What force must be applied to keep the ice with a volume of 500 cm3 in alcohol. Show the strength in the picture.

What force must be applied to keep the ice with a volume of 500 cm3 in alcohol. Show the strength in the picture. The density of ice is 900 kg / m3, the density of alcohol is 800 kg / m3.

Initial data: V (ice volume) = 500 cm3 (500 * 10 ^ -6 m3).

Reference data: g (acceleration due to gravity) ≈ 10 m / s2; by condition ρl (ice density) = 900 kg / m3; ρс (alcohol density) = 800 kg / m3.

The force applied to hold ice in kerosene is determined from the equality: F (resultant force, upward) = Ft (gravity, downward) – Fa (Archimedes force, upward) = m * g – ρс * g * V = ρl * V * g – ρс * g * V = (ρl – ρс) * g * V.

Calculation: F = (900 – 800) * 10 * 500 * 10 ^ -6 = 0.5 N.