Answer:
The absolute pressure on me increases to 130,900 Pa.
Explanation:
Since I dive to a depth of 2.0 m, there is a depth change of 2.0 m which will cause a corresponding pressure change. The pressure change is thus ΔP = ρgΔh where ρ = density of water = 1000 kg/m³, g = acceleration due to gravity = 9.8m/s² and Δh = depth change = 2.0 m
ΔP = ρgΔh = 1000 kg/m³ × 9.8m/s² × 2.0 m = 19,600 Pa.
Since the absolute pressure P at 1.0 m is the pressure due to atmosphere plus the pressure of 1.0 m of water, we have
P = P₀ + ρgΔh' where P₀ = atmospheric pressure = 101,500 Pa and Δh' = 1.0 m
P = 101,500 Pa + 1000 kg/m³ × 9.8m/s² × 1.0 m = 101,500 Pa + 9800 Pa = 111,300 Pa
So, the absolute pressure at 2.0 m is P' = P + ΔP = 111,300 Pa + 19,600 Pa = 130,900 Pa.
So the absolute pressure on me increases to 130,900 Pa.
