Calculate the tensile stress in a rod with a diameter of 80 mm subjected to a tensile force of 900 N. Express the answer in MPa.

The diameter of the rod is $D = 80 \text{ mm}$.
First, convert the diameter to meters: $D = 80 \times 10^{-3} \text{ m} = 0.08 \text{ m}$.
The radius of the rod is $r = D/2 = 0.08 \text{ m} / 2 = 0.04 \text{ m}$.
The cross-sectional area $A$ of a circular rod is given by the formula $A = \pi r^2$.

$$A = \pi (0.04 \text{ m})^2 = \pi \times 0.0016 \text{ m}^2 \approx 0.0050265 \text{ m}^2$$

Tensile stress ($\sigma$) is defined as the tensile force ($F$) divided by the cross-sectional area ($A$).
The tensile force is given as $F = 900 \text{ N}$.

$$\sigma = \frac{F}{A} = \frac{900 \text{ N}}{0.0050265 \text{ m}^2}$$ $$\sigma \approx 179047.6 \text{ Pa}$$

Since $1 \text{ MPa} = 10^6 \text{ Pa}$, we divide the stress in Pascals by $10^6$.

$$\sigma \approx \frac{179047.6}{10^6} \text{ MPa} \approx 0.179 \text{ MPa}$$

The tensile stress is $\text{0.179 MPa}$.