Правка разделов «Statement», «Solution», «Answer»
en/13.4.24.md
+4 −4
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $13.4.24.$ [Insert the problem statement] | ||
| + | $13.4.24.$ In a smoke screen of opaque particles with a radius of $5$ microns, with a content of $0.004$ g of substance in 1 m$^3$ of air, visibility is $50$ m. How much substance in 1 m$^3$ of air is sprayed by a curtain source that creates particles with a radius of $10$ microns, if visibility is reduced to $20$ m? | ||
| ### Solution | |||
| Since each thin layer of air with uniform density of polluting particles obstructs the same fraction of light travelling perpendicular to it, the intensity of light decays exponentially over distance, i.e. $I=I_0e^{-x/L}$, where $L$ is a characteristic distance depending on the polluting particles. Over a short distance $x$, we have $I\approx I_0(1-x/L)$, so $x/L$ is the fraction of area that light is obstructed, i.e. | |||
| \[\frac{x}{L}=\frac{\pi r^2\rho x}{m},\] | |||
| @@ -10,10 +10,10 @@Solution | |||
| where $\rho$ is the density of the pollution and $m$ and $r$ are the mass and the radius of each particle. Thus, in order for different pollutions to cause the same reduction of light intensity over different distances, we must have | |||
| − | \[\frac{r_1^2\rho_1x1}{m_1}=\frac{r_2^2\rho_2x_2}{m_2}\] | ||
| + | \[\frac{r_1^2\rho_1x_1}{m_1}=\frac{r_2^2\rho_2x_2}{m_2}\] | ||
| − | If a larger particle has twice the radius of a smaller particle ($r_2=2r_1$) with the same composition, then its mass is eightfold ($m_2=8m_1$). Using $x_1=50$ m and $x_2=20$ m, we have $\rho_2=5\rho_1=5 | ||
| + | If a larger particle has twice the radius of a smaller particle ($r_2=2r_1$) with the same composition, then its mass is eightfold ($m_2=8m_1$). Using $x_1=50$ m and $x_2=20$ m, we have $\rho_2=5\rho_1=5\times(0.004$ g/m$^3)=0.02$ g/m$^3$. | ||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $0.02$ g/m$^3$ | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $13.4.24.$ [Insert the problem statement] | $13.4.24.$ In a smoke screen of opaque particles with a radius of $5$ microns, with a content of $0.004$ g of substance in 1 m$^3$ of air, visibility is $50$ m. How much substance in 1 m$^3$ of air is sprayed by a curtain source that creates particles with a radius of $10$ microns, if visibility is reduced to $20$ m? | ||
| ### Solution | ### Solution | ||
| Since each thin layer of air with uniform density of polluting particles obstructs the same fraction of light travelling perpendicular to it, the intensity of light decays exponentially over distance, i.e. $I=I_0e^{-x/L}$, where $L$ is a characteristic distance depending on the polluting particles. Over a short distance $x$, we have $I\approx I_0(1-x/L)$, so $x/L$ is the fraction of area that light is obstructed, i.e. | Since each thin layer of air with uniform density of polluting particles obstructs the same fraction of light travelling perpendicular to it, the intensity of light decays exponentially over distance, i.e. $I=I_0e^{-x/L}$, where $L$ is a characteristic distance depending on the polluting particles. Over a short distance $x$, we have $I\approx I_0(1-x/L)$, so $x/L$ is the fraction of area that light is obstructed, i.e. | ||
| \[\frac{x}{L}=\frac{\pi r^2\rho x}{m},\] | \[\frac{x}{L}=\frac{\pi r^2\rho x}{m},\] | ||
| @@ -10,10 +10,10 @@Solution | |||
| where $\rho$ is the density of the pollution and $m$ and $r$ are the mass and the radius of each particle. Thus, in order for different pollutions to cause the same reduction of light intensity over different distances, we must have | where $\rho$ is the density of the pollution and $m$ and $r$ are the mass and the radius of each particle. Thus, in order for different pollutions to cause the same reduction of light intensity over different distances, we must have | ||
| \[\frac{r_1^2\rho_1x1}{m_1}=\frac{r_2^2\rho_2x_2}{m_2}\] | \[\frac{r_1^2\rho_1x_1}{m_1}=\frac{r_2^2\rho_2x_2}{m_2}\] | ||
| If a larger particle has twice the radius of a smaller particle ($r_2=2r_1$) with the same composition, then its mass is eightfold ($m_2=8m_1$). Using $x_1=50$ m and $x_2=20$ m, we have $\rho_2=5\rho_1=5 |
If a larger particle has twice the radius of a smaller particle ($r_2=2r_1$) with the same composition, then its mass is eightfold ($m_2=8m_1$). Using $x_1=50$ m and $x_2=20$ m, we have $\rho_2=5\rho_1=5\times(0.004$ g/m$^3)=0.02$ g/m$^3$. | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $0.02$ g/m$^3$ | ||