If the hose mentioned in the previous task is lifted by $0.4 \mathrm{~m}$, the water will flow out from under the water. What will be its average speed? If the surface of the lake does not sink due to the outflow, so that the speed remains constant, how many kilograms of water will flow out per second? If this energy is utilized through turbines, how many indicated horsepower will be available?
The speed at the upper part of the orifice:

$$
v_{1}=\sqrt{2 \cdot g \cdot 5} \frac{\mathrm{m}}{\mathrm{sec}}
$$

at the lower part of the orifice:

$$
v_{2}=\sqrt{2 \cdot g \cdot 5.4} \frac{\mathrm{m}}{\mathrm{sec}}
$$

the average speed $v=\frac{1}{2}\left(v_{1}+v_{2}\right) \frac{\mathrm{m}}{\mathrm{sec}}=10.1 \frac{\mathrm{m}}{\mathrm{sec}}$

The volume of water flowing out per second:

$$
k=0.4 \cdot 1.5 \cdot v \mathrm{~m}^{3}=6.060 \mathrm{~m}^{3}
$$

its mass

$$
\mathrm{m}=6060 \mathrm{~kg}
$$

The power is the energy produced per second, which is

$$
E=\frac{1}{2} m v^{2}=\frac{1}{2} 6060 \cdot 10^{3} \cdot 1010^{2} \mathrm{erg}=\frac{1}{2} 6060 \cdot 10.1^{2} \cdot \frac{1}{g} \mathrm{~kg} \mathrm{~m}
$$

the power:

$$
H=\frac{6060 \cdot 10.1^{2}}{2 \cdot 9.81 \cdot 75}(H P)_{i}=420(H P)_{i}
$$

(Imre Csada, Pápa.)

The problem was also solved by: L. Bánó, P. Ehrenstein, K. Epstein, A. Frank, E. Freund, R. Földes, Gy. Jánosy, P. Merse, O. Morvai, T. Pözel, A. Stagl, O. Szilas.