapter 4.3, Problem 43IP

a.

To determine

To find: that how many times as intense is a rock concert as a normal conversation.

Answer to Problem 43IP

The rock concert as a normal conversation is ${10}^5$ times.

Explanation of Solution

Given information:

sound	decibels	Intensity
Rock concert	110	${10}^{11}$
Vacuum cleaner	80	${10}^8$
Normal conversation	60	${10}^6$
whispering	20	${10}^2$

Calculation : to find that how many times as intense is a rock concert as a normal conversation, compare the intensity of rock concert and a normal conversation.

Let the rock concert is x time as a normal conversation,

  $\begin{array}{l}\text{intensity of rock concert}=x·\text{intensity of normal conversation}\\ {\text{10}}^{11}=x·{10}^6\text{ [substitute]}\\ x=\frac{{10}^{11}}{{10}^6}\text{ [divide]}\\ x={10}^{11-6}\text{ [}{a}^m÷a^n=a^{m-n}]\\ x={10}^5\text{ [simplify the exponents]}\end{array}$

Thus, The rock concert as a normal conversation is ${10}^5$ times.

b.

To determine

To find: that how many times as intense is a vacuum cleaner as a person whispering.

Answer to Problem 43IP

The vacuum cleaner as a person whispering is ${10}^4$ times.

Explanation of Solution

Given information:

sound	decibels	Intensity
Rock concert	110	${10}^{11}$
Vacuum cleaner	80	${10}^8$
Normal conversation	60	${10}^6$
whispering	20	${10}^2$

Calculation : to find that how many times as intense is a vacuum cleaner as a person whispering, compare the intensity of vacuum cleaner and person whispering.

Let the vacuum cleaner is x time as a person whispering,

  $\begin{array}{l}\text{intensity of vacuum cleaner}=x·\text{intensity of person whispering}\\ {\text{10}}^8=x·{10}^2\text{ [substitute]}\\ x=\frac{{10}^8}{{10}^2}\text{ [divide]}\\ x={10}^{8-2}\text{ [}{a}^m÷a^n=a^{m-n}]\\ x={10}^4\text{ [simplify the exponents]}\end{array}$

Thus, The vacuum cleaner as a person whispering is ${10}^4$ times.