Answer:
The current through the coil is 2.05 A
Explanation:
Given;
number of turns of the coil, N = 1
radius of the coil, r = 9.8 cm = 0.098 m
magnetic moment of the coil, P = 6.2 x 10⁻² A m²
The magnetic moment is given by;
P = IA
Where;
I is the current through the coil
A is area of the coil = πr² = π(0.098)² = 0.03018 m²
The current through the coil is given by;
I = P / A
I = (6.2 x 10⁻² ) / (0.03018)
I = 2.05 A
Therefore, the current through the coil is 2.05 A
Answer:
The strength coefficient is K = 591.87 MPa
Explanation:
We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by

where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.
Solving for strength coefficient
From the strain hardening equation we can solve for K

And we can replace values

Thus we get that the strength coefficient is K = 591.87 MPa
Yes it does have a large spot
Answer:
a) heat gain per unit tube length = 
b) heat gain per unit tube length = 
Explanation:
Assumptions:
- Constant properties
- Steady state conditions
- Negligible effect of radiation
- Negligible constant resistance between tube and insulation
- one dimensional radial conduction
a) What is the heat gain per unit tube length

Therefore 








heat gain per unit tube length = 
b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

and
are the same, but
changes.
Therefore:


The total resistance 
heat gain per unit tube length = 
Answer:
nothing will happen
Explanation:
except for whatever that wire feeds will lose power
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