Answer:
Pressure = 11.38 psi
Force = 13.981 Ibf
Explanation:
Step by step solution is in the attached document.
Explanation:
<u>Filler:</u>
Filler is the material rod is used when we are joining two material by using welding process.If thickness of work piece is more so it will become compulsory to provide some filler material for making the welding join to withstand high stresses.
<u>Electrode:</u>
Electrode is the element which is used to complete the electric circuit in welding .Some time electrode is connected with positive terminal and some time with negative terminal ,it depends on the requirement of welding process.In Tungsten inert gas welding electrode is connected negative terminal but on the other hand Metal inert gas welding electrode is connected with positive terminal.Electrode can be consumable non-consumable depends on the condition.
Yes electrode can be work as filler material ,in Metal inert gas welding wire is used as electrode as well as filler material.In Metal inert gas welding consumable electrode is used on the other hand Tungsten inert gas welding non-consumable electrode is used.In Tungsten inert gas welding if thickness of work pieces is less than 5 mm then no need to used any filler material but if thickness is more than 5 mm then we have to use filler material.
Answer: (a) 36.18mm
(b) 23.52
Explanation: see attachment
Answer:
metals, composite, ceramics and polymers.
Explanation:
The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.
i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.
ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.
iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.
iv) Polymers: They have low weight and are poor conductors of electricity and heat
Answer:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
