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castortr0y [4]
3 years ago
5

CUP 6. A wire has a diameter of 0.032 inches. The AWG rating of this wire is most likely to be O A. 20 O B. 14. O C. 12. O D. 18

.​
Physics
1 answer:
never [62]3 years ago
7 0

Answer:

A. 20

Explanation:

The AWG rating of the wire can be determined by applying the formula;

d_{i} = 0.005 x 92^{\frac{(36 - s)}{39}} }

where d_{i} is the diameter of wire in inches, and s is the diameter of wire in AWG.

Given that d_{i} = 0.032 inches, then;

0.032 = 0.005 x 92^{\frac{(36 - s)}{39}} }

\frac{0.032}{0.005} =  92^{\frac{(36 - s)}{39}} }

6.4 = 92^{\frac{(36 - s)}{39}} }

Find the log of both sides to have,

log 6.4 = log 92^{\frac{(36 - s)}{39}} }

log 6.4 = (\frac{36 -s}{39}) log 92

\frac{log 6.4}{log92} = (\frac{36 -s}{39})

0.410523 = (\frac{36 -s}{39})

36 - s = 0.410523 x 39

         = 16.0104

⇒ s = 36 - 16.0104

      = 19.9896

s = 20

Therefore, the AWG of the wire is 20.

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