A data bus can be visualized as a multilane highway
A. and each component is located at an intersection where it will turn or go straight
B. with each component having an individual address
C. and each component located in a curve where the data will slow down
D. with each component acting as a traffic light, stop-go

Answers

Answer 1

Answer:

B. with each component having an individual address

Explanation:

A data bus is a system within a computing system, that consists of a  set of wires or connectors, that provides transportation for data. A data bus can transfer data and information through a computing system, or between two or more computing systems. Each component in the computer has its own unique address by which data or information is sent to it, to or fro the central processing unit, and the data or information travelling through the data bus, which serves as the highway, reaches each component using this address. This is analogous to the postal service system of moving letters and packages, through a series of networks that locates the address of the receiver.


Related Questions

a mercury barometer located in a lab has a heigh tof 750 mm. what is the atmosphereic pressur ein kPa

Answers

Answer:

The atmospheric pressure from a mercury barometer equals to 99.992 kilopascals.

Explanation:

760 milimeters of mercury column is equal to 101.325 kilopascals. The atmospheric pressure is found by simple rule of three:

[tex]p_{atm} = \frac{750\,mm\,Hg\times 101.325\,kPa}{760\,mm\,Hg}[/tex]

[tex]p_{atm} = 99.992\,kPa[/tex]

The atmospheric pressure from a mercury barometer equals to 99.992 kilopascals.

Please select the word from the list that best fits the definition I love horses and want to be a veterinarian

Answers

The answer is Intrinsic

Answer:

the right word is Intrinsic

5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025

Answers

Answer:

LMC wall thickness= 5.05

Explanation:

Given:

Minimum inner diameter = 25 - 0.05 = 24.95

Maximum outer diameter = 35 + 0.05 = 35.05

Find:

LMC wall thickness

Computation:

LMC wall thickness = (maximum outer diameter - minimum inner diameter) / 2

LMC wall thickness = (35.05 - 24.95) / 2

LMC wall thickness= 5.05

Tetrahymena thermophila protozoa have a minimum doubling time of 6.5 hours when grown using bacteria as the limiting substrate. The yield of protozoal biomass is 0.33 g per g of bacteria and the substrate constant is 12 mg/ l. The protozoa are cultured at steady state in a chemostat using a feed stream containing 10 g/ l of nonviable bacteria. What is the maximum dilution rate for operation of the chemostat?

Answers

Answer:

The answer is "[tex]\bold {3.3 \ \frac{g}{L}}[/tex]"

Explanation:

If the endogenous metabolic rate wasn’t substantial after, which [tex]\mu_{net} = \mu_{max} = \frac{0.693}{ \text{ doubling period}}[/tex] .

Throughout the calculating of doubling the time is set at 6.5 hours, consequently [tex]\mu_{max} = \frac{0.693}{6.5} = \frac{0.1}{hours}.[/tex]

Know we calculate the two-equation,  

Initially, the maximum dilution frequency for [tex]D_{max}[/tex] that is:  

[tex]D_{max}= \mu _{max}\times \frac{[S_0]}{{K_s + [S_0]}} .....(a)[/tex]

In secondary the steady concentration of state cells X,  

[tex]X = \frac{Y_{\frac{x}{s}} (s[S_0]-K_sD)}{(\mu _{max}-D)}...... (b)[/tex]

In this section, we will display the [tex]Y_{\frac{x}{s}}[/tex] , that is equal to [tex]0.33\ \frac{g}{g-substrate}[/tex], and the value of the [tex]K_s = 12 \times 10^{-3} \ \frac{g}{L}[/tex].  

For equation (a):

[tex]D_{max} = 0.1 \times \frac{10}{ (12 \times 10^{-3} + 10)}[/tex]

         [tex]= 0.1 \times \frac{10}{ (\frac{12}{10^{3}} + 10)}\\\\= 0.1 \times \frac{10}{ 0.012 + 10)}\\\\= 0.1 \times \frac{10}{ 10.012}\\\\=0.1 \times 0.9988\\\\= 0.09988\\\\= 0.1 \ \ \frac{1}{h}[/tex]

In the Operating value is equal to [tex]D =\frac{1}{2}[/tex] of [tex]D_{max}[/tex], so D is =[tex]\frac{0.05}{h}[/tex] in our case.  

