Question

A circular swimming pool has a flat bottom and a 9.45 m diameter. it is filled with water to a depth of 2.02 m. there is 0.709 atm of pressure on the top surface. what is the absolute pressure at the bottom? the acceleration of gravity is 9.8 m/s 2 . answer in units of pa. 007 (part 2 of 2) 10.0 points two people with a combined mass of 158 kg get into the pool and float there quietly. what is the resulting increase in the absolute pressure at the bottom? answer in units of pa.

Answer 1

Part a Answer: $9.16 \times 10^4 Pa$ Absolute pressure (P) = atmospheric pressure $P_o$ + pressure due to weight of fluid above it $\rho g h$. $P=P_o+ \rho g h$ g is acceleration due to gravity $= 9.8 m/s^2$ h is height of the water column above it =2.02 m $\rho$ is density of water $= 1000 kg/m^3$

$P_o$ is atmospheric pressure = 0.709 atm.  

Substitute the values:

$P= 0.709 \times 101325 Pa + 1000 \times 9.8 \times 2.02 Pa$ Because: $1 atm = 101325 Pa$ Hence, absolute pressure at the bottom is $9.16 \times 10^4 Pa$ Part b Answer: 104.34 Pa Radius of the pool = 4.73 m Increase in the absolute pressure at the bottom = Absolute pressure with two people in the pool - Absolute pressure without them = Pressure due to two people $P'-P =\frac {Mg}{A} = \frac {Mg}{\pi r^2}$ Where M is the combined mass of the person, g is the acceleration of gravity and r is the radius of the pool.  $\Delta P=\frac {158 \times 9.8}{\pi {4.73}^2} Pa = 104.34 Pa$