A wire 10 cm long with a diameter of 0.1 mm breaks under the action of a load weighing 1 kg. What mass will the load break a wire 20 cm long with a diameter of 0.2 mm?
Given: l1 (length of the first wire) = 10 cm; d1 (diameter of the first wire) = 0.1 mm; m1 (mass of the first load) = 1 kg; l2 (length of the second wire) = 20 cm; d2 (second wire diameter) = 0.2 mm.
To find the mass of the load that will break the second wire, we will use the formula for calculating the ultimate tensile strength: Ϭw = P1 / F1 = P2 / F2; m1 * g / (0.25 * Π * d1 ^ 2) = m2 * g / (0.25 * Π * d2 ^ 2) and m2 = m1 * d2 ^ 2 / d1 ^ 2.
Calculation: m2 = 1 * 0.2 ^ 2 / 0.12 = 4 kg.
Answer: The wire will be torn by a weight of 4 kg.
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