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Sladkaya [172]
3 years ago
12

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Engineering
1 answer:
vodka [1.7K]3 years ago
5 0

Answer:

(⌐■-■) (⌐■-■) (⌐■-■) (⌐■-■)

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Ultimate tensile strength is: (a) The stress at 0.2% strain (b) The stress at the onset of plastic deformation (c) The stress at
MissTica

Answer:

By definition the ultimate tensile strength is the maximum stress in the stress-strain deformation. The stress at 0.2% strain, the stress at the onset of plastic deformation, the stress at the end of the elastic deformation and the stress at the fracture correspond, by definition, to other points of the stress-strain curve.

Explanation:

4 0
3 years ago
Another name for your computer, running the web browser program is: Web user The client The mainframe Browsing agent
timofeeve [1]

Answer:

Browsing agent

Explanation:

Hope this helps!

6 0
3 years ago
What is the braks mean effictive pressure?
OverLord2011 [107]

Engine cylinder pressure

<u>Explanation:</u>

  1. Brake mean effective pressure is a method to calculate  the engine cylinder pressure which  would give the measured brake horsepower. Brake mean effective pressure is used to identify engine efficiency regardless of capacity or engine speed.
  2. It is used to identify engine efficiency.It is measured by means of transducers or pressure gauges.
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7 0
3 years ago
Based on experimental observations, the acceleration of a particle is defined by the relationa = -( 0.1 + sin(x/b) ),where a and
yKpoI14uk [10]

Answer:

a) v = +/- 0.323 m/s

b) x = -0.080134 m

c) v = +/- 1.004 m/s

Explanation:

Given:

                             a = - (0.1 + sin(x/b))

b = 0.8

v = 1 m/s @ x = 0

Find:

(a) the velocity of the particle when x = -1 m

(b) the position where the velocity is maximum

(c) the maximum velocity.

Solution:

- We will compute the velocity by integrating a by dt.

                           a = v*dv / dx =  - (0.1 + sin(x/0.8))

- Separate variables:

                           v*dv = - (0.1 + sin(x/0.8)) . dx

-Integrate from v = 1 m/s @ x = 0:

                          0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5

                          0.5v^2 =  0.8cos(x/0.8) - 0.1x - 0.3

- Evaluate @ x = -1

                          0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3

                          v = sqrt (0.104516)

                          v = +/- 0.323 m/s

- v = v_max when a = 0:

                           -0.1 = sin(x/0.8)

                             x = -0.8*0.1002

                             x = -0.080134 m

- Hence,

                            v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134

                            v = sqrt (0.504)

                            v = +/- 1.004 m/s

4 0
3 years ago
True or false? the arc welding power voltage used for GMAW and FCAW produces a constant voltage​
7nadin3 [17]

Answer:

False

Explanation:

3 0
3 years ago
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