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Artyom0805 [142]
3 years ago
8

The densities of the elements K, Ca, Sc, and Ti are 0.86, 1.5, 3.2, and 4.5 g/cm³, respectively. One of these elements crystalli

zes in a body-centered cubic structure; the other three crystallize in a face-centered cubic structure.
Which one crysta llizes in the body-centered cubic structure? Justify your answer
Physics
1 answer:
IgorC [24]3 years ago
4 0

Answer:

K

Explanation:

Given that

Elements                                    Density

K                                                  0.86 g/cm³

Ca                                                 1.5 g/cm³

Sc                                                  3.2 g/cm³

Ti                                                    4.5 g/cm³

As we know that density of the BCC ( body centered cubic )element is lower than the FCC(Face centered cubic )element.

In the given problem only K have lower density all of these element that is why is in BCC and all others are in FCC.

The packing efficiency of the FCC structure element is higher than the BCC element that is why FCC is more denser than BCC.And also FCC having good packing of element.

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Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = q_{int} /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

       

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

   

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

 

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

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