Answer: λ2= 2.34 * 10^-6 C/m
Explanation: In order to calculate the value of the linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so
Volume of cylinder:2*π*b*L *(b-a) where (b-a) is the thickness, then
λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m
Answer : The heat change of the cold water in Joules is, 
Explanation :
First we have to calculate the mass of cold water.
As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.


Now we have to calculate the heat change of cold water.
Formula used :

where,
Q = heat change of cold water = ?
m = mass of cold water = 45 g
c = specific heat of water = 
= initial temperature of cold water = 
= final temperature = 
Now put all the given value in the above formula, we get:


Therefore, the heat change of cold water is 
Answer:
Flow rate 2.34 m3/s
Diameter 0.754 m
Explanation:
Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.
The area at the well head is

So the volume flow rate along the pipe is

We can use the similar logic to find the cross-section area at the refinery

The radius of the pipe at the refinery is:



So the diameter is twice the radius = 0.38*2 = 0.754m