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Answer:
1) R₁ > R₃ > R₂ correct B , 2) the wire that dissipates the most is wire 2
Explanation:
1) The resistance of a wire is given by the expression
R =
where ρ is the resistivity of the material, l the length of the wire and A the area of the wire
The area is given by
A = π r² = π d² / 4
we substitute
R = (ρ 4 /π)
the amount in parentheses is constant for this case
let's analyze the situation presented, to find the resistance of each wire
* indicate l₁ = l₂ and d₂ = 2 d₁
the resistance of wire 1 is
R₁ = (ρ 4 /π)
the resistance of wire 2 is
R₂ = (ρ 4 /π) \frac{l_2}{d_2^2}
R₂ = (ρ 4 /π)
R₂ = (ρ 4 /π ) ¼
R₂ = ¼ R₁
* indicate that d₂ = d₃ and l₃ = 2 l₂
R2 = (ρ 4 /π)
the resistance of wire 3 is substituting the indicated condition
R3 = (ρ 4 /π 2) \frac{l_3}{d_3^2}
R3 = (ρ 4 /π)
R3 = 2 R₂
let's write the relations obtained
R₁ = (ρ 4 /π)
R₂ = ¼ R₁
R₃ = 2 R₂
let's write everything as a function of R1
R₁ =(ρ 4 /π)
R₂ = ¼ R₁
R₃ = ½ R₁
the resistance of the wire in decreasing order is
R₁ > R₃ > R₂
2) The power dissipated by a wire is
P = V I
the voltage is
V = I R
I = V / R
substituting
P = V² / R
therefore the power dissipated by each wire is
wire 1
P₁ = V² / R₁
wire 2
P₂ = V² / R₂
P₂ =
P₂ = 4 P₁
wire 3
P₃ = V² / R₃
P₃ =
P₃ = 2 P₁
Therefore, the wire that dissipates the most is wire 2