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DiKsa [7]
3 years ago
7

Just some random stufff

Engineering
1 answer:
NeTakaya3 years ago
5 0

Answer:

Nice!

Explanation:

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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

//main() function.

return 0;

}

4 0
3 years ago
In a 1D compression test with double drainage, the pore pressure readings are practically zero after 8 minutes for a clay sample
frosja888 [35]

Answer:

The duration of the consolidation process for the same clay is 32 min

Explanation:

for clay 1:

t1=0

H1=thickness=2 cm

for the clay 2:

t2=?

H2=2 cm

The time factor is equal to:

T=(\frac{Cv}{d^{2} })t

where Cv is the coefficient of consolidation

(\frac{Cvt}{d^{2} })_{1}=  (\frac{Cvt}{d^{2} })_{2}

if Cv is constant, we have:

(\frac{t1}{(\frac{H1}{2}) ^{2} })_{1}=(\frac{t2}{H2^{2} })_{2}\\\frac{0}{(\frac{2}{2})^{2})  }=\frac{t2}{2^{2} }

Clearing t2:

t2=32 min

3 0
3 years ago
A pressure gage and a manometer are connected to a compressed air tank to measure its pressure. If the reading on the pressure g
mixer [17]

Answer:

h=1.122652m

Explanation:

Assuming density of air = 1.2kg/m³

the differential pressure is given by:

h^{i} =h(\frac{density of manometer}{density of flowing air}-1)\\h^{i} =h(\frac{1000}{1.2}-1)\\ h^{i}=832.33h...(1)\\\\but\\ h^{i} =\frac{change in pressure}{air density*g} \\\\h^{i} =\frac{11*10^3}{1.2*9.81}\\\\h^{i}= 934.42...(2)\\\\equating, \\\\934.42=832.33h\\\\h=1.122652m

7 0
4 years ago
A company specification calls for a steel component to have a minimum tensile strength of 1240 MPa. Tension tests are conducted
polet [3.4K]

Answer:

HV = 372.4 KGf/mm^2

HR_C = 39.445

Explanation:

given data:

minimum tensile strength = 1240 MPa

we knwo that \sigma_{vrs}  and brinell hardness number (HB} relation given as

\sigma_{vrs} = 3.4 \times H.B

H.B = \frac{1240}{3.4} = 364.7

1) relation betwen H.B and Rockwell is

 HR_C = 6.96 + 0.089 HB

 HR_C = 6.96 + (0.089\times 365} = 39.445

2) relation between HB  and Vickers hardnes HV is given as

 HB \times 1.05 = HV + 10.5

HV = HB\times 1.05 - 10.5

HV = 372.4 KGf/mm^2

4 0
4 years ago
Is santa real or nah is santa real or nah
Elena L [17]

Answer:

nah

Explanation:

3 0
3 years ago
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