Chapter 1, Problem 5PQ

(a)

To determine

Represent $0.53\text{cm}$ in SI units.

Answer to Problem 5PQ

In SI units, $0.53\text{cm}$ is $5.3\times {10}^{-3}\text{m}$.

Explanation of Solution

The SI unit for length is $\text{m}$. Here, $\text{m}$ is meters and $1\text{cm}={10}^{-2}\text{m}$.

Conclusion:

Multiply the $0.53\text{cm}$ by $\frac{{10}^{-2}\text{m}}{1\text{cm}}$

  $\begin{array}{c}0.53\text{cm}=0.53\text{cm}\cdot \frac{{10}^{-2}\text{m}}{1\text{cm}}\\ =5.3\times {10}^{-3}\text{m}\end{array}$

Thus, $0.53\text{cm}$ in SI unit is $5.3\times {10}^{-3}\text{m}$.

(b)

To determine

Represent $128.92\text{g}$ in SI units.

Answer to Problem 5PQ

In SI units, $128.92\text{g}$ is $0.12892\text{kg}$.

Explanation of Solution

The SI unit for mass is $\text{kg}$. Here, $\text{kg}$ is kilograms and $1\text{g}={10}^{-3}\text{kg}$.

Conclusion:

Multiply the $128.92\text{g}$ by $\frac{{10}^{-3}\text{kg}}{1\text{g}}$

  $\begin{array}{c}128.92\text{g}=128.92\text{g}\cdot \frac{{10}^{-3}\text{kg}}{1\text{g}}\\ =0.12892\text{kg}\end{array}$

Thus, $128.92\text{g}$ in SI unit is $0.12892\text{kg}$.

(c)

To determine

Represent $35.7{\text{cm}}^3$ in SI units.

Answer to Problem 5PQ

In SI units, $35.7{\text{cm}}^3$ is $3.57\times {10}^{-5}{\text{m}}^3$.

Explanation of Solution

The SI unit for length is $\text{m}$. Here, $\text{m}$ is meters and $1{\text{cm}}^3={10}^{-6}{\text{m}}^3$.

Conclusion:

Multiply the $35.7{\text{cm}}^3$ by $\frac{{10}^{-6}{\text{m}}^3}{1{\text{cm}}^3}$

  $\begin{array}{c}35.7{\text{cm}}^3=35.7{\text{cm}}^3\cdot \frac{{10}^{-6}{\text{m}}^3}{1{\text{cm}}^3}\\ =3.57\times {10}^{-5}{\text{m}}^3\end{array}$

Thus, $35.7{\text{cm}}^3$ in SI unit is $3.57\times {10}^{-5}{\text{m}}^3$.

(d)

To determine

Represent $65.7{\text{g/cm}}^3$ in SI units.

Answer to Problem 5PQ

In SI units, $65.7{\text{g/cm}}^3$ is $6.57\times {10}^4{\text{kg/m}}^3$.

Explanation of Solution

The SI unit for length is $\text{m}$ and for mass is $\text{kg}$.

Conclusion:

Multiply $65.7{\text{g/cm}}^3$ by $\frac{{10}^{-3}\text{kg}}{1\text{g}}\cdot \frac{1{\text{cm}}^3}{{10}^{-6}{\text{m}}^3}$

  $\begin{array}{c}65.7{\text{g/cm}}^3=65.7\frac{\text{g}}{{\text{cm}}^{\text{3}}}\cdot \frac{{10}^{-3}\text{kg}}{1\text{g}}\cdot \frac{1{\text{cm}}^3}{{10}^{-6}{\text{m}}^3}\\ =6.57\times {10}^4{\text{kg/m}}^3\end{array}$

Thus, $65.7{\text{g/cm}}^3$ in SI unit is $6.57\times {10}^4{\text{kg/m}}^3$.