Why do union members generally earn higher wages than non-union workers?

Describe in your own words the three strengthening mechanisms

TEA [102]

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

3 0

3 years ago

Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic j

elixir [45]

Answer:

A) Flow depth = 2.46 m, Froude number after jump = 0.464

B) head loss = 0.572 m

C) dissipation ratio = 0.173

Explanation:

Given data :

Velocity before jump ( v1 ) = 7 m/s

flow depth before jump ( y1 ) = 0.8 m

g = 9.81 m/s

Esi = 3.3 m ( calculated )

attached below is a detailed solution of the problem

3 0

3 years ago

Identify the prefixes used in the International System of

lesya [120]

Answer:

i need points 425677

Explanation:

yurrrrrr awnser C

3 0

3 years ago

A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank

Bad White [126]

Given:

mass of water, m = 2000 kg

temperature, T = $30^{\circ}C$ = 303 K

extacted mass of water = 100 kg

Atmospheric pressure, P = 101.325 kPa

Solution:

a) Using Ideal gas equation:

PV = m $\bar{R}$ T (1)

where,

V = volume

m = mass of water

P = atmospheric pressure

$\bar{R} = \frac{R}{M}$

R= Rydberg's constant = 8.314 KJ/K

M = molar mass of water = 18 g/ mol

Now, using eqn (1):

$V = \frac{m\bar{R}T}{P}$

$V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}$

$V = 2762.44 m^{3}$

Therefore, the volume of the tank is $V = 2762.44 m^{3}$

b) After extracting 100 kg of water, amount of water left, m' = m - 100

m' = 2000 - 100 = 1900 kg

The remaining water reaches thermal equilibrium with surrounding temperature at T' = $30^{\circ}C$ = 303 K

At equilibrium, volume remain same

So,

P'V = m' $\bar{R}$ T'

$P' = \frac{1900\times \frac{8.314}{18}\times 303}{2762.44}$

Therefore, the final pressure is P' = 96.258 kPa

4 0

3 years ago

). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially

Delicious77 [7]

Answer:

$\frac{e'_z}{e_z} = 0.87142$

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

εx = - [ σx - ν( σy + σz ) ] / E

εy = - [ σy - ν( σx + σz ) ] / E

εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

εx = [ ν*σz ] / E

εy = [ ν*σz ] / E

εz = - [ σz ] / E .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

εx' = - [ σx' - ν( σy' + σz ) ] / E = [ ν*σz ] / 2E

εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

εx' = - [ σy' - ν( σx' + σz ) ] / E = [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note: σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

- [ σx' - ν( σx'+ σz ) ] = [ ν*σz ] / 2

σx' ( 1 - v ) = [ ν*σz ] / 2

σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

εz' = - [ σz - ν( σy' + σx' ) ] / E

εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

$\frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ] }{-\frac{s_z}{E}} \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142$ ... Answer

5 0

3 years ago