a) The energy stored in a C—C bond is 348 10.mola and its length is 0.154 nm. Assume that the bond stretches twice its equilibrium length before rupture. Calculate the lifting capacity of a single C—C bond and its tensile strength.

b) How many (12, 0) zigzag nanotubes are required to lift an average person (wt. = 150 pounds, F _c.c = 3.73 x 10 ^-9 N). There are 12 carbon-carbon bonds aligned vertically along the long axis of the nanotube and the diameter of this carbon nanotube is 1.019 nm.

c) What is the diameter of the cable? Assume that there is tight packing (i.e. tubes are touching with packing efficiency in two-dimensions of 0.91).

d) What is the tensile strength of this cable?

Answer rating: 100% (QA)

a I Force F E bond x strech 348 10 3 1 mol6022 10 23 0154 10 9 375 ... View the full answer