Finally, the amount of protozoa cells in equation (b):

[tex]X = 0.33 \times (10 - 12 \times 10^{-3} \times \frac{0.05}{(0.1 - 0.05)})\\\\X = 0.33 \times (10 - 12 \times 10^{-3} \times \frac{0.05}{-0.05})\\\\X = 0.33 \times (10 - 12 \times 10^{-3} \times -1 )\\\\X = 0.33 \times (10 + \frac{12}{10^{-3}} )\\\\X = 0.33 \times ( 10.012 )\\\\X= 3.303\\[/tex]

what is
entrained
the difference between Air-
and Air- entrapped,
2​

Answers

Answer:

sometimes small air bubbles are intentionally incorporated (entrained) into the mix using admixtures; other times larger bubbles are entrapped during mixing. When the bubbles are smaller than 0.04 inch, the air is called entrained; larger, and it's called entrapped.

Explanation:

What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn?
a. 2.6 at% Pb and 97.4 at% Sn.
b. 7.6 at% Pb and 92.4 at% Sn.
c. 97.4 at% Pb and 2.6 at% Sn.
d. 92.4 at% Pb and 7.6 at% Sn.

Answers

Answer: Option A is correct -- 2.6 at% Pb and 97.4 at% Sn.

Explanation:

Option A is the only correct option -- 2.6 at% Pb and 97.4 at% Sn. While option B, which is 7.6 at% Pb and 92.4 at% Sn. and option C, which is 97.4 at% Pb and 2.6 at% Sn. and option D, which is 92.4 at% Pb and 7.6 at% Sn. are wrong.

An open channel has a trapezoidal cross section with a bottom width of 5 m and side slopes of 2 : 1 (H : V). If the depth of flow is 2 m and the average velocity in the channel is 1 m/s, calculate the discharge in the channel.

Answers

Answer:

Q = 18 m^3/sec

Explanation:

In trapezoidal channel

Area  [tex]A = \frac{1}{2}(a+b)h[/tex]

where a, b are parallel side and h is distance b/w them.

Also, as per question

[tex]tan\theta = \frac{1}{2}\\\Rightarrow \theta = 26.56^{\circ}[/tex]

a = 5 m , b =5+2(2/tanθ) = 13 m

Therefore, A = 0.5(5+13)2 = 18 m^2

Discharge Q = AV  = 18×1 = 18 m^3/sec

V= velocity = 1 m/s (given)

Why should low and high side access valves be installed when recovering refrigerant from a household refrigerator

Answers

Answer:

Because it helps to speed up recovery and required recovery efficiency When recovering refrigerant from a household refrigerator who's compressor does not run

What is the purpose of making internal resistance of milli-ammeter very low?A) High accuracy.B) Minimum voltage drop across meter.C) Maximum voltage drop across meter.D) High sensitivity.

Answers

Answer:

B) Minimum voltage drop across meter

Explanation:

An ammeter is a measuring instrument device that is used to measure small currents flowing through a circuit in milli-ampere. The internal resistance of the milli ammeter is very low so that there would be a minimum voltage drop across the meter causing a minimum effect of the current in the circuit, this results to a measurement closer to the original value. If a high resistance is place, there would be large voltage drop therefore the current measured would be lesser than the original value.

When two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the shaft, the arrangement is known as spiral gearing.
a) true
b) false

Answers

Answer:

False

Explanation:

The spiral gearing is the application that is commonly used in any vehicle where the drive from the shaft will turn perpendicular to the drive of the wheel.

The two parallel and coplanar shafts are connected by gears having helical teeth perpendicular to the axis of the shaft.

Spiral gearing application is appropriate for any machine or vehicle with a demand of great velocity and large torque power.

When two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the shaft, the arrangement is known as spiral gearing is therefore false.

Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb which acts up to the left, in the same direction as from B towards A. Take θ = 30 ∘.

Answers

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = [tex]\sqrt{400^2 + 650^2 - 2(400)(650)cos30}[/tex]

      = [tex]\sqrt{160000 + 422500 - 80210}[/tex]

      = 708.72 Ib

using the sine law to determine the design angle

[tex]\frac{sin}{400} = \frac{sin 30}{Fac}[/tex]

hence ∅ = [tex]sin^{-1} (\frac{sin 30 *400}{708.72} )[/tex]

              = [tex]sin^{-1} 0.0871[/tex] =  4.9968 ≈ 5⁰

The design angle of the struts system given is;

ϕ = 38.26°

Parallelogram law of vector addition

The diagram showing the struts system is missing and so i have attached it.

