Component(s) that only allow(s) electrons to flow in one direction. Mark all that apply

If a material is found to be in the tertiary phase of creep, the following procedure should be implemented:

kirill [66]

When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Therefore, Option C is correct.

<h3>What do you mean by a tertiary degree of creep?</h3>

Tertiary Creep has an extended creep rate and terminates when the material breaks or ruptures. It is related to each necking and formation of grain boundary voids. The wide variety of possible stress-temperature- time combos is infinite.

Therefore, When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Option C is correct.

Learn more about creep:

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8 0

2 years ago

Consider a cubic crystal with the lattice constant a. Complete the parts (a)-(c) below. (a) Sketch the crystallographic planes w

Anna [14]

Answer:

(a) See attachment

(b) The two planes are parallel because the intercepts for plane [220] are X = 0,5 and Y = 0,5 and for plane [110] are X = 1 and Y = 1. When the planes are drawn, they keep the same slope in a 2D plane.

(c) $d = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}} = \frac{1}{\sqrt{2}} = 0,707$

Explanation:

(a) To determine the intercepts for an specific set of Miller indices, the reciprocal intercepts are taken as follows:

For [110]

$X = \frac{1}{1} = 1; Y = \frac{1}{1} = 1; Z = \frac{1}{0} = \inf.$

For [220]

$X = \frac{1}{2} = 0,5;Y = \frac{1}{2} = 0,5;Z = \frac{1}{0} = \inf.$

The drawn of the planes is shown in the attachments.

(b) Considering the planes as two sets of 2D straight lines with no intersection to Z axis, then the slope for these two sets are:

For (1,1):

$K_1 = \frac{1}{1} = 1$

For (0.5, 0.5):

$K_2 = \frac{0.5}{0.5} = 1$

As shown above, the slopes are exactly equal, then, the two straight lines are considered parallel and for instance, the two planes are parallel also.

(c) To calculate the d-spacing between these two planes, the distance is calculated as follows:

The Miller indices are already given in the statement. Then, the distance is:

$\frac{1}{d^{2}} = \frac{h^{2} + k^{2} + l^{2}}{a^{2}}$

$d = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}} = \frac{1}{\sqrt{2}} = 0,707$

7 0

3 years ago

Consider a cylinder of height h, diameter d, and wall thickness t pressurized to an internal pressure P_0 (gauge pressure, relat

Serggg [28]

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

Eq 3

Finally, solving for unknown pressure as a function of hoop strain:

Eq 4

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

Eq 8

For Wheat stone bridge:

Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

part b

Since, axial strain(1-2v) < hoop strain (2-v). V out axial < V out hoop.

Hence, dV hoop < dV axial.

part c

Given data:

P = 253313 Pa

D = d + 2t = 0.09013 m

t = 65 um

GF = 2

E = 75 GPa

v = 0.33

Use the data above and compute Vout using Eq3

Eq 11

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3 0

3 years ago

What do electrons have to do with electrical current?

erastova [34]

Answer:

Electrons in atoms can act as our charge carrier, because every electron carries a negative charge. If we can free an electron from an atom and force it to move, we can create electricity.

6 0

4 years ago

A well-insulated, rigid vessel contains 3 kg of saturated liquid water at 40°C. The vessel also contains an electrical resistor

fenix001 [56]

Answer:

The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C

Explanation:

given information:

mass, m = 3 kg

initial temperature, T₁ = 40°C

current, I = 10 A

voltage, V = 50 V

time, t = 30 min = 1800 s

Heat for the system because of the resistance is

Q = V I t

where

V = voltage (V)

I = current (A)

t = time (s)

Q = heat transfer to the system (J)

so,

Q = V x I x t

= 50 x 10 x 1800

= 900000

= 9 x 10⁵ J

the heat transfer in the closed system is

Q = ΔU + W

where

U = internal energy

W = work done by the system

thus,

Q = ΔU + W

9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.

ΔU = 9 x 10⁵ J = 900 kJ

then, the energy change in the system is

ΔU = m c ΔT

ΔT = ΔU / m c, c = 4.186 J/g°C

= 900 / (3 x 4.186)

= 71.67°C

so,the final temperature (T₂)

ΔT = T₂ - T₁

T₂ = ΔT + T₁

= 71.67°C + 40°C

= 111.67°C

7 0

3 years ago