Question

Hi guys...can anyone answer this? i can give a medal:)

How much lead is needed to coat a cable, 55mm in diameter, if the cable is 5 km long and the lead coat is 4 mm thick? Use 11.4x10^3 kg/m^3 for the density of the lead.

Answer 1

^^

Answer 2

this is what i've got from your equation ...
m=135432

Answer 3

no... i just plug it in my calcu

Answer 4

where can we use the 4mm thickness?

Answer 5

So the total diameter will be 59 mm

So,

The total volume of lead will be the volume of the cable plus the thickness of the coat minus the volume of the cable .

So, (.059^2)(pi)(5000) - (.055^2)(pi)(5000) = 7.16 m^3

then m =(11.4 x10^3)(7.16) = 81624 kg

Sorry did not see the 4 mm at the beginning hopefully this makes more sense to you now :)

Sorry for any confusion :P

Answer 6

Does this make sense are you there?