Question

a can hold water with a constant depth of 0.6 m. the surface of the water is exposed to the atmosphere. what is the pressure on the bottom of the can? (g = 10n/kg and atm pressure = 101.3 kpa)

Explanation:

Step1: Calculate hydrostatic pressure

The formula for hydrostatic pressure is $P_{hydro} =
ho gh$. Assuming the density of water $
ho = 1000\ kg/m^{3}$, $g = 10\ N/kg$ and $h = 0.6\ m$. So $P_{hydro}=1000\times10\times0.6 = 6000\ Pa$.

Step2: Calculate total pressure

The total pressure at the bottom of the can is the sum of the atmospheric pressure $P_{atm}$ and the hydro - static pressure $P_{hydro}$. Given $P_{atm}=101.3\times10^{3}\ Pa$. Then $P = P_{atm}+P_{hydro}=101.3\times 10^{3}+6000=107300\ Pa = 107.3\ kPa$.

Answer:

$107.3\ kPa$