Chapter 9, Problem 2SP

(a)

To determine

The volume of water column.

Answer to Problem 2SP

The volume of water column is $1.25{\text{ m}}^3$.

Explanation of Solution

Given info: The density of water is $1000\text{kg}/{\text{m}}^3$. The depth of the pool is $2.5\text{m}$ and its cross-sectional area is $0.5\text{}{\text{m}}^2$.

Write the expression to find the volume of the water column.

$V=A\times t$

Here,

$V$ is the volume of the water column

$A$ is the area of the water column

$t$ is the depth of the water column

Substitute $2.5\text{m}$ for $t$ and $0.5\text{}{\text{m}}^2$ for $A$ in equation to find $V$

$\begin{array}{c}V=\left(2.5\text{m}\right)\left(0.5\text{}{\text{m}}^2\right)\\ =1.25{\text{ m}}^3\end{array}$

Conclusion:

Therefore, the volume of water column is $1.25{\text{ m}}^3$.

(b)

To determine

The mass of the water column.

Answer to Problem 2SP

The mass of the water column is $1.25\times {10}^3\text{ kg}$.

Explanation of Solution

Write the equation to find density of the water column.

$\rho =\frac{M}{V}$

Here,

$\rho$ is the density of water

$M$ is the mass of the water column

$V$ is the volume of the water column

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $1.25{\text{ m}}^3$ for $V$ to find the mass of the water column.

$\begin{array}{c}M=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(1.25{\text{ m}}^3\right)\\ =1.25\times {10}^3\text{kg}\end{array}$

Conclusion:

Therefore, the mass of the water column is $1.25\times {10}^3\text{ kg}$.

(c)

To determine

The weight of the water column in newtons.

Answer to Problem 2SP

The weight of the water column in newtons is $12.25\times {10}^4\text{ N}$.

Explanation of Solution

Write the equation to find the weight of the water column.

$W=Mg$

Here,

$W$ is the weight of the water column

$M$ is the mass of the water column

$g$ is the acceleration due to gravity

Substitute $1.25\times {10}^3\text{ kg}$ for $M$ and $9.8\text{m}/{\text{s}}^2$ for $g$ to find the weight of the water column.

$\begin{array}{c}W=\left(1.25\times {10}^3\text{ kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =12.25\times {10}^4\text{ N}\end{array}$

Conclusion:

Therefore, the weight of the water column in newtons is $12.25\times {10}^4\text{ N}$.

(d)

To determine

The excess pressure exerted by the column of water on the bottom of the pool.

Answer to Problem 2SP

The excess pressure exerted by the column of water on the bottom of the pool is $24500\text{ Pa}$.

Explanation of Solution

Write the expression to find the excess pressure.

$P=h\rho g$

Here,

$P$ is the pressure

$h$ is the depth of the water column

$\rho$ is the density of water

$g$ is the acceleration due to gravity

Substitute $2.5\text{m}$ for $h$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find the excess pressure.

$\begin{array}{c}P=\left(2.5\text{m}\right)\left(\text{1000 }\text{kg}/{\text{m}}^3\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =24500\text{ Pa}\end{array}$

Conclusion:

Therefore, the excess pressure exerted by the column of water on the bottom of the pool is $24500\text{ Pa}$.

(e)

To determine

The value of pressure of the water column compared with the atmospheric pressure.

Answer to Problem 2SP

The value of pressure of the water column is about $\frac{1}{4}\text{th}$ of the atmospheric pressure.

Explanation of Solution

Write the expression to find the ratio of pressure of the water column and the atmospheric pressure.

$ratio=\frac{P_w}{P_a}$

Here,

$P_w$ is the pressure of the water column

$P_a$ is the atmospheric pressure

Substitute $24500\text{ Pa}$ for $P_w$ and $101325\text{ Pa}$ for $P_a$ in the above equation to find the ratio.

$\begin{array}{c}\text{ratio}=\frac{24,500\text{Pa}}{101,325\text{Pa}}\\ =\frac{1}{4}\end{array}$

Conclusion:

Therefore, the value of pressure of the water column is about $\frac{1}{4}\text{th}$ of the atmospheric pressure.