Новое решение
en/4.2.1.md
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| + | ### Statement | ||
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| + | $4.2.1.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | ### Statement | ||
| + | |||
| + | $4.2.1.$ Determine the liquid pressure on the bottom surface of a floating puck of cross-section $S$ and mass $m$, if the atmospheric pressure is $P_0$. | ||
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| + | ### Solution | ||
| + | |||
| + | The Archimedes force is equal to the force of gravity: | ||
| + | |||
| + | $$F_A = mg$$ | ||
| + | |||
| + | The Archimedes force arises from the difference in pressure from different sides on the same area, therefore: | ||
| + | |||
| + | $$(P - P_0)S = mg$$ | ||
| + | |||
| + | We get the answer: | ||
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| + | $$P = \frac{mg}{S} + P_0$$ | ||
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| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $4.2.1.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $4.2.1.$ Determine the liquid pressure on the bottom surface of a floating puck of cross-section $S$ and mass $m$, if the atmospheric pressure is $P_0$. | |||
| ### Solution | |||
| The Archimedes force is equal to the force of gravity: | |||
| $$F_A = mg$$ | |||
| The Archimedes force arises from the difference in pressure from different sides on the same area, therefore: | |||
| $$(P - P_0)S = mg$$ | |||
| We get the answer: | |||
| $$P = \frac{mg}{S} + P_0$$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||