Edit to “Statement”
en/1.3.12.md
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| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $1.3.12^*.$ From a hose lying on the ground, water shoots at an angle of $45^\circ$ to the horizon with an initial velocity of $10~\mathrm{\frac{m}{s}}$. The cross-sectional area of the hose opening is $5~\text{ | ||
| + | $1.3.12^*.$ From a hose lying on the ground, water shoots at an angle of $45^\circ$ to the horizon with an initial velocity of $10~\mathrm{\frac{m}{s}}$. The cross-sectional area of the hose opening is $5~\text{cm}^2$. Determine the mass of the jet in the air. | ||
| ### Solution | |||
| First, let's determine the flight time of each water drop. This time is: | |||
| $$ t = \frac{2v_0 \cdot \sin\alpha}{g} $$ | |||
| There is water in the air, flowing out of the hose in exactly time $t$. The mass of this water is easy to determine: it is the mass of water in a cylinder of height $v_0 t$ and base area $S$. | |||
| $$ m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot \sin\alpha}{g} $$ $$ m = \frac{2 \rho v_0^2 S \cdot \sin\alpha}{g} = 7.2 \text{ kg}$$ | |||
| #### Answer | |||
| $$x = \frac{2\nu^{2}}{g(\tan\beta +\tan\alpha )}$$ | |||
| unchanged lines 11 | |||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $1.3.12^*.$ From a hose lying on the ground, water shoots at an angle of $45^\circ$ to the horizon with an initial velocity of $10~\mathrm{\frac{m}{s}}$. The cross-sectional area of the hose opening is $5~\text{ |
$1.3.12^*.$ From a hose lying on the ground, water shoots at an angle of $45^\circ$ to the horizon with an initial velocity of $10~\mathrm{\frac{m}{s}}$. The cross-sectional area of the hose opening is $5~\text{cm}^2$. Determine the mass of the jet in the air. | ||
| ### Solution | ### Solution | ||
| First, let's determine the flight time of each water drop. This time is: | First, let's determine the flight time of each water drop. This time is: | ||
| $$ t = \frac{2v_0 \cdot \sin\alpha}{g} $$ | $$ t = \frac{2v_0 \cdot \sin\alpha}{g} $$ | ||
| There is water in the air, flowing out of the hose in exactly time $t$. The mass of this water is easy to determine: it is the mass of water in a cylinder of height $v_0 t$ and base area $S$. | There is water in the air, flowing out of the hose in exactly time $t$. The mass of this water is easy to determine: it is the mass of water in a cylinder of height $v_0 t$ and base area $S$. | ||
| $$ m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot \sin\alpha}{g} $$ $$ m = \frac{2 \rho v_0^2 S \cdot \sin\alpha}{g} = 7.2 \text{ kg}$$ | $$ m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot \sin\alpha}{g} $$ $$ m = \frac{2 \rho v_0^2 S \cdot \sin\alpha}{g} = 7.2 \text{ kg}$$ | ||
| #### Answer | #### Answer | ||
| $$x = \frac{2\nu^{2}}{g(\tan\beta +\tan\alpha )}$$ | $$x = \frac{2\nu^{2}}{g(\tan\beta +\tan\alpha )}$$ | ||
| unchanged lines 11 | |||