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nadya68 [22]
2 years ago
9

In case of damaged prestressed concrete I girders which are used for restoring strength?

Engineering
1 answer:
Ahat [919]2 years ago
7 0

Answer: The use of post-tensioning rods is useful in this situation.

Explanation:

The prestressed concrete has not been accepted as a good building material. The high tensile strength of steel is combined with the concrete to give the compressive strength to the concrete.

Post-tensioning rods of steel along with jacking can be used to restore the strength of the prestressed girder and it can begin with the calculated preload. This way the damaged concrete can be repaired and restored.  

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SUBJECT : SCIENCE
juin [17]

because not all substances react the same to temperature changes. If you heat a metal and an organic substance and you turn the temperature up, the organic substance like water will react (boil) while most metals need higher temperature to react.

3 0
2 years ago
(40 points) Program the following sorting algorithms: InsertionSort, MergeSort, and QuickSort. There are 9 test les uploaded for
babunello [35]

Answer:

Explanation:

MERGE SORT

#include<stdlib.h>

#include<stdio.h>

#include<string.h>

void merge(int arr[], int l, int m, int r)

{

int i, j, k;

int n1 = m - l + 1;

int n2 = r - m;

 

int L[n1], R[n2];

for (i = 0; i < n1; i++)

L[i] = arr[l + i];

for (j = 0; j < n2; j++)

R[j] = arr[m + 1+ j];

i = 0;

j = 0;

k = l;

while (i < n1 && j < n2)

{

if (L[i] <= R[j])

{

arr[k] = L[i];

i++;

}

else

{

arr[k] = R[j];

j++;

}

k++;

}

while (i < n1)

{

arr[k] = L[i];

i++;

k++;

}

while (j < n2)

{

arr[k] = R[j];

j++;

k++;

}

}

void mergeSort(int arr[], int l, int r)

{

if (l < r)

{

int m = l+(r-l)/2;

mergeSort(arr, l, m);

mergeSort(arr, m+1, r);

merge(arr, l, m, r);

}

}

void printArray(int A[], int size)

{

int i;

for (i=0; i < size; i++)

printf("%d ", A[i]);

printf("\n");

}

int main()

{

int arr[1000] = {0};

int arr_size =0;

int data;

char file1[20];

strcpy(file1,"data.txt");

FILE *fp;

fp = fopen(file1,"r+");

if (fp == NULL) // if file not opened return error

{

perror("Unable to open file");

return -1;

}

else

{

fscanf (fp, "%d", &data);    

arr[arr_size]=data;

arr_size++;

while (!feof (fp))

{  

fscanf (fp, "%d", &data);  

arr[arr_size]=data;

arr_size++;    

}

}

printf("Given array is \n");

printArray(arr, arr_size);

mergeSort(arr, 0, arr_size - 1);

printf("\nSorted array Using MERGE SORT is \n");

printArray(arr, arr_size);

return 0;

}

3 0
3 years ago
An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current trave
german

Answer:

true

Explanation:

true An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE

3 0
2 years ago
(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows
maxonik [38]

Answer:

diameter of the sprue at the bottom is 1.603 cm

Explanation:

Given data;

Flow rate, Q = 400 cm³/s

cross section of sprue: Round

Diameter of sprue at the top d_{top} = 3.4 cm

Height of sprue, h = 20 cm = 0.2 m

acceleration due to gravity g = 9.81 m/s²

Calculate the velocity at the sprue base

V_{base} = √2gh

we substitute

V_{base} = √(2 × 9.81 m/s² × 0.2 m )

V_{base} = 1.98091 m/s

V_{base} = 198.091 cm/s

diameter of the sprue at the bottom will be;

Q = AV = (πd_{bottom}^2/4) × V_{base}

d_{bottom} = √(4Q/πV_{base})

we substitute our values into the equation;

d_{bottom} = √(4(400 cm³/s) / (π×198.091 cm/s))

d_{bottom}  = 1.603 cm

Therefore, diameter of the sprue at the bottom is 1.603 cm

6 0
2 years ago
A heating system must maintain the interior of a building at 20°C during a period when the outside air temperature is 5°C and th
Anettt [7]

Answer:

a. W = 51,194.54 kJ

b. W = 102,390 kJ

c. W = 153,585 kJ

Explanation:

(COP)_{HP} =\frac{Desired-effectx}{Work-done}= \frac{Q_{1} }{W} \\\\(COP)_{HP} =(COP)_{Ideal}\\\\\frac{Q1}{W} =\frac{T_{1} }{T_{1} -T_{2} }

W=Q_{1} \frac{T_{1}-T_{2}  }{T_{1} }

a. the ground at 15°C.

T_{1}=20°C = 273 K + 20 = 293 K

T_{2}=15°C = 273 K + 15 = 288 K

Q_{1}=3x10^{6} kJ

W=3x10^{6} kJ \frac{293 K-288 K}{293 K}=3x10^{6} kJ \frac{5 K}{293 K}=3x10^{6} kJ x 0.017065}

W = 0.051195x10^{6} kJ

W = 51,194.54 kJ

b. a pond at 10°C.

T_{2}=10°C = 273 K + 10 = 283 K

W=3x10^{6} kJ \frac{293 K-283 K}{293 K}=3x10^{6} kJ \frac{10 K}{293 K}=3x10^{6} kJ x 0.034130}

W = 0.102390x10^{6} kJ

W = 102,390 kJ

c. the outside air at 5°C.

T_{2}=5°C = 273 K + 5 = 278 K

W=3x10^{6} kJ \frac{293 K-278 K}{293 K}=3x10^{6} kJ \frac{15 K}{293 K}=3x10^{6} kJ x 0.051195}

W = 0.153585x10^{6} kJ

W = 153,585 kJ

Hope this helps!

3 0
2 years ago
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