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Mazyrski
8 days ago
12

Where are revolved sections placed in a print? A) in between break lines B) cutting planes are used to identify their locations

C) in between section lines D) stand alone
Engineering
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A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7
mote1985 [299]

Answer:

r=0.31

Ф=18.03°

Explanation:

Provided:

Original diameter of bar = 75 mm

Diameter post-cutting = 73 mm

Average diameter of the bar d= (75+73)/2=74 mm

Average length of uncut chip = πd

Average length of uncut chip = π x 74 =232.45 mm

Thus, cutting ratio r

Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}

r=\dfrac{73.5}{232.45}   r=0.31

Therefore, the cutting ratio equals 0.31.

Now, the shearing angle is given as

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

Next by substituting the values

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }\

Ф=18.03°

Concluding, the shearing angle is 18.03°.

4 0
3 months ago
Consider the following program:
Daniel [329]

Answer:

The resolution for this question is below in the explanation section.

Explanation:

The right response to this question is A, which is 112002.

The correct code relating to this question is presented below

#include  <stdio.h>

#include  <string.h>

#include  <sys/types.h>

//#include "csapp.h"

void end(void)

{

printf("2");

}

int main()

{

if (fork() == 0)

atexit(end);

if (fork() == 0)

printf("0");

else

printf("1");

exit(0);

}

/* $end forkprob2 */

When this program is executed, the fork function will yield varying results.

However, it should be noted that running it multiple times will produce inconsistent values. The most common output the program will generate is A.

Attached is an image of the program's execution for further clarification.

5 0
1 month ago
Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe
iogann1982 [368]

An item of protective gear that shields individuals passing by from stray sparks or metal during the welding process performed by another worker is known as: E. Welding Screens.

An operator is a person tasked with joining two or more metals using a technique called wielding.

In the course of wielding, both sparks and tiny metallic fragments are released, which pose a danger to the operator and others working nearby.

As a result, the equipment outlined below should be worn or utilized directly by a worker actuating the wielding process:

  • Visors.
  • Goggles.
  • Protective clothing.
  • Dark walls.

Nonetheless, a type of protective gear that defends other workers nearby from stray sparks or metallic fragments while the operator (worker) is in the act of welding is called welding screens.

Find more information:

4 0
1 month ago
The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and re
mote1985 [299]

Answer:

The power of the pump is 23.09 kW.

Explanation:

Parameters

gravitational constant, g = 9.81 m/s^2

mass flow rate, \dot{m} = 60.6 kg/s

flow density, \rho = 1000 kg/m^3

efficiency of the pump, \eta = 0.74

output gauge pressure, p_o = 344.75 kPa

input gauge pressure, p_i = 68.95 kPa

cross-sectional area of output pipe, A_o = 0.069 m^2

cross-sectional area of input pipe, A_i = 0.093 m^2

height of discharge, z_o = 1.22 m - 0.61 m = 0.61 m (evaluated at pump’s maximum height of 1.22 m)

input height, z_i = 0 m

hydraulic power of the pump,P =? kW

Initially, the volumetric flow (Q) needs to be determined

Q = \frac{\dot{m}}{\rho}

Q = \frac{60.6 kg/s}{1000 kg/m^3}

Q = 0.0606 m^3/s

Next, compute the velocity (v) for both input and output

v_o = \frac{Q}{A_o}

v_o = \frac{0.0606 m^3/s}{0.069 m^2}

v_o = 0.88 m/s

v_i = \frac{Q}{A_i}

v_i = \frac{0.0606 m^3/s}{0.093 m^2}

v_i = 0.65 m/s

Subsequently, the total head (H) can be calculated

H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}

H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}

H = 28.74m

Finally, the computation of pump power is done as follows

P = \frac{Q \, \rho \, g \, H}{\eta}

P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}

P = 23.09 kW

6 0
2 months ago
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