V1 = 1 liter = 0.001 m ^ 3 – volume of water;
ro1 = 1000 kg / m ^ 3 – water density;
V2 = 250 cm ^ 3 = 0.00025 m ^ 3 – the volume of the lead body;
ro2 = 11340 kg / m ^ 3 – lead density;
g = 10 N / kg – acceleration of gravity.
It is required to determine the force of gravity acting on the water F1 and the lead body F2 (Newton).
Let’s find the force of gravity acting on the water:
F1 = m1 * g = ro1 * V1 * g = 1000 * 0.001 * 10 = 1 * 10 = 10 Newtons.
Let us find the force of gravity acting on the lead body:
F2 = m2 * g = ro2 * V2 * g = 11340 * 0.00025 * 10 = 2.835 * 10 = 28.4 Newton.
Answer: gravity equal to 10 Newtons acts on 1 liter of water, and gravity equal to 28.4 Newtons acts on 250 cm ^ 3 of a lead body.
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