91影视

The dimensions of the base area are $14.0\mathrm{cm}脳17.0\mathrm{cm}$

The volume of each candy, $\mathrm{v}=0.50{\mathrm{mm}}^3$.

The mass of each candy, $\mathrm{m}=0.0200\mathrm{g}$.

The density of a material is given by mass per unit volume. The rate of increase in mass indicates the amount of increase in mass per unit time.

The expression for the density is given as:

$\mathrm{p}=\frac{\mathrm{m}}{\mathrm{v}}$ 鈥� (i)

Here, $\mathrm{p}$is the density, m is the mass and V is the volume.

Using equation (i), the density is calculated as:

$$\begin{gathered} \mathrm{p}=\frac{\mathrm{m}}{\mathrm{v}} \\ =\frac{0.0200\mathrm{g}}{50.0{\mathrm{mm}}^3} \\ =4脳{10}^{鈥�4}\frac{\mathrm{g}}{{\mathrm{mm}}^3} \end{gathered}$$

Now, convert the density into $\mathrm{kg}/{\mathrm{cm}}^3$

$$\begin{gathered} \mathrm{p}=4脳{10}^{鈥�4}\frac{\mathrm{g}}{{\mathrm{mm}}^3}\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)鈥�\left(\frac{1000{\mathrm{mm}}^3}{1{\mathrm{cm}}^3}\right) \\ =4脳{10}^{鈥�4}\frac{\mathrm{kg}}{{\mathrm{cm}}^3} \end{gathered}$$

Thus, the density of candy is $4脳{10}^{鈥�4}\mathrm{kg}/{\mathrm{cm}}^3$

As the volume of the empty spaces between the candies is negligible, the mass of the candies in the container with the heighthwill be

$\mathrm{M}=\mathrm{p}脳\mathrm{A}脳\mathrm{h}$ 鈥� (ii)

The area A is calculated as:

$$\begin{gathered} \mathrm{A}=14.0\mathrm{cm}脳17.0\mathrm{cm} \\ =238{\mathrm{cm}}^3 \end{gathered}$$

With this, the rate of change of mass will be given by differentiating equation (ii) with respect to time. This gives,

$$\begin{gathered} \frac{\mathrm{dM}}{\mathrm{dt}}=\frac{\mathrm{g}\left(\mathrm{p}脳\mathrm{A}脳\mathrm{h}\right)}{\mathrm{dt}} \\ =\mathrm{p}脳\mathrm{A}脳\frac{\mathrm{dh}}{\mathrm{dt}} \end{gathered}$$

Substitute the values in the above equation.

$$\begin{gathered} \frac{\mathrm{dM}}{\mathrm{dt}}=4脳{10}^{鈥�4}\frac{\mathrm{kg}}{{\mathrm{cm}}^3}脳238{\mathrm{cm}}^2脳0.250\frac{\mathrm{cm}}{\mathrm{s}} \\ =0.0238\frac{\mathrm{kg}}{\mathrm{s}} \end{gathered}$$

Convert this rate from kg/s to kg/min.

$$\begin{gathered} \frac{\mathrm{dM}}{\mathrm{dt}}=0.0238\frac{\mathrm{kg}}{\mathrm{s}}鈥�\left(\frac{1\mathrm{s}}{0.01667\min }\right) \\ =1.43\frac{\mathrm{kg}}{\min } \end{gathered}$$

Thus, the rate of increase in mass is $1.43\mathrm{kg}/\min$