Question

a 24 m³ incompressible box is completely submerged in a fluid. what volume of fluid is displaced by the box if the fluid is...
a.)...air? the density of air is 1.2 kg/m³.

m³
b.)...mercury? the density of mercury is is 13.6 g/ml.

m³
c.)...helium? the density of helium is 0.1786 g/l.

m³
d.)...water? the density of water is 1 g/cm³.

m³

Explanation:

To solve this problem, we use the principle of Archimedes, which states that the volume of fluid displaced by a completely submerged object is equal to the volume of the object itself, regardless of the fluid's density. The volume of the box is given as \(24 \, \text{m}^3\).

Part A: Air

Step 1: Apply Archimedes' Principle

The volume of fluid displaced is equal to the volume of the submerged object. Since the box is completely submerged in air, the volume of air displaced is equal to the volume of the box.

$$ \text{Volume of air displaced} = 24 \, \text{m}^3 $$

Part B: Mercury

Step 1: Apply Archimedes' Principle

The volume of fluid displaced is equal to the volume of the submerged object. Since the box is completely submerged in mercury, the volume of mercury displaced is equal to the volume of the box.

$$ \text{Volume of mercury displaced} = 24 \, \text{m}^3 $$

Part C: Helium

Step 1: Apply Archimedes' Principle

The volume of fluid displaced is equal to the volume of the submerged object. Since the box is completely submerged in helium, the volume of helium displaced is equal to the volume of the box.

$$ \text{Volume of helium displaced} = 24 \, \text{m}^3 $$

Part D: Water

Step 1: Apply Archimedes' Principle

The volume of fluid displaced is equal to the volume of the submerged object. Since the box is completely submerged in water, the volume of water displaced is equal to the volume of the box.

$$ \text{Volume of water displaced} = 24 \, \text{m}^3 $$

Final Answers:

A. \(\boxed{24}\)
B. \(\boxed{24}\)
C. \(\boxed{24}\)
D. \(\boxed{24}\)

Answer:

To solve this problem, we use the principle of Archimedes, which states that the volume of fluid displaced by a completely submerged object is equal to the volume of the object itself, regardless of the fluid's density. The volume of the box is given as \(24 \, \text{m}^3\).

Part A: Air

Step 1: Apply Archimedes' Principle

The volume of fluid displaced is equal to the volume of the submerged object. Since the box is completely submerged in air, the volume of air displaced is equal to the volume of the box.

$$ \text{Volume of air displaced} = 24 \, \text{m}^3 $$

Part B: Mercury

Step 1: Apply Archimedes' Principle

The volume of fluid displaced is equal to the volume of the submerged object. Since the box is completely submerged in mercury, the volume of mercury displaced is equal to the volume of the box.

$$ \text{Volume of mercury displaced} = 24 \, \text{m}^3 $$

Part C: Helium

Step 1: Apply Archimedes' Principle

The volume of fluid displaced is equal to the volume of the submerged object. Since the box is completely submerged in helium, the volume of helium displaced is equal to the volume of the box.

$$ \text{Volume of helium displaced} = 24 \, \text{m}^3 $$

Part D: Water

Step 1: Apply Archimedes' Principle

The volume of fluid displaced is equal to the volume of the submerged object. Since the box is completely submerged in water, the volume of water displaced is equal to the volume of the box.

$$ \text{Volume of water displaced} = 24 \, \text{m}^3 $$

Final Answers:

A. \(\boxed{24}\)
B. \(\boxed{24}\)
C. \(\boxed{24}\)
D. \(\boxed{24}\)