Новое решение
en/3.2.37.md
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| + | ### Statement | ||
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| + | $3.2.37.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
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| + | When the ship is displaced by the distance $x$ deeper into the water, the extra buoyancy force $\rho_wgSx$ serves as the restoring force. Consequently, $\omega^2=\rho_wgS/m$, and | ||
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| + | \[m=\frac{\rho_wgS}{\omega^2}=\frac{\rho_wgST^2}{4\pi^2}.\] | ||
| + | |||
| + | Thus, | ||
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| + | \[\Delta m\approx\frac{\rho_wgS\cdot2T\Delta T}{4\pi^2}.\] | ||
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| + | Using $T=7$ s and $\Delta T=0.5$ s, we find that the mass of the additional cargo is approximately $900$ tons. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $3.2.37.$ [Insert the problem statement] | |||
| ### Solution | |||
| When the ship is displaced by the distance $x$ deeper into the water, the extra buoyancy force $\rho_wgSx$ serves as the restoring force. Consequently, $\omega^2=\rho_wgS/m$, and | |||
| \[m=\frac{\rho_wgS}{\omega^2}=\frac{\rho_wgST^2}{4\pi^2}.\] | |||
| Thus, | |||
| \[\Delta m\approx\frac{\rho_wgS\cdot2T\Delta T}{4\pi^2}.\] | |||
| Using $T=7$ s and $\Delta T=0.5$ s, we find that the mass of the additional cargo is approximately $900$ tons. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||