Answer to Question #85308 in Mechanics | Relativity for Amelia

Question

A deep sea diver is at a depth of 125m below sea level. If the density of sea water is 1025kgm-³ where g=10Nkg-¹ determine the pressure on the diver due to the sea water. If atmospheric pressure is 101000Pa, determine the total pressure on the diver

Answer 1

We can determine the pressure on the diver due to the sea water from the formula:

$$P = \rho_{sea \thinspace water} gh,$$

here, $P$ is the pressure on the diver due to the sea water, $\rho_{sea \thinspace water}$ is the density of the sea water, $g$ is the acceleration due to gravity and $h$ is the depth.

Then, we get:

$$P_{sea \thinspace water} = 1025 \dfrac{kg}{m^3} \cdot 10 \dfrac{N}{kg} \cdot 125 m = 1.28 \cdot 10^6 Pa.$$

We can find the total pressure on the diver from the formula:

$$P_{total} = P_{atm} + P_{sea \thinspace water} = 1.01 \cdot 10^5 Pa + 1.28 \cdot 10^6 Pa = 1.38 \cdot 10^6 Pa.$$

Answer:

$P_{sea \thinspace water} = 1.28 \cdot 10^6 Pa.$

$P_{total} = 1.38 \cdot 10^6 Pa.$