Answer :
Final answer:
The specific weight of the unleaded gasoline is 46.5 lbs/ft³. The density is 1.443 lbs/ft³. The specific volume is 0.692 ft³/lb. The specific gravity is 0.0231.
Explanation:
To determine the specific weight, density, specific volume, and specific gravity of the unleaded gasoline, we can use the given weight of 46.5 lbs and the fact that one cubic foot of gasoline occupies one cubic foot of volume.
The specific weight is the weight of a substance per unit volume. In this case, the specific weight of the unleaded gasoline is 46.5 lbs/ft³.
The density is the mass of a substance per unit volume. To calculate the density, we need to convert the weight to mass. Since weight is a force, we can use the equation:
Weight = Mass × Acceleration due to gravity
Since the weight is given as 46.5 lbs, we can rearrange the equation to solve for mass:
Mass = Weight / Acceleration due to gravity
Using the value of acceleration due to gravity as 32.2 ft/s², we can calculate the mass:
Mass = 46.5 lbs / 32.2 ft/s² = 1.443 lbs·s²/ft
Now, we can calculate the density:
Density = Mass / Volume
Since the volume is given as one cubic foot, the density of the unleaded gasoline is:
Density = 1.443 lbs·s²/ft / 1 ft³ = 1.443 lbs/ft³
The specific volume is the volume of a substance per unit mass. To calculate the specific volume, we can use the reciprocal of the density:
Specific Volume = 1 / Density
Therefore, the specific volume of the unleaded gasoline is:
Specific Volume = 1 / 1.443 lbs/ft³ = 0.692 ft³/lb
The specific gravity is the ratio of the density of a substance to the density of a reference substance, usually water. Since the density of water is 62.4 lbs/ft³, we can calculate the specific gravity:
Specific Gravity = Density of Gasoline / Density of Water
Specific Gravity = 1.443 lbs/ft³ / 62.4 lbs/ft³ = 0.0231
Learn more about properties of unleaded premium gasoline here:
https://brainly.com/question/32666273
#SPJ14