We appreciate your visit to 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. 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!
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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