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:
25 seconds
Explanation:
Assuming the woman is accelerating at a constant rate of
from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.
Let she takes t seconds to cover the distance, s=115 m.
As acceleration, 

Now, from the equation of motion


[ from equation (i)]

seconds.
Hence, she takes 25 seconds to walk the distance.
F should be 10. If The Whole top is 50cm, then we should subtract 10n and 30n which gives us 10.
Or it could be 15 if both top and bottom are 25. 10+15= 25.