Need help Please 12.5 Two 5-m-high telephone poles are separated by 45m. How hig

Question

Need help Please

12.5 Two 5-m-high telephone poles are separated by 45m. How high off the ground is the center of the cable if the sag allowed is 1 percent of the span length between poles?

12.6 When cables are installed in ducts, the ratio of the sum of the cross-sectional areas of the cables to the inside cross-sectional area of the duct is called the fill ratio. Suppose that three optical fiber cables which have outer diameters of 6.0, 7.5, and 9.0 mm are blown into a duct that has a 20-mm inner diameter. What is the fill ratio for this case?

Explanation / Answer

12.5:

centre of cable comes height h down then ,

h = 1% of 45m = 45 x 1 /100 = 0.45 m

height from ground, H = 45 - 0.45 = 44.55 m ......Ans

12.6:

sum of cross sectional area = pi [ (6/2)^2 + (7.5/2)^2 + (9/2)^2]

= 43.3125pi mm^2

cross section area of duct = pi ( 20/2)^2 = 100pi mm^2

Ration = 43.3125 / 100 = 0.433

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