Answer:
a) the average length per chip is 3 mm
b) the metal removal rate MR is 90 mm³/sec
c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min
Explanation:
Given that;
wheel diameter = 150 mm
infeed = 0.06 mm
wheel speed = 1600 m/min = 16000000 mm/s
work speed = 0.30 m/s = 300 mm/s
cross feed = 5 mm
active grits per area = 50 grits/cm²
a)
Average length per chip
Average chip length is given by
Lc = √fd
f is infeed and d is diameter of the wheel
so we substitute
Lc = √( 0.06 × 150
Lc = √ 9
Lc = 3 mm
Therefore the average length per chip is 3 mm
b)
metal removal rate MR is expressed as;
MR = fvc
v is work speed, c is cross feed and f is infeed
so we substitute
MR = 0.06 × 300 × 5
MR = 90 mm³/sec
Therefore the metal removal rate MR is 90 mm³/sec
c)
number of chips formed per unit time is expressed as
No = Vw × c × G
Vw is wheel speed and G is active grits per area
so we substitute
No = 1600000 × 5 × 50/10²
= 4,000,000 chips/min
Therefore number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min
What are some of the trade-offs of a move to an enterprise-level analytics solution for individual end users who might have grown accustomed to working with their own customized solutions for generating data?
Answer:
THE thing is 435
Explanation:
Some of the trade-offs that must be made while transitioning from an individually tailored solution to such an enterprise-level solution are about as continues to follow.
Enterprise-level analytics Different individuals must be given some amount of control over their data in corporate-level systems.To counteract opposition from individuals who seem to be comfortable working utilizing customized statistics, training needs to be provided so that they can utilize cutting-edge large enterprise technologies.Adequate orientation training, as well as perks, should have been provided so that employees may become acquainted using the changing approach.
As a result, the aforementioned response seems to be correct.
Find out more information about enterprise-level analytics here:
https://brainly.com/question/25071524
Consider an ideal turbojet at flight conditions: To =220 K, PO = 20 kPa, and Mo= 0.8. Calculate and plot: 1) the specific thrust, 2) f, and 3) ISP versus the compressor stagnation pressure ratio (πc). Perform calculations for Tt,A=1600 K and 1800 K and for πc from 1 to 50. The fuel heat of combustion is hf= 42,800 kJ/kg. What πc maximizes thrust at the two peak stagnation temperatures?
Answer:
i) specific thrust = 11.065
ii) F = attached below
iii) Isp vs [tex]\pi[/tex]c = attached below
iv) [tex]\pi c1 = 16.38* 10^8\\\pi c2 = 24.47 * 10^8[/tex]
Explanation:
Given data :
To = 220k
Po = 20 kPa
Mo = 0.8
Hf = 42800 KJ/kg
Describe how the fracture behavior would be different for a fiber-reinforced tape such as duct tape.
Answer:
A Normal tape is very weak under tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape
Explanation:
The fracture behavior would be different for a fiber-reinforced tape in the following way :
* It's behavior during tensile stress and its fracture behavior.
A Normal tape is very weak under tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape
Water flows at 10 m3/s in a 5-m-wide channel. What is the height of a suppressed rectangular (sharp-crested) weir that will cause the depth of flow in the channel to be 2 m
Answer:
Hw = 1.01 meters
Explanation:
Given data:
flow rate = 10 m^3
depth of flow in channel = 2 m
Determine the height of a suppressed rectangular weir ( Hw ) using the following expressions
expression for the elevation of of water surface above crest of weir
H = 2 - [tex]H_{w}[/tex] ------ ( 1 )
expression for the height of the weir ( Hw )
Hw = 2 - [tex]( \frac{Q}{C_{w} b}) ^{\frac{3}{2} }[/tex] ---------- ( 2 )
expression for the weir coefficient ( Cw )
Cw = [tex]\frac{2}{3} C_{d} \sqrt{2g}[/tex] -------------- ( 3 )
expression for the coefficient of discharge ( Cd )
Cd = 0.611 + 0.075 [tex]\frac{H}{Hw}[/tex] ---------- ( 4 )
Finally to determine the value of Hw we apply the trial and error method
in the trial and error method the value of LHS = RHS for the number chosen to be true
How can technology interfere with good study habits?
Answer:
the third anwser is right on edgeinuity
Explanation:
Answer:
Technology can pull students’ focus away from their tasks.
