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.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~