High School

We appreciate your visit to A small rectangular tank measuring 5 00 in by 9 00 in is filled with mercury a If the total force on the bottom of. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

A small rectangular tank measuring 5.00 in. by 9.00 in. is filled with mercury.

(a) If the total force on the bottom of the tank is 165 lb, how deep is the mercury? (Weight density of mercury = 0.490 lb/in³)

Answer :

The depth of the mercury in the tank is approximately [tex]\( 0.093 \)[/tex] inches.

To find the depth of the mercury in the tank, we first need to calculate the pressure exerted by the mercury on the bottom of the tank using the given force.

The pressure exerted by a fluid at a certain depth is given by the formula:

[tex]\[ P = \rho \cdot g \cdot h \][/tex]

Where:

- [tex]\( P \)[/tex] is the pressure (in pounds per square inch, psi)

- [tex]\( \rho \)[/tex] is the density of the fluid (in pounds per cubic inch, lb/in³)

- [tex]\( g \)[/tex] is the acceleration due to gravity (in inches per second squared, in/s²)

- [tex]\( h \)[/tex] is the depth of the fluid (in inches, in)

We're given:

- Total force on the bottom of the tank [tex]\( F = 165 \)[/tex] lb

- Density of mercury [tex]\( \rho = 0.490 \)[/tex] lb/in³

We need to find [tex]\( h \),[/tex] the depth of the mercury.

First, let's convert the given force into pressure using the formula:

[tex]\[ P = \frac{F}{A} \][/tex]

where [tex]\( A \)[/tex] is the area of the bottom of the tank.

Given that the tank is rectangular and has dimensions [tex]\( 5.00 \)[/tex] in by [tex]\( 9.00 \)[/tex] in, the area [tex]\( A \)[/tex] is:

[tex]\[ A = \text{length} \times \text{width} = (5.00 \times 9.00) \, \text{in}^2 \][/tex]

Now, we find the pressure [tex]\( P \):[/tex]

[tex]\[ P = \frac{165}{5.00 \times 9.00} \, \text{psi} \][/tex]

Next, we equate this pressure to the pressure exerted by the mercury:

[tex]\[ \frac{165}{5.00 \times 9.00} = 0.490 \times g \times h \][/tex]

To find [tex]\( h \),\\[/tex] we rearrange the equation:

[tex]\[ h = \frac{\frac{165}{5.00 \times 9.00}}{0.490 \times g} \][/tex]

Now, we need the value of the acceleration due to gravity [tex]\( g \)[/tex]. This is usually taken as [tex]\( 32.174 \)[/tex] ft/s², but since we're working in inches, we convert it to [tex]\( 386.174 \)[/tex] in/s².

Now, let's plug in the values and solve for [tex]\( h \):[/tex]

[tex]\[ h = \frac{\frac{165}{5.00 \times 9.00}}{0.490 \times 386.174} \, \text{in} \]\[ h \approx \frac{165}{5.00 \times 9.00 \times 0.490 \times 386.174} \, \text{in} \]\[ h \approx \frac{165}{1775.11} \, \text{in} \]\[ h \approx 0.093 \, \text{in} \][/tex]

So, the depth of the mercury in the tank is approximately [tex]\( 0.093 \)[/tex] inches.

Thanks for taking the time to read A small rectangular tank measuring 5 00 in by 9 00 in is filled with mercury a If the total force on the bottom of. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada