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1 year ago

What are the disadvantages of military shovels?

Engineering

1 answer:

Xelga [282]1 year ago

4 0

Answer:

being prepared and having sufficient military resources available can be great IF there is an attack although it can be viewed as intimidating and scare off allies. also, it can cause other resources too thin.

Explanation:

it is great to have a sense of national pride and togetherness although if it gets out of hand it can be almost brainwashing and used to control citizens.

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(a) Consider a germanium semiconductor at T 300 K. Calculate the thermal equilibrium electron and hole concentrations for (i) Nd

padilas [110]

Answer:

i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2 × 10¹¹ cm⁻³

ii. electron concentration, n₀ = 1.33 × 10¹¹ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³

i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2.205 × 10⁻³ cm⁻³

ii. electron concentration, n₀ = 1.47 × 10⁻³ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³

c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.

Explanation:

a. For Germanium, intrinsic concentration n₁ = 2 × 10¹³ cm⁻³.

i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 0,

n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ] since N₁ > N₂

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ - 0) +√[(2 × 10¹⁵ cm⁻³ - 0)² + 4(2 × 10¹³ cm⁻³)²] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4.0016 × 10³⁰ cm⁻⁶] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ + 2.0004 × 10¹⁵ cm⁻³ ]

n₀ = 1/2[4.0004 × 10¹⁵ cm⁻³ ]

n₀ = 2.0002 × 10¹⁵ cm⁻³ ≅ 2 × 10¹⁵ cm⁻³

The hole concentration p₀ is gotten from

n₀p₀ = n₁²

p₀ = n₁²/n₀ = (2 × 10¹³ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/2 × 10¹⁵ cm⁻³

p₀ = 2 × 10¹¹ cm⁻³

ii. For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 10¹⁶ cm⁻³,

p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ] since N₂ > N₁

p₀ = 1/2[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³) +√[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³)² + 4(2 × 10¹³ cm⁻³)²] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9.0016 × 10³⁰ cm⁻⁶] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ + 3.0003 × 10¹⁵ cm⁻³ ]

p₀ = 1/2[6.0003 × 10¹⁵ cm⁻³ ]

p₀ = 3.00015 × 10¹⁵ cm⁻³ ≅ 3 × 10¹⁵ cm⁻³

Te electron concentration n₀ is gotten from

n₀p₀ = n₁²

n₀ = n₁²/p₀ = (2 × 10¹³ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/3 × 10¹⁵ cm⁻³

n₀ = 1.33 × 10¹¹ cm⁻³

b. For GaAs, intrinsic concentration n₁ = 2 × 10⁶ cm⁻³.

i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 0,

n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ] since N₁ > N₂ and N₁ - N₂ = 2 × 10¹⁵ cm⁻³ >> n₁ = 2 × 10⁶ cm⁻³

n₀ = (N₁ - N₂) = 2 × 10¹⁵ cm⁻³ - 0 = 2 × 10¹⁵ cm⁻³

The hole concentration p₀ is gotten from

n₀p₀ = n₁²

p₀ = n₁²/n₀ = (2.1 × 10⁶ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/2 × 10¹⁵ cm⁻³

p₀ = 2.205 × 10⁻³ cm⁻³

ii. For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 10¹⁶ cm⁻³,

p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ] since N₂ > N₁ and N₂ - N₁ = 10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³ = 3 × 10¹⁵ cm⁻³ >> n₁ = 2.1 × 10⁶ cm⁻³

p₀ ≅ N₂ - N₁ = 3 × 10¹⁵ cm⁻³

The electron concentration n₀ is gotten from

n₀p₀ = n₁²

n₀ = n₁²/p₀ = (2.1 × 10⁶ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/3 × 10¹⁵ cm⁻³

n₀ = 1.47 × 10⁻³ cm⁻³

c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.

4 0

3 years ago

Yasir is trying to build an energy-efficient wall and deciding what materials to use. How can he calculate the thermal resistanc

777dan777 [17]

Answer:

Add the thermal R values of each wall layer.

Explanation:

Fiberglass insulation R value, sheetrock, sheathing, siding etc. All sum together to one composite opposition to heat transfer/loss.

Q= U x A x Delta T

Q= heat transfer btuh

U= 1/R inverse of resistance

A= area of surface

Delta T is temerature difference across the wall.

8 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