The dynamics of the spray gun follow the principle of conservation of mass or the continuity equation from fluid dynamics. This principle states that the mass flow rate of a fluid entering a system must equal the mass flow rate leaving a system. For incompressible fluids like liquids, or compressed air where changes in density are negligible, this is often simplified to the rate of volume flow or the continuity equation:

$$ A_1v_1 = A_2v_2 $$

where

$$A_1$$ is the cross-sectional area of the piston,

$$v_1$$ is the speed of the piston,

$$A_2$$ is the cross-sectional area of the nozzle, and

$$v_2$$ is the velocity of the fluid (air) leaving the nozzle.

We can use this equation to find the speed of air leaving the nozzle. Given that the piston's speed is 5 mm/s and its radius is 20 mm, we have

$$ A_1 = \pi (20\, \text{mm})^2 = 400\pi\, \text{mm}^2 $$

and

$$ v_1 = 5\, \text{mm/s} $$

The nozzle's radius is 1 mm, thus

$$ A_2 = \pi (1\, \text{mm})^2 = \pi\, \text{mm}^2 $$

Substituting the values into the continuity equation, we can solve for $$v_2$$:

$$ A_1v_1 = A_2v_2 \rightarrow 400\pi\, \text{mm}^2 \times 5\, \text{mm/s} = \pi\, \text{mm}^2 \times v_2 \rightarrow v_2 = \frac{400\pi\, \text{mm}^2 \times 5\, \text{mm/s}}{\pi\, \text{mm}^2} = 2000\, \text{mm/s} $$

Converting this to meters per second, we get

$$ v_2 = 2000\, \text{mm/s} \times \frac{1\, \text{m}}{1000\, \text{mm}} = 2\, \text{m/s} $$

Therefore, the air comes out of the nozzle with a speed of 2 m/s, which corresponds to Option C.