An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?

Answers

Answer 1

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

[tex]specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3[/tex]

Fraction of the object's weight below the surface of water is calculated as;

[tex]= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}[/tex]

Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀


Related Questions

When 26400j of energy is supplied to a 2.0kg bloom of aluminum it temperature rise from 20oc to 35oc.The block is well so there is no energy lost to sorround determine the specific heat capacity of aluminum

Answers

Answer:

880J/kelvin

Explanation:

Q =MC ×change in t

c =C/m

C=Q/change in t

c= Q/ m× change in t

c = 26400 / 2.0 × 15

c = 880 J/kelvin

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