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en/1.2.11.md
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| + | <meta name="author" content="Aliaksandr Melnichenka"> | ||
| + | <meta name="date" content="2023-10" scheme="YYYY-MM"> | ||
| + | <meta property="og:title" content="a. In a conical vessel, the water level rises at a constant rate v_0. How does the rate of water entering a vessel through a tube of section s depend on time? At time zero, the vessel is empty.b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness h. How does the speed of movement of the spot boundary depend on time, if the volume of oil q enters per unit of time? At the initial time, the spot radius is zero."> | ||
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| + | <title>a. In a conical vessel, the water level rises at a constant rate v_0. How does the rate of water entering a vessel through a tube of section s depend on time? At time zero, the vessel is empty.b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness h. How does the speed of movement of the spot boundary depend on time, if the volume of oil q enters per unit of time? At the initial time, the spot radius is zero.</title> | ||
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| + | <body style=""> | ||
| + | <header style="text-align:center;"> | ||
| + | <h2>Solutions of Savchenko Problems in Physics</h2> | ||
| + | <p class="author"> | ||
| + | Aliaksandr Melnichenka <br/> | ||
| + | October 2023 | ||
| + | </p> | ||
| + | </header> | ||
| + | |||
| + | <h3 id="back-link"><a href="../#1.2">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $1.2.11.$ a. In a conical vessel, the water level rises at a constant rate $v_0$. How does the rate of water entering a vessel through a tube of section $s$ depend on time? At time zero, the vessel is empty. | ||
| + | </p> | ||
| + | |||
| + | <p> | ||
| + | b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness $h$. How does the speed of movement of the spot boundary depend on time, if the volume of oil $q$ enters per unit of time? At the initial time, the spot radius is zero. | ||
| + | |||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="statement.png" | ||
| + | loading="lazy" width="160" /> | ||
| + | <figcaption> | ||
| + | For problem $1.2.11$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | |||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="drawing1.png" | ||
| + | loading="lazy" width="200" /> | ||
| + | <figcaption> | ||
| + | Cone vessel | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | |||
| + | $a)$ By time $t$, the water level will be $v_0t$. And the rate of change of volume will be equal to: | ||
| + | $$\frac{dV}{dt} = \frac{\pi r(t)^2 dx}{dt}$$ | ||
| + | Where $dx$ is the change in water level: | ||
| + | |||
| + | $$\frac{dV}{dt} = \pi r(t)^2 v_0$$ | ||
| + | |||
| + | From Geometry, | ||
| + | $$r(t) = v_0 t \cdot tg(\alpha)$$ | ||
| + | |||
| + | By definition, the velocity of incoming water is equal to | ||
| + | $$v = \frac{dV}{sdt}$$ | ||
| + | |||
| + | Substituting the previous expressions: | ||
| + | $$v = {\pi v_0^3 t^2 \cdot tg^2(\alpha)}/s$$ | ||
| + | |||
| + | $b)$ For a small time interval $dt$, the volume changes by | ||
| + | $$dV = q dt$$ | ||
| + | |||
| + | Also the volume increment can be written as | ||
| + | $$dV = 2\pi r dr \cdot h $$ | ||
| + | |||
| + | Thus: | ||
| + | |||
| + | $$ q dt = 2\pi r dr \cdot h$$ | ||
| + | |||
| + | Considering $v = \frac{dr}{dt}$, | ||
| + | $$ \fbox{$v = \frac{q}{2\pi r h}$}$$ | ||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$v=\frac{\pi v_0^3t^2\operatorname{tg}^2\alpha}{s}$$ | ||
| + | $$v=\frac{1}{2}\sqrt{\frac{q}{\pi ht}}$$ | ||
| + | </p> | ||
| + | |||
| + | |||
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| + | <p> | ||
| + | <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | ||
| + | </p> | ||
| + | <p> | ||
| + | <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small> | ||
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| <meta charset="utf-8"> | |||
| <meta name="viewport" content="width=device-width, initial-scale=1.0"> | |||
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| <meta name="description" content="a. In a conical vessel, the water level rises at a constant rate v_0. How does the rate of water entering a vessel through a tube of section s depend on time? At time zero, the vessel is empty.b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness h. How does the speed of movement of the spot boundary depend on time, if the volume of oil q enters per unit of time? At the initial time, the spot radius is zero."> | |||
| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
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| </head> | |||
| <body style=""> | |||
| <header style="text-align:center;"> | |||
| <h2>Solutions of Savchenko Problems in Physics</h2> | |||
| <p class="author"> | |||
| Aliaksandr Melnichenka <br/> | |||
| October 2023 | |||
| </p> | |||
| </header> | |||
| <h3 id="back-link"><a href="../#1.2">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $1.2.11.$ a. In a conical vessel, the water level rises at a constant rate $v_0$. How does the rate of water entering a vessel through a tube of section $s$ depend on time? At time zero, the vessel is empty. | |||
| </p> | |||
| <p> | |||
| b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness $h$. How does the speed of movement of the spot boundary depend on time, if the volume of oil $q$ enters per unit of time? At the initial time, the spot radius is zero. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="statement.png" | |||
| loading="lazy" width="160" /> | |||
| <figcaption> | |||
| For problem $1.2.11$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| <center> | |||
| <figure> | |||
| <img src="drawing1.png" | |||
| loading="lazy" width="200" /> | |||
| <figcaption> | |||
| Cone vessel | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| $a)$ By time $t$, the water level will be $v_0t$. And the rate of change of volume will be equal to: | |||
| $$\frac{dV}{dt} = \frac{\pi r(t)^2 dx}{dt}$$ | |||
| Where $dx$ is the change in water level: | |||
| $$\frac{dV}{dt} = \pi r(t)^2 v_0$$ | |||
| From Geometry, | |||
| $$r(t) = v_0 t \cdot tg(\alpha)$$ | |||
| By definition, the velocity of incoming water is equal to | |||
| $$v = \frac{dV}{sdt}$$ | |||
| Substituting the previous expressions: | |||
| $$v = {\pi v_0^3 t^2 \cdot tg^2(\alpha)}/s$$ | |||
| $b)$ For a small time interval $dt$, the volume changes by | |||
| $$dV = q dt$$ | |||
| Also the volume increment can be written as | |||
| $$dV = 2\pi r dr \cdot h $$ | |||
| Thus: | |||
| $$ q dt = 2\pi r dr \cdot h$$ | |||
| Considering $v = \frac{dr}{dt}$, | |||
| $$ \fbox{$v = \frac{q}{2\pi r h}$}$$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$v=\frac{\pi v_0^3t^2\operatorname{tg}^2\alpha}{s}$$ | |||
| $$v=\frac{1}{2}\sqrt{\frac{q}{\pi ht}}$$ | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
| </html> | |||