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2 years ago

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What is the major drawback in nanocrystalline alloys? a)- high brittleness b)-low hardness c)-rapid grain growth upon heating d)

-low thermal conductivity

1 answer:

patriot [66]2 years ago

8 0

Answer: a) High brittleness

Explanation: Nano crystalline alloys are the alloys that have the packing of nano crystal that is the size of the crystal is very minute or small. The size of the nano crystalline alloys usually ranges around 100 nm. Due to such arrangement of the crystal the nano crystalline alloy becomes hard in nature and because of this property it becomes difficult to make them ductile. Therefore the major drawback in nano crystalline alloys are that it has high brittleness or hardness.

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1. Which type of fit implies that a piece will never fit? a. interference fit b. construction fit c. transition fit d. impeding

VladimirAG [237]

Answer:

d

Explanation:

interference fit: the dimensions are larger and high force like heating is required to fit in the object

transition fit: the dimensions are fractionally larger and slight force like hammering is required to fit in the object

impeding fit: the dimensions are larger and the object will not fit at all

construction fit: activities making a commercial space sutiable for tenant occupation.

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2 years ago

If you’re still enrolled in school, but are looking for a job, on your resume you should highlight:

Ann [662]

Answer:

c

Explanation:

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2 years ago

The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur.

OlgaM077 [116]

Answer:

a) the velocity of the implant immediately after impact is 20 m/s

b) the average resistance of the implant is 40000 N

Explanation:

a) The impulse momentum is:

mv1 + ∑Imp(1---->2) = mv2

According the exercise:

v1=0

∑Imp(1---->2) = F(t2-t1)

m=0.2 kg

Replacing:

$0+F(t_{2} -t_{1} )=0.2v_{2}$

if F=2 kN and t2-t1=2x10^-3 s. Replacing

$0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s$

b) Work and energy in the system is:

T2 - U(2----->3) = T3

where T2 and T3 are the kinetic energy and U(2----->3) is the work.

$T_{2} =\frac{1}{2} mv_{2}^{2} \\T_{3} =0\\U_{2---3} =-F_{res} x$

Replacing:

$\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N$

3 0

2 years ago

A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar

jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading 114.06 m^3/d/m

Explanation:

calculation for single clarifier

$sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d$

$surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2$

$surface area = 314.16 m^2$

volume of tank $V = A\times side\ water\ depth$

$=314.16\times 2 = 628.32m^3$

$Length\ of\ weir = \pi \times diameter of weir$

$= \pi \times 18 = 56.549 m$

overflow rate = $v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2$

Detention time $t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr$

weir loading $= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m$

6 0

2 years ago

One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$

$T_{1}$ = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

$v_{2}$ = $v_{1}$ = 0.15 $m^{3}$ / Kg

In saturated water labels for $T_{2}$ = 40°C.

$v_{f}$ = 0.001008 $m^{3}$ / Kg

$v_{g}$ = 19.515 $m^{3}$ / Kg

The final state is two phase region $v_{f}$ < $v_{2}$ < $v_{g}$ .

In saturated water labels for $T_{2}$ = 40°C.

$P_{2}$ = $P_{Sat}$ = 7.3851 kPa

$P_{2}$ = 7.3851 kPa

7 0

2 years ago