Iron has a mass of 7.87 g per cubic centimeter of volume, and the mass of an iron atom is 9.27 × 10-26 kg. If you simplify and treat each atom as a cube, (a) what is the average volume (in cubic meters) required for each iron atom and (b) what is the distance (in meters) between the centers of adjacent atoms?

Answers

Answer 1

Answer: known values; mass per / volume and mass of each atom

average volume= volume per atom.
density*mass per atom = volume per atom

1cm^3 / 7.87 gram * 9.27 *10^-26 kg /atom* 1000 gram / kg=
=1.178 * 10^-22 cm^3 per atom

Convert that to mass
1 cm^3 * (1 m / 100 cm)^3
average volume= 1.178 *10^-29 m^3 per atom

distance:

Volume of a cube is L^3

so L = (1.178*10^-29)^(1/3)

Half way through the cube is distance 1/2 L.
Second cube is 1/2 L
So distance between the center of 2 cubes = L

=2.28 *10^-10 m

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Answers

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Answers

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Answers

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Answers

I'll tell you how I look at this, although I may be missing something important.

Position = x(t) = 0.5 sin(pt + p/3)

Speed = position' = x'(t) = 0.5 p cos(pt + p/3)

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In order to evaluate this, don't I still have to know what 'p' is ? ?

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Answers

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Answers

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