(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why? (b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?

Answers

Answer 1
Answer:

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.


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A cube with 1 m on a side is located in the positive x-y-z octant in a Cartesian coordinate system, with one of its points located at the origin. Find the total charge contained in the cube if the charge is given by p_v = x^2 ye^-z mC/m^3

Answers

Answer:

4.61 mC

Explanation:

The cube has 1 m side in the positive x-y-z octant in a Cartesian coordinate system, with one of its points located at the origin. The charge density is given as:

\rho_v=x^2ye^(-z) \ mC/m^3

Charge density is the charge per unit length or area or volume. It is the amount of charge in a particular region.

The charge Q is given as:

Q=\int\limits_v {\rho_v} \, dv  \nQ=\int\limits_v {\rho_v} \, dv=\int\limits^2_(x=0)\int\limits^2_(y=0)\int\limits^2_(z=0) {x^2ye^(-z)} \, dxdydz\n

Q=\int\limits^2_(x=0) {x^2} \, dx \int\limits^2_(y=0) {y} \, dy \int\limits^2_(z=0) {e^(-z)} \, dz \n\nQ=((1)/(3) [x^3]^2_0)((1)/(2) [y^2]^2_0)(-1 [e^(-z)]^2_0)\n\nQ=(-1)/(6) ([x^3]^2_0)( [y^2]^2_0)( [e^(-z)]^2_0)\n\nQ=(-1)/(6)[2^3-0^3][2^2-0^2][e^(-2)-e^0]\n\nQ=(-1)/(6)(8)(4)(0.1353-1)=(-1)/(6)(8)(4)(-0.8647)\n\nQ=4.61\ mC

A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool 5 large watermelons, 10 kg each, to 8 C. If the watermelons are initially at 28 C, determine how long it will take for the refrigerator to cool them.

Answers

Answer:

\Delta t = 5866.667\,s\,(97.778\,m)

Explanation:

The specific heat for watermelon above freezing point is 3.96\,(kJ)/(kg\cdot K). The heat liberated by the watermelon to cool down to 8°C is:

Q_(cooling) = (5)\cdot (10\,kg)\cdot (3.96\,(kJ)/(kg\cdot K) )\cdot (20\,K)

Q_(cooling) = 3960\,kJ

The heat absorbed by the household refrigerator is:

\dot Q_(L) = COP\cdot \dot W_(e)

\dot Q_(L) = 1.5\cdot (0.45\,kW)

\dot Q_(L) = 0.675\,kW

Time needed to cool the watermelons is:

\Delta t = (Q_(cooling))/(\dot Q_(L))

\Delta t = (3960\,kJ)/(0.675\,kW)

\Delta t = 5866.667\,s\,(97.778\,m)

 

Water needs to be turned into steam in a high altitude lab where the atmospheric pressure is 84.6 KPa. Computte the heat energy (in calories) required to evaporate 900g of water at 15 degree C under these conditions.

Answers

Answer:

558.1918 kilocalories = 558191.8 calories

Explanation:

Data provided in the question:

Atmospheric pressure = 84.6 KPa

Mass of water, m = 900 g = 0.90 kg

Temperature = 15°C

Now,

Temperature at 84.6 KPa = 94.77°C

Therefore,

Heat energy required = m(CΔT + L)

here,

C is the specific heat of the water = 4.2 KJ/kg.°C

L = Latent heat of water = 2260 KJ/kg

Thus,

Heat energy required = 0.90[ 4.2 × (94.77 - 15) + 2260 ]

= 2335.53 KJ

also,

1 KJ = 0.239  Kilocalories

Therefore,

2335.53 KJ = 0.239 × 2335.53 Kilocalories

= 558.1918 kilocalories = 558191.8 calories

The substance called olivine may have any composition between Mg2SiO4 and Fe2SiO4, i.e. the Mg atoms can be replaced by Fe atoms in any proportion without altering the crystal structure except by expanding it slightly: this is an example of a binary solid solution series. For different compositions, the lines in the powder diffraction patterns are in slightly different positions, because of the cell expansion, but the overall pattern remains basically the same. The spacing of the lattice planes varies linearly with composition, and this can be used in a rapid and non- destructive method of analysis. a. The (062) reflection from olivine is strong and well resolved from other lines. Calculate d062 for an olivine that displays its (062) reflection at a Bragg angle of 37.21° (i.e., a diffraction angle of 74.42°) when x-rays with a wavelength of 0.1790 nm are used. b. The d062 spacing as measured accurately for synthetic materials is 0.14774 nm for Mg2SiO4 and 0.15153 nm for Fe2SiO4. What would be the approximate composition, expressed in mol.% Mg2SiO4, of an olivine material for which do62 has the value obtained in part 2.1 above?

Answers

Answer:

The answer is "0.147 nm and  99.63 mol %"

Explanation:

In point (a):

\to nk1 = 062

\to \text{Bragg angle}\theta =37.21^(\circ)

\to \text{diffraction angle}2 \theta = 74.42^(\circ)

\to \lambda = 0.1790 nm

find:

d(062)=?

formula:

\to nx = 2d \sin  \theta

\to  d(062) = (1 * 0.1790^(\circ))/(2 * \sin 37.21^(\circ))\n

               = (0.1790^(\circ))/(2 * 0.604738126)\n\n= (0.29599589)/(2)\n\n= 0.147 \n

In point (b):

\to Mg_2SiO_4\longleftrightarrow  Fe_2SiO_4

d= 0.14774  \ \ \ \ \ olivine = 0.147 \ \ \ \  \ 0.15153

formula:

\to d=(a)/(√(n^2+k^2+i^2))\n

that's why the composition value equal to 99.63 %

What is 29.95 inHg in kPa?

Answers

Answer:

101.42235 kPa

Explanation:

The unit inHg means "inches of mercury", Its a pressure unit commonly used  by the US aviators.

The conversion value to KPa (kilopascal) is

1 inHg= 3.386389 kPa

So now we only have to multiply:

29.95 inHg * 3.386389 kPa/in Hg =101.42235 kPa

Have a nice day and Good Luck!

Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim working on now? A. design development B. schematic design C. mechanical D. structural

Answers

Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.

What is a schematic design?

A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.

The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.

The schematic design consists of a rough sketch with markings and measurements.

Therefore, the correct option is B. schematic design.

To learn more about schematic design, refer to the link:

brainly.com/question/14959467

#SPJ5

Answer:

B. schematic design

Explanation:

This correct for Plato/edmentum

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