Explanation:
when you reach a yield sign yield to cross traffic and ___ before you enter the intersection
Answer:
Wain for a safe gap
Explanation:
Find the complex power, average power, and reactive power for the following:
v(t) = 339.4 sin (377t + 90°) V, i(t) = 5.657 sin (377t + 60°) A
Answer:
Explanation:
Given that:
v(t) = 339.4 sin(377t + 90°) V
i(t) = 5.657 sin (377t + 60°) A
v = 339.4 ∠ 90° [tex]v_m \angle\phi_1[/tex]
i = 5.657∠60° [tex]I_m \angle \phi_2[/tex]
The phase difference [tex]\phi = 90 -60[/tex] = 30
The average power [tex]P_{avg}[/tex] can be expressed as:
[tex]P_{avg} = \dfrac{v_m}{\sqrt{2}}\dfrac{I_m}{\sqrt{2}} \times cos (30)[/tex]
[tex]P_{avg} = \dfrac{339.4}{\sqrt{2}}*\dfrac{5.657}{\sqrt{2}} \times cos (30)[/tex]
[tex]\mathbf{P_{avg} = 831.38 \ watts}[/tex]
The reactive power Q is as follow;
[tex]Q = \dfrac{v_m}{\sqrt{2}} * \dfrac{I_m}{\sqrt{2}} \ sin \phi\\[/tex]
[tex]Q = \dfrac{339.4}{\sqrt{2}}*\dfrac{5.657}{\sqrt{2}} \times sin (30)[/tex]
Q = 479.99 VAR
The complex power S = P + jQ
The complex power S = 831.38 W + j479.99 VAR
Writing Prompt 2: Split Decisions PLEASE HELP
Answer:
Explanation:
is there a book or smg
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?
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
which of the following is a function of a safety device
Answer:
what are the options available?
Equal moles of pure liquid 1-Butanol, Benzene, and Phenol form an ideal solution system at 353K. Determine Yi for each component at vapor-liquid equilibrium (VLE) ofthe mixtureat 353K.
Answer:
[tex]\mathbf{y_1 =0.1750}[/tex]
[tex]\mathbf{y_2 = 0.8088}[/tex]
[tex]\mathbf{y_3 = 0.0161}[/tex]
Explanation:
Given that:
The temperature of the ideal solution formed by the compounds = 353K
= (353 - 273)° C
= 80° C
The data below shows the Antoine constant obtained for 1-Butanol, Benzene, and Phenol.
Compound A B C
1 - Butanol 15.3144 3212.43 182.739
Benzene 13.7819 2726.81 217.572
Phenol 14.4397 3507.80 175.400
By the application of the Antoine constant, we can find the vapor pressure of each corresponding component at the given temperature of 80° C.
The Antoine equation is expressed as:
[tex]In (P^*) = A - \dfrac{B}{T+C}[/tex]
[tex]P_1 ^* = exp \bigg [15.314 - \dfrac{3212.43}{80+182.739} \bigg ][/tex]
[tex]P_1 ^* = exp \bigg [15.314 - \dfrac{3212.43}{262.739} \bigg ][/tex]
[tex]P_1 ^* = 21.9267 \ kPa[/tex]
[tex]P_2^* = exp \bigg [13.7819 - \dfrac{2726.81}{80+217.572} \bigg ][/tex]
[tex]P_2^* = exp \bigg [13.7819 - \dfrac{2726.81}{297.572} \bigg ][/tex]
[tex]P_2^* = 101.3287 \ kPa[/tex]
[tex]P_3^* = exp \bigg [14.4387 - \dfrac{3507.80}{80+175.400} \bigg ][/tex]
[tex]P_3^* = exp \bigg [14.4387 - \dfrac{3507.80}{255.4} \bigg ][/tex]
[tex]P_3^* =2.0221 \ kPa[/tex]
However, since the ideal solution has equimolar composition.
Then:
component (1) = 1 mole ,
component (2) = 1 mole,
component (3) = 1 mole,
The total mole = 1 + 1 + 1 = 3,
Thus, mole fraction = mole of component/total mole,
mole fraction of component (1) [tex]x_1[/tex] = 1/3,
mole fraction of component (2) [tex]x_2[/tex] = 1/3,
mole fraction of component (3) [tex]x_3[/tex] = 1/3,
i.e.
