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(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

Student A says hazardous waste can take the form of solid, liquid, or gas. Student B says hazardous waste can only take the form

lina2011 [118]

Answer:

Student A

Explanation:

hope this helps have a great day

4 0

3 years ago

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

3 years ago

An oxygen–nitrogen mixture consists of 35 kg of oxygen and 40 kg of nitrogen. This mixture is cooled to 84 K at 0.1 MPa pressure

Tamiku [17]

Answer:

The mass of oxygen in liquid phase = 14.703 kg

The mass of oxygen in the vapor phase = 20.302 kg

Explanation:

Given that:

The mass of the oxygen $m_{O_2}$ = 35 kg

The mass of the nitrogen $m_{N_2}$ = 40 kg

The cooling temperature of the mixture T = 84 K

The cooling pressure of the mixture P = 0.1 MPa

From the equilibrium diagram for te-phase mixture od oxygen-nitrogen as 0.1 MPa graph. The properties of liquid and vapor percentages are obtained.

i.e.

Liquid percentage of $O_2$ = 70% = 0.70

Vapor percentage of $O_2$ = 34% = 0.34

The molar mass (mm) of oxygen and nitrogen are 32 g/mol and 28 g/mol respectively

Thus, the number of moles of each component is:

number of moles of oxygen = 35/32

number of moles of oxygen = 1.0938 kmol

number of moles of nitrogen = 40/28

number of moles of nitrogen = 1.4286 kmol

Hence, the total no. of moles in the mixture is:

$N_{total} = 1.0938+1.4286$

$N_{total} = 2.5224 \ kmol$

So, the total no of moles in the whole system is:

$N_f + N_g = 2.5224 --- (1)$

The total number of moles for oxygen in the system is

$0.7 \ N_f + 0.34 \ N_g = 1.0938 --- (2)$

From equation (1), let N_f = 2.5224 - N_g, then replace the value of N_f into equation (2)

∴

0.7(2.5224 - N_g) + 0.34 N_g = 1.0938

1.76568 - 0.7 N_g + 0.34 N_g = 1.0938

1.76568 - 0.36 N_g = 1.0938

1.76568 - 1.0938 = 0.36 N_g

0.67188 = 0.36 N_g

N_g = 0.67188/0.36

N_g = 1.866

From equation (1)

$N_f + N_g = 2.5224$

N_f + 1.866 = 2.5224

N_f = 2.5224 - 1.866

N_f = 0.6564

Thus, the mass of oxygen in the liquid and vapor phases is:

$m_{fO_2} = 0.7 \times 0.6564 \times 32$

$m_{fO_2} = 14.703 \ kg$

The mass of oxygen in liquid phase = 14.703 kg

$m_{g_O_2} = 0.34 \times 1.866 \times 32$

$m_{g_O_2} = 20.302 \ kg$

The mass of oxygen in the vapor phase = 20.302 kg

8 0

3 years ago

Describe how gene therapy can be a social issue and give rise to moral, ethical and legal debates. Justify your response in two

den301095 [7]

Answer: stigmatism and discrimination

Explanation:

Knowledge of genetic risks can lead to potential social and psychological consequences for the individual. Socially, knowledge from genetic tests may lead to stigmatization and discrimination within the community.

8 0

3 years ago

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