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3 years ago

Construction lines are thick lines true false

Engineering

2 answers:

n200080 [17]3 years ago

5 0

True hope this helps

ANTONII [103]3 years ago

4 0

Answer:

true

Explanation:

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2. When performing an alignment, what action should be taken immediately after putting a vehicle on the rack?

mash [69]

D...................

3 0

2 years ago

A 2-cm-diameter vertical water jet is injected upward by a nozzle at a speed of 15 m/s. Determine the maximum weight of a flat p

Ede4ka [16]

Answer:58.28 N

Explanation:

Given data

$dia. of nozzle \left ( d\right )=2 cm$

$initial velocity\left ( u\right )=15 m/s$

$height\left ( h\right )=2m$

Now velocity of jet at height of 2m

$v^2-u^2=2gh$

$v^2=15^2-2\left ( 9.81\right )\left ( 2\right )$

$v=\sqrt{185.76}=13.62 m/s$

$Now\ forces\ on\ plate\ are\ weight\left ( Downward\right ) and jet\ force\left ( upward\right )$

equating them

W= $\left ( \rho Av\right )v$

W= $10^{3}\times \frac{\pi}{4}\left ( 0.02\right )^2\times 13.62^2$

W=58.28 N

7 0

3 years ago

Georgia Tech is committed to creating solutions to some of the world’s most pressing challenges. Tell us how you have improved o

Kaylis [27]

Answer:

Georgia Tech is committed to WGAR 53566 THE ANSWER IS JELLY IS KING AND THE JELLY IS KING AND hope to improve the human condition in your community.

Explanation:

6 0

3 years ago

A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the

Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + $\frac{u^2}{30(\frac{a}{g}\pm G)}$ ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5 + $\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}$

solve it we get

SSD = 644 ft

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1 - $cos (\frac{28.65 SSD}{Rv})$ ) .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1 - $cos (\frac{28.65 \times 664}{2294})$ )

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R = $\frac{u^2}{15(e+f)}$ ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = $\frac{65^2}{15(e+0.10)}$

solve it we get

e = 2%

3 0

3 years ago

Given a matrix, clockwise-rotate elements in it. Please add code to problem3.cpp and the makefile. Use the code in p3 to test yo

rusak2 [61]

Answer:

/* C Program to rotate matrix by 90 degrees */

#include<stdio.h>

int main()

{

int matrix[100][100];

int m,n,i,j;

printf("Enter row and columns of matrix: ");

scanf("%d%d",&m,&n);

/* Enter m*n array elements */

printf("Enter matrix elements: \n");

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

{

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

{

scanf("%d",&matrix[i][j]);

}

/* matrix after the 90 degrees rotation */

printf("Matrix after 90 degrees roration \n");

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

{

for(j=m-1;j>=0;j--)

{

printf("%d ",matrix[j][i]);

}

printf("\n");

}

return 0;

}

5 0

3 years ago