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zhuklara [117]
2 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]2 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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Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m
Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0
2 years ago
Prompt the user to input an integer, a double, a character, and a string, storing each into separate variables. Then, output tho
Likurg_2 [28]

Answer:

See explanation

Explanation:

//Include the

//required header files.

#include <stdio.h>

//Define the

//main() function.

int main(void) {

//Declare the

//required variables.

char input_char;

int input_int;

double input_double;

char input_string[100];

//Prompt the user

//to enter an integer.

printf("Enter integer: ");

//Read and store

//the integer.

scanf("%d", &input_int);

//Prompt the user

//to enter a double value.

printf("Enter double: ");

//Read and store

//the double value.

scanf("%lf", &input_double);

//Prompt the user

//to enter a character.

printf("Enter character: ");

//Read and store

//the character.

scanf(" %c", &input_char);

//Prompt user to

//enter the string

printf("Enter string: ");

//Read and

//store the string.

scanf("%s", input_string);

//(1)

//Display the values.

printf("%d %lf %c %s\n",

input_int, input_double,

input_char, input_string);

//(2)

//Display the values

//in reverse order.

printf("%s %c %lf %d\n",

input_string, input_char,

input_double, input_int);

//(3)

//Cast the double to

//an integer and display it.

printf("%lf cast to an integer is %d",

input_double, (int)(input_double));

//Return from the

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return 0;

}

4 0
3 years ago
What is the importance of knowing the parts and function of the ignition system in servicing your own vehicle
lisov135 [29]

Answer:

El conocer la función del sistema nervioso permite comprender la estructura del cuerpo. Puesto que, toda acción que se se realice en ella dependerá del sistema nervioso para que sea eficaz

4 0
1 year ago
Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and
Alborosie

Answer:

V_{pp}=2V

Explanation:

Source Voltage V= 120V

Frequency f=60Hz

Peak output voltage Vp=15V

Peak Output Voltage with filter V_p'=14V

Generally the equation for Peak to peak voltage is mathematically given by

V_p'=V_p-\frac{V_{pp}}{2}

Therefore

V_{pp}=2(V_p-v_p')

V_{pp}=2(15-14)

V_{pp}=2V

5 0
2 years ago
A compressible clay layer has a thickness of 3.8 m. After 1.5 yr, when the clay is 50% consolidated, 7.3 cm of settlement has oc
grigory [225]

The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.

<h3>How to determine the amount of settlement?</h3>

For a layer of 3.8 m thickness, we were given the following parameters:

U = 50% = 0.5.

Sc = 7.3 cm.

For Sf, we have:

Sf = Sc/U

Sf = 7.3/0.5

Sf = 14.6

Therefore, Sf for a layer of 38 m thickness is given by:

Sf = 14.6 × 38/3.8

Sf = 146 cm.

At 50%, the time for a layer of 3.8 m thickness is: t_{50} = 1.5 year.

At 50%, the time for a layer of 38 m thickness is:

t_{50} = 1.5 × (38/3.8)²

t_{50} = 150 years.

For the thickness of 38 m, U₂ is given by:

\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25  \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05

The new settlement after 1.5 year is:

Sc = U₂Sf

Sc = 0.05 × 146

Sc = 7.3 cm.

For time, t₂ = 5 year:

U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25  \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09

The new settlement after 5 year is:

Sc = U₂Sf

Sc = 0.09 × 146

Sc = 13.14 cm.

Read more on clay layer here: brainly.com/question/22238205

8 0
2 years ago
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