There is water in the tank of a street watering machine, how much will its weight decrease if it spills 200 liters of water

Given:

V = 200 liters – volume of poured water;

ro = 1000 kg / m ^ 3 – water density;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to determine P (Newton) – the weight of the spilled water.

Let’s bring the volume to SI units:

V = 200 liters = 200 * 0.001 = 0.2 m ^ 3.

Find the mass of spilled water:

m = ro * V = 1000 * 0.2 = 200 kilograms.

Then the weight of the spilled water will be equal to:

P = m * g = 200 * 10 = 2000 Newtons.

Answer: the weight of the tank will decrease by 2000 Newtons, and the mass by 200 kilograms.