To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105 mm^-2 . Suppose that all the dislocations in 1000 mm^3 (1 cm^3) were somehow removed and linked end to end.

Required:
a. How far (in miles) would this chain extend?
b. Now suppose that the density is increased to 1010 mm^-2 by cold working. What would be the chain length of dislocations in 1000 mm^3 of material?

Answers

Answer 1

Answer:

[tex]62.14\ \text{miles}[/tex]

[tex]6213727.37\ \text{miles}[/tex]

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = [tex]10^5\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}[/tex]

[tex]1\ \text{mile}=1609.34\ \text{m}[/tex]

[tex]\dfrac{10^5}{1609.34}=62.14\ \text{miles}[/tex]

The chain would extend [tex]62.14\ \text{miles}[/tex]

Dislocation density = [tex]10^{10}\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}[/tex]

[tex]\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}[/tex]

The chain would extend [tex]6213727.37\ \text{miles}[/tex]


Related Questions

You are playing guitar on a stool that is 22" tall. How tall is the stool if it is expressed as a combination of feet and inches?

Answers

Answer:

1 foot 10 inches

Explanation:

1 foot = 12 inches + 10 inches = 22 inches

irhagoaihfw

A spherical Gaussian surface of radius R is situated in space along with both conducting and insulating charged objects. The net electric flux through the Gaussian surface is:______

Answers

Answer:

Ф = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]

Explanation:

Radius of Gaussian surface = R

Charge in the Sphere ( Gaussian surface ) = Q

lets take the radius of the sphere to be equal to radius of the Gaussian surface i.e. R

To determine the net electric flux through the Gaussian surface

we have to apply Gauci law

Ф = 4[tex]\pi r^2 E[/tex]

Ф = [tex]\frac{Q_{enc} |}{e_{0} }[/tex]

    = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]


A 550 kJ of heat quantity needed to increase water temperature from 32°C to 80°C. Calculate the mass
of the water when the specific heat capacity of water is 4200 J/kg °C.​

Answers

Answer:

  2.728 kg

Explanation:

The units help you keep the calculation straight.

  [tex]\dfrac{550\text{ kJ}}{(80^\circ\text{C}-32^\circ\text{C})(4.200\text{ kJ/kg\,$^\circ$C})}=\dfrac{550}{48\cdot4.2}\text{ kg}\approx\boxed{2.728\text{ kg}}[/tex]

the pressure rise, across a pump can be expressed as where D is the impeller diameter, p, is the fluid density, w is the rotational speed, adn q is the flowrate. determine a suitable set of dimensionless parameters

Answers

Answer:

hello your question is incomplete below is the complete question

The pressure rise Δp across a pump can be expressed as Δp = f(D, p, w, Q) where D is the impeller diameter, p is the fluid density, w is the rotational speed, and Q is the flowrate. determine a suitable set of dimensionless parameters

answer : Δp / D^2pw^2 = Ф (Q / D^3w )

Explanation:

k ( number of variables ) = 5

r ( number of reference dimensions ) = 3

applying the pi theorem

hence the number of pi terms = k - r = 5 - 3 = 2

A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistances are both 1 kΩ?

Answers

Answer:

required feedback resistance ( R2 ) = 100 k Ω

Explanation:

Given data :

Voltage gain = 100

input resistance ( R1 ) = 1 k ohms

calculate feedback resistance required

voltage gain of differential amplifier

[tex]\frac{Vout}{V2 - V1 } = \frac{R2}{R1}[/tex]

= Voltage gain =  R2/R1

= 100 = R2/1

hence required feedback resistance ( R2 ) = 100 k Ω

. In the U.S. fuel efficiency of cars is specified in miles per gallon (mpg). In Europe it is often expressed in liters per 100 km. Write a MATLAB userdefined function that converts fuel efficiency from mpg to liters per 100 km. For the function name and arguments, use Lkm

Answers

Answer:

MATLAB Code is written below with comments in bold, starting with % sign.

MATLAB Code:

function L = Lkm(mpg)

 L = mpg*1.60934/3.78541;  %Conversion from miles per gallon to km per   liter

 L = L^(-1);  %Conversion to liter per km

 L = L*100;   %Conversion to liter per 100 km

end

Explanation:

A function named Lkm is defined with an output variable "L" and input argument "mpg". So, in argument section, we give function the value in miles per gallon, which is stored in mpg. Then it converts it into km per liter by following formula:

L = (mpg)(1.60934 km/1 mi)(1 gallon/3.78541 liter)

Then this value is inverted to convert it into liter per km, in the next line. Then to find out liter per 100 km, the value is multiplied by 100 and stored in variable "L"

Test Run:

>> Lkm(100)

ans =

   2.3522

Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.

Answers

Hope this helps...........

1. Consider a solid cube of dimensions 1ft x 1ft x 1ft (=0.305m x 0.305m x 0.305m). Its top surface is 10
ft (=3.05 m) below the surface of the water. The density of water is pf=1000 kg/m3.
Consider two cases:
a) The cube is made of cork (pB=160.2 kg/m3)
b) The cube is made of steel (pB=7849 kg/m3)
In what direction does the body tend to move?​

Answers

Answer:

  a) up

  b) down

Explanation:

When the cube is less dense than water, it will tend to float (move upward). When it is more dense, it will sink (move downward).

a) 160.2 kg/m^3 < 1000 kg/m^3. The cube will move up.

__

b) 7849 kg/m^3 > 1000 kg/m^3. The cube will move down.

What test should be performed on abrasive wheels

Answers

Answer:

before wheel is put on it should be looked at for damage and a sound or ring test should be done to check for cracks, to test the wheel it should be tapped with a non metallic instrument (I looked it up)

The  test that should be performed on abrasive wheels is the ring test.

What is the purpose of the ring test on the  abrasive wheels?

The ring test can be regarded as one of the mechanical test that is used to know whether the wheel is cracked or damaged.

To carry out this test , the wheel will be arranged to be in the 45 degrees each side and it is then aligned to be at a specific diameter, this can be done by the expert in this field to know the state of that wheel.

Learn more about  ring test  on:

https://brainly.com/question/4621112

#SPJ9

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