physics question

Question

The density of seawater is 1023 kg/m^3, what is the change in pressure ∆P for a diver going from a depth of 70.7 m to a depth of 48.3 m? Enter units of Pa.

Answer 1

Given

Density = 1023 kg/m^3

h2 = 48.3m

h1 = 70.7 m

Find

∆P

Solution

∆P = density * g * (h2-h1)

∆P = 1023 * 9.81 * (48.3 - 70.7) = -22.4 *10^3 Pa