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Marta_Voda [28]

3 years ago

9

The inner surface of a hollow cylinder is subjected to tangential and axial stresses of 40,000 and 24,000 psi, respectively. Det

ermine the maximum shear stress at the inner surface, if the cylinder is pressurized to 10,000 p

1 answer:

Furkat [3]3 years ago

5 0

Answer:

15,000 psi

Explanation:

The solution / solving is attach below.

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Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

Allisa [31]

Answer:

theoretical fracture strength = 16919.98 MPa

Explanation:

given data

Length (L) = 0.28 mm = 0.28 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa

solution

we get here theoretical fracture strength s that is express as

theoretical fracture strength = $s_{0} \times \sqrt{\frac{L}{r} }$ .............................1

put here value and we get

theoretical fracture strength = $1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }$

theoretical fracture strength = $16919.98 \times 10^6$

theoretical fracture strength = 16919.98 MPa

3 0

3 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

13 Which pump is the most appropriate for low head high discharge capacity installations?

gavmur [86]

Options are missing and they are;

A. Propeller pump

B. Jet (mixed flow) pump

C. Centrifugal pump

D. Positive displacement pump

Answer:

C. Centrifugal pump

Explanation:

Dynamic pumps which are also known as centrifugal pumps are the ones widely used when we require high discharge. This is because they deal with transfer of fluids with low viscosity in flow rates that are high. Another reason is that they are also very good with low pressure installation.

5 0

3 years ago

If a sky diver decides to jump off a jet in Arkansas

andrew-mc [135]

Answer:

The answer to this question can be defined as follows:

Explanation:

The sky driver began his sky journey from Arkansas, drove across the Tennessee River then landed in North Carolina. He returned to both the north in the very same direction. He began with NC, traveled through Tennessee, eventually lands in Arkansas. But North Carolina has been in the third state on which skydiver was traveling over, and It's also more than 700 miles from Arkansas to the NC.

6 0

3 years ago

Consider a magnetic disk consisting of 16 heads and 400 cylinders. This disk has four 100-cylinder zones with the cylinders in d

e-lub [12.9K]

Answer:

Disk Capacity = 45056000 bytes

Optimal skew = 26.455 ≈ 27

Maximum Data transfer rate = 262.5 GB per second

Explanation:

given data

heads = 16

cylinders = 400

cylinder zones = 100

each sector contains = 512 bytes

average seek time = 1 msec

disk rotates = 7200 RPM

to find out

disk capacity, optimal track skew, and maximum data transfer rate

solution

first we get total number of sectors that is

total number of sector = number of zones × (number of sectors in different zones

total number of sector = 100 × (160+200+240+280)

total number of sector = 88000

so

Disk Capacity = total number of sectors × size of each sector

Disk Capacity = 88000 × 512

Disk Capacity = 45056000 bytes

and

Rotation time = $\frac{60}{7200}$

Rotation time = 8.33 milli seconds

so Optimal number of sectors in a track = average of ( 160,200,240,280 )

Optimal number of sectors in a track = 220

now New sector is read every $\frac{8.33}{220}$ i.e = 0.0378 ms

here Optimal skew = seek time ÷ new sector read time

Optimal skew = $\frac{1}{0.0378}$

Optimal skew = 26.455 ≈ 27

and

here we know that for maximum transfer rate we will select cylinder with maximum number of sectors i.e here 280 sectors

so

capacity of one track with maximum = 280 × 512

capacity of one track with maximum = = 143360 bytes

and Number of rotations in 1 second is = $\frac{7200}{60}$

Number of rotations in 1 second is = 120

so Data transfer rate = Number of heads × Capacity of one track × Number of rotations in one second

Data transfer rate = 16 × 143360 × 120

Data transfer rate = 275251200 bytes per second

Data transfer rate = 262.5 GB per second

3 0

3 years ago