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| Решение задачи: скорость звука в различных средах |
| Speed of sound in different media |
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| Дано: |
| Given: |
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| $$\beta_{\mathrm{рт}}=3\cdot10^{-5}\,\mathrm{атм}^{-1},\qquad |
| $$\beta_{\mathrm{Hg}}=3\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad |
| \rho_{\mathrm{рт}}=13.6\cdot10^3\,\mathrm{кг/м^3}$$ |
| \rho_{\mathrm{Hg}}=13.6\cdot10^3\,\mathrm{kg/m^3}$$ |
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| $$\beta_{\mathrm{вод}}=5\cdot10^{-5}\,\mathrm{атм}^{-1},\qquad |
| $$\beta_{\mathrm{water}}=5\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad |
| \rho_{\mathrm{вод}}=1.0\cdot10^3\,\mathrm{кг/м^3}$$ |
| \rho_{\mathrm{water}}=1.0\cdot10^3\,\mathrm{kg/m^3}$$ |
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| $$\beta_{\mathrm{возд}}=0.71\,\mathrm{атм}^{-1},\qquad |
| $$\beta_{\mathrm{air}}=0.71\,\mathrm{atm}^{-1},\qquad |
| \rho_{\mathrm{возд}}=1.2\,\mathrm{кг/м^3}$$ |
| \rho_{\mathrm{air}}=1.2\,\mathrm{kg/m^3}$$ |
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| Основная формула: |
| Main formula: |
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| $$v=\sqrt{\frac{1}{\beta\rho}}$$ |
| $$v=\sqrt{\frac{1}{\beta\rho}}$$ |
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| где $\beta$ — сжимаемость среды, а $\rho$ — плотность среды. |
| where $\beta$ is the compressibility of the medium, and $\rho$ is the density of the medium. |
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| Переведём сжимаемость из $\mathrm{атм}^{-1}$ в $\mathrm{Па}^{-1}$. |
| We convert the compressibility from $\mathrm{atm}^{-1}$ to $\mathrm{Pa}^{-1}$. |
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| $$1\,\mathrm{атм}=101325\,\mathrm{Па}$$ |
| $$1\,\mathrm{atm}=101325\,\mathrm{Pa}$$ |
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| Следовательно, |
| Therefore, |
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| $$1\,\mathrm{атм}^{-1}=\frac{1}{101325}\,\mathrm{Па}^{-1}$$ |
| $$1\,\mathrm{atm}^{-1}=\frac{1}{101325}\,\mathrm{Pa}^{-1}$$ |
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| 1. Ртуть |
| 1. Mercury |
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| $$\beta_{\mathrm{рт}}= |
| $$\beta_{\mathrm{Hg}}= |
| \frac{3\cdot10^{-5}}{101325} |
| \frac{3\cdot10^{-5}}{101325} |
| \approx2.96\cdot10^{-10}\,\mathrm{Па}^{-1}$$ |
| \approx2.96\cdot10^{-10}\,\mathrm{Pa}^{-1}$$ |
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| Используем формулу: |
| We use the formula: |
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| $$v_{\mathrm{рт}}= |
| $$v_{\mathrm{Hg}}= |
| \sqrt{\frac{1}{\beta_{\mathrm{рт}}\rho_{\mathrm{рт}}}}$$ |
| \sqrt{\frac{1}{\beta_{\mathrm{Hg}}\rho_{\mathrm{Hg}}}}$$ |
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| Подставляем значения: |
| Substituting the values: |
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| $$v_{\mathrm{рт}}= |
| $$v_{\mathrm{Hg}}= |
| \sqrt{\frac{1} |
| \sqrt{\frac{1} |
| {(2.96\cdot10^{-10})(13.6\cdot10^3)}}$$ |
| {(2.96\cdot10^{-10})(13.6\cdot10^3)}}$$ |
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| Сначала вычислим знаменатель: |
| First, we calculate the denominator: |
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| $$(2.96\cdot10^{-10})(13.6\cdot10^3) |
| $$(2.96\cdot10^{-10})(13.6\cdot10^3) |
| \approx4.0256\cdot10^{-6}$$ |
| \approx4.0256\cdot10^{-6}$$ |
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| Поэтому: |
| Therefore: |
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| $$v_{\mathrm{рт}}= |
| $$v_{\mathrm{Hg}}= |
| \sqrt{\frac{1}{4.0256\cdot10^{-6}}}$$ |
| \sqrt{\frac{1}{4.0256\cdot10^{-6}}}$$ |
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| $$v_{\mathrm{рт}}\approx498\,\mathrm{м/с}$$ |
| $$v_{\mathrm{Hg}}\approx498\,\mathrm{m/s}$$ |
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| Ответ для ртути: |
| Answer for mercury: |
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| $$\boxed{v_{\mathrm{рт}}\approx500\,\mathrm{м/с}}$$ |
| $$\boxed{v_{\mathrm{Hg}}\approx500\,\mathrm{m/s}}$$ |
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| 2. Вода |
| 2. Water |
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| $$\beta_{\mathrm{вод}}= |
