A force of 20 N is applied to a material with a cross-sectional area of 5 cm2. Calculate the stress.

The force applied, $F = 20\, \text{N}$.
The cross-sectional area, $A = 5\, \text{cm}^2$.
The formula for stress ($\sigma$) is:

$$\sigma = \frac{F}{A}$$

Since $1\, \text{m} = 100\, \text{cm}$, then $1\, \text{m}^2 = (100\, \text{cm})^2 = 10000\, \text{cm}^2$.
Therefore, to convert $\text{cm}^2$ to $\text{m}^2$, we divide by 10000.

$$A = 5\, \text{cm}^2 \times \frac{1\, \text{m}^2}{10000\, \text{cm}^2} = 5 \times 10^{-4}\, \text{m}^2$$

Substitute the force and the converted area into the stress formula:

$$\sigma = \frac{20\, \text{N}}{5 \times 10^{-4}\, \text{m}^2}$$ $$\sigma = \frac{20}{5} \times 10^4\, \text{N/m}^2$$ $$\sigma = 4 \times 10^4\, \text{Pa}$$

The stress is $\mathbf{4 \times 10^4\, Pa}$.