Правка разделов «Statement», «Solution», «Answer»
en/4.5.19.md
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| @@ -1,13 +1,32 @@ | |||
| ### Statement | |||
| − | $4.5.19.$ The outer radius of a soap bubble is | ||
| + | $4.5.19.$ The outer radius of a soap bubble is $R$, the thickness of its wall is $h$. Find the air pressure inside the bubble. The air pressure outside the bubble is $P_0$, the surface tension of water is $\sigma$. | ||
| ### Solution | |||
| − |  | ||
| + | The pressure inside the bubble wall is: | ||
| + | \[ | ||
| + | P_1 = P_0 + P_L = P_0 + \frac{2\sigma}{R} | ||
| + | \] | ||
| + | The pressure inside the bubble is: | ||
| + | \[ | ||
| + | P = P_1 + \frac{2\sigma}{R-h} | ||
| + | \] | ||
| + | Combining the two equations, we get: | ||
| + | \[ | ||
| + | P = P_0 + \frac{2\sigma}{R} + \frac{2\sigma}{R - h} | ||
| + | \] | ||
| + | |||
| + | \[ | ||
| + | P = P_0 + 2\sigma \left( \frac{1}{R} + \frac{1}{R - h} \right) | ||
| + | \] | ||
| + | |||
| #### Answer | |||
| − | P = P | ||
| + | \[ | ||
| + | P = P_0 + 2\sigma \left( \frac{1}{R} + \frac{1}{R - h} \right) | ||
| + | \] | ||
| @@ -1,13 +1,32 @@ | |||
| ### Statement | ### Statement | ||
| $4.5.19.$ The outer radius of a soap bubble is |
$4.5.19.$ The outer radius of a soap bubble is $R$, the thickness of its wall is $h$. Find the air pressure inside the bubble. The air pressure outside the bubble is $P_0$, the surface tension of water is $\sigma$. | ||
| ### Solution | ### Solution | ||
|  | ||
| The pressure inside the bubble wall is: | |||
| \[ | |||
| P_1 = P_0 + P_L = P_0 + \frac{2\sigma}{R} | |||
| \] | |||
| The pressure inside the bubble is: | |||
| \[ | |||
| P = P_1 + \frac{2\sigma}{R-h} | |||
| \] | |||
| Combining the two equations, we get: | |||
| \[ | |||
| P = P_0 + \frac{2\sigma}{R} + \frac{2\sigma}{R - h} | |||
| \] | |||
| \[ | |||
| P = P_0 + 2\sigma \left( \frac{1}{R} + \frac{1}{R - h} \right) | |||
| \] | |||
| #### Answer | #### Answer | ||
| P = P |
\[ | ||
| P = P_0 + 2\sigma \left( \frac{1}{R} + \frac{1}{R - h} \right) | |||
| \] | |||