Question

Чему равна архимедова сила действующая в воде на тела объёмом 50см3 из алюминия

Answer 1

$F_a$ – ?
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ρ воды = 1000 кг/м³
V = 50 см³ = 0,00005 м³
ρ тела = 2700 кг/м³
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$F_a$ = g × ρ жидкости × V тела
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$F_a$ = 10H/кг × 1000 кг/м³ × 0,00005 м³ = 0,5H

Answer 2

Archimedes' Force on an Aluminum Body with a Volume of 50 cm³ in Water

The Archimedes' force acting on a body submerged in a fluid is equal to the weight of the fluid displaced by the body. This force can be calculated using the formula:

F = ρ * g * V

Where: - F is the Archimedes' force, - ρ is the density of the fluid, - g is the acceleration due to gravity, and - V is the volume of the fluid displaced by the body.

For an aluminum body with a volume of 50 cm³ submerged in water, we can calculate the Archimedes' force by considering the density of water and the volume of the body.

Calculation

The density of water is approximately 1000 kg/m³ or 1 g/cm³ The volume of the aluminum body is 50 cm³.

Using the formula F = ρ * g * V, we can calculate the Archimedes' force:

F = 1 g/cm³ * 9.81 m/s² * 50 cm³

Converting the units to be consistent: - 1 g/cm³ = 1000 kg/m³ - 50 cm³ = 0.00005 m³

F = 1000 kg/m³ * 9.81 m/s² * 0.00005 m³

Result

Calculating the above expression gives us the Archimedes' force acting on the aluminum body with a volume of 50 cm³ in water.

F ≈ 0.4905 N

Therefore, the Archimedes' force acting on the aluminum body with a volume of 50 cm³ in water is approximately 0.4905 Newtons.