R01= 14.1 Ω
R02= 0.03525Ω
<h3>Calculations and Parameters</h3>
Given:
K= E2/E1 = 120/2400
= 0.5
R1= 0.1 Ω, X1= 0.22Ω
R2= 0.035Ω, X2= 0.012Ω
The equivalence resistance as referred to both primary and secondary,
R01= R1 + R2
= R1 + R2/K2
= 0.1 + (0.035/9(0.05)^2)
= 14.1 Ω
R02= R2 + R1
=R2 + K^2.R1
= 0.035 + (0.05)^2 * 0.1
= 0.03525Ω
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Answer:
B) 5.05
Explanation:
The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:
Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2
Given that:
Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05
Maximum outer diameter = 35 + 0.05 = 35.05
Minimum inner diameter = 25 - 0.05 = 24.95
Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05
or
Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05
Therefore the LMC wall thickness is 5.05
Hopefully that helps you out and is this for history or science?
Answer:
± 0.003 ft
Explanation:
Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.
Let L' = our distance and L = our tape measure
So, L' = 100L
Now by error determination ΔL' = 100ΔL
Now ΔL' = ± 0.30 ft
ΔL = ΔL'/100
= ± 0.30 ft/100
= ± 0.003 ft
So, the maxim error per tape is ± 0.003 ft
