College

For problems 55 and 56:

A cylindrical container is completely filled with a fluid that has a density of [tex]91.7 \, \text{lbm/ft}^3[/tex]. If the filled container weighs [tex]127 \, \text{lbf}[/tex], and the container weighs [tex]2.00 \, \text{lbf}[/tex] when empty, what is the interior volume of the container?

A. [tex]0.734 \, \text{ft}^3[/tex]
B. [tex]1.38 \, \text{ft}^3[/tex]
C. [tex]43.9 \, \text{ft}^3[/tex]
D. [tex]44.6 \, \text{ft}^3[/tex]
E. [tex]1.36 \, \text{ft}^3[/tex]

56. What is the specific weight of the fluid in the container?

A. [tex]2950 \, \text{lbf/ft}^3[/tex]
B. [tex]3.02 \, \text{lbf/ft}^3[/tex]
C. [tex]97.4 \, \text{lbf/ft}^3[/tex]
D. [tex]91.7 \, \text{lbf/ft}^3[/tex]
E. Not enough information given

Answer :

The interior volume of the container is 0.0423 ft^3, which corresponds to answer choice (a).

The specific weight of the fluid in the container is 2950 lbf/ft^3, which corresponds to answer choice (a).

Let's start with problem 55.

The weight of the filled container is 127lbf, and the weight of the empty container is 2.00lbf. Therefore, the weight of the fluid in the container is:

127 lbf - 2.00 lbf = 125 lbf

We know that the density of the fluid is 91.7lbm/ft3. To find the volume of the fluid, we can use the formula:

density = mass / volume

Solving for volume, we get:

volume = mass / density

The mass of the fluid can be found using its density and the weight of the fluid:

mass = weight / gravitational acceleration

mass = 125 lbf / (32.2 ft/s^2)

mass = 3.88 lbm

Now we can find the volume of the fluid:

volume = mass / density

volume = 3.88 lbm / 91.7 lbm/ft^3

volume = 0.0423 ft^3

Therefore, the interior volume of the container is 0.0423 ft^3, which corresponds to answer choice (a).

Moving on to problem 56, the specific weight of a fluid is defined as the weight per unit volume of the fluid. We can find it using the formula:

specific weight = weight / volume

We already know the weight of the fluid (125 lbf) and its volume (0.0423 ft^3), so we can plug these values into the formula:

specific weight = 125 lbf / 0.0423 ft^3

specific weight = 2950 lbf/ft^3

Therefore, the specific weight of the fluid in the container is 2950 lbf/ft^3, which corresponds to answer choice (a).

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