a material has r = 1, r = 10, and = 0.02 s/m at 20 ghz. problem [1] <5 points> what kind of a lossy material is this? a) lossless b) low loss c) quasi-conductor (medium loss) d) good conductor

Answers

Answer 1

Based on the given values of r and  at 20 GHz, we can determine that the material is a quasi-conductor or a medium loss material. This is because a material with r values ranging from 1 to 10 indicates that it has moderate resistance to the flow of electric current.

Additionally, a conductivity value of 0.02 s/m suggests that the material is not a good conductor of electric current, which further supports the idea that it is a medium loss material. In summary, the material can be classified as a quasi-conductor or medium loss material due to its moderate resistance and low conductivity values.
Based on the given information, this material can be classified as a quasi-conductor (medium loss).

Here's a step-by-step explanation:
1. The material's resistivity (r) values are given as r = 1 and r = 10.
2. The material's loss tangent is given as 0.02 s/m at 20 GHz frequency.
3. Comparing these values to standard classifications, we can conclude:
  a) Lossless: This material is not lossless because it has a non-zero loss tangent.
  b) Low loss: A low loss material typically has a much smaller loss tangent than 0.02 s/m.
  c) Quasi-conductor (medium loss): This material fits the description of a quasi-conductor since it has a moderate loss tangent.
  d) Good conductor: A good conductor usually has lower resistivity values and a higher loss tangent than this material.

So, the correct answer is c) quasi-conductor (medium loss).

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Related Questions

The pressure less than atmospheric pressure is known as:

Suction pressure

Negative gauge pressure

Vacuum pressure

All of the above

Answers

Answer:

answer is option (d) all of the above

how skateboards works?

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Answer

The skateboarder applies pressure to the trucks and gives/releases pressure on the levers. Second, the wheels and the axles are also examples of simple machines. They help the skater ride, spin, grind, and do a bunch of other radical movements on a skateboard.:

Explanation:

I don’t know the answer to this question

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Answer:

I dont know the answer either

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Answer:

flux

Explanation:

what is the distance in term of wavelengh between successive minima in the standing wave ratio​

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Answer:

hi there

Explanation:

Convert an acceleration of 12m/s² to km/h²​

Answers

Here is your answer , hope you like it

What current works best when the operator
encounters magnetic arc blow?

•DCEP

•ACEN

•CC

•AC

Answers

Answer:

AC

Explanation:

One situation when alternating current would work better than direct current is if the operator is encountering magnetic arc blow.

Current works best when the operator  encounters magnetic arc blow is AC

Magnetic arc blow is simply defined as the arc deflection due to the warping of the magnetic field that is produced by electric arc current.

This is caused as a result of the following;

- if the material being welded has residual magnetism at an intolerable level

- When the weld root is being made, and the welding current is direct current which indicates constant direction and maintains constant polarity (either positive or negative).

Since it is caused by DC(Direct Current) which means constant polarity , it means the opposite will be better which is AC(alternating current) because it means that electricity direction will be switching to and fro and as such the polarity will also be revered in response to this back and forth switch manner.

Thus, Current works best when the operator  encounters magnetic arc blow is AC

Read more at; brainly.in/question/38789815?tbs_match=1

What are the top 4 solar inventions, how they are used, and how they are better than the original way of powering them

Answers

Yes I will answer soon

During a shrinkage limit test, a 19.3 cm3 saturated clay sample with a mass of 37 g was placed in a porcelain dish and dried in the oven. The oven-dried sample had a mass of 28 g with a final volume of 16 cm3 . Determine the shrinkage limit and the shrinkage ratio.

Answers

Answer:

shrinkae limit = 20.35%

shrinkage ratio = 1.45

Explanation:

1. to get the shrinkage limit we would first calculate the moisture content w.

w = (37-28)/28

= 9/28

= 0.3214

then the formula for shrinkage limit is

[tex][w-\frac{(V-Vd}{wd} ]*100[/tex]

w = 0.3214

V = 19.3

Vd = 16

Wd = 28

when we put these values into the formula:

[tex][0.3214-\frac{(19.3-16)}{28} ]*100\\[/tex]

= 20.35%

2. the shrinkage limit = Wd/V

= 28/19.3

= 1.45

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 2.00 MPa, calculate the magnitude of applied stress necessary to cause slip to occur on the (111) plane in the direction.

