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3 years ago

By using hydraulic cylinders of

Engineering

1 answer:

JulsSmile [24]3 years ago

3 0

Answer:

Explanation:

Hydraulic systems can increase or decrease the force applied to them. To make the force larger, the pressure is applied to a larger area. For example, if a 100-N force is applied to the left cylinder in (Figure) and the right cylinder has an area five times greater, then the output force is 500 N.

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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

Q7. A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cros

umka2103 [35]

Answer:

11.2mm or 0.45in

Explanation:

The percent cold work, attendant tensile strength and ductility if drawing is carried out without interruption is given by the equation you will find in the attached file.

Please go through the attached file for a step by step solution to this question.

5 0

3 years ago

Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average valu

Varvara68 [4.7K]

The question is not complete. We are supposed to find the average value of v_o.

Answer:

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

3/(T/4) = 0.7/t1

So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)

v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)

v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

8 0

3 years ago

A freezer is maintained at 20°F by removing heat from it at a rate of 75 Btu/min. The power input to the freezer is 0.70 hp, and

Igoryamba

Answer:

Explanation:

Cop of reversible refrigerator = TL / ( TH - TL)

TL = low temperature of freezer = 20 °F

TH = temperature of air around = 75 °F

Heat removal rate QL = 75 Btu/min

W actual, power input = 0.7 hp

conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9

COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))

COP reversible = 480 / 55 = 8.73

irreversibility expression, I = W actual - W rev

COP r = QL / Wrev

W rev = QL / COP r where 75 Btu/min = 1.76856651 hp where W actual = 0.70 hp

a) W rev = 1.76856651 hp / 8.73 = 0.20258 hp is reversible power

I = W actual - W rev

b) I = 0.7 hp - 0.20258 hp = 0.4974 hp

c) the second-law efficiency of this freezer = W rev / W actual = 0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %

8 0

4 years ago

Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5

pochemuha

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

$Aa.Va=Ab.Vb=Q$

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

$Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3$

$Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3$

Using the volume rate:

$Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s$

$Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s$

Assuming no losses, the energy equation for fluids can be written as:

$Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb$

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

$Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za$

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

$Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m$

$Pb = 70000\ Pa-27303\ Pa - 14715\ Pa$

$Pb = 27,996\ Pa = 28\ kPa$

6 0

3 years ago