Answer: the modulus of elasticity of the aluminum is 75740.37 MPa
Explanation:
Given that;
Length of Aluminum bar L = 125 mm
square cross section s = 16 mm
so area of cross section of the aluminum bar is;
A = s² = 16² = 256 mm²
Tensile load acting the bar p = 66,700 N
elongation produced Δ = 0.43
so
Δ = PL / AE
we substitute
0.43 = (66,700 × 125) / (256 × E)
0.43(256 × E) = (66,700 × 125)
110.08E = 8337500
E = 8337500 / 110.08
E = 75740.37 MPa
Therefore, the modulus of elasticity of the aluminum is 75740.37 MPa
Answer:C 0.12 V
Explanation:
Given
Concentration of 
Concentration of 
Standard Potential for Ni and Fe are 
and 
Answer:
no
Explanation:
it's not a dead load because when load is put on the pillars it's not fully straining it's been slowly getting to be heavier in that period of time before it falls
