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| <meta name="author" content="Aliaksandr Melnichenka"> | | <meta name="author" content="Aliaksandr Melnichenka"> |
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| <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.38.$ The air resistance force acting on fog drops is proportional to the product of the radius and velocity: $f = \gamma rv$. Drops of radius $r = 0.1$ mm, falling from a great height, have a speed of about $1$ $\frac{m}{s}$ near the ground. What speed will drops have if their radius is half as large? ten times less? | | $2.1.38.$ The air resistance force acting on fog drops is proportional to the product of the radius and velocity: $f = \gamma rv$. Drops of radius $r = 0.1$ mm, falling from a great height, have a speed of about $1$ $\frac{m}{s}$ near the ground. What speed will drops have if their radius is half as large? ten times less? |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| <figure> | | <figure> |
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| loading="lazy" width="80" /> | | loading="lazy" width="80" /> |
| <figcaption> | | <figcaption> |
| Forces acting on a drop | | Forces acting on a drop |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <p> | | <p> |
| A falling drop is acted upon by two forces: the constant force of gravity, accelerating the drop's movement, and the force of air resistance, slowing its movement and increasing with the drop's speed. The force of air resistance increases until it becomes equal to the force of gravity. Then the speed stops changing, and the drop falls at a constant speed. | | A falling drop is acted upon by two forces: the constant force of gravity, accelerating the drop's movement, and the force of air resistance, slowing its movement and increasing with the drop's speed. The force of air resistance increases until it becomes equal to the force of gravity. Then the speed stops changing, and the drop falls at a constant speed. |
| </p> | | </p> |
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| <p> | | <p> |
| Let's write the equation after a long period of time: | | Let's write the equation after a long period of time: |
| $$ mg=γrv\;(1) $$ | | $$ mg=γrv\;(1) $$ |
| Let's find $m$ through the volume $V$: | | Let's find $m$ through the volume $V$: |
| $$ m=ρV=\frac{4}{3} ρ \pi r^3 $$ | | $$ m=ρV=\frac{4}{3} ρ \pi r^3 $$ |
| And we substitute into $(1)$: | | And we substitute into $(1)$: |
| $$ \frac{4}{3} ρ \pi r^3 g=γrv $$ | | $$ \frac{4}{3} ρ \pi r^3 g=γrv $$ |
| From here: | | From here: |
| $$ v = \frac{4}{3} \frac{ρ \pi g}{γ} \cdot r^2 =\alpha r^2\;(2) $$ | | $$ v = \frac{4}{3} \frac{ρ \pi g}{γ} \cdot r^2 =\alpha r^2\;(2) $$ |
| | | |
| $$ \alpha = \frac{4}{3} \frac{ρ \pi g}{γ} =\frac{v}{r^2}=10^8 \,\frac{1}{\text{m}\cdot\text{s}} $$ | | $$ \alpha = \frac{4}{3} \frac{ρ \pi g}{γ} =\frac{v}{r^2}=10^8 \,\frac{1}{\text{m}\cdot\text{s}} $$ |
| We substitute and find the answer | | We substitute and find the answer |
| $$ v(\frac{r}{2}) = \alpha \frac{r^2}{4}=0.25~\text{m/s} $$ | | $$ v(\frac{r}{2}) = \alpha \frac{r^2}{4}=0.25~\text{m/s} $$ |
| | | |
| $$ v(\frac{r}{10}) = \alpha \frac{r^2}{100}=0.01~\text{m/s} $$ | | $$ v(\frac{r}{10}) = \alpha \frac{r^2}{100}=0.01~\text{m/s} $$ |
| </p> | | </p> |
| </p> | | </p> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$v_1 ≈ 0.25 ~\text{m/s}$$ | | $$v_1 ≈ 0.25 ~\text{m/s}$$ |
| $$v_2 ≈ 0.01 ~\text{m/s}$$ | | $$v_2 ≈ 0.01 ~\text{m/s}$$ |
| </p> | | </p> |
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