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3 years ago

Describe in your own words the three strengthening mechanisms

Engineering

1 answer:

TEA [102]3 years ago

3 0

Answer:

a) Decreasing grain size decreases the amount of possible pile up at the boundary, increasing the amount of applied stress necessary to move a dislocation across a grain boundary. ... Grain sizes can range from about 100 μm (0.0039 in) (large grains) to 1 μm (3.9×10−5 in) (small grains).

b) Solid solution strengthening is a type of alloying that can be used to improve the strength of a pure metal. The technique works by adding atoms of one element (the alloying element) to the crystalline lattice of another element (the base metal), forming a solid solution.

c) Work hardening, also known as strain hardening, is the strengthening of a metal or polymer by plastic deformation. ... This strengthening occurs because of dislocation movements and dislocation generation within the crystal structure of the material.

Explanation:

I hope this will help you bro/sis

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3 years ago

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Explanation:

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Is a compass a analog or a digital sensor?

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2 years ago

Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and

lina2011 [118]

Answer:

y[n] = x[n] x[n-1] x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is $\frac{1}{x[n] . x[n-1] x[n+1]}$

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

for x[n] = 2 $\pi$ , y[n] = cos(2 $\pi$ ) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

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2 years ago

Convert A'B'C'D' + A'B'C'D + A'B'CD' + A'BC'D + AB'C'D' + AB'C'D+ AB'CD' to SOP form

bazaltina [42]

Answer:

thats really hard how could you answerthis hhhhhhh

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2 years ago

Read 2 more answers