Answer:
A renewable electricity generation technology harnesses a naturally existing energy. But they have other features that a few fringe customers value.
Explanation:
The free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.
The first image shows the diagram of the beam and the second image shows the free-body diagram of the beam.
The resolution of forces in the system is well understood by the principle of equilibrium where a stationary body will remain balanced when subject to parallel forces provided that the total sum of the overall external forces is zero.
The free-body diagram is a graphical representation used to visualize the forces applied to an object.
The equilibrium of forces on the x-axis is:
The equilibrium of forces on the y-axis is:
The equilibrium condition at any point is:
From the free body diagram attached in the second image below,
- the horizontal reaction is located at point A as
- the vertical reaction is located at point A as
- the tension = T
- the weight = W
Therefore, we can conclude that the free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.
Learn more about the free-body diagram here:
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Answer:
a) The minimum acceptable value is 387.5 HV using Vickers hardness test.
b) The minimum acceptable value is 39.4 HRC using Rockwell C hardness test.
Explanation:
To get the tensile strength of a material from its hardness, we multiply it by an empirical constant that depends on things like yield strength, work-hardening, Poisson's ratio and geometrical factors. The incidence of cold-work varies this relationship.
According to DIN 50150 (a conversion table for hardness), the constant for Vickers hardness is ≈ 3.2 (an empirical approximate):
According to DIN 50150, the constant for Rockwell C hardness test is ≈31.5 around this values of tensile strength:
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
The length of the arm is the main part of natur