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:
true
Explanation:
Equipment that are "locked-out/tagged-out" <em>prevent the electrician from being electrocuted</em> or attaining a serious injury in relation to it. Locking out an equipment prevents it from releasing its energy because such energy can be <em>hazardous</em> to the electrician. There are instances when the equipment accidentally starts up, thus, it is essential that the equipment's source of energy is<em> isolated.</em>
Answer:
The expected settlement for the pile group using the given information is 19.92mm or 0.79 inch
Explanation:
In this question, we are asked to calculate the expected settlement for the pole group given some information.
Please check attachment for complete solution and step by step explanation
Answer:
The volume flow rate of air is 
Explanation:
A random duct is shown in the below attached figure
The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time
Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters
From the attached figure we can see that
The volume of the prism that the flow occupies in 1 second equals

Hence the volume flow rate is 