Chapter 2, Problem 2.1P

Interpretation Introduction

(a)

Interpretation:

: To convert 2 wk to microseconds.

Concept introduction:

The international system of unit (SI units):

Day is represented by the symbol.	d
Week is represented by the symbol	wk
Hour is represented by the symbol	h
Minutes is represented by the symbol	min
Seconds is represented by the symbol	s
Microsecond is represented by the symbol	µs
Meter is represented by the symbol	m
Feet is represented by the symbol	ft
Kilometer is represented by the symbol	km
Kilogram is represented by the symbol	kg
Pound is represented by the symbol	lb

Answer to Problem 2.1P

In two weeks there are $1.2\times {10}^{12}\mu s$.

Explanation of Solution

In one week there are seven days and in one day there are twenty-four hours. In one hour there are sixty minutes and in one minute there are sixty seconds. In one second there are 10⁶ microseconds.

Convert week to microseconds as follows:

$\begin{array}{c}2\text{wk}=\left(2\text{wk}\right)\left(\frac{7\text{d}}{1\text{wk}}\right)\left(\frac{24\text{h}}{1\text{d}}\right)\left(\frac{60\min }{1\text{h}}\right)\left(\frac{60\text{s}}{1\min }\right)\left(\frac{{10}^6\text{μs}}{1\text{s}}\right)\\ =1.2\times {10}^{12}\text{μs}\end{array}$.

Interpretation Introduction

(b)

Interpretation:

To convert 38.1 ft/s to kilometers/h.

Concept introduction:

The international system of unit (SI units):

Day is represented by the symbol	d
Week is represented by the symbol	wk
Hour is represented by the symbol	h
Minutes is represented by the symbol	min
Seconds is represented by the symbol	s
Microsecond is represented by the symbol	µs
Meter is represented by the symbol	m
Feet is represented by the symbol	ft
Kilometer is represented by the symbol	km
Kilogram is represented by the symbol	kg
Pound is represented by the symbol	lb

Answer to Problem 2.1P

In $38.1\frac{\text{ft}}{\text{s}}$ there are $41.8\frac{\text{km}}{\text{h}}$.

Explanation of Solution

In one day there are twenty-four hours and in one hour there are sixty minutes.

In one pound there are 0.453593 kg and in one feet there are 0.3048 m. Convert meter per day into kilogram to feet per minute into pound as follows:

$\begin{array}{c}\text{554}\frac{{\text{m}}^{\text{4}}}{\text{day}\cdot \text{kg}}\text{=}\left(\text{554}\frac{{\text{m}}^{\text{4}}}{\text{day}\cdot \text{kg}}\right)\left(\frac{\text{1}\text{day}}{24\text{ h}}\right)\left(\frac{1\text{h}}{60\text{}\min }\right)\left(\frac{\text{0}\text{.453593}\text{kg}}{\text{1}{\text{lb}}_m}\right)\left(\frac{1{\text{ft}}^{\text{4}}}{{\left(0.3048\text{ m}\right)}^4}\right)\\ \text{=20}\text{.2}\frac{{\text{ft}}^{\text{4}}}{\text{min}\cdot {\text{lb}}_{\text{m}}}\end{array}$.

Interpretation Introduction

(c)

Interpretation:

To convert 554 m⁴/(day.kg) to ft⁴/(min.lb_m ).

Concept introduction:

The international system of unit (SI units):

Day is represented by the symbol	d
Week is represented by the symbol	wk
Hour is represented by the symbol	h
Minutes is represented by the symbol	min
Seconds is represented by the symbol	s
Microsecond is represented by the symbol	µs
Meter is represented by the symbol	m
Feet is represented by the symbol	ft
Kilometer is represented by the symbol	km
Kilogram is represented by the symbol	kg
Pound is represented by the symbol	lb

Answer to Problem 2.1P

In $\text{554}\frac{{\text{m}}^{\text{4}}}{\text{day}\cdot \text{kg}}$ there are $\text{20}\text{.2}\frac{{\text{ft}}^{\text{4}}}{\min \cdot {\text{lb}}_{\text{m}}}$.

Explanation of Solution

In one day there are twenty-four hours and in one hour there are sixty minutes.

In one pound there are 0.453593 kg and in one feet there are 0.3048 m. Convert meter per day into kilogram to feet per minute into pound as follows:

$\begin{array}{c}\text{554}\frac{{\text{m}}^{\text{4}}}{\text{day}\cdot \text{kg}}\text{=}\left(\text{554}\frac{{\text{m}}^{\text{4}}}{\text{day}\cdot \text{kg}}\right)\left(\frac{\text{1}\text{day}}{24\text{ h}}\right)\left(\frac{1\text{h}}{60\text{}\min }\right)\left(\frac{\text{0}\text{.453593}\text{kg}}{\text{1}{\text{lb}}_m}\right)\left(\frac{1{\text{ft}}^{\text{4}}}{{\left(0.3048m\right)}^4}\right)\\ \text{=20}\text{.2}\frac{{\text{ft}}^{\text{4}}}{\min \cdot {\text{lb}}_{\text{m}}}\end{array}$.