A person swims from a depth of 0.50 m to a depth of 1.70 m below the surface of the sea. Density of the sea water = 1030 kg/m^3. Gravitational field strength = 9.8 N/kg. Calculate the increase in pressure on the swimmer. Give the unit.

We may use from the equation sheet, that p = h ρ g. Here, we are looking for a change in pressure, so one or more of the terms on the other side of the equation must be changing too. The question tells us that the depth, which we can consider as h, is in fact changing. Therefore, our equation becomes:Δp = Δh ρ g,Let us say that the swimmer is going from h1 (0.50m) to h2 (1.70m). By substituting this into our equation, we obtain:Δp = (h2 - h1) ρ gWe know from the question that the density, ρ, of the sea water is 1030kg/m^3 and that the gravitational field strength, g, is 9.8N/kg, so we can plug those into the equation like so:Δp = (1.70 - 0.50)10309.8From which we obtain that the increase in pressure on the swimmer is Δp = 12112.8Pa, which we must round to 2 sig. figs to give Δp = 12000Pa.