Answer to Question #129363 in Physics for kimhab

Question

The density of seawater is 1030kg per cubic meter. The pressure exerted in the depth H is 201880Pa. What is the depth of seawater?Take g=9.8N/kg.

Answer 1

The pressure due to a liquid in liquid columns of constant density is represented by the following formula:

$$p = \rho gh$$

where $p = 201880\space Pa$, $\rho = 1030 \space kg/m^3$ is the density of seawater, $g = 9.8 \space N/kg$ and $h$ is the depth. Expressing $h$ obtain:

$$h = \dfrac{p}{\rho g} = \dfrac{201880}{1030 \cdot 9.8} = 20 m$$

Answer. 20 m.