What is the force on the top surface of a $10.5 \mathrm{~cm}$-diameter soup can that is submerged in seawater (density $1,017 \mathrm{~kg} / \mathrm{m}^{3}$ ) at a depth of $350 \mathrm{~m}$ ? The air pressure at the surface is $101.5 \mathrm{kPa}$.
A) $30.2 \mathrm{kN}$
B) $121 \mathrm{kN}$
C) $31.1 \mathrm{kN}$
D) $124 \mathrm{kN}$

Question

Quizplus · Accepted Answer

The force on the top surface of the can is determined by the pressure at the depth of 350 m, which includes both the pressure due to the water column above it and the atmospheric pressure at the surface. The pressure due to the water column is calculated using the formula $P = \rho gh$, where $\rho$ is the density of seawater (1017 kg/m³), $g$ is the acceleration due to gravity (approximately 9.81 m/s²), and $h$ is the depth (350 m). The total pressure at the depth is the sum of this pressure and the atmospheric pressure at the surface (101.5 kPa). The force is then calculated by multiplying this total pressure by the area of the top surface of the can, which is a circle with a diameter of 10.5 cm (or 0.105 m). The correct calculation leads to a force of approximately 31.1 kN.

What Is the Force on the Top Surface of A 10.5