Answer:
1. SG
true
=2.689
2. The object is probably some sort of minerals and rocks such as Feldspar, Corals, Beryl, etc.
Explanation:
Given:
mass in the air= 0.0675 kg
mass in water= 0.0424 kg
The specific gravity of the object will be 2.6892. It is the ratio of the density of the given fluid and the standard fluid.
Density is specified as the mass divided by the volume. It is represented by the unit of measurement as kg/m³.
The mass of the object in air;
m=Vρ₀
m=0.0675 kg
Buoyant force on the object;
B= Vρₐg
For equilibrium;
N+B=m₀g
n=m₀g-Vρₓg
N/g=m₀-Vρₓ
N/g=0.0424 kg
[tex]\rm \frac{V\rho_0}{V\rho_x} =\frac{0.0675 }{m_0-0.0424 \ kg} \\\\ \frac{\rho_0}{\rho_x} =\frac{0.0675}{0.0675-0.0424} \\\\ \frac{\rho_0}{\rho_x} =2.6892[/tex]
Hence, the specific gravity of the object will be 2.6892.
To learn more about the density refers to the link;
brainly.com/question/952755
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