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2 years ago

Can you screen record on WOW Presents Plus on iPhone?

Engineering

1 answer:

faust18 [17]2 years ago

8 0

Answer:

what do u mean.‎‎‎‎‎‎‎‎‎‎‎

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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

Makovka662 [10]

Answer:

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

6 0

3 years ago

What are the nine Historical periods?

Rufina [12.5K]

Answer:

https://quizlet.com/148993376/the-nine-distinct-periods-of-time-flash-cards/

Explanation:

you can find them all : )

7 0

2 years ago

If a steel cable is rated to take 800-lb and the steel has a yield strength of 90,000psi, what is the diameter of the cable?

goldfiish [28.3K]

Answer:

d = 2.69 mm

Explanation:

Assuming the cable is rated with a factor of safety of 1.

The stress on the cable is:

σ = P/A

Where

σ = normal stress

P: load

A: cross section

The section area of a circle is:

A = π/4 * d^2

Then:

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

$d = \sqrt{4*P / (\pi*\sigma)}$

Replacing:

$d = \sqrt{4*800 / (\pi*\90000)} = 0.106 inches$

0.106 inches = 2.69 mm

5 0

3 years ago

1. You should

vladimir2022 [97]

D
Step by step explanation

3 0

2 years ago

Read 2 more answers

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago