A glass full of water has a bottom area of 20 cm^2 , a height of 20...

The given parameters are $A = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$, $h = 20 \text{ cm} = 0.2 \text{ m}$, $V = 0.5 \text{ L} = 0.5 \times 10^{-3} \text{ m}^3$, and $\rho = 1000 \text{ kg/m}^3$.

Since the air is evacuated, the pressure at the free surface of the water is zero. The pressure at the bottom of the glass is:
$P = \rho g h = 1000 \times 10 \times 0.2 = 2000 \text{ N/m}^2$

The force applied by the water on the bottom of the glass is:
$F_{\text{bottom}} = P \times A = 2000 \times 20 \times 10^{-4} = 4 \text{ N}$

The total weight of the water in the glass is:
$W = \rho V g = 1000 \times (0.5 \times 10^{-3}) \times 10 = 5 \text{ N}$

For the water to be in equilibrium, the net vertical force must be zero. The forces acting on the water are the downward weight $W$, the upward normal force from the bottom $F_{\text{bottom}}$, and the net vertical force from the sides $F_{\text{sides}}$.

Balancing the vertical forces:
$F_{\text{bottom}} + F_{\text{sides}} = W$
$4 + F_{\text{sides}} = 5$
$F_{\text{sides}} = 1 \text{ N}$

Therefore, the net vertical force exerted by the sides of the glass on the water is $1 \text{ N}$ upward.