What is the force of gravity acting on a steel part with a volume of 480 cm³. density of steel 7800 kg / m³
August 28, 2021 | education|
Initial data: V (volume of steel part) = 480 cm ^ 3 = 480 * 10 ^ -6 m ^ 3.
Reference data: g (acceleration of gravity) = 10 m / s ^ 2; according to the condition ρс (steel density) = 7800 kg / m ^ 3.
1) Determine the mass of the steel part: m = V * ρс = 480 * 10 ^ -6 * 7800 = 3.744 kg.
2) Calculate the force of gravity that acts on the steel part: Ft = m * g = 3.744 * 10 = 37.44 N.
Answer: The force of gravity that acts on the steel part is 37.44 N.
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