Answers

Answer:

Explanation:

From the given information:

The equation for applied stress can be expressed as:

[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda}[/tex]

where;

[tex]\phi[/tex] = angle between the applied stress [100] and [111]

To determine the [tex]\phi[/tex] and [tex]\lambda[/tex] for the system

Using the equation:

[tex]\phi= cos^{-1}\Big [\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}}\Big][/tex]

for [100]

[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]

for [111]

[tex]l_1 = 1 , m_1 = 1, n_1 = 1[/tex]

Thus;

[tex]\phi= cos^{-1}\Big [\dfrac{1*1+0*1+0*1}{\sqrt{(1^2+0^2+0^2)(1^2+1^2+1^2)}}\Big][/tex]

[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(3)}}\Big][/tex]

[tex]\phi= 54.74^0[/tex]

To determine  [tex]\lambda[/tex]  for [tex][1 \overline 1 0][/tex]

where;

for [100]

[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]

for [tex][1 \overline 1 0][/tex]

[tex]l_1 = 1 , m_1 = -1, n_1 = 0[/tex]

Thus;

[tex]\lambda= cos^{-1}\Big [\dfrac{1*1+0*1+0*0}{\sqrt{(1^2+0^2+0^2)(1^2+(-1)^2+0^2)}}\Big][/tex]

[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(2)}}\Big][/tex]

[tex]\phi= 45^0[/tex]

Thus, the magnitude of the applied stress can be computed as:

[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda }[/tex]

[tex]\sigma_{app} = \dfrac{2.00}{cos (54.74) \ cos (45) }[/tex]

[tex]\mathbf{\sigma_{app} =4.89 \ MPa}[/tex]

An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter D=1 mm, electrical resistivity rhoe=10−6Ω⋅m, thermal conductivity k=25W/m⋅K, and emissivity ε=0.20. The heater is designed to deliver air at a temperature of T[infinity]=50∘C under flow conditions that provide a convection coefficient of h=250W/m2⋅K for the wire. The temperature of the housing that encloses the wire and through which the air flows is Tsur=50∘C. If the maximum allowable temperature of the wire is Tmax=1200∘C, what is the maximum allowable electric current I? If the maximum available voltage is ΔE=110V, what is the corresponding length L of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.

Answers

Solution :

Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :

[tex]$h_T= \epsilon \sigma (T_s+T_{surr})(T_s^2 +T^2_{surr})$[/tex]

[tex]$h_T= 0.20 \times 5.67 \times 10^{-8} (1473+323)(1473^2 +323^2)$[/tex]

[tex]$h_T=46.3 \ W/m^2 .K$[/tex]

The total heat transfer coefficient will be :

[tex]$h_T=(250+46.3) \ W/m^2 .K$[/tex]

    [tex]$=296.3 \ W/m^2 .K$[/tex]

Now calculating the maximum volumetric heat generation :

[tex]$\dot {q}_{max}=\frac{2h_t}{r_0}(T_s-T_{\infty})$[/tex]

[tex]$\dot {q}_{max}=\frac{2\times 296.3}{0.0005}(1200-50)$[/tex]

        [tex]$= 1.362 \times 10^{9} \ W/m^3$[/tex]

The heat generation inside the wire is given as :

[tex]$\dot{q} = \frac{I^2R}{V}$[/tex]

Here, R is the resistance of the wire

         V is the volume of the wire

∴ [tex]$\dot{q} = \frac{I^2\left( \rho \times \frac{L}{A} \right)}{A \times L}$[/tex]

     [tex]$=\frac{I^2 \rho}{\left(\frac{\pi}{4}D^2 \right)}$[/tex]

where, ρ is the resistivity.

[tex]$I_{max}= \left(\frac{\dot{q}_{max}}{\rho} \right)^{1/2} \times \frac{\pi}{4}D^2$[/tex]

[tex]$I_{max}= \left(\frac{1.36 \times 10^9}{10^{-6}} \right)^{1/2} \times \frac{3.14}{4}(1 \times 10^{-3})^2$[/tex]

        = 28.96 A

Now considering the relation for the current flow through the finite potential difference.

[tex]$E=I_{max} \times R$[/tex]

[tex]$E=I_{max} \times \rho \times \frac{L}{A}$[/tex]

[tex]$L=\frac{AE}{I_{max} \ \rho}$[/tex]

[tex]$L=\frac{\frac{\pi}{4} \times (1 \times 10^{-3})^2 \times 110}{28.96 \times 10^{-6}}$[/tex]

   = 2.983 m

Now calculating the power rating of the heater:

[tex]$P= E \times I_{max}$[/tex]

[tex]$P= 110 \times 28.96}$[/tex]

