You are originally 1.0 m beneath the surface of a pool. If you dive to 2.0 m beneath the surface, what happens to the absolute pressure on you

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.

do you have a formula possibly