Now, from the forces given to us acting on the system attached, I have drawn a force diagram in form of a parallelogram showing the direction of the forces action. The second attached image shows this force diagram called diagram a.

From the diagram, we can see that it follows parallelogram law of vector addition.

Now, I have narrowed down the force system to the upper part of the strut as depicted in the third image attached.

From the third image attached, we can use the concept of parallelogram law of vector addition and the law of cosine to get the force on the strut AC which is;

F_AC = √(400² + 600² - 2(400 × 600)cos 30)

F_AC = 322.97 lb

Now, to get the design angle, we will solve as;

(sinϕ)/(sin 30) = 400/322.97

Where; ϕ is the design angle.

Thus;

(sinϕ) = (400/322.97) × sin 30

(sinϕ) = 1.2385 × 0.5

(sinϕ) = 0.61925

ϕ = sin^(-1) 0.61925

ϕ = 38.26°

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In case of damaged prestressed concrete I girders which are used for restoring strength?

Answers

Answer: The use of post-tensioning rods is useful in this situation.

Explanation:

The prestressed concrete has not been accepted as a good building material. The high tensile strength of steel is combined with the concrete to give the compressive strength to the concrete.

Post-tensioning rods of steel along with jacking can be used to restore the strength of the prestressed girder and it can begin with the calculated preload. This way the damaged concrete can be repaired and restored.  

Exceeding critical mach may result in the onset of compressibility effects such as:______.

Answers

Answer:

Sound barrier.

Explanation:

Sound barrier is a sudden increase in drag and other effects when an aircraft travels faster than the speed of sound. Other undesirable effects are experienced in the transonic stage, such as relative air movement creating disruptive shock waves and turbulence. One of the adverse effect of this sound barrier in early plane designs was that at this speed, the weight of the engine required to power the aircraft would be too large for the aircraft to carry. Modern planes have designs that now combat most of these undesirable effects of the sound barrier.

Which of the three predominant equations for friction loss in fluid flow can be used only for water and is primarily used for open-channel (gravity) flow?
A. Darcy-Weisbach
B. Manning
C. Hazen-Williams
D. All of the above

Answers

Answer:

B. Manning

Explanation:

The Manning formula is an empirical formula for velocity estimation of a liquid flowing in a open channel, only used for water. The Manning formula is:

[tex]V = \frac{k}{n} \cdot R_{h}^{2/3}\cdot S^{1/2}[/tex]

Where:

[tex]V[/tex] - Cross sectional average velocity, measured in meters per second.

[tex]k[/tex] - Conversion factor, dimensionless. ([tex]k = 1[/tex] for SI units)

[tex]n[/tex] - Gauckler-Manning constant, measured in [tex]\frac{s}{m^{1/3}}[/tex].

[tex]R_{h}[/tex] - Hydraulic radius, measured in meters.

[tex]S[/tex] - Slope of the hydraulic grade line, dimensionless.

Both Darcy-Weisbach and Hazen-Williams formulas are used only for fluid flowing in pipes.

In a nutshell, the correct answer is B.

Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.

Answers

Answer:

Explanation:

Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.

Low angle grain boundaries (quasi-coherent) are formed by the dislocation network positioned along the geometric plane with small tilt angle differences between successive peers that is tilt boundary made up edge dislocations therefore it may only divert the slip direction of the incoming gliding dislocation with very little frictional stresses. And on the other hand, a high angle grain boundary region because of their disordered almost liquid like structure which acts as a strong barrier against dislocation slip motion and causes actually formation of dislocations file-up against it by arresting their motion unless that the stress concentration at the leading dislocation becomes high enough to go though the barrier.

The term route of entry on an sds refers to the way a ___ enters the body

Answers

The correct answer is A) Chemical

Explanation:

The abbreviations SDS means Safety Data Sheet, which refers to a very complete document about chemicals and safety related to these, including their properties and potential hazards, as well as, procedures or steps to avoid hazards and accidents.

This means the main focus of this document is chemicals; moreover, in this, there is information about the route of entry, which refers to the way chemical can enter the body and includes through inhalation, ingestion, contact with skin, among others. Also, depending on the chemical some routes of entry represent a major hazard. According to this, the correct answer is A.

Answer: Chemical!!!

