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3 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]3 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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What is the De Broglie wavelength of an electron under 150 V acceleration?

yanalaym [24]

Answer:

0.1 nm

Explanation

Potential deference of the electron is given as V =150 V

Mass of electron $m=9.1\times 10^{-31}$

Let the velocity of electron = v

Charge on the electron $=1.6\times 10^{-19}C$

plank's constant h = $6.67\times 10^{-34}$

According to energy conservation $eV =\frac{1}{2}mv^2$

$v=\sqrt{\frac{2eV}{M}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 150}{9.1\times 10^{-31}}}=7.2627\times 10^{-6}m/sec$

Now we know that De Broglie wavelength $\lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm$

4 0

3 years ago

Read 2 more answers

A 30 mm thick AISI 1020 steel plate is sandwiched between two 10 mm thick 2024-T3 aluminum plates and compressed with a bolt and

denis-greek [22]

Answer:

275 MPa

Explanation:

Regardless of what it is holding, the stiffness of a bolt depends on its own material properties and geometry.

The stiffness is:

$k = E * \frac{A}{l}$

I assume this one is made of steel, because regular bolts are steel.

The Young's modulus for steel is E = 210 GPa

The longitude is given. (But note that in a real application you have to consider the length up to the nut.)

The section is (using the nominal diameter of 10 mm)

$A = \frac{\pi * d^2}{4} = \frac{\pi * 0.01^2}{4} = 7.85e-5 m^2$

Then:

$k = 2.1e11 * \frac{7.85e-5}{0.06} = 275e6 Pa = 275 MPa$

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3 years ago

A(n) ______ is used to measure fluid flow in engineering

Kazeer [188]

Explanation:

instrument of engineering and management skills that can be used to identify a device on a network of devices

6 0

3 years ago

A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip

JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

$DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})$

Above equation can be written as:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})$

First Consider no wire i.e d/D=0

Above expression will become:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}$

Part A: (d/D=0.1)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574$

$DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.574}{1}*100$

DeltaV percent=42.6%

Part B:(d/D=0.01)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783$

$DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.783}{1}*100$

DeltaV percent=21.7%

5 0

3 years ago

9. An embankment having a volume of 320,000 yd is to be constructed from local borrow. The dry unit weight and moisture content

tatuchka [14]

Answer:

correct option is (B) 315,500

Explanation:

given data

volume = 320,000 yd³ = 8640000 ft³

dry unit weight = 106 pcf

moisture content = 18.2%

total unit weight = 122 pcf

moisture content = 16.7%

to find out

volume of borrow lyd needed

solution

first we get here weight of material that is

weight = volume × unit weight

weight = 8640000 × 122

weight = 1054080000 lb

that weight is weight of water + weight of solid so

0.167 × weight is weight of water + weight of solid ) = 1054080000 lb

and weight of solid = $\frac{1054080000}{1.167}$

weight of soil solid is = 903239075 pound

and weight of water = 150840925 pound

so volume of soil = 903239075 ÷ 106 lb/ft³ = 8521123.34 ft³

and volume required = 8521123.34 ft³ ÷ 27 ft³ = 315597.161 yd³

volume required = 315500 yd³

so correct option is (B) 315,500

3 0

3 years ago