A tractor (4 tonnes, 4000 cm² tyre contact area) and a car (1 tonne, 150 cm² tyre contact area) are on soft ground. Explain the physical features enabling the tractor to move easily and the car to sin

Given: $1 \, \text{tonne} = 1000 \, \text{kg}$, $1 \, \text{m}^2 = 10000 \, \text{cm}^2$.
Assume acceleration due to gravity $g = 10 \, \text{m/s}^2$.

For the tractor:
Mass $m_T = 4 \, \text{tonnes} = 4 \times 1000 \, \text{kg} = 4000 \, \text{kg}$
Area $A_T = 4000 \, \text{cm}^2 = 4000 \times 10^{-4} \, \text{m}^2 = 0.4 \, \text{m}^2$

For the car:
Mass $m_C = 1 \, \text{tonne} = 1 \times 1000 \, \text{kg} = 1000 \, \text{kg}$
Area $A_C = 150 \, \text{cm}^2 = 150 \times 10^{-4} \, \text{m}^2 = 0.015 \, \text{m}^2$

Force $F = mg$.
For the tractor:

$$F_T = m_T g = 4000 \, \text{kg} \times 10 \, \text{m/s}^2 = 40000 \, \text{N}$$

For the car:

$$F_C = m_C g = 1000 \, \text{kg} \times 10 \, \text{m/s}^2 = 10000 \, \text{N}$$

Pressure $P = \frac{F}{A}$.
For the tractor:

$$P_T = \frac{F_T}{A_T} = \frac{40000 \, \text{N}}{0.4 \, \text{m}^2} = 100000 \, \text{Pa}$$

For the car:

$$P_C = \frac{F_C}{A_C} = \frac{10000 \, \text{N}}{0.015 \, \text{m}^2} \approx 666666.67 \, \text{Pa}$$

Comparison:
The pressure exerted by the tractor is $100000 \, \text{Pa}$.
The pressure exerted by the car is approximately $666667 \, \text{Pa}$.
The car exerts significantly more pressure on the ground than the tractor.
The pressure exerted by the tractor is $\boxed{\text{100000 Pa}}$ and by the car is approximately $\boxed{\text{666667 Pa}}$.