   = 3185.6 W

   = 3.1856 kW

A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction. The axial load is 30 kips. Soil conditions consist of a normally consolidated clay layer with following properties; su=800 psf, andγsat= 110 pcf. Design for a FOS of 3. Assume frost depth to be 3ft below grade

Answers

Solution :

Finding the cohesion of the soil(c) using the relation:

[tex]$c = \frac{q_u}{2}$[/tex]

Here, [tex]$q_u$[/tex] is the unconfined compression strength of the soil;

[tex]$c = \frac{800}{2}$[/tex]

   = 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction [tex]$0^\circ$[/tex]

    [tex]$N_c = 5.14$[/tex]

   [tex]$N_q = 1.0$[/tex]

   [tex]$N_r = 0$[/tex]

Therefore,

[tex]$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$[/tex]

     =  2386 psf

∴ Allowable bearing capacity [tex]$q_{a} = \frac{Q_{allow}}{A}$[/tex]

                                                     [tex]$=\frac{30}{B^2}$[/tex]

∴ [tex]$q_a = \frac{q_{ult}}{F.O.S}$[/tex]

  [tex]$\frac{30}{B^2} = \frac{2386}{3}$[/tex]

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

                                                                                 [tex]$=0.04 \ ft^2$[/tex]

Find the derivative of x ​

Answers

Answer:

this is your answer. if mistake don't mind.

I have an AC waveform with the voltage peak value of 100 V. I'm trying to find the RMS value of the voltage​

Answers

Answer:

98 x 100=9,00

Explanation:

Now, you get a turn to practice writing a short program in Scratch. Try to re-create the program that was shown that turns the sprite in a circle. After you have completed that activity, see if you can make one of the improvements suggested. For example, you can try adding a sound. If you run into problems, think about some of the creative problem-solving techniques that were discussed.

When complete, briefly comment on challenges or breakthroughs you encountered while completing the guided practice activity.


Pls help im giving 100 points for this i have this due in minutes

Answers

Answer:

u need to plan it out

Explanation:

u need to plan it out

Answer:

use the turn 1 degrees option and put a repeat loop on it

Explanation:

u can add sound in ur loop

Tech A says that LED brake lights illuminate faster than incandescent bulbs. Tech B says that LED brake lights have
more visibility and last longer. Who is correct?

Answers

Answer:

Both

Explanation:

Describe how to contribute to
zero/low carbon work outcomes
within the built environment.

Answers

Answer:

day if you workout without Zero billing that means you're not sweating. Sweating you're not losing anything that means you have zero outcomes

Explanation:

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16.5 s later. The descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude. Assume that g = 32.2 ft/s2.
Determine
(a) the speed v1 of the rocket at the end of powered flight,
(b) the maximum altitude reached by the rocket.

Answers

Answer:

[tex]u = 260.22m/s[/tex]

[tex]S_{max} = 1141.07ft[/tex]

Explanation:

Given

[tex]S_0 = 89.6ft[/tex] --- Initial altitude

[tex]S_{16.5} = 0ft[/tex] -- Altitude after 16.5 seconds

[tex]a = -g = -32.2ft/s^2[/tex] --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

[tex]S = ut + \frac{1}{2}at^2[/tex]

The final altitude after 16.5 seconds is represented as:

[tex]S_{16.5} = S_0 + ut + \frac{1}{2}at^2[/tex]

Substitute the following values:

[tex]S_0 = 89.6ft[/tex]       [tex]S_{16.5} = 0ft[/tex]     [tex]a = -g = -32.2ft/s^2[/tex]    and [tex]t = 16.5[/tex]

So, we have:

[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2[/tex]

[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45[/tex]

[tex]0 = 89.6 + 16.5u- 4383.225[/tex]

Collect Like Terms

[tex]16.5u = -89.6 +4383.225[/tex]

[tex]16.5u = 4293.625[/tex]

Make u the subject

[tex]u = \frac{4293.625}{16.5}[/tex]

[tex]u = 260.21969697[/tex]

[tex]u = 260.22m/s[/tex]

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

[tex]v=u + at[/tex]

At the maximum height:

[tex]v =0[/tex] --- The final velocity

[tex]u = 260.22m/s[/tex]

[tex]a = -g = -32.2ft/s^2[/tex]

So, we have:

[tex]0 = 260.22 - 32.2t[/tex]

Collect Like Terms

[tex]32.2t = 260.22[/tex]

Make t the subject

[tex]t = \frac{260.22}{ 32.2}[/tex]

[tex]t = 8.08s[/tex]

The maximum height is then calculated as:

[tex]S_{max} = S_0 + ut + \frac{1}{2}at^2[/tex]

This gives:

[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2[/tex]

