A metal piece weighing 500 g was suspended from a dynamometer and lowered into gas.

A metal piece weighing 500 g was suspended from a dynamometer and lowered into gas. What is the volume of the part if the dynamometer reading is 4.4N?

Density of engine oil BY THE REFERENCE = 0.910 kg / liter
Part weight in oil = m * g – V * P * g = 4, 6 H
0, 5 * 9, 8 – V * 0, 910 * 9, 8 = 4, 6
we transfer the known ones to one side, and the unknowns to the other side, when transferring values, the signs change to the opposite, then we get
– V * 0, 910 * 9, 8 = 4, 6 – 0, 5 * 9, 8
V * 0.910 * 9.8 = – 4.6 + 0.5 * 9.8
V = (0, 5 * 9, 8 – 4, 6) / (0, 9 1 0 * 9, 8) = (4, 9 – 4, 6) / (0, 9 1 0 * 9, 8) = 0.3 / (0.9 1 0 * 9.8) = 0.3 / (8.918) = 0.0336 liters = 33.6 CM IN CUBE