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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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While discussing VIN numbers, Technician A says that the first digit of the VIN identifies the country where the vehicle was man

ruslelena [56]

Usually the first digit of the vin id’s the country it was built. So technician A would be correct. That’s usually how it is. Hope this helps. Please let me know if this is incorrect

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3 years ago

Determine the work input and entropy generation during the compression of steam from 100 kPa to 1 MPa in (a) an adiabatic pump a

Goshia [24]

Answer:

See attachment below

Explanation:

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3 years ago

Milk has a density of as much as 64.6 lb/ft3. What is the gage pressure at the bottom of the straw 6.1 inches deep in the milk?

gregori [183]

Answer:

Explanation:

1 inch is 0.0833333feet

6.1 inches is 0.5083 feet

Density = mass/volume

64.6 = mass/0.50833

mass = 64.6 x 0.5083 =32.83618lb

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3 years ago

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and

k0ka [10]

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

$z1=$ $\frac{\left(150-137\right)}{27.7}$ $=0.4693$

$z2=\frac{\left(201-137\right)}{27.7}=2.3105$

$P(0.4693$

B.)

$z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289$

$\\P(2.9309$

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

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3 years ago

Read 2 more answers

An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e

Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in $2.25 * 10^4 BTU$ . It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - $1 BTU = 778.17 lbf*ft$

$2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft$

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:

$Ep = m*g*(h_2 - h_1)\\ W = m*g$