[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22[/tex]

[tex]S_{max} = 89.6 + 260.22 * 8.08 - 1051.11[/tex]

[tex]S_{max} = 1141.0676[/tex]

[tex]S_{max} = 1141.07ft[/tex]

Hence, the maximum height is 1141.07ft

Suppose you are choosing between four different desktop computers: one is an Apple Mac Intosh and the other three are PC-compatible computers that use a Pentium 4, an AMD processor (using the same compiler as the Pentium 4), and a Pentium 5 (which does not yet exist in 2004 but has the same architecture as the Pentium 4 and uses the same compiler). Which of the following statements are true?
a. The fastest computer will be the one with the highest clock rate.
b. Since all PCs use the same Intel-compatible instruction set and execute the same number of instructions for a program, the fastest PC will be the one with the highest clock rate.
c. Since AMD uses different techniques than Intel to execute instructions,they may have different CPIs. But, you can still tell which of the two Pentium-based PCs is fastest by looking at the clock rate.
d. Only by looking at the results of benchmarks for tasks similar to your workload can you get an accurate picture of likely performance.

Answers

Answer:

d.

Explanation:

Which of the following choices accurately contrasts a categorical syllogism with a conditional syllogism?


An argument constructed as a categorical syllogism uses deductive reasoning whereas an argument constructed as a conditional syllogism uses inductive reasoning.

A categorical syllogism contains two premise statements and one conclusion whereas a conditional syllogism contains one premise statement and one conclusion.

A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

An argument constructed as a categorical syllogism is valid whereas an argument constructed as a conditional syllogism is invalid.

Answers

Answer:

The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

Explanation:

As,

Categorical syllogisms follow an "If A is part of C, then B is part of C" logic.

Conditional syllogisms follow an "If A is true, then B is true" pattern of logic.

So,

The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

If the hypotenuse of a right triangle is 12 and an acute angle is 37 degrees find leg a and leg b lengths

Answers

scrity añao devid codicie

Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per year. On the second site, the wind blows steadily at 10 m/s for 2000 hours per year. The density of air on the both sites is 1.25 kg/m3 . Assuming the wind power generation is negligible during other times.Calculate the maximum power of wind on each site per unit area, in kW/m2 .

Answers

Solution :

Given :

[tex]$V_1 = 7 \ m/s$[/tex]

Operation time, [tex]$T_1$[/tex] = 3000 hours per year

[tex]$V_2 = 10 \ m/s$[/tex]

Operation time, [tex]$T_2$[/tex] = 2000 hours per year

The density, ρ = [tex]$1.25 \ kg/m^3$[/tex]

The wind blows steadily. So, the K.E. = [tex]$(0.5 \dot{m} V^2)$[/tex]

                                                             [tex]$= \dot{m} \times 0.5 V^2$[/tex]

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = [tex]$\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$[/tex]

Regarding that [tex]$\dot m \propto V$[/tex]. Then,

Power [tex]$ \propto V^3$[/tex] → Power = constant x [tex]$V^3$[/tex]

Since, [tex]$\rho_a$[/tex] is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, [tex]$P_1= \text{const.} \times V_1^3$[/tex]

            [tex]$P_1 = \text{const.} \times 343 \ W$[/tex]

For the second site,

Power, [tex]$P_2 = \text{const.} \times V_2^3 \ W$[/tex]

           [tex]$P_2 = \text{const.} \times 1000 \ W$[/tex]

This is silence I couldent find the tab... 30 points plus marked brainliest if corrects!


The most recent evidence supporting the theory of plate tectonics would include
es )
A)
GPS monitoring of plate speeds and movements.
B)
the WWII discovery of paleomagnetic reversals.
Elimi
O
the 1963 mapping of the tectonic plate boundaries.
D
C-14 dating of marine fossils found in the Himalayas.

Answers

Yeah the answer is A

who was part of dempwolf his firm when he first started

Answers

Explanation:

Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with  brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.

With _____, only one criterion must evaluate true in order for a record to be selected and with _____, all criteria must be evaluate true in order for a record to be selected.

a. parameter criteria, double criteria
b. function criteria, IF criteria
c. simple criteria, complex criteria
d. OR criteria, AND criteria

Answers

Answer:

d

Explanation:

OR criteria, AND criteria

In an OR criteria, it doesn't need all the records to be true. Just one record is enough and all other criterion becomes true.

In an AND criteria, it's unlike the OR criteria and works in opposite. It needs every member of the record to be true to be able to adjudge the whole record as true.

