3.she wore the wrong outfit for the tree plantation drive.
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:
Force per unit plate area is 0.1344 
Solution:
As per the question:
The spacing between each wall and the plate, d = 10 mm = 0.01 m
Absolute viscosity of the liquid, 
Speed, v = 35 mm/s = 0.035 m/s
Now,
Suppose the drag force that exist between each wall and plate is F and F' respectively:
Net Drag Force = F' + F''
where
= shear stress
A = Cross - sectional Area
Therefore,
Net Drag Force, F = 
Also
F = 
where
= dynamic coefficient of viscosity
Pressure, P = 
Therefore,
Answer:
The Employee
Explanation:
Because it is there responsibility
Answer:
d. is applied after the ceiling joists are
installed.