[tex]x_1 =x_2=x_3 = \dfrac{1}{3}[/tex]
Using Raoult's law, The total pressure is computed as:
[tex]P = x_1P_1^* + x_2 P_2^* + x_3P_3^*[/tex]
[tex]P = \dfrac{1}{3}(21.9267) + \dfrac{1}{3} (101.3287) + \dfrac{1}{3} (2.0221)[/tex]
P = 41.7592 kPa
and;
[tex]P_1 *x_1 = y_1P[/tex]
[tex]\implies y_1 = \dfrac{P_1\times x_1}{P}[/tex]
Thus:
[tex]y_1 = \dfrac{21.9267 \times \dfrac{1}{3}}{41.7592}[/tex]
[tex]\mathbf{y_1 =0.1750}[/tex]
[tex]y_2 = \dfrac{P_2\times x_2}{P}[/tex]
[tex]y_2 = \dfrac{101.3287 \times \dfrac{1}{3}}{41.7592}[/tex]
[tex]\mathbf{y_2 = 0.8088}[/tex]
[tex]y_3 = \dfrac{P_3\times x_3}{P}[/tex]
[tex]y_3 = \dfrac{2.0221\times \dfrac{1}{3}}{41.7592}[/tex]
[tex]\mathbf{y_3 = 0.0161}[/tex]
What type of circles have two or more circles with different center points?
Answer:
Concentric circles
Explanation:
Concentric circles are two or more circles which have the same center point. The region between two concentric circles is called an annulus.
A method that uses low temperature heat-treating that imparts toughness without reduction in hardness is called:_______
Answer:
"Tempering Process" seems to be the appropriate choice.
Explanation:
Tempering seems to be a method of heat preparation which is mostly used in completely hard materials to increase consistency, strength, durability, and also some decreasing brittleness. The tempering method is used to examine good functionality as well as flexural by reducing stiffness again after the substance has indeed been quenched towards its toughest state.A rigid tank contains 1 kg of N2 at 25°C and 300 kPa is connected to another rigid tank that contains 3 kg of O2 at 25°C and 500 kPa. The valve connecting the two tanks is opened, and the two gases are allowed to mix. If the final mixture temperature is 25°C.
Required:
Determine the volume of each tank and the final mixture pressure.
Answer:
Following are the solution to this question:
Explanation:
let,
[tex]\text{m= mass of the gas}\\\text{M= molar weight of the gas}\\ m_{N_2} = 1 \kg \\T_{N_2}= 298 \ K\\ M_{N_2} = 28 \ \frac{kg}{kmol}\\ P_{N_2}= 300 \ kPa\\m_{O_2} = 3 \ kg \\T_{O_2} = 298 \ K \\M_{O_2}=32 \ \frac{kg}{kmol}\\ P_{O_2} = 500 \ kPa \\T_m= 298 \ K[/tex]
Calculating the mole of nitrogen:
[tex]N_{N_2} = \frac{m_{N_2}}{M_{N_2}}[/tex]
[tex]=\frac{1}{28} \\\\= 0.0357 \ kmol[/tex]
Calculating the mole oxygen:
[tex]N_{O_2} = \frac{m_{O_2}}{M_{O_2}}\\\\=\frac{3}{32} \\\\= 0.09375 \ kmol[/tex]
Calculating the total mole:
[tex]N_m=N_{N_2}+N_{O_2}\\\\= 0.0357+0.09375\\\\= 0.1295 \ kmol[/tex]
Calculating the volume of nitrogen:
[tex]V_{N_2} =\frac{N_{N_2} \times R_{u}\times T_{N_2}}{P_{N_2}}\\\\= \frac{0.0357 \times 8.314 \times 298}{300}\\\\= 0.295 m^3[/tex]
Calculating the volume of oxygen:
[tex]V_{O_2} =\frac{N_{O_2} \times R_{u}\times T_{O_2}}{P_{O_2}}\\\\= \frac{0.09375 \times 8.314 \times 298}{500}\\\\= 0.465 m^3[/tex]
Calculating the total volume:
[tex]V_m=V_{N_2}+V_{O_2}\\\\=0.295+0.465 \\\\=0.76 \ m^3[/tex]
Calculating the mixture pressure:
[tex]P_m = \frac{N_{m} \times R_u \times T_m}{V_m} \\\\= \frac{0.1295 \times 8.314 \times 298}{0.76} \\\\= 422.2\ kPa[/tex]
PLSSSSS Help !!!!!!!!!!!. It's due today.
