There is water in the cistern of the street watering machine. How much will the weight of the machine decrease if it spills 200 liters of water?
The weight of a body is the force with which this body acts on a support or stretches the suspension. The initial weight of the machine is P = m * g, where m is the mass of the machine with water, g is the acceleration of gravity 10 m / s2. When the machine spills 200 liters of water, its mass will decrease by the mass of the poured water, which means that the weight of the entire machine will also decrease by the weight of the poured water. Let’s find the mass of 200 liters of water m = ρ * V, where ρ is the density of water 1 kg / l, V is the volume of water. m = 1kg / l * 200l = 200 kg. Find the weight of spilled water weighing 200 kg, P = m * g = 200kg * 10m / s2 = 2000N = 2kN. Machine weight will be reduced by 20,000N or 2kN nga.
Answer: ≈ by 2 kN
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