155Ω
Explanation:
R = R ref ( 1 + ∝ ( T - Tref)
where R = conduction resistance at temperature T
R ref = conductor resistance at reference temperature
∝ = temperature coefficient of resistance for conductor
T = conduction temperature in degrees Celsius
T ref = reference temperature that ∝ is specified at for the conductor material
T = 600 k - 273 k = 327 °C
Tref = 300 - 273 K = 27 °C
R = 50 Ω ( 1 + 0.007 ( 327 - 27) )
R = 155Ω
Answer:

where E = electric field intensity
Explanation:
As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical
So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field
while weight of the ball is vertically downwards
so here we have


since string makes 30 degree angle with the vertical so we will have





where E = electric field intensity
An atom. Hope this helped
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

where ‘h’ is the length of the imaginary Gaussian surface.

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

3. At the boundary where r = R:

As can be seen from above, two E-field values are equal as predicted.