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4 years ago

What factors are likely to promote the formation of stress corrosion cracks?

Engineering

1 answer:

Naddika [18.5K]4 years ago

7 0

Answer:

Stress corrosion cracking

Explanation:

This occurs when susceptible materials subjected to an environment that causes cracking effect by the production of folds and tensile stress. This also depends upon the nature of the corrosive environment.

Factors like high-temperature water, along with Carbonization and chlorination, static stress, and material properties.

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A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri

babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are 73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

$T_{BE}-W=0$ hence

$T_{BE}=W=20*9.81=196.2 N$

Therefore, tension in the cable, $T_{BE}=196.2 N$

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

$196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0$

$24.525-39.24+0.2D_x=0$

$D_x=73.575 N$

Similarly,

$A_x-D_y=0$

$A_x=73.575 N$

Therefore, both reactions at A and D are 73.575 N

7 0

3 years ago

The position of a particle is given by s = 0.27t

Natali [406]

Sorry bro people do this22.2 pls

8 0

2 years ago

Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton

SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude $q_{1},q_{2}$ the magnitude of force between them is given by

$F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}$

where

$k_{e}$ is constant

$r$ is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

$F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208$

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

$F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}$

Applying value of the constant we get

$F_{1}=\frac{62.208}{0.7^{2}}$

Thus $F_{2}=126.955Newtons$

5 0

3 years ago

A thin-film gold conductor for an integrated circuit in a satellite application is deposited from a vapor, and the deposited gol

ahrayia [7]

Answer:

Explanation:

Considering the relation of the equilibrium vacancy concentration ;

nv/N = exp (-ΔHv/KT)

Where T is the temperature at which the vacancy sites are formed

K = Boltzmaan constant

ΔHv = enthalpy of vacancy formation

Rearranging the equation and expressing in term of the temperature and plugging the values given to get the temperature. The detailed steps is as shown in the attached file

5 0

3 years ago

Rank the magnitudes of the diffusion coefficients from greatest to least for the following systems:(a) Cr in Fe at 600°C; (b) C

jek_recluse [69]

Answer:

Explanation:

The rank of the magnitude of the diffusion coefficient from greatest to least is as follows:

C in Fe at 900°C > Cr in Fe at 900°C > Cr in Fe at 600°C

Reason

C in Fe is an interstitial impurity while Cr in Fe is a substutional impurity.Therefore interstitial impurity occurs in C in Fe systems,while substutitional diffusion occurs in Cr in Fe system.Interstitial is much faster than substitutional diffusion hence the order

Also with increasing temperature magnitude of diffusion coefficient increases,due to the relation.

D = D₀exp(-Qd/RT)

Where D₀=Temperature independent per exponential

Qd= The activation energy for diffusion

R= Universal gas constant

T=absolute temperature

3 0

3 years ago

Read 2 more answers