Where is Eq.(28) ?? You should show it to find the result
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We are given a CSP with only binary can concentrate assume we run backtrackingSearch with ARC
Answer:
the welding gun liner regulates the shielding gas.
Explanation:
The purpose of the welding gun liner is to properly position the welding wire from the wire feeder till it gets to the nozzle or contact tip of the gun. <em>Regulation of the shielding gas depends on factors such as the speed, current, and type of gas being used. </em>In gas metal arc welding, an electric arc is used to generate heat which melts both the electrode and the workpiece or base metal.
The electric arc produced is shielded from contamination by the shielding gas. The heat generated by the short electric arc is low.
Answer:
1561.84 MPa
Explanation:
L=20 cm
d1=0.21 cm
d2=0.25 cm
F=5500 N
a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa
lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16
longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3
(assuming a poisson's ration of 0.3)
ε_l =0.16/0.3 = 0.5333
b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)
σ_true = 1561.84 MPa
ε_true = ln( 1+ε_l)= ln(1+0.5333)
ε_true= 0.4274222
The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.
