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zhuklara [117]
3 years ago
13

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. Calculate the number of atoms in t

he unit cell and diameter of the metal atom.

Engineering
1 answer:
uysha [10]3 years ago
8 0

Answer:288 pm

Explanation:

Number of atoms(s) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- whereas

For an FCC lattices √2a =4r =2d

Therefore d = a/√2a = 408pm/√2a= 288pm

I think with this step by step procedure the, the answer was clearly stated.

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Chemical manufacturers must present which information on the product's label?
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hazardous chemicals leaving the workplace is labeled, tagged or marked with the following information: product identifier; signal word; hazard statement

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The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam
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Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

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Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

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Answer:

public static int average(int j, int k) {

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Method declared: average of integer data type

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