Explanation:

In a high-quality coaxial cable, the power drops by a factor of 10 approximately every 5 km. If the original signal power is 0.25 W (=2.5 x 10-1), how far will a signal be transmitted before the power is attenuated to 25 μW? As part of your answer, include a Table showing the signal power vs. distance in 5 km intervals. If optical fibre is used instead of the coaxial cable, briefly explain how you would expect the above calculated distance value to change. You are not required to include another Table.

Answers

Answer:

20 km for 40 dB loss80 km for 40 dB loss, or 10 dB loss for 20 km

Explanation:

Here's your table of (distance, power level):

  (0 km, 250 mW), (5 km, 25 mW), (10 km, 2.5 mW),

  (15 km, 250 μW), (20 km, 25 μW)

The signal can be transmitted 20 km before being attenuated to 25 μW.

__

Reportedly, the loss in fiber optic cable is about 0.5 dB/km. This compares to 10 dB/5 km = 2 dB/km for the coaxial cable. Loss in dB/km is a factor of 4 less for fiber optic cable, so the distance for the same loss would be multiplied by 4.

The amount of time an activity can be delayed and yet not delay the project is termed:_________
A. Total slack.
B. Free slack.
C. Critical float.
D. Float pad.
E. Slip pad.

Answers

The correct answer is A. Total slack

Explanation:

Projects include multiple tasks or activities; moreover, each activity and the project itself have a specific due date. Besides this, activities can be delayed and this might affect or not the due date of the project. Indeed, some activities can have extra time without any effect on the date of the project, which is known as total slack or total float (delay in activities that does not delay the project.) Additionally, total slack is possible if activities or tasks are flexible, for example, if each task in a project should take 1 day and one of the tasks takes one day and a half, this can be compensated if another task takes less time. According to this, the correct answer is A.

Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam is superheated to a higher temperature, (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease.

Answers

Answer:

Option D - the moisture content at turbine exit will decrease

Explanation:

In an ideal rankine system, the phenomenon of superheating occurs at a state where the vapor state of the fluid is heated above its saturation temperature and the phase of the fluid is changed from the vapor phase to the gaseous phase.

Now, a vapour phase has two different substances at room temperature, whereas a gas phase consists of just a single substance at a defined thermodynamic range, at standard room temperature.

At the turbine exit, since it's just a single substance in gaseous phase, it means it will have less moisture content.

Thus, the correct answer is;the moisture content at turbine exit will decrease

In which order are the following measuring instruments listed from lowest precision to highest precision:
Select one:
a. Outside Calipers.
Ruler.
Micrometer.
Digital Calipers.
b. Outside Calipers.
Digital Calipers.
Ruler.
Micrometer.
c. Outside Calipers.
Digital Calipers.
Micrometer.
Ruler.

Answers

Wouldn’t it be C tho?

Answer:

Answer is C

Explanation:

Outside Calipers.

Digital Calipers.

Micrometer.

Ruler.

Kieran and Kurt spend most of their day performing physical labor, aligning materials, and inspecting projects for quality and safety. However, Kieran has no fear of heights, while Kurt works with his feet on the ground. Which most likely explains the jobs of Kieran and Kurt? Kieran is a Roofer, and Kurt is a Plumber. Kieran is a Plumber, and Kurt is a Roofer. Kieran is a Carpenter, and Kurt is a Drafter. Kieran is a Drafter, and Kurt is a Carpenter.

Answers

Answer:

Kieran is a Carpenter, and Kurt is a Drafter

Explanation:

Among all the professions listed, Kurt can only possibly be a Drafter.  A Drafter is an engineering technician who makes detailed technical drawings or plans for machinery, buildings, electronics, infrastructure, sections, etc. Drafters make use of computer software and manual sketches to convert the designs, plans, and layouts of engineers and architects into a set of technical drawings. All the other professions will at some point need for the professional to stand and work at an height above the ground except for a drafter, whose feet will remain on the ground at almost all times.

Answer:

A is the right answer.