| $$\beta_{\mathrm{water}}= |
| \frac{5\cdot10^{-5}}{101325} |
| \frac{5\cdot10^{-5}}{101325} |
| \approx4.93\cdot10^{-10}\,\mathrm{Па}^{-1}$$ |
| \approx4.93\cdot10^{-10}\,\mathrm{Pa}^{-1}$$ |
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| Используем формулу: |
| We use the formula: |
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| $$v_{\mathrm{вод}}= |
| $$v_{\mathrm{water}}= |
| \sqrt{\frac{1}{\beta_{\mathrm{вод}}\rho_{\mathrm{вод}}}}$$ |
| \sqrt{\frac{1}{\beta_{\mathrm{water}}\rho_{\mathrm{water}}}}$$ |
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| Подставляем значения: |
| Substituting the values: |
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| $$v_{\mathrm{вод}}= |
| $$v_{\mathrm{water}}= |
| \sqrt{\frac{1} |
| \sqrt{\frac{1} |
| {(4.93\cdot10^{-10})(1.0\cdot10^3)}}$$ |
| {(4.93\cdot10^{-10})(1.0\cdot10^3)}}$$ |
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| Вычисляем знаменатель: |
| We calculate the denominator: |
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| $$(4.93\cdot10^{-10})(1.0\cdot10^3) |
| $$(4.93\cdot10^{-10})(1.0\cdot10^3) |
| =4.93\cdot10^{-7}$$ |
| =4.93\cdot10^{-7}$$ |
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| Следовательно: |
| Therefore: |
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| $$v_{\mathrm{вод}}= |
| $$v_{\mathrm{water}}= |
| \sqrt{\frac{1}{4.93\cdot10^{-7}}}$$ |
| \sqrt{\frac{1}{4.93\cdot10^{-7}}}$$ |
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| $$v_{\mathrm{вод}}\approx1424\,\mathrm{м/с}$$ |
| $$v_{\mathrm{water}}\approx1424\,\mathrm{m/s}$$ |
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| Ответ для воды: |
| Answer for water: |
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| $$\boxed{v_{\mathrm{вод}}\approx1420\,\mathrm{м/с}}$$ |
| $$\boxed{v_{\mathrm{water}}\approx1420\,\mathrm{m/s}}$$ |
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| 3. Воздух |
| 3. Air |
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| $$\beta_{\mathrm{возд}}= |
| $$\beta_{\mathrm{air}}= |
| \frac{0.71}{101325} |
| \frac{0.71}{101325} |
| \approx7.01\cdot10^{-6}\,\mathrm{Па}^{-1}$$ |
| \approx7.01\cdot10^{-6}\,\mathrm{Pa}^{-1}$$ |
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| Используем формулу: |
| We use the formula: |
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| $$v_{\mathrm{возд}}= |
| $$v_{\mathrm{air}}= |
| \sqrt{\frac{1}{\beta_{\mathrm{возд}}\rho_{\mathrm{возд}}}}$$ |
| \sqrt{\frac{1}{\beta_{\mathrm{air}}\rho_{\mathrm{air}}}}$$ |
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| Подставляем значения: |
| Substituting the values: |
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| $$v_{\mathrm{возд}}= |
| $$v_{\mathrm{air}}= |
| \sqrt{\frac{1} |
| \sqrt{\frac{1} |
| {(7.01\cdot10^{-6})(1.2)}}$$ |
| {(7.01\cdot10^{-6})(1.2)}}$$ |
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| Вычисляем знаменатель: |
| We calculate the denominator: |
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| $$(7.01\cdot10^{-6})(1.2) |
| $$(7.01\cdot10^{-6})(1.2) |
| =8.412\cdot10^{-6}$$ |
| =8.412\cdot10^{-6}$$ |
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| Следовательно: |
| Therefore: |
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| $$v_{\mathrm{возд}}= |
| $$v_{\mathrm{air}}= |
| \sqrt{\frac{1}{8.412\cdot10^{-6}}}$$ |
| \sqrt{\frac{1}{8.412\cdot10^{-6}}}$$ |
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| $$v_{\mathrm{возд}}\approx345\,\mathrm{м/с}$$ |
| $$v_{\mathrm{air}}\approx345\,\mathrm{m/s}$$ |
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| Ответ для воздуха: |
| Answer for air: |
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| $$\boxed{v_{\mathrm{возд}}\approx345\,\mathrm{м/с}}$$ |
| $$\boxed{v_{\mathrm{air}}\approx345\,\mathrm{m/s}}$$ |
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| Итоговый ответ: |
| Final answer: |
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| $$\boxed{ |
| $$\boxed{ |
| v_{\mathrm{рт}}\approx500\,\mathrm{м/с},\qquad |
| v_{\mathrm{Hg}}\approx500\,\mathrm{m/s},\qquad |
| v_{\mathrm{вод}}\approx1420\,\mathrm{м/с},\qquad |
| v_{\mathrm{water}}\approx1420\,\mathrm{m/s},\qquad |
| v_{\mathrm{возд}}\approx345\,\mathrm{м/с} |
| v_{\mathrm{air}}\approx345\,\mathrm{m/s} |
| }$$ |
| }$$ |
