Question

Медный брусок размером 4х5х15 см на 1/3 часть погружен в керосин. чему равна выталкивающая сила,действующая на брусок? Полное решение(с дано,с формулами) 15 баллов

Answer 1

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F Архимеда=0,8 Н

Answer 2

Given Information

The copper block has dimensions of 4x5x15 cm, and one-third of it is submerged in kerosene.

Formula for Buoyant Force

The buoyant force acting on an object submerged in a fluid can be calculated using the formula: Fb = ρ * V * g Where: - Fb is the buoyant force, - ρ is the density of the fluid, - V is the volume of the fluid displaced by the submerged part of the object, and - g is the acceleration due to gravity.

Calculation of Buoyant Force

Given that one-third of the copper block is submerged in kerosene, we can calculate the volume of kerosene displaced and then use the formula to find the buoyant force.

The volume of the copper block is: V_block = 4 * 5 * 15 = 300 cm^3

As one-third of the block is submerged, the volume of kerosene displaced is: V_kerosene = (1/3) * V_block = (1/3) * 300 = 100 cm^3

The density of kerosene is approximately 0.8 g/cm^3.

Using the formula for buoyant force: Fb = ρ * V * g Fb = 0.8 g/cm^3 * 100 cm^3 * 9.81 m/s^2 Fb ≈ 784.8 g m/s^2 ≈ 784.8 N

Answer

The buoyant force acting on the copper block submerged in kerosene is approximately 784.8 N.