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One mole of iron (6 × 1023 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-center distance between atoms is 2.28 × 10-10 m. You have a long thin bar of iron, 2.7 m long, with a square cross section, 0.07 cm on a side. You hang the rod vertically and attach a 100 kg mass to the bottom, and you observe that the bar becomes 2.70 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in iron. 1) What is the spring stiffness of the entire wire, considered as a single macroscopic (large scale), very stiff spring? ks= N/m 2) How many side-by-side atomic chains (long springs) are there in this wire? This is the same as the number of atoms on the bottom surface of the iron wire. Note that the cross-sectional area of one iron atom is (2.28× 10-10)2 m2. Number of side-by-side long chains of atoms =

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Answer:

a) k[tex]_{s}[/tex] = 3.6 N/m ; b) 9.43 * 10¹²; c) 1.18 * 10¹¹; d) 75.08 N/m

Step-by-step explanation:

Length of iron bar = L = 2.7m;

side length of cross section = a = 0.07cm = 0.0007m;

x = 2.7 cm = 0.027m;

m = 100kg;

ρ = 7.87gm/cm³;

da = 2.28 * 10⁻¹⁰m;

a)

Fnet = F - mg

where Fnet = 0

So,

F = mg where F=k[tex]_{s}[/tex]x

k[tex]_{s}[/tex] = mg/x = 100*9.8/0.027

k[tex]_{s}[/tex] = 3.6 N/m

b)

Nchain = Aw/Aa =a²/(da)²

                           = (o.ooo7)²/(2.28 * 10⁻¹⁰)²

                           = 9.43 * 10¹²

c)

Nbond = L/da = 2.7/2.28 * 10⁻¹⁰

                       = 1.18 * 10¹¹

d)

Spring stiffness of wire = ksi = (Nbondk[tex]_{s}[/tex])/Nchain

                                      = [(1.18* 10¹¹)(6*10⁴)]/(9.43 * 10¹²)

                                      = 75.08 N/m

The spring stiffness of the wire is 75.08 N/m and the total number of side-by-side atomic chains in the wire are [tex]9.43\times 10^{12}[/tex] and trhis can be determined by using the given data.

Given :

  • One mole of iron (6 × 1023 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-center distance between atoms is 2.28 × 10-10 m.
  • You have a long thin bar of iron, 2.7 m long, with a square cross section, 0.07 cm on a side.
  • You hang the rod vertically and attach a 100 kg mass to the bottom, and you observe that the bar becomes 2.70 cm longer.

1) The net force is given by:

[tex]\rm F_N=F-mg[/tex]

The net force is zero so, the above expression becomes:

F = mg

where the value of F = Kx

Kx = mg

[tex]\rm K=\dfrac{mg}{x}=\dfrac{100\times 9.8}{0.027}[/tex]

K = 3.6 N/m

2) The total number of side by side atomic chains in the wire is:

[tex]\rm N_{chain}=\dfrac{A_w}{A_a}[/tex]

[tex]\rm N_{chain}=\dfrac{(0.0007)^2}{(2.28\times10^{-10})^2}[/tex]

[tex]\rm N_{chain}=9.43\times 10^{12}[/tex]

3) Now, the value of N bond is given by:

[tex]\rm N_{bond}=\dfrac{L}{d_a}=\dfrac{2.7}{2.28\times 10^{-10}}[/tex]

[tex]\rm N_{bond}=1.18\times 10^{11}[/tex]

4) Now, the value of the spring stiffness is given by:

[tex]\rm Stiffness = \dfrac{N_{bond}K_s}{N_{chain}}[/tex]

[tex]\rm Stiffness = \dfrac{1.18\times 10^{11}\times 6\times 10^{4}}{9.43\times 10^{12}}[/tex]

Stiffness = 75.08 N/m

For more information, refer to the link given below:

https://brainly.com/question/13095331

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