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Elenna [48]
3 years ago
9

In a website browser address bar, what does “www” stand for?

Engineering
1 answer:
Ludmilka [50]3 years ago
3 0

Answer:

www stands for world wide web

Explanation:

It will really help you thank you.

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A vertical cylinder (Fig. P3.227) has a 61.18-kg piston locked with a pin, trapping 10 L of R-410a at 10◦C with 90% quality insi
Whitepunk [10]

Hey! How are you? My name is Maria, 19 years old. Yesterday broke up with a guy, looking for casual sex.

Write me here and I will give you my phone number - *pofsex.com*

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4 0
2 years ago
What is hardness and how is it generally tested?
drek231 [11]

Answer:

Hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

Explanation:

Hardness of a material is understood as the resistance that the material opposes to its permanent surface plastic deformation by scratching or penetration. It is always true that the hardness of a material is inversely proportional to the footprint that remains on its surface when a force is applied.

In this sense, the hardness of a material can also be defined as that property of the surface layer of the material to resist any elastic deformation, plastic or destruction due to the action of local contact forces caused by another body (called indenter or penetrator), harder, of certain shape and dimensions, which does not suffer residual deformations during contact.

That is, hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

The following conclusions can be drawn from the previous definition of hardness:  

  1) hardness, by definition, is a property of the surface layer of the material, and is not a property of the material itself;  

  2) the methods of hardness by indentation presuppose the presence of contact efforts, and therefore, the hardness can be quantified within a scale;

  3) In any case, the indenter or penetrator must not undergo residual deformations during the test of hardness measurement of the body being tested.

To determine the hardness of the materials, durometers with different types of tips and ranges of loads are used on the various materials. Below are the most commonly used tests to determine the hardness of the materials.

   Rockwell hardness :

It refers to the Rockwell hardness test, a method with which the hardness or resistance of a material to be penetrated is calculated. It is characterized by being a fast and simple method that can be applied to all types of materials. An optical reader is not required.

    Brinell hardness :

Brinell hardness is a scale that is used to determine the hardness of a material through the indentation method, which consists of penetrating with a hardened steel ball tip into the hard material, a load and for a certain time.  

This test is not very precise but easy to apply. It is one of the oldest and was proposed in 1900 by Johan August Brinell, a Swedish engineer.

    Vickers hardness:

Vickers hardness is a test that is used in all types of solid and thin or soft materials. In this test, a square-shaped pyramid-shaped diamond and a   136° vertex angle are placed on the penetrating equipment.

In this test the hardness measurement is performed by calculating the diagonal penetration lengths.

However, its result is not read directly on the equipment used, therefore, the following formula must be applied to determine the hardness of the material: HV = 1.8544 · F / (dv2).

3 0
3 years ago
Ughhh my cramps hurt sm
kozerog [31]

Answer:

Explanation:Come

tomate

5 0
2 years ago
Read 2 more answers
Find the time-domain sinusoid for the following phasors:_________
sattari [20]

<u>Answer</u>:

a.  r(t) = 6.40 cos (ωt + 38.66°) units

b.  r(t) = 6.40 cos (ωt - 38.66°) units

c.  r(t) = 6.40 cos (ωt - 38.66°) units

d.  r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = \sqrt{a^2 + b^2}

Ф = direction = tan⁻¹ (\frac{b}{a})

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{-5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{-5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

3 0
3 years ago
Which of the following is a possible consequence of poor measurement in construction as stated in the segment?
Kamila [148]

Answer:

Decreased risk of structure failure

6 0
3 years ago
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