Q43P - Fundamentals Of Physics

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.

Step-by-Step Solution

Answer

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

Step 1: Given data

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

Step 2: Understanding the unit conversion

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

Step 3: 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,

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

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 ${}^{\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{mg}\mathbf{/}\mathbf{s}}$ .