Answer:
a)
(Ω-m)^{-1}
b) Resistance = 121.4 Ω
Explanation:
given data:
diameter is 7.0 mm
length 57 mm
current I = 0.25 A
voltage v = 24 v
distance between the probes is 45 mm
electrical conductivity is given as

![\sigma = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%5Cfrac%7B0.25%20%5Ctimes%2045%5Ctimes%2010%5E%7B-3%7D%7D%7B24%20%5Cpi%20%5B%5Cfrac%7B7%20%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5D%5E2%7D)
(Ω-m)^{-1}[/tex]
b)


![= \frac{57 \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B57%20%20%5Ctimes%2010%5E%7B-3%7D%7D%7B12.2%20%5Ctimes%20%5Cpi%20%5B%5Cfrac%7B7%20%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5D%5E2%7D)
Resistance = 121.4 Ω
Answer:
maximum value of the power delivered to the circuit =3.75W
energy delivered to the element = 3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750
Explanation:
V =75 - 75e-1000t V
l = 50e -IOOOt mA
power = IV = 50 * 10^-3 e -IOOOt * (75 - 75e-1000t)
=50 * 10^-3 e -IOOOt *75 (1 - e-1000t)
=
maximum value of the power delivered to the circuit =3.75W
the total energy delivered to the element = 

Answer:
The level of the service is loss and the density is 34.2248 pc/mi/ln
Explanation:
the solution is attached in the Word file