Electricity is the controlled flow of electrons. They are made to travel in the same direction at the same time. If you have an insulator, it stops or severely inhibits the flow of electrons. Such a device is found on telephone poles. Those are made of glass like materials.
Answer:
Hello There!
It is Speed.
Because in the question it is said that at which speed car is travelling.
Speed is a scalar quantity.
If it is velocity then first know this that it is a vector quantity.
It must include both magnitude and direction.
If it is velocity then it would be 40 mph east or west or north or south.
<h2>I hope it is helpful to you...</h2><h3>Cheers!_____________</h3>
Answer:
<em>The internal energy change is 330.01 J</em>
Explanation:
Given
the initial volume = 5.75 L
the final volume = 1.23 L
is the external pressure = 1.00 atm
q the heat energy removed = -128 J (since is removed from the system)
expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;
W = -ΔV
W = -1.00 x(1.23 - 5.75)
W = -1.00 x -4.52
W = 4.52 L atm
converting to joules we have
W = 4.52 L atm x 101.33 J/ L atm = 458.01 J
The internal energy change during compression can be calculated thus;
ΔU = q + W
ΔU = -128 J + 458.01 J
ΔU = 330.01 J
Therefore the internal energy change is 330.01 J
I have solved parts A and B by doing the following:
A) R = ρL/A First I found the constants a and b by solving for them, which I found to be: a = 2.25x10^-8 Ωm b = 2.78x10^-8 Ωm A = πr^2 A = 3.8013x10-4 m2 Taking the integral of the Resistance equation gave me: 1/A*(ax + bx3/3) I then took the integral from 0 to 1.50m, thereby giving me: (1/3.801299x10-4)*3.38x10-8*1.50 + 2.78x10-8*1.53/3 = 1.71x10-4 This I know to be correct.
B) ∫E.dl = V dV/dx = d/dx(IR) 1.75/A*(a + bx2) Plugging in 0.75 for x gave me E = 1.76x10^-4 V/m This is also correct
C) This is where I'm stuck. I had thought that if I used the integral in part A to find the integral from 0 to 0.75 then that would be my resistance in the left hand side. Then I thought that I could subtract the left hand value from the value I found in part A to give me the right hand value, but this hasn't worked. My workings are as follow: R = ρL/A A = 3.8x10^-4 m2 Rleft hand1/A*(2.25x10^-8)(0.75) + (2.78x10^-8)(0.75^3/3) = 4.43966x10^-5 Then Rright hand = 1.71x10^-4 - 4.43966x10^-5 = 1.266x10^-4