Chapter 5.4, Problem 35IP

To determine

To find: The order of group of rates from least to greatest.

Answer to Problem 35IP

The order of rates from least to greatestare $\text{18}\text{lb/min}<\text{500}\text{kg/z}<\text{5}\text{oz/s}$ .

Explanation of Solution

Given information:The given differentrates are $\text{500}\text{kg/h}$ , $\text{5}\text{oz/s}$ and $\text{18}\text{lb/min}$ .

Calculation:

To compare different weight per time, convert the rates to the same basic unit.

To convert kilogram and pound to ounce, use $\text{1}\text{kg}=\text{35}\text{.274}\text{oz}$ and $\text{1}\text{lb}=\text{16}\text{oz}$ .

To convert hours and minutes to seconds, use $\text{1}\text{hr}=\text{3600}\text{s}$ and $\text{1}\text{min}=\text{60}\text{s}$ .

The rate from kg per hour to oz per second can be converted as:

  $\begin{array}{c}\text{500}\text{kg/h}=\frac{\text{500}\text{kg}}{\text{1}\text{h}}\cdot \frac{\text{35}\text{.274}\text{oz}}{\text{1}\text{kg}}\cdot \frac{\text{1}\text{h}}{\text{3600}\text{s}}\\ =\text{4}\text{.9}\text{oz/s}\end{array}$

The rate from lb per minute to oz per second can be converted as:

  $\begin{array}{c}\text{18}\text{lb/min}=\frac{\text{18}\text{lb}}{\text{1}\text{min}}\cdot \frac{\text{16}\text{oz}}{\text{1}\text{lb}}\cdot \frac{\text{1}\text{min}}{\text{60}\text{s}}\\ =\text{4}\text{.8}\text{oz/s}\end{array}$

Compare the calculated value with gram per minute.

  $\text{4}\text{.8}\text{oz/s}<\text{4}\text{.9}\text{oz/s}<\text{5}\text{oz/s}$

Therefore, the order of rates from least to greatestis $\text{18}\text{lb/min}<\text{500}\text{kg/z}<\text{5}\text{oz/s}$ .