Explanation:

Hope that helps

A data bus can be visualized as a multilane highway
A. and each component is located at an intersection where it will turn or go straight
B. with each component having an individual address
C. and each component located in a curve where the data will slow down
D. with each component acting as a traffic light, stop-go

Answers

Answer:

B. with each component having an individual address

Explanation:

Data bus is a system within a computer or device, consisting of a connector or set of wires, that provides transportation for data. Data bus needs an address unique to each component in order to deliver the right data to the right place. Every memory location has a unique binary address. A microprocessor architecture is mainly composed of two main buses: The data bus and the address bus.

which of the following devices are used in networking? check all that apply?
switches, hubs,TCP or routers​

Answers

Answer:

Hub , switch port , transmission media , hope I helped

Explanation:

The resultant force is directed along the positive x axis and has a magnitude of 1330 N.
Determine the magnitude of F_A. Express your answer to three significant figures and include the appropriate units. Determine the direction theta of F_A. Express your answer using three significant figures.

Answers

Answer:

the magnitude of F_A is 752 N

the direction theta of F_A is 57.9°

Explanations:

Given that,

Resultant force = 1330 N in x direction

∑Fx = R

from the diagram of the question which i uploaded along with this answer

FB = 800 N

FAsin∅ + FBcos30 = 1330 N

FAsin∅ = 1330 - (800 × cos30)

FA = 637.18 / sin∅

Now ∑Fx = 0

FAcos∅ - FBsin30 = 0

we substitute for FA

(637.18 / sin∅)cos∅ = 800 × sin30

637.18 / 800 × sin30 = sin∅/cos∅

and we know that { sin∅/cos∅ = tan∅)

so tan∅ = 1.59295

∅ = 57.88° ≈ 57.9°

THEREFORE FROM THE EQUATION

FA = 637.18 / sin∅

we substitute ∅

so FA = 637.18 / sin57.88

FA = 752 N

Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
a. 2.17000000 A-2.258571429 A
b. 2.20000000 A-2.29000000 A
c. 2.211445 A-2.30000000 A
d. 2.20144927 A-2.29130435 A
e. 2.00000000 A-2.30000000 A

Answers

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

         x_average = ∑ [tex]x_{i}[/tex] / n

The tolerance or error is the current value over the mean value per 100

         Δx₁ = x₁ / x_average

         tolerance = | 100 -Δx₁  100 |

bars indicate absolute value

let's look for these values ​​for each case

a)

    x_average = (2.1700000+ 2.258571429) / 2

    x_average = 2.2142857145

fluctuation for x₁

        Δx₁ = 2.17000 / 2.2142857145

        Tolerance = 100 - 97.999999991

        Tolerance = 2.000000001%

fluctuation x₂

        Δx₂ = 2.258571429 / 2.2142857145

        Δx2 = 1.02

        tolerance = 100 - 102.000000009

        tolerance 2.000000001%

b)

    x_average = (2.2 + 2.29) / 2

    x_average = 2,245

fluctuation x₁

         Δx₁ = 2.2 / 2.245

         Δx₁ = 0.9799554

         tolerance = 100 - 97,999

         Tolerance = 2.00446%

fluctuation x₂

          Δx₂ = 2.29 / 2.245

          Δx₂ = 1.0200445

          Tolerance = 2.00445%

c)

   x_average = (2.211445 +2.3) / 2

   x_average = 2.2557225

       Δx₁ = 2.211445 / 2.2557225 = 0.9803710

       tolerance = 100 - 98.0371

       tolerance = 1.96%

       Δx₂ = 2.3 / 2.2557225 = 1.024624

       tolerance = 100 -101.962896

       tolerance = 1.96%

d)

   x_average = (2.20144927 + 2.29130435) / 2

   x_average = 2.24637681

       Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

       tolerance = 100 - 98.000043

       tolerance = 2.000002%

       Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

       tolerance = 2.0000002%

e)

   x_average = (2 +2,3) / 2

   x_average = 2.15

   Δx₁ = 2 / 2.15 = 0.93023

   tolerance = 100 -93.023

   tolerance = 6.98%

   Δx₂ = 2.3 / 2.15 = 1.0698

   tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values ​​may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

Based on the information given, the correct option will be C. 2.211445 A-2.30000000 A.

It should be noted that the tolerance range simply means the maximum limit of variation that can be allowed for a particular quantity.

It can be noted that the range that depicts the 2% tolerance range to the full 9 digits provided is 2.211445 A-2.30000000 A. It illustrates a high precision regarding the question.

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What is the approximate probability that no people in a group of seven have the same birthday?
(A) 0.056
(B) 0.43
(C) 0.92
(D) 0.94

Answers

Answer: (D) 0.94

P(7 distinct birthdays) = 363/365 = 0.9438 (0.94)

The probability that no people in a group of seven have the same birthday is; D: 0.94

How to find the probability?

