Question

A person on a diet might lose ${}^{\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{kg}\mathbf{}\mathbf{per}\mathbf{}\mathbf{week}}$. Express the mass loss rate in milligrams per second, as if the dieter could sense the second-by-second loss.

Short answer

The mass-loss rate is ${}^{\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$.

Step-by-step solution

Given data

The given rate of loss of mass is${}^{\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{kg}\mathbf{/}\mathbf{week}}$.

Understanding the unit conversion

Unit conversion is the conversion between different units of measure for the same quantity, usually through multiplier conversion factors.

Determination of mass-loss rate in required units

Using the conversion of mass (Kg) in milligram and time (week) into seconds, find the mass-loss rate.

It is known that,

$$\begin{gathered} \mathbf{1}\mathbf{}\mathbf{kg}\mathbf{}\mathbf{=}{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{gm} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{mg} \end{gathered}$$

Therefore, the rate of loss in milligrams per week is ${}^{\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}}\mathbf{}\mathbf{mg}\mathbf{/}\mathbf{week}$.

The number of seconds per week is,

$$\begin{gathered} \mathbf{Number}\mathbf{}\mathbf{seconds}\mathbf{}\mathbf{per}\mathbf{}\mathbf{week}\mathbf{}\mathbf{=}\left(\frac{7\mathrm{day}}{\mathrm{wek}}\right)\left(\frac{24\mathrm{hr}}{\mathrm{day}}\right)\left(\frac{3600s}{\mathrm{hr}}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{604}\mathbf{,}\mathbf{800}\mathbf{}\mathbf{s}\mathbf{/}\mathbf{week} \end{gathered}$$

So, the rate of loss of mass is,

$$\begin{gathered} \mathbf{Mass}\mathbf{}\mathbf{loss}\mathbf{}\mathbf{rate}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{mg}}{\mathbf{604800}\mathbf{}\mathbf{s}}\mathbf{}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{mg}\mathbf{/}\mathbf{s} \end{gathered}$$

Thus, the rate of loss of mass is${}^{\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{mg}\mathbf{/}\mathbf{s}}$ .