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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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I need help with this question please
solniwko [45]

Answer:

The resultant moment is 477.84 N·m

Explanation:

We note that the resultant moment is given by the moment about a given point

The length of the sides of the formed triangles are;

l = sin(40°) × 4/sin(110°) ≈ 2.736

Taking the moment about the lower left hand corner of the figure, with the convention that clockwise moments are positive, we have;

The resultant moment, ∑m, is given as follow;

∑M = 250 N × 4 m + 400 N × cos(40°) × 4 m - 400 N × cos(40°) × 2 m + 400 N × sin(40°) × 2 m × tan(40°) - 600 N × cos(40°) × 2 m - 600 N× sin(40°) × 2 m × tan(40°) = 477.837084 N·m

Therefore, the resultant moment, ∑m ≈ 477.84 N·m clockwise.

6 0
3 years ago
134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.
vichka [17]

Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = - 10^{\circ}C = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = 60^{\circ}C = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T =60^{\circ}C:

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at  P = 140 kPa, T =- 10^{\circ}C:

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = s'_{s} and h'_{s} = 278.06 kJ/kg.K at 700kPa

(a) Isentropic efficiency of the compressor, \eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}

\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%

(b) The temperature of the environment, T_{e} = 27^{\circ}C = 273 + 27 = 300 K

Availability at state 1, \Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg

Similarly for state 2, \Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg

Now, the efficiency of the compressor as per the second law;

\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757 = 67.57%

4 0
4 years ago
Given an unsorted array of distinct positive integers A[1..n] in the range between 1 and 10000 and an integer i in the same rang
AnnZ [28]

Answer:

Explanation:

Arbitrary means That no restrictions where placed on the number rather still each number is finite and has finite length. For the answer to the question--

Find(A,n,i)

for j =0 to 10000 do

frequency[j]=0

for j=1 to n do

frequency[A[j]]= frequency[A[j]]+1

for j =1 to n do

if i>=A[j] then

if (i-A[j])!=A[j] and frequency[i-A[j]]>0 then

return true

else if (i-A[j])==A[j] and frequency[j-A[j]]>1 then

return true

else

if (A[j]-i)!=A[j] and frequency[A[j]-i]>0 then

return true

else if (A[j]-i)==A[j] and frequency[A[j]-i]>1 then

return true

return false

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timurjin [86]

Answer:

Explanation:

Attached is the solution

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3 years ago
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nataly862011 [7]

Answer:

You need to use option '–author' along with option '-l' to print the author name of each file.

Explanation:

hopefully this helps you sorry if it doesn’t help you

4 0
3 years ago
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