Chapter 9, Problem 2SP

(a)

To determine

The volume of water column.

Answer to Problem 2SP

The volume of water column is $1.44{\text{m}}^3$

Explanation of Solution

Given info: Density of water is $1000\text{kg}/{\text{m}}^{\text{3}}$ , depth of water is $2.4\text{m}$ , area is $0.6{\text{m}}^2$

Write the equation to find the volume.

$V=A\times t$

Here,

$V$ is the volume

$A$ is the area

$t$ is the depth

Substitute $2.4\text{m}$ for $t$ and $0.6{\text{m}}^2$ for $A$ in equation to find $V$

$\begin{array}{c}V=2.4\text{m}\times \text{0}\text{.6}{\text{m}}^2\\ =1.44{\text{m}}^3\end{array}$

Conclusion:

The volume of water column is $1.44{\text{m}}^3$

(b)

To determine

Mass of water column.

Answer to Problem 2SP

Mass of the water column is $1.44\times {10}^3\text{kg}$

Explanation of Solution

Write the equation to find density.

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

Here,

$\rho$ is the density

$M$ is the mass

$V$ is the volume

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $1.44{\text{m}}^3$ for $V$ to get $M$ .

$\begin{array}{c}M=1000\text{kg}/{\text{m}}^{\text{3}}\times 1.44{\text{m}}^3\\ =1.44\times {10}^3\text{kg}\end{array}$

Conclusion:

Mass of the water column is $1.44\times {10}^3\text{kg}$

(c)

To determine

Weight of water column.

Answer to Problem 2SP

The weight of water column is $14\text{kN}$

Explanation of Solution

Write the equation to find the weight.

$W=Mg$

Here,

$W$ is the weight

$M$ is the mass

$g$ is the acceleration due to gravity

Substitute $1.44\times {10}^3\text{kg}$ for $M$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ to get $W$.

$\begin{array}{c}W=1.44\times {10}^3\text{kg}\times \text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}\\ =14\text{kN}\end{array}$

Conclusion:

The weight of water column is $14\text{kN}$

(d)

To determine

The excess pressure of the water column on bottom layer.

Answer to Problem 2SP

The excess pressure is $23,500\text{Pa}$

Explanation of Solution

Write the equation to find the pressure.

$P=h\rho g$

Here,

$P$ is the pressure

$h$ is the depth

$\rho$ is the density

$g$ is the acceleration due to gravity

Substitute $2.4\text{m}$ for $h$ , $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ to get $P$

$\begin{array}{c}P=2.4\text{m}\times \text{1000}\text{kg}/{\text{m}}^{\text{3}}\times 9.8\text{m}/{\text{s}}^{\text{2}}\\ =23,520\text{Pa}\end{array}$

Conclusion:

The excess pressure is $23,500\text{Pa}$

(e)

To determine

How this value of pressure compare to atmospheric pressure.

Answer to Problem 2SP

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

Explanation of Solution

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

$\begin{array}{c}ratio=\frac{23,500\text{Pa}}{101325\text{Pa}}\\ =0.23\end{array}$

Conclusion:

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