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kenny6666 [7]
3 years ago
10

Technician A says that excessive end play causes rapid wear of the sealing rings and possibly the sealing ring groove. Technicia

n B says that seal does not rotate in the groove; therefore, the groove could not wear. Which technician is correct
Engineering
1 answer:
DerKrebs [107]3 years ago
7 0

Answer:

Technician A only

Explanation:

Rapid wear of the sealing ring and the seal ring groove are result of excessive end play

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A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which th
Inessa [10]

Answer:

attached below

Explanation:

4 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 particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is imp
Nikolay [14]

Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

Mass = density × volume

Mass = 1200 × 6.5 × 10^-14

Mass M = 7.9 × 10^-11 kg

Using the formula

V = sqrt( 2Mg/ pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 × 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 × 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s

6 0
3 years ago
g A circular oil slick of uniform thickness is caused by a spill of one cubic meter of oil. The thickness of the oil slick is de
Anika [276]

Answer:

the rate of increase of radius is dR/dt = 0.804 m/hour = 80.4 cm/hour

Explanation:

the slick of oil can be modelled as a cylinder of radius R and thickness h, therefore the volume V is

V = πR² * h

thus

h = V / (πR²)

Considering that the volume of the slick remains constant, the rate of change of radius will be

dh/dt = V d[1/(πR²)]/dt

dh/dt = (V/π) (-2)/R³ *dR/dt

therefore

dR/dt = (-dh/dt)* (R³/2) * (π/V)

where dR/dt = rate of increase of the radius , (-dh/dt)= rate of decrease of thickness

when the radius is R=8 m , dR/dt is

dR/dt = (-dh/dt)* (R³/2) * (π/V) = 0.1 cm/hour *(8m)³/2 * π/1m³ *(1m/100 cm)= 0.804 m/hour = 80.4 cm/hour

4 0
3 years ago
In this problem set, you will implement multidimensional scaling (MDS) from scratch. You may use standard matrix/vector librarie
EleoNora [17]

Features of Multidimensional scaling(MDS) from scratch is described below.

Explanation:

Multidimensional scaling (MDS) is a way to reduce the dimensionality of data to visualize it.  We basically want to project our (likely highly dimensional) data into a lower dimensional space and preserve the distances between points.  

If we have some highly complex data that we project into some lower N dimensions, we will assign each point from our data a coordinate in this lower dimensional space, and the idea is that these N dimensional coordinates are ordered based on their ability to capture variance in the data.  Since we can only visualize things in 2D, this is why it is common to assess your MDS based on plotting the first and second dimension of the output.  

If you look at the output of an MDS algorithm, which will be points in 2D or 3D space, the distances represent similarity. So very close points = very similar, and points farther away from one another = less similar.

Working of MDS

The input to the MDS algorithm is our proximity matrix.  There are  two kinds of classical MDS that we could use:  Classical (metric) MDS is for data that has metric properties, like actual distances from a map or calculated from a vector .Nonmetric MDS is for more ordinal data (such as human-provided similarity ratings) for which we can say a 1 is more similar than a 2, but there is no defined (metric) distance between the values of 1 and 2.

Uses

Multidimensional scaling (MDS) is a means of visualizing the level of similarity of individual cases of a dataset. MDS is used to translate "information about the pairwise 'distances' among a set of n objects or individuals" into a configuration of n points mapped into an abstract Cartesian space.

8 0
3 years ago
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