Answer:
D. 30 Ω
Explanation:
R₁ + R₂ = 33
-0.0005 R₁ + 0.005 R₂ = 0
Solve the system of equations for R₁.
-0.0005 R₁ + 0.005 R₂ = 0
0.005 R₂ = 0.0005 R₁
R₂ = 0.1 R₁
R₁ + R₂ = 33
R₁ + 0.1 R₁ = 33
1.1 R₁ = 33
R₁ = 30
Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.
While outside the shell, the electric field is given by: k(q + Q)/r²
Where;
K= is a constant which is given as, 8.99 x 10^9 N m² / C².
Q= source charge which creates the electric field
q= is the test charge which is used to measure the strength of the electric field at a given location.
r= is the radius
Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.
It is D, you can not replace minerals. This makes them valuable.
