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aalyn [17]
4 years ago
10

In natural convection, the fluid is forced to flow over a surface or in a tube by an external mean such as a pump or a fan. a)-

True b)- False
Engineering
1 answer:
lesantik [10]4 years ago
6 0

Answer:

(b) FALSE

Explanation:

Here in the statement it is given that fluid is force to flow over a surface means it is a forced convention in natural convention we do not need any force for fluid flow. In natural convention fluid flow itself on the other hand in forced convention we need force for fluid flow so this given statement is false statement as the fluid is force to flow

So the given statement is false statement  

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Explain why surface temperature increases when two bodies are rubbed against each other. What is the significance of temperature
Ronch [10]

Answer:

The surface temperature increases when two bodies are rubbed against each other due to friction.

Explanation:

No object has a perfectly even surface. So, when two bodies with uneven surfaces are rubbed against each other, they experience friction.

Friction is a resistance experienced by the two bodies when they are moved against each other.

The friction between the two surfaces, converts the kinetic energy of the movement to the thermal energy.

Thus, resulting in rise in the surface temperature of the two bodies.

Therefore, when two bodies are rubbed against each other, the surface temperature increases due to friction.

7 0
4 years ago
Gina is about to use a fire extinguisher on a small fire. What factor determines the
Effectus [21]

Answer:

Each portable fire extinguisher should only be used for its specific type of fire. Class A extinguishers could cause a Class B fire to spread or electrocution in a Class C fire. A Class B extinguisher could fail to completely extinguish a Class A fire, causing the flame to re-ignite later.

7 0
3 years ago
You are given a noninverting 741 op-amp with a dc-gain of 23.6 dB. The input signal to this amplifier is;Vin(t) = (0.18)∙cos(2π(
Vsevolod [243]

Answer:

Output voltage equation is V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

Explanation:

Given:

dc gain A = 23.6 dB

Input signal V_{in} (t) = 0.18 \cos (2\pi (57000)t +18.3)

Now convert gain,

A = 10^{\frac{23.6}{20} } = 15.13

DC gain at frequency f = 0 is given by,

  A = \frac{V_{out} }{V_{in} }

V_{out} =AV_{in}

V_{out} = 15.13 \times   0.18 \cos (2\pi (57000)t +18.3)

At zero frequency above equation is written as,

V_{out} = 2.72 \times \cos 18.3

V_{out} = 2.72

Now we write output voltage as input voltage,

V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

Therefore, output voltage equation is V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

7 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
(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.3 mm in diameter in a steel alloy when a load of 1000
vampirchik [111]

Answer:

(a)  We are asked to compute the Brinell hardness for the given indentation. for HB, where P= 1000 kg, d= 2.3 mm, and D= 10 mm.  

Thus, the Brinell hardness is computed as

HB=2P/\pi D{D-\sqrt{D^2-d^2}

=2*1000hg/\pi (10mm)[10mm-\sqrt{(1000^2-(2.3mm)^2} ]

(b)    This  part  of  the  problem  calls  for  us  to  determine  the  indentation diameter d which  will  yield  a  270  HB  when P=  500  kg.  

d=\sqrt{D^2-[D-\frac{2P}{(HB)\pi D} } ]^2\\=\sqrt{(10mm)^2-[10mm-\frac{2*500}{450( \pi10mm)} } ]^2

6 0
4 years ago
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