Chapter 2, Problem 1E

Convert the following to engineering notation:

1. (a) 0.045 W
2. (b) 2000 pJ
3. (c) 0.1 ns
4. (d) 39,212 as
5. (e) 3 Ω
6. (f) 18,000 m
7. (g) 2,500,000,000,000 bits
8. (h) 10¹⁵ atoms/cm³

To determine

Convert $\text{0}\text{.045 W}$ into engineering notation.

Answer to Problem 1E

The conversion of $\text{0}\text{.045 W}$ into engineering notation is $\text{45 mW}$.

Explanation of Solution

Given data:

The power given is $\text{0}\text{.045 W}$.

Calculation:

The conversion of $\text{0}\text{.045 W}$ in engineering notation is as follows.

$\begin{array}{c}\text{0}\text{.045 W}=0.045\times {10}^3\text{ mW }\left\{\because 1\text{ W}={\text{10}}^3\text{ mW}\right\}\\ =45\text{ mW}\end{array}$

Conclusion:

Thus, the conversion of $\text{0}\text{.045 W}$ into engineering notation is $\text{45 mW}$.

To determine

Convert $\text{2000 pJ}$ into engineering notation.

Answer to Problem 1E

The conversion of $\text{2000 pJ}$ into engineering notation is $2\text{ nJ}$.

Explanation of Solution

Given data:

The energy given is $\text{2000 pJ}$.

Calculation:

The conversion of $\text{2000 pJ}$ in engineering notation is as follows.

$\begin{array}{c}\text{2000 pJ}=2000\times {10}^{-3}\text{ nJ }\left\{\because 1\text{ pJ}={\text{10}}^{-3}\text{ nJ}\right\}\\ =2\text{ nJ}\end{array}$

Conclusion:

Thus, the conversion of $\text{2000 pJ}$ into engineering notation is $2\text{ nJ}$.

To determine

Convert $0.1\text{ ns}$ into engineering notation.

Answer to Problem 1E

The conversion of $0.1\text{ ns}$ into engineering notation is $100\text{ ps}$.

Explanation of Solution

Given data:

The time given is $0.1\text{ ns}$.

Calculation:

The conversion of $0.1\text{ ns}$ in engineering notation is as follows.

$\begin{array}{c}0.1\text{ ns}=0.1\times {10}^3\text{ ps }\left\{\because 1\text{ ns}={\text{10}}^3\text{ ps}\right\}\\ =100\text{ ps}\end{array}$

Conclusion:

Thus, the conversion of $0.1\text{ ns}$ into engineering notation is $100\text{ ps}$.

To determine

Convert $\text{39212 as}$ into engineering notation.

Answer to Problem 1E

The conversion of $\text{39212 as}$ into engineering notation is $39.212\text{ fs}$.

Explanation of Solution

Given data:

The time given is $\text{39212 as}$.

Calculation:

The conversion of $\text{39212 as}$ in engineering notation is as follows.

$\begin{array}{c}\text{39212 as}=\text{39212}\times {10}^{-3}\text{ fs }\left\{\because 1\text{ as}={\text{10}}^{-3}\text{ fs}\right\}\\ =39.212\text{ fs}\end{array}$

Conclusion:

Thus, the conversion of $\text{39212 as}$ into engineering notation is $39.212\text{ fs}$.

To determine

Convert $3\Omega$ into engineering notation.

Answer to Problem 1E

The conversion of $3\Omega$ into engineering notation is $3\Omega$.

Explanation of Solution

Given data:

The resistance given is $3\Omega$.

The given data is already in engineering notation.

Conclusion:

Thus, the conversion of $3\Omega$ into engineering notation is $3\Omega$.

To determine

Convert $18000\text{ m}$ into engineering notation.

Answer to Problem 1E

The conversion of $18000\text{ m}$ into engineering notation is $18\text{ km}$.

Explanation of Solution

Given data:

The distance given is $18000\text{ m}$.

Calculation:

The conversion of $18000\text{ m}$ in engineering notation is as follows.

$\begin{array}{c}18000\text{ m}=18000\times {10}^{-3}\text{ km }\left\{\because 1\text{ m}={\text{10}}^{-3}\text{ km}\right\}\\ =18\text{ km}\end{array}$

Conclusion:

Thus, the conversion of $18000\text{ m}$ into engineering notation is $18\text{ km}$.

To determine

Convert $2,500,000,000,000\text{ bits}$ into engineering notation.

Answer to Problem 1E

The conversion of $2,500,000,000,000\text{ bits}$ into engineering notation is $2.5\text{ terabits}$.

Explanation of Solution

Given data:

The memory given is $2,500,000,000,000\text{ bits}$.

Calculation:

The conversion of $2,500,000,000,000\text{ bits}$ in engineering notation is as follows.

$\begin{array}{c}2,500,000,000,000\text{ bits}=\left(\begin{array}{l}2,500,000,000,000\\ \times {10}^{-12}\text{ terabits}\end{array}\right)\left\{\because 1\text{ bit}={\text{10}}^{-12}\text{ terabits}\right\}\\ =2.5\text{ terabits}\end{array}$

Conclusion:

Thus, the conversion of $2,500,000,000,000\text{ bits}$ into engineering notation is $2.5\text{ terabits}$.

To determine

Convert ${10}^{15}\frac{\text{atoms}}{{\text{cm}}^3}$ into engineering notation.

Answer to Problem 1E

The conversion of ${10}^{15}\frac{\text{atoms}}{{\text{cm}}^3}$ into engineering notation is ${10}^{21}\frac{\text{atoms}}{{\text{ m}}^3}$.

Explanation of Solution

Given data:

The volume density of atom given is ${10}^{15}\frac{\text{atoms}}{{\text{cm}}^3}$.

Calculation:

The conversion of ${10}^{15}\frac{\text{atoms}}{{\text{cm}}^3}$ in engineering notation is as follows.

$\begin{array}{c}{10}^{15}\frac{\text{atoms}}{{\text{cm}}^3}={10}^{15}\frac{\text{atoms}}{{\text{10}}^{-6}{\text{ m}}^3}\left\{\because 1{\text{ cm}}^3={\text{10}}^{-6}{\text{ m}}^3\right\}\\ ={10}^{21}\frac{\text{atoms}}{{\text{ m}}^3}\end{array}$

Conclusion:

Thus, the conversion of ${10}^{15}\frac{\text{atoms}}{{\text{cm}}^3}$ into engineering notation is ${10}^{21}\frac{\text{atoms}}{{\text{ m}}^3}$.