Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
I believe the correct answer from the choices listed above is option C. The instrument that is <span>best suited for measuring the dimensions of a shoebox would be a ruler. A triple-beam balance is for measuring mass. A volumetric flask is for volume. A caliper is measuring lengths of small objects.</span>
1.c
2.b
3.a
4.c
5.b
I hop this helps
1 coulomb of electric charge is carried by 6.25 x 10^18 electrons
1 Ampere = 1 coulomb per second
10 A = 10 coulombs per second
(2.0 x 10^20 electrons) x (coul / 6.25 x 10^18 electrons) / (10 coul/sec) =
(2.0 x 10^20) / (6.25 x 10^18 x 10) sec = <em>3.2 seconds</em>
Answer:
<em><u>In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.</u></em>