I will give brainliest to correct answer
Answer:
b
Explanation:
Answer:
b
Explanation:
A specimen of aluminum having a rectangular cross section 10 mm x 12.7 mm is pulled in tension with 35,500 N (8000 lbf) force, producing only elastic deformation. Calculate the resulting strain.
Answer:
The resulting strain is 4.05 x 10⁻³
Explanation:
Given;
dimension of the specimen = 10 mm x 12.7 mm
Cross sectional area of the aluminum specimen = 10 mm x 12.7 mm = 127 mm² = 1.27 x 10⁻⁴ m²
applied force, F = 35,500 N
Young's modulus is given by;
[tex]E = \frac{Stress}{Strain}\\\\E = \frac{F}{A(strain)}\\\\E = \frac{F}{A(\epsilon)}\\\\\epsilon = \frac{F}{A E}\\\\[/tex]
Where;
ε is the resulting strain
E is Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²
[tex]\epsilon = \frac{35500}{(1.27*10^{-4}) (69*10^{9})}\\\\ \epsilon = 4.05*10^{-3}[/tex]
Therefore, the resulting strain is 4.05 x 10⁻³
Tech A says that 18 AWG wire is larger than 12 AWG wire. Tech B says that the larger the diameter of the conductor, the more electrical resistance it has. Who is correct?
Answer:
Both of them are wrong
Explanation:
The two technicians have given the wrong information about the wires.
This is because firstly, a higher rating of AWG means it is smaller in diameter. Thus, the diameter of a 18 AWG wire is smaller than that of a 12 AWG wire and that makes the assertion of the technician wrong.
Also, the higher the resistance, the smaller the cross sectional area meaning the smaller the diameter. A wire with bigger cross sectional area will have a smaller resistance
So this practically makes the second technician wrong too
A Carnot heat engine absorbs 235 KW of heat from a heat source and rejects 164 KW to the atmosphere. Determine the thermal efficiency of the heat engine.
Answer:
43.2%
Explanation:
Given that,
Heat absorbed by a carnot heat engine, [tex]Q_1=235\ kW[/tex]
Heat rejected to the atmosphere, [tex]Q_2=164\ kW[/tex]
We need ti find the thermal efficiency of the heat engine. It is equal to the ratio of output work to the energy supplied. Its mathematical form is given by :
[tex]\eta=1-\dfrac{Q_1}{Q_2}\\\\\eta=1-\dfrac{235}{164}\\\\\eta=-0.432[/tex]
or
[tex]\eta=-43.2\%[/tex]
The egative value of efficiency shows work is done by the engine.
look told you i made it a meme
Answer:
Dam they be thinking we racist
Explanation:
Answer:
XDDDD
Explanation:
XD I LOVE THIS MEME IM SAVING THIS TO MY GALLERY XD
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it through the resistors where it is heated. If the density of air at the inlet is 1.20 kg/m^3 and 0.95kg/m^3 at the exit, determine the percent increase in the velocity of the air as it flows through the dryer.
Answer:
% increase = 26.32%
Explanation:
From conservation of mass, we can say that;
Mass flow rate at inlet = mass flow rate at exit.
Thus;
m'1 = m'2
Formula for mass flow rate is;
m' = ρV'
Where V' is volumetric flow rate = Av
Thus;
m' = ρAv
Where;
ρ is density
A is area
v is velocity
Therefore from m'1 = m'2, we can say that;
ρ1•A1•v1 = ρ2•A2•v2
Since the duct has a constant diameter, then A1 = A2
Thus, we now have;
ρ1•v1 = ρ2•v2
Making v2 the subject, we have;
v2 = ρ1•v1/ρ2
Now, since we want to find the percent increase in the velocity of the air as it flows through the dryer,we would use;
% increase = ((v2 - v1)/v1) × 100%
We have v2 = ρ1•v1/ρ2
Thus;
% increase = ((ρ1•v1/ρ2) - v1)/v1) × 100%
Factorizing v1 out, we have;
% increase = ((ρ1/ρ2) - 1)/1) × 100%
We are given;
ρ1 = 1.2 kg/m³
ρ2 = 0.95 kg/m³
Thus;
% increase = ((1.2/0.95) - 1)/1) × 100%
% increase = 26.32%
. 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
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
Technician A states that air tools generally produce more noise than electric tools, so wear ear protection when using air tools. Technician B states that you should always use impact sockets with impact guns. Who is correct?
