Answer:
The maximum potential difference is 186.02 x 10¹⁵ V
Explanation:
formula for calculating maximum potential difference

where;
Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²
k is the dielectric constant = 2.3
b is the outer radius of the conductor = 3 mm
a is the inner radius of the conductor = 0.8 mm
λ is the linear charge density = 18 x 10⁶ V/m
Substitute in these values in the above equation;

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V
Answer:
CALCIUM IDENINE ADEININE AND PHOSPHATE
Explanation:
Answer:
143 °
Explanation:
a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then
d sinθ = ( 2n+1) λ/2
for first dark fringe
d sinθ = λ/2
d /λ = 1/ 2 sinθ
1 / 2 sin15
= 1.93
b )
For intensity of fringe at angle θ, the relation is
I = I₀ cos²θ
I / I₀ = cos²θ/2
Given I / I₀ =0. 1
0.1 = cos²θ/2
θ/2 = 71.5
θ = 143 °