The probability that no people in a group of seven have the same birthday is;

P(no two people in a group of seven) = P(1) * P(2) * P(3) * P(4) * P(5) * P(6) *P(7)

⇒ 365/365 * 364/365 * 363/365 * 362/365 * 361/365 * 360/365 * 359/365 = 0.9438 ≈ 0.94

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What is the biggest expectation when engineers test out designs?

A.
Designs need to be totally unique.

B.
There are no expectations about the results, since all ideas are considered.

C.
Designs need to be feasible.

D.
Designs should be implemented within a certain time frame.

Answers

The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.

Answer: Designs need to be feasible.

Explanation: don’t know, guessed got it right

Argon is compressed in a polytropic process with n = 1.2 from 100 kPa and 30°C to 1200 kPa in a piston–cylinder device. Determine the final temperature of argon.

Answers

Answer:

181 °C

Explanation:

Initial pressure [tex]P_{1}[/tex] = 100 kPa

Initial temperature [tex]T_{1}[/tex] = 30 °C = 30 + 273 K = 303 K

Final pressure [tex]P_{2}[/tex] = 1200 kPa

Final temperature [tex]T_{2}[/tex] = ?

n = 1.2

For a polytropic process, we use the relationship

([tex]T_{2}[/tex]/[tex]T_{1}[/tex] ) = ([tex]P_{2}[/tex]/[tex]P_{1}[/tex])^γ

where γ = (n-1)/n

γ = (1.2-1)/1.2 = 0.1667

substituting into the equation, we have

([tex]T_{2}[/tex]/303) = (1200/100)^0.1667

[tex]T_{2}[/tex]/303 = 12^0.1667

[tex]T_{2}[/tex]/303 = 1.513

[tex]T_{2}[/tex] = 300 x 1.513 = 453.9 K

==> 453.9 - 273 = 180.9 ≅ 181 °C

The final temperature of Argon in polytropic process is T = 181°C

Given data:

The initial pressure  [tex]P_{1}=100kPa[/tex]

The final pressure  [tex]P_{2}=1200kPa[/tex]

The initial temperature  [tex]T_{1}=303K[/tex]

The value of n = 1.2

For a polytropic process:

[tex]\frac{T_{2}}{T_{1}} =\frac{P_{1}}{P_{2}}^{\beta }[/tex]

The value of  [tex]\beta = \frac{(n-1)}{n}[/tex]

So, [tex]\beta =0.1667[/tex]

Substituting the values in the equation:

[tex]\frac{T_{2}}{303} =\frac{1200}{100}^{0.1667 }[/tex]

So, the value of  [tex]T_{2}=453.9K[/tex]

The temperature in degree Celsius is  [tex]T_{2}=181C[/tex].

Hence, the final temperature is T = 181°C.

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A circular hoop sits in a stream of water, oriented perpendicular to the current. If the area of the hoop is doubled, the flux (volume of water per unit time) through it:___________

Answers

Answer:

The flux (volume of water per unit time) through the hoop will also double.

Explanation:

The flux = volume of water per unit time = flow rate of water through the hoop.

The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.

This means that

Flow rate = AV

where A is the area of the hoop

V is the velocity of the water through the hoop

This flow rate = volume of water per unit time = Δv/Δt =Q

From all the above statements, we can say

Q = AV

From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2

You are driving on a roadway with multiple lanes of travel in the same direction, and are approaching an emergency vehicle parked along the roadway. You must:a. Leave the lane closest to the emergency as soon as it is safe to do so, or slow down to a speed of 20 MPH below the posted speed limit.b. Yield to the emergency vehicle.c. If you are in the lane closest to the emergency vehicle, immediately come to a complete stop and ask if anyhelp is needed.d. All of the above.

Answers

Answer: a. Leave the lane closest to the emergency as soon as it is safe to do so, or slow down to a speed of 20 MPH below the posted speed limit.

Explanation:

Giving a way to the law enforcement vehicle and a medical emergency vehicle is necessary. If one approaches an emergency vehicle parked along the roadway one should change the lane as the vehicle may not move and the driver may also waste his or her time also one should also slow down his or her speed while approaching the vehicle as most of the emergency vehicle are in rush to reach the hospital so the driver should maintain some distance with the medical emergency vehicle.

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