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laila [671]
2 years ago
6

What is the function of engineering

Engineering
1 answer:
ra1l [238]2 years ago
6 0

Answer:

The interpretation including its given question has always been discussed in the following sections on the elaboration.

Explanation:

In engineering, a function or a feature has always been perceived as a particular process, active engagement, or workload a system must perform.

  • Generally differentiated consequently throughout the life span of infrastructure projects: Specifications as well as Functional technical specifications.
  • Typically the Specifications define the most critical device attributes demanded.
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A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
Amanda [17]

Question in order:

A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

Mirror Radius (mm) Bending Failure Stress (MPa)

0.603                         225

0.203                         368

0.162                         442

Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

Bending Stress = \frac{1}{(Mirror Radius)^{n}}

At mirror radius 1 = 0.603 mm   Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm  Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm   Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

\frac{Stress 1}{Stress 2} = ({\frac{Radius 2}{Radius 1}})^{n_1}

\frac{225}{368} = ({\frac{0.203}{0.603}})^{n_1}

0.6114 = (0.3366)^{n_1}

Taking the natural logarithm of both side

ln(0.6114) = n ln(0.3366)

n₁ = ln(0.6114)/ln(0.3366)

n₁ = 0.452

comparing case 2 and 3 using the above equation

\frac{Stress 2}{Stress 3} = ({\frac{Radius 3}{Radius 2}})^{n_2}

\frac{368}{442} = ({\frac{0.162}{0.203}})^{n_2}

0.8326 = (0.7980)^{n_2}

Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

n₂ = 0.821

comparing case 1 and 3 using the above equation

\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}

\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}

0.5090 = (0.2687)^{n_3}

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

n₃ = ln(0.5090)/ln(0.2687)

n₃ = 0.514

average for n

n = \frac{n_1 + n_2 + n_3}{3}

n = \frac{0.452 +0.821 + 0.514}{3}

n = 0.596

Hence to get bending stress x at mirror radius 0.796

\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}

\frac{Stress x}{225} = ({\frac{0.603}{0.796}})^{0.596}

\frac{Stress x}{225} = 0.8475

stress x = 191 MPa

3 0
3 years ago
Which one is suitable for industries petrol engine or diesel engine and why?
klio [65]

Answer:

diesel engine

Explanation:

because diesel is stronger than petrol

3 0
3 years ago
Read 2 more answers
Bob and Alice are solving practice problems for CSE 2320. They look at this code: for(i = 1; i <= N; i = (i*2)+17 ) for(k = i
MissTica

Answer:

Alice is correct.

The loop are dependent.

Explanation:

for(i = 1; i <= N; i = (i*2)+17 )

for(k = i+1; k <= i+N; k = k+1) // notice i in i+1 and i+N

printf("B")

This is a nested for-loop.

After the first for-loop opening, there is no block of statement to be executed rather a for-loop is called again. And the second for-loop uses the value of i from the first for-loop. The value of N is both called from outside the loop.

So, the second for-loop depend on the first for loop to get the value of i. For clarity purpose, code indentation or use of curly brace is advised.

8 0
3 years ago
Read 2 more answers
What is a microwave transmitter?a) A technology that uses active or passive tags in the form of chips or smart labels that can s
Blababa [14]

Answer:

b) Commonly used to transmit network signals over great distances.

Explanation:

The transmission of information or data by using microwave radio waves is known as microwave transmission. Microwave transmitter is commonly used to transmit network signals over great distances. It is an electronic device that transmits and receives radio frequency signals ranging from 1GHz to 100GHz.

The microwave transmitter has a wide range of applications and these includes, radio stations, television stations, mobile phones, radio astronomy, radar,

5 0
2 years ago
Read 2 more answers
A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con
Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question  

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters  = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

6 0
3 years ago
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