# What force must be applied to keep a steel rail with a volume of 0.7 m3 in water.

What force must be applied to keep a steel rail with a volume of 0.7 m3 in water. The density of water is 1000 kg / m3. The density of the steel is 7800 kg / m3.

Initial data: V (steel rail volume) = 0.7 m3.

Reference data: g (acceleration due to gravity) ≈ 10 m / s2; according to the condition ρw (water density) = 1000 kg / m3; ρс (steel density) = 7800 kg / m3.

The resultant force for holding a steel rail in water is determined by the formula: F = Ft – Fa = m * g – ρw * g * V = ρw * V * g – ρw * g * V = (ρw – ρw) * V * g.

Calculation: F = (7800 – 1000) * 0.7 * 10 = 47600 N (47.6 kN).

Answer: A force of 47.6 kN is required to hold the steel rail in water.

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