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lbvjy [14]
3 years ago
14

Calculate the number of vacancies per cubic meter at 1000∘C for a metal that has an energy for vacancy formation of 1.22 eV/atom

, a density of 6.25g/cm^3, and an atomic weight of 37.4 g/mol.
(A) 1.49×10^18 m^−3
(B) 7.18×10^22 m^−3
(C) 1.49×10^24 m^−3
(D) 2.57×10^24 m^−3
Engineering
1 answer:
Makovka662 [10]3 years ago
6 0

Answer:

C

Explanation:

N = Na.P/A------(1)

Na = avogadro's number = 6.02210²³

P = density

A = atomic weight of metal

When we substitute into equation 1 above we get

1.0x10²⁹atoms/m³

From here we calculate the number of vacancies

T = 1000⁰c = 1273K

The formula to use is

Nv= Nexo(Qt/K.T) -----(2)

Qt = 1.22eV

K = Boltzmann's constant = 8.6210x10^-5

When we substitute values into equation 2

We get Nv = 1.49 x 10²⁴m-3

Therefore option c is correct

Check attachment for a more detailed calculation of this question

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Answer: a. 0.4667

b. 0.4667 and C 0.0667

Explanation:

Given Data:

N = population size (10)

n = random selection (2)

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When y = 1

f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )

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When y = 2

f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )

= 7 / 15

= 0.4667

When y = 0

f(0) = ( 7/0 ) ( 10 - 7 / 2 -0) / ( 10/2 )

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The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

K = \sigma Y \sqrt{\pi a}

where;

fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

applied stress \sigma = ???

crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

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Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

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P = 1728.289 × 4.5

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If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp
victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

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N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

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Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

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d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

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