Answer:
Specific Weight = 76.19 lb/ft³
mass density = 34.56
specific gravity = 1.22
Explanation:
given data
weighs = 38,300 pounds
diameter = 8 ft
high = 10 ft
solution
we get here Specific Weight that is
Specific Weight = [tex]\frac{weight}{volume}[/tex] .................1
Specific Weight = [tex]\frac{38300}{\pi r^2h }[/tex]
Specific Weight = [tex]\frac{38300}{\pi \times 4^2 \times 10}[/tex]
Specific Weight = 76.19 lb/ft³
and
mass density = [tex]\frac{m}{v}[/tex] .............2
mass density = [tex]\frac{38300}{\pi \times 4^2 \times 10 \times 2.2046}[/tex]
mass density = 34.56
and
specific gravity = [tex]\frac{density (fluid)}{density(st.fluid)}[/tex] ............3
specific gravity = [tex]\frac{76.19}{62.43}[/tex]
specific gravity = 1.22