Answer:
Both Technician A and Technician B are correct
Explanation:
Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.
Technician B is also correct because it is always advisable to use impact sockets while using impact guns due to the ability of the impact sockets to withstand the force caused by operating impact guns and make work neater when nuts and bolts are being loosened or tightened.
When jump-starting a car, always remember ___________
Answer:
to keep the cables away from any belts so that when starting the car back up, they do not get tangled or caught up into the engine
Explanation:
When jump-starting a car, always remember jumper cables typically have two clamps, one with the label “positive” in red and “negative” in black.
What is a jump-start?Jump-starting a car means starting a car when it is off for a long time, and it should be started by giving heat to a battery. The cables of the car are attached to the battery box and the current is given.
The steps to jump-starting a car are:
With a functional battery, start the vehicle. Start a working car's engine, then move it near to the vehicle with the dead battery.Attach jumper cords to the batteries.Start the vehicle.Disconnect the two batteries with care.After the jump start, check the cables' condition.Thus, when jump-starting a car, always remember the cables are in the right position.
To learn more about jump-start, refer to the link:
https://brainly.com/question/16148146
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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?
Answer:
1 foot 10 inches
Explanation:
1 foot = 12 inches + 10 inches = 22 inches
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A cylindrical specimen of a brass alloy having a length of 104 mm (4.094 in.) must elongate only 5.20 mm (0.2047 in.) when a tensile load of 101000 N (22710 lbf) is applied. Under these circumstances what must be the radius of the specimen? Consider this brass alloy to have the stress–strain behavior
Answer:
The radius of the specimen is assumed to be 9.724 mm
Explanation:
Given that:
For a cylindrical specimen of a brass alloy;
The length = 104 mm, Elongation = 5.20 mm and the tensile load = 101000 N
Let's first determine the radius of the cylindrical brass alloy from the knowledge of the cross-sectional area of a cylinder.
[tex]A_0 = \pi r ^2[/tex]
[tex]r = \sqrt{\dfrac{A_o}{\pi}}[/tex]
[tex]r = \sqrt{\dfrac{\bigg (\dfrac{F}{\sigma} \bigg )}{\pi}}[/tex]
[tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]
To estimate the tensile stress:
We need to first determine the strain relating to elongation at 5.20 mm
[tex]Strain \ \ \varepsilon= \dfrac{\Delta l}{l_o}[/tex]
[tex]Strain \ \ \varepsilon= \dfrac{5.20}{104}[/tex]
Strain ε = 0.05
Using the stress-strain plot; let assume that under the circumstances; [tex]\sigma[/tex] = 340 MPa for stress corresponding to 0.05 strain
Thus;
The cylindrical brass alloy radius [tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]
[tex]r =\sqrt{ \dfrac{101000}{(340\times 10^{6})\pi}[/tex]
r = 0.009724 m
r = 9.724 mm
Find the value of P(-1.5≤Z≤2)
Answer:
0.9104
Explanation:
Suitable technology can tell you the probability.
P(-1.5≤Z≤2) ≈ 0.9104
__
A phone app gives the probability as 0.9104426667829628.
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a feed of 0.02 ipr? What is the cutting time?
Answer:
a) the metal removal rate is 14.4 in³/min
b) the cutting time is 0.98 min
Explanation:
Given the data from the question
first we find the rpm for the spindle of the drilling tool, using the equation
Ns = 12V/πD
V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)
so we substitute
Ns = 12 × 120 / π2
Ns = 1440 / 6.2831
Ns = 229.18 rmp
Now we find the metal removal rate using the equation
MRR = (πD²/4) Fr × Ns
Fr is the feed rate( 0.02 ipr ),
so we substitute
MRR = ((π × 2²)/4) × 0.02 × 229.18
MRR = 14.3998 ≈ 14.4 in³/min
Therefore the metal removal rate is 14.4 in³/min
Next we find the allowance for approach of the tip of the drill
A = D/2
A = 2/2
= 1 in
now find the time required to drill the hole
Tm = (L + A) / (Fr × Ns)
Lis the the depth of the hole( 3.5 in)
so we substitute our values
Tm = (3.5 + 1) / (0.02 × 229.18 )
Tm = 4.5 / 4.5836
Tm = 0.98 min
Therefore the cutting time is 0.98 min
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:______
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]
why do u have to have certain limits for questions
Answer:
ok
Explanation:
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?
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]
