Answer:
A. Juna's answer is correct because both the number of feet and the number of inches are correct
Explanation:
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:
Ammeters must always be connected in series with the circuit under test. Always start with the highest range of an ammeter. Deenergize and discharge the circuit completely before you connect or disconnect the ammeter. In dc ammeters, observe the proper circuit polarity to prevent the meter from being damaged.
Explanation:
