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Luba_88 [7]
3 years ago
13

What is the primary responsibility of ABET?

Engineering
1 answer:
nalin [4]3 years ago
5 0

Answer:

C

Explanation:

The ABET (Accreditation Board for Engineering and Technology) is a non-governmental organization that accredits programs in applied science, computing, engineering, and engineering technology, both in the United States and elsewhere.

Give Brainliest pls

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Dampness or moisture introduces ____ into the weld, which causes cracking when some metals are welded.
N76 [4]

Answer: Dampness or moisture introduces hydrogen into the weld, which causes cracking when some metals are welded.

Explanation:

<em>This moisture (hydrogen) is a major cause of weld cracking and porosity. </em>

5 0
2 years ago
- Viscoelastic stress relaxation
My name is Ann [436]

Explanation:

The correct answers to the fill in the blanks would be;

1. Viscoelastic stress relaxation refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).

2. Viscoelastic creep refers to scenarios for which a polymer will permanently flow over time in response a constant applied stress.

The polymer whose properties have been mentioned above is commonly known as Kevlar.

It is mostly used in high-strength fabrics and its properties are because of several hydrogen bonds between polymer molecules.

5 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
A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a
vladimir2022 [97]

Answer:

diameter is 14 mm

Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

to find out

minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60    .................1

put here value

15 ×10^{3} = 2π×1750×T / 60

so

T = 81.84 Nm

and

torsion = T / Z                        ..........2

here Z is section modulus i.e = πd³/ 16

so from equation 2

torsion = 81.84 / πd³/ 16

so torsion = 416.75 / / d³     .................3

so from shear stress theory

torsion = σy / factor of safety

so here σy = 530 for 1020 steel

so

torsion = σy / factor of safety

416.75 / d³ = 530 × 10^{6} / 3

so d = 0.0133 m

so diameter is 14 mm

3 0
3 years ago
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
3 years ago
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