And as such, we have

With OR criteria, only one criterion must evaluate true in order for a record to be selected and with AND criteria, all criteria must be evaluate true in order for a record to be selected.

The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a particle) is 2 m above the road. Calculate the speed v of the truck.

Answers

Answer:

19.81 m/s

Explanation:

The total acceleration of the truck (a) is due to the centripetal acceleration and as a result of the linear acceleration. Therefore the total acceleration (a) is given by:

[tex]a^2=a_n^2+a_t^2\\\\where\ a_n=centripetal\ acceleration=\frac{v^2}{r},a_t=linear \ acceleration\\\\But\ since\ the \ speed\ is \ constant, the \ linear \ acceleration(a_t)\ would\ be\ 0.\\\\a^2=a_n^2+a_t^2\\\\a^2=a_n^2\\\\a=a_n=\frac{v^2}{r} \\\\v^2=ar\\\\v=\sqrt{ar} \\\\a=0.4g=0.4*9.81,r=98\ m+2\ m=100\ m\\\\v=\sqrt{0.4*9.81*100} \\\\v=19.81\ m/s[/tex]

A cylindrical bar of metal having a diameter of 15.7 mm and a length of 178 mm is deformed elastically in tension with a force of 49100 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34 respectively, Determine the following:(a) The amount by which this specimen will elongate (in mm) in the direction of the applied stress. The entry field with incorrect answer now contains modified data.(b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answers

Answer:

0.6727

-0.02017

Explanation:

diameter = 15.7

l = 178

E =elastic modulus = 67.1 Gpa

poisson ratio = 0.34

p = force = 49100N

first we calculate the area of the cross section

[tex]A=\frac{\pi }{4} d^{2}[/tex]

[tex]A=\frac{\pi }{4} (15.7)^{2} \\A = \frac{774.683}{4} \\[/tex]

A = 193.6mm²

1. Change in directon of the applied stress

[tex]= \frac{pl}{AE}[/tex]

= 49100*178/193.6*67.1*10³

= [tex]=\frac{8739800}{12990560}[/tex]

δl = 0.6727  mm

2. change in diameter of the specimen

equation for poisson distribution =

m = -(δd/d) / (δl/l)

0.34 = (δd/15.7) / (0.6727/178)

0.34 = (-δd * 178) / 15.7 * 0.6727

0.34 = -178δd / 10.56139

we cross multiply

10.56139*0.34 =-178δd

3.5908726 = -178δd

δd = 3.5908726/-178

δd = -0.02017 mm

the change in dimeter has a negative sign. it decreases

1. What is the productivity rate using cycle time for the following information:
I
Type of Work – Hauling
Average Cycle Time – 35 Minutes
Truck Capacity – 25 Tons
Crew - One Driver
Productivity Factor - 0.85
System Efficiency – 55 Minutes
per
Hour

Answers

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If the sum of the two numbers is 4 and the sum of their squares minus three times their product is 76,find the number

Answers

Answer:

-2 and 6

Explanation:

Let "x" and "y" be 2 numbers.

The sum of the two numbers is 4. The mathematical expression is:

x + y = 4

y = 4 - x   [1]

The sum of their squares minus three times their product is 76. The mathematical expression is:

x² + y² - 3 x y = 76   [2]

If we substitute [1] in [2], we get:

x² + (4 - x)² - 3 x (4 - x) = 76

x² + 16 - 8 x + x² - 12 x + 3 x² = 76

5 x² - 20 x - 60 = 0

We apply the solving formula for second order equations and we get x₁ = 6 and x₂ = -2.

If we replace these x values in [1], we get:

y₁ = 4 - x₁ = 4 - 6 = -2

y₂ = 4 - x₂ = 4 - (-2) = 6

As a consequence, one of the numbers is 6 and the other is -2.

Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 399 MPa causes the specimen to plastically elongate to a length of 54.4 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 47.7 mm to a length of 57.8 mm.

Answers

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

[tex]\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}[/tex]

Solve the following for the real stress and pressure for the stable.[tex]\sigma_{r1}=K(\varepsilon_{r1})^{n}[/tex]

[tex]K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}[/tex]

Solve the following for the true state stress and stress2.

[tex]\sigma_{r2}=K(\varepsilon_{r2})^n[/tex]

     [tex]=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa[/tex]

... is an actual sequence of interactions (i.e., an instance) describing one specific situation; a ... is a general sequence of interactions (i.e., a class) describing all possible ... associated with a situation. ... are used as examples and for clarifying details with the client. ... are used as complete descriptions to specify a user task or a set of related system features.

Answers

Answer:

ScenarioUse caseScenariosScenariosUse case

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.

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