Question

A mass of 1000 kg hangs at the end of a steel rod 5.0 m in length. The diameter of the rod is 0.80 cm and Young's modulus for the rod is 210,000 MN/m^2. What is the elongation of the rod? 1.8 cm 1.2 cm 0.46 cm 0.047 cm 0.12 cm A rock is suspended from a scale reads 20.0 N. A beaker of water is raised up so the rock is totally submerged in the water. The scale now reads 12.5 N. What is the density of the rock? 2.67 times 10^3 kg/m^3 2.50 times 10^3 kg/m^3 2.33 times 10^3 kg/m^3 3.00 times 10^3 kg/m^3 1.60 times 10^3 kg/m^3

Answer 1

the elongation of the rod due to this object is cal culated as follows FL &L=- AY mgL 72 4 (1000 kg) (9.8 m/s2) (5.0 m) 08:10 リ!(210,000×106 Nm *) 4 = 0.00464 m = 0.464 cm 0.46 cm

The buoyant force is given by F Vock Pwaterg and equal to the difference between the two scale readings. Thus, ρ water g The density of the rock is given by mrock 200N 2.67x 10'kg/m 1000kg/m rock ρ water g

While submerged in water, the rock has displaced (20.0 N - 12.5 N =) 7.5 N