Consider a uniformly charged sphere (an insulating sphere of radius R) and a spherical Gaussian surface with radius R 2 concentr
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Answer:
d_{b} = 2 d_{a}
Explanation:
The electrical resistance for a cylindrical wire is described by the expression
R = ρ L / A
The area of a circle is
A = π r²
r = d / 2
A = π d²/4
We substitute
R = ρ L 4 /π d²
Let's apply this expression to our case, they indicate that the resistance of wire A is 4 times the resistance of wire B
= 4 R_{b}
We substitute
ρ 4/π ² = 4 (ρ 4/π d_{b}²)
1 / d_{a}² = 4 / d_{b}²
d_{a} = d_{b} / 2
Answer:
C.) 1.5 kg
Explanation:
Start with the equation:
Plug in what you know, and solve:
Find matching soluation:
C.) 1.5 kg