Answer:
Explanation:
Given data:
initial construction co = 0.286 wt %
concentration at surface position cs = 0 wt %
carbon concentration cx = 0.215 wt%
time = 7 hr
for 0.225% carbon concentration following formula is used
where, erf stand for error function
from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815
from given table
x = 0.002395 mm
Answer:
R=1923Ω
Explanation:
Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.
Coil length(L) of the wire=37.0m
Cross-sectional area of the conductor or wire (A) = πr^2
A= π * (2.053/1000)/2=3.31*10^-6
To calculate for the resistance (R):
R=ρ*L/A
R=(1.72*10^8)*(37.0)/(3.31*10^-6)
R=1922.65Ω
Approximately, R=1923Ω