Respuesta :
Explanation:
The given data is as follows.
Mass of gold = 645 lb, Volume = 0.5348 ft3
Density of water = 62.4 lbs/ft3
It is known that specific gravity is defined as density of substance divided by the density of standard fluid.
Mathematically, Specific gravity = [tex]\frac{\text{density of gold}}{\text{density of water}}[/tex]
Specific gravity = [tex]\frac{\text{density of gold}}{62.4 (lbs.ft^{-3})}[/tex]
Now, calculate the density of gold then from density we will calculate specific gravity as follows
Since, Density = [tex]\frac{mass}{volume}[/tex]
Density = [tex]\frac{645 lbs}{0.5348 ft^{3}}[/tex]
= 1206.06 [tex]lbs/ft^{3}[/tex]
As, Specific gravity = [tex]\frac{\text{density of gold}}{\text{density of water (standard fluid)}}[/tex]
= [tex]\frac{1206.06 (lbs/ft^{3})}{62.4 (lbs/ft^{3})}[/tex]
= 19.32
Therefore, the value of specific gravity is 19.32.
Specific gravity has no units as it is density divided by density. Hence, all the units get canceled out.
