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3 years ago

5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of

Engineering

1 answer:

marta [7]3 years ago

8 0

Answer:

(a) $3.185*10^{21} cells$

(b) $6.37*10^{21} atoms$

Explanation:

(a)

Volume, V of unit cell

$V=(2.866*10^{-8})^{3}=2.354*10^{-23}$

Number of unit cells, N

$N=\frac {W_{mat}}{V\rho_{mat}}$ Where $W_{mat}$ is weight of material and $\rho_{mat}$ is density of material

$N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells$

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron= $3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms$

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Rafel knows that lessons learned is a valuable aid to future projects. When should he and his team address

Arada [10]

Answer: Create lessons learned at the end of the project.

Explanation:

Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.

The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.

5 0

3 years ago

A 0.5m diameter sphere containing pollution monitoring equipment is dragged through the Charles River at a relative velocity of

Dmitriy789 [7]

Answer:

$\phi = 155.57$

Explanation:

from figure

taking summation of force in x direction be zero

$\sum x = 0$

$F_D = Tsin \theta$ .....1

$\frac{c_d \rho v^2 A}{2} =Tsin \theta$

taking summation of force in Y direction be zero

$F_B - W- Tcos \theta$

$T = \frac{F_B -W}{cos \theta}$ .........2

putting T value in equation 1

$F_D - \frac{F_B -W}{cos \theta} sin\theta$

$F_D = \rho g V ( 1 -Sg) tan \theta$ .........3

$F_D = \rho g [\frac{\pi d^3}{6}] ( 1 -Sg) tan \theta$

$tan \theta = \frac{6 c_D \rho v&2 A}{ 2 \rho g V \pi D^3 (1- Sg)}$

Water at 10 degree C has kinetic viscosity v = 1.3 \times 10^{-6} m^2/s

Reynold number

$Re = \frac{ VD}{\nu} = \frac{10\times 0.5}{1.3 \times 10^{-6}} = 3.84 \times 10^6$

so for Re = $3.84 \times 10^6$ cd is 0.072

$tan \theta = \frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}$

$\theta = tan^{-1} [\frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}]$

$\theta = - 65.57 degree$

$\phi = 90 - (-65.57) = 1557.57$ degree

8 0

3 years ago

Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0.846 a

MAXImum [283]

Answer:

Yes it is possible.

Explanation:

This problem is about to possibility to have alloy of iron-carbon for which mass fraction of ferrite, $W_{\alpha} = 0.846$ and proeutectoid cementite, $W_{Fe_3C}=0.049$