U=0
<span>t=10 </span>
<span>a=9.8m/s/s </span>
<span>v is velocity (the tower must be very high to be able to fall for 10 seconds!!!) </span>
<span>you work out the result now</span>
Answer:
a
Solid Wire
Stranded Wire 
b
Solid Wire
Stranded Wire
Explanation:
Considering the first question
From the question we are told that
The radius of the first wire is 
The radius of each strand is 
The current density in both wires is 
Considering the first wire
The cross-sectional area of the first wire is

= >
= >
Generally the current in the first wire is

=> 
=>
Considering the second wire wire
The cross-sectional area of the second wire is

=> 
=> 
Generally the current is

=> 
=> 
Considering question two
From the question we are told that
Resistivity is 
The length of each wire is 
Generally the resistance of the first wire is mathematically represented as
=>
=>
Generally the resistance of the first wire is mathematically represented as
=>
=>
Answer:
The minimum wall thickness required for the spherical tank is 0.0189 m
Explanation:
Given data:
d = inside diameter = 8.1 m
P = internal pressure = 1.26 MPa
σ = 270 MPa
factor of safety = 2
Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?
The allow factor of safety:

The minimun wall thickness:

Answer:
V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s average speed
t = 7.2 / 7.25 = .993 sec time to cross patch
a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2 or 1.5 m/s^2