ENGINEERING COLLEGE

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. Calculate the number of atoms in the unit cell and diameter of the metal atom.

Answers

Answer 1

Answer:

Answer:288 pm

Explanation:

Number of atoms(s) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- whereas

For an FCC lattices √2a =4r =2d

Therefore d = a/√2a = 408pm/√2a= 288pm

I think with this step by step procedure the, the answer was clearly stated.

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Young students show a preference for which modality?

Answers

Answer. In the present study, it was found that 61% students had multimodal learning style preferences and that only 39% students had unimodal preferences. Amongst the multimodal learning styles, the most preferred mode was bimodal, followed by trimodal and quadrimodal respectively

Explanation:

Write a Python program that does the following. Create a string that is a long series of words separated by spaces. The string is your own creative choice. It can be names, favorite foods, animals, anything. Just make it up yourself. Do not copy the string from another source. Turn the string into a list of words using split. Delete three words from the list, but delete each one using a different kind of Python operation. Sort the list. Add new words to the list (three or more) using three different kinds of Python operation. Turn the list of words back into a single string using join. Print the string.

Answers

Answer:

The code is attached.

Explanation:

I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.

$\n$ Then I used methods append(), insert() and extend() for adding elements to the list.

$\n$ Finally I converted list into a string using join() and adding space in between the elements of the list.

A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.

Answers

Answer:

$\sigma _1=10.33MPa$

$\sigma _2=5.16MPa$

Explanation:

Given that

Diameter(d)=62 mm

Thickness(t)= 300 μm=0.3 mm

Internal pressure(P)=100 KPa

Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .

The hoop stress

$\sigma _h=(Pd)/(2t)$

Longitudinal stress

$\sigma _l=(Pd)/(4t)$

Now by putting the values

$\sigma _h=(Pd)/(2t)$

$\sigma _h=(100* 62)/(2* 0.3)$

$\sigma _h=10.33MPa$

$\sigma _l=5.16MPa$

So the principle stress are

$\sigma _1=10.33MPa$

$\sigma _2=5.16MPa$

The reservoir pressure of a supersonic wind tunnel is 5 atm. A static pressure probe is moved along the centerline of the nozzle, taking measurements at various stations. For these probe measurements, calculate the local Mach number and area ratio: a. 4 atm; b. 2.64 atm; c. 0.5 atm.

Answers

Answer

Given,

Reservoir pressure of a supersonic wind tunnel = 5 atm

Local Mach number = ?

Area ration = ?

a) 4 atm.

Pressure ratio = $(4)/(5)$

= 0.8

From Isentropic Flow Tables

M = 0.58 A/A* = 1.213

b) 2.64 atm

Pressure ratio = $(2.64)/(5)$

= 0.528

From Isentropic Flow Tables

M = 1 A/A* = 1

c) 0.5 atm

Pressure ratio = $(0.5)/(5)$

= 0.1

From Isentropic Flow Tables

M =2.10 A/A* = 1.8369

With the aid of a labbled diagram describe the operation of a core type single phase transformer

Answers

Answer:

A simple single-phase transformer has each winding being wound cylindrically on a soft iron limb separately to provide a necessary magnetic circuit, which is commonly referred to as “transformer core”. It offers a path for the flow of the magnetic field to induce voltage between two windings.

An activated sludge plant is being designed to handle a feed rate of 0.438 m3 /sec. The influent BOD concentration is 150 mg/L and the cell concentration (MLVSS) is 2,200 mg/L. If you wish to operate the plant with a food-to-microorganism ratio of 0.20 day-1 , what volume of aeration tank should you use? Please give your answer in m3 .

Answers

Answer:

Volume of aeration tank = 1.29 x 10^4 m³

Explanation:

Food/Micro- organism Ratio = 0.2/day

Feed Rate (Q) = 0.438 m³/s

Influent BOD = 150 mg/L

MLVSS = 2200 mg/L

The above mentioned parameters are related by the equation

F/M = QS₀/VX

where S₀ is the influent BOD and X is the cell concentration, with V being the Volume of the tank. Re- arranging the above equation for V and putting the values in, we get

V = 0.4380 x 150/0.2 x 2200

V = 0.1493 (m³/s) x day

V = 0.1493 x 24 x 60 x 60

V = 1.29 x 10^4 m³

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