Physics – 1.8 Pressure | e-Consult
1.8 Pressure (1 questions)
Login to see all questions.
Click on a question to view the answer
We are given the equation Δp = ρ g Δh. We need to calculate the increase in pressure (Δp) as the diving bell descends 20m (Δh = 20m). We need to assume a density (ρ) for seawater. A reasonable value for the density of seawater is 1025 kg/m3.
Therefore, Δp = (1025 kg/m3) * (9.8 m/s2) * (20 m)
Δp = 195000 Pa = 1.95 x 105 Pa
The pressure inside the diving bell at 20m is 1.2 x 105 Pa. The atmospheric pressure at the surface is 1.01 x 105 Pa. The increase in pressure inside the diving bell is therefore:
Δp = 1.2 x 105 Pa - 1.01 x 105 Pa = 1.9 x 104 Pa
Therefore, the increase in pressure inside the diving bell as it descends to 20m is 1.9 x 104 Pa.