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+### Statement
+
+$3.6.24.$ [Insert the problem statement]
+
+### Solution
+
+Condition:
+
+The compressibility of mercury, water, and air is equal to $3 \cdot 10^{-5}$, $5 \cdot 10^{-5}$, and $0.71\,\mathrm{atm}^{-1}$, respectively, and their density is $13.6 \cdot 10^3$, $1 \cdot 10^3$, and $1.2\,\mathrm{kg/m}^3$, respectively.
+Determine the speed of sound in these media.
+
+
+Given:
+\[
+\beta_{\mathrm{Hg}}=3\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{Hg}}=13.6\cdot10^3\,\mathrm{kg/m^3}
+\]
+\[
+\beta_{\mathrm{H_2O}}=5\cdot10^{-5}\,\mathrm{atm}^{-1},\qquad \rho_{\mathrm{H_2O}}=1.0\cdot10^3\,\mathrm{kg/m^3}
+\]
+\[
+\beta_{\text{air}}=0.71\,\mathrm{atm}^{-1},\qquad \rho_{\text{air}}=1.2\,\mathrm{kg/m^3}
+\]
+
+Main formula:
+\[
+v=\sqrt{\frac{1}{\beta\rho}}
+\]
+where $\beta$ is the compressibility of the medium, and $\rho$ is the density of the medium.
+
+We convert the compressibility from $\mathrm{atm}^{-1}$ to $\mathrm{Pa}^{-1}$:
+\[
+1\,\mathrm{atm}=101325\,\mathrm{Pa} \implies 1\,\mathrm{atm}^{-1}=\frac{1}{101325}\,\mathrm{Pa}^{-1}
+\]
+
+1. Mercury
+
+\[
+\beta_{\mathrm{Hg}}=\frac{3\cdot10^{-5}}{101325} \approx 2.96\cdot10^{-10}\,\mathrm{Pa}^{-1}
+\]
+
+We use the formula:
+\[
+v_{\mathrm{Hg}}=\sqrt{\frac{1}{\beta_{\mathrm{Hg}}\rho_{\mathrm{Hg}}}}
+\]
+
+Substituting the values:
+\[
+v_{\mathrm{Hg}}=\sqrt{\frac{1}{(2.96\cdot10^{-10})(13.6\cdot10^3)}}
+\]
+
+We calculate the denominator:
+\[
+(2.96\cdot10^{-10})(13.6\cdot10^3) \approx 4.0256\cdot10^{-6}
+\]
+
+Therefore:
+\[
+v_{\mathrm{Hg}}=\sqrt{\frac{1}{4.0256\cdot10^{-6}}} \approx 498\,\mathrm{m/s}
+\]
+
+Answer for mercury:
+\[
+v_{\mathrm{Hg}}\approx 500\,\mathrm{m/s}
+\]
+
+2. Water ($\mathrm{H_2O}$)
+
+\[
+\beta_{\mathrm{H_2O}}=\frac{5\cdot10^{-5}}{101325} \approx 4.93\cdot10^{-10}\,\mathrm{Pa}^{-1}
+\]
+
+We use the formula:
+\[
+v_{\mathrm{H_2O}}=\sqrt{\frac{1}{\beta_{\mathrm{H_2O}}\rho_{\mathrm{H_2O}}}}
+\]
+
+Substituting the values:
+\[
+v_{\mathrm{H_2O}}=\sqrt{\frac{1}{(4.93\cdot10^{-10})(1.0\cdot10^3)}}
+\]
+
+We calculate the denominator:
+\[
+(4.93\cdot10^{-10})(1.0\cdot10^3) = 4.93\cdot10^{-7}
+\]
+
+Therefore:
+\[
+v_{\mathrm{H_2O}}=\sqrt{\frac{1}{4.93\cdot10^{-7}}} \approx 1424\,\mathrm{m/s}
+\]
+
+Answer for water:
+\[
+v_{\mathrm{H_2O}}\approx 1420\,\mathrm{m/s}
+\]
+
+3. Air
+
+\[
+\beta_{\text{air}}=\frac{0.71}{101325} \approx 7.01\cdot10^{-6}\,\mathrm{Pa}^{-1}
+\]
+
+We use the formula:
+\[
+v_{\text{air}}=\sqrt{\frac{1}{\beta_{\text{air}}\rho_{\text{air}}}}
+\]
+
+Substituting the values:
+\[
+v_{\text{air}}=\sqrt{\frac{1}{(7.01\cdot10^{-6})(1.2)}}
+\]
+
+We calculate the denominator:
+\[
+(7.01\cdot10^{-6})(1.2) = 8.412\cdot10^{-6}
+\]
+
+Therefore:
+\[
+v_{\text{air}}=\sqrt{\frac{1}{8.412\cdot10^{-6}}} \approx 345\,\mathrm{m/s}
+\]
+
+Answer for air:
+\[
+v_{\text{air}}\approx 345\,\mathrm{m/s}
+\]
+
+Answer:
+\[
+v_{\mathrm{Hg}}\approx500\,\mathrm{m/s},\qquad v_{\mathrm{H_2O}}\approx1420\,\mathrm{m/s},\qquad v_{\text{air}}\approx345\,\mathrm{m/s}
+\]
+
+#### Answer
+
+[Insert a concise answer or boxed result]