Considering the sign's symmetrical setup and equal cable length, the tension in each cable holding a 50 kg sign is closest to 250 N, which represents half the sign's total weight.
To determine the tension in each of the cables supporting a hanging sign, we need to consider the weight of the sign and how it is distributed between the cables. The weight of the sign acts downward due to gravity and is equal to the mass of the sign times the acceleration due to gravity (50 kg × 9.8 m/s² = 490 N). With two cables, this weight is distributed equally between them if they are symmetrical.
For each cable to support the sign, the tension must equal half the weight of the sign (490 N / 2 = 245 N). However, if the sign weighs 50 kg and the weight is equally distributed, the closest provided option to the calculated tension of 245 N is 250 N. Therefore, the most accurate provided tension for each cable is 250 N, given that the cables are of equal length and the sign is hung symmetrically.
So, assuming symmetrical setup, the tension in each of the cables supporting the 50 kg sign is 250 N.