Define pressure.
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Pressure is the force per unit area.
Pressure
A woman stands on soft ground. She then puts on snowshoes, which have a much larger area than her boots. Explain why she sinks less into the ground when wearing snowshoes.
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Her weight (force) remains the same; the snowshoes increase the area over which this force acts; because pressure is inversely proportional to area, the larger area results in a lower pressure on the ground; therefore the ground is less compressed and she sinks less.
A builder uses a wide, flat foundation beneath a wall. State how this affects the pressure on the ground and explain why.
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The wide foundation increases the area over which the weight of the wall acts; because pressure is inversely proportional to area (p = F/A), a larger area results in a lower pressure on the ground; this prevents the wall from sinking.
Calculate the force exerted by a gas at a pressure of 200 000 Pa acting on a piston of area 0.0050 m².
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Rearranging for force $$F=p\times A$$ Given $$p=200 000\text{ Pa}$$ $$A=0.0050{\text{ m}}^{2}$$ Substitution $$F=200 000\times 0.0050$$ $$F=1000\text{ N}$$ The force exerted on the piston is 1000 N.
State two factors that affect the pressure beneath the surface of a liquid.
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The depth beneath the surface of the liquid; the density of the liquid.
A diver swims from a depth of 5 m to a depth of 20 m in the sea. Explain what happens to the pressure on the diver and why.
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The pressure on the diver increases; because at a greater depth there is a greater weight of water above the diver; this greater weight exerts a greater force per unit area on the diver; therefore the pressure increases with depth.
Calculate the change in pressure at a depth of 15 m below the surface of a lake. The density of water is 1000 kg m⁻³ and g = 9.8 N kg⁻¹.
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Equation used $$\Delta p=\rho g\Delta h$$ Given $$\rho =1000{\text{ kg m}}^{-3}$$ $$g=9.8{\text{ N kg}}^{-1}$$ $$\Delta h=15\text{ m}$$ Substitution $$\Delta p=1000\times 9.8\times 15$$ $$\Delta p=147 000\text{ Pa}$$ The change in pressure is 147 000 Pa.
A diver experiences a pressure change of 206 000 Pa at a certain depth in seawater of density 1030 kg m⁻³. Determine the depth of the diver below the surface. Use g = 10 N kg⁻¹.
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Rearranging for depth change $$\Delta h=\frac{\Delta p}{\rho g}$$ Given $$\Delta p=206 000\text{ Pa}$$ $$\rho =1030{\text{ kg m}}^{-3}$$ $$g=10{\text{ N kg}}^{-1}$$ Substitution $$\Delta h=\frac{206 000}{1030\times 10}$$ $$\Delta h=\frac{206 000}{10 300}$$ $$\Delta h=20\text{ m}$$ The diver is at a depth of 20 m below the surface.