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2 years ago

Describe in your own words the three strengthening mechanisms

Engineering

1 answer:

TEA [102]2 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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A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

2 years ago

Which answer best describes the relationship between the left-hand and right-hand panels in both File Explorer and Finder?

Art [367]

Answer: I think The left-hand panel shows like a high level list of folders. And Right-hand panel shows the contents of selected folder.

4 0

2 years ago

What is the primary function of NCEES?

joja [24]

National Council of Examiners for engineering and surveying a nonprofit organization

7 0

3 years ago

A heating system must maintain the interior of a building at 20°C during a period when the outside air temperature is 5°C and th

Anettt [7]

Answer:

a. W = 51,194.54 kJ

b. W = 102,390 kJ

c. W = 153,585 kJ

Explanation:

$(COP)_{HP} =\frac{Desired-effectx}{Work-done}= \frac{Q_{1} }{W} \\\\(COP)_{HP} =(COP)_{Ideal}\\\\\frac{Q1}{W} =\frac{T_{1} }{T_{1} -T_{2} }$

$W=Q_{1} \frac{T_{1}-T_{2} }{T_{1} }$

a. the ground at 15°C.

$T_{1}$ =20°C = 273 K + 20 = 293 K

$T_{2}$ =15°C = 273 K + 15 = 288 K

$Q_{1}=3x10^{6} kJ$

$W=3x10^{6} kJ \frac{293 K-288 K}{293 K}=3x10^{6} kJ \frac{5 K}{293 K}=3x10^{6} kJ x 0.017065}$

$W = 0.051195x10^{6} kJ$

W = 51,194.54 kJ

b. a pond at 10°C.

$T_{2}$ =10°C = 273 K + 10 = 283 K

$W=3x10^{6} kJ \frac{293 K-283 K}{293 K}=3x10^{6} kJ \frac{10 K}{293 K}=3x10^{6} kJ x 0.034130}$

$W = 0.102390x10^{6} kJ$

W = 102,390 kJ

c. the outside air at 5°C.

$T_{2}$ =5°C = 273 K + 5 = 278 K

$W=3x10^{6} kJ \frac{293 K-278 K}{293 K}=3x10^{6} kJ \frac{15 K}{293 K}=3x10^{6} kJ x 0.051195}$

$W = 0.153585x10^{6} kJ$

W = 153,585 kJ

Hope this helps!

3 0

2 years ago

Can a heat engine operating with only one thermal reservoir?

Harlamova29_29 [7]

Answer:

Explanation:

Heat engine is a device which operates between two thermal reservoir.One thermal reservoir supply the thermal energy to the heat engine and another thermal reservoir absorb heat rejected by heat engine.These thermal reservoir are infinite amount of heat capacity system.Temperature of thermal reservoir remain constant always.

So we can say that heat engine can not operate by using only one thermal reservoir.

5 0

3 years ago