For a n-JFET CS amplifier circuit with the following values: VDD 18V, RL -20 ks2, R₁ = 60 ks2, R₂ = 80 k2, Rp 12k2, Rss = 1 k2, Rs = 10052 (source internal resistance). Assume Ipss=20mA and V₂ - 4.0 V. Assume Rss is fully bypassed. Given the equation for A, as following: a. Find the operating points Ip, Vos and VDs b. Find the ac voltage gain A,: [ The equation is: [A] = gm Ra (RD|R₁)/(Rs+RG)] c. The input Resistance Ri d. Draw the ac equivalent circuit using a JFET ac model

You get a result immediately from Euler's formula:

e ^(i π/2) = cos(π/2) + i sin(π/2) = 0 + i * 1 = i

Answer:

What ! you are an Engineer and I am a High school why are You in my feed..!!

The characteristics of the fifth component are:

Resistance = 15.36 ohms

Reactance = -7.692 ohms

(a) The arrangement of the described load can be illustrated as follows:

    +---R1---+

    |        |

 |-----|      |

 |     L1     |

 |-----|      |

    |        |

    +---R2---+

    |        |

 |-----|      |

 |     L2     |

 |-----|      |

    |        |

    +---R3---+

    |        |

 |-----|      |

 |     L3     |

 |-----|      |

    |        |

    +---R4---+

where R1, X1, R2, X2, R3, and X3 are the resistance and reactance values of components 1, 2, and 3 respectively, and R4 and X4 are the resistance and reactance values of component 4.

(b) (i) For each of the four components of the total load, we can use the following formulas to determine the current and power factor:

Apparent Power = VI

Real Power = VIcos(θ)

Reactive Power = VIsin(θ)

Power Factor = cos(θ)

where V is the voltage of the supply, I is the current in the component, and θ is the phase angle between the voltage and current.

Using these formulas, we can obtain the following results:

Component 1: Current = 20.39A, Power Factor = 0.9769, Resistance = 13.33 ohms, Reactance = 8.88 ohms

Component 2: Current = 22.55A, Power Factor = 0.8959 lagging, Resistance = 23.048 ohms, Reactance = -11.51 ohms

Component 3: Current = 8.13A, Power Factor = 0.9532 lagging, Resistance = 9.216 ohms, Reactance = 6.912 ohms

Component 4: Current = 30.25A, Power Factor = 0.909 lagging

To determine the characteristics (resistance and reactance) of component 4, we can use the following formula:

Apparent Power = VI

Reactive Power = VIsin(θ)

Impedance Z = V/I

Resistance R = Real Power / I^2

Reactance X = sqrt(Z^2 - R^2)

where Z is the impedance of the component.

Using these formulas with the values given in the problem, we get:

Component 4: Apparent Power = 13.2kVA Reactive Power = 10.81kVAR Impedance Z = 7.937 ohms Resistance R = 3.913 ohms Reactance X = 6.200 ohms

(ii) Now that we know the characteristics of the three components and the total load, we can determine the characteristics of the new component 5 by subtracting the apparent power and reactive power of the total load from the values for components 1-4. Using the same formulas as above, we get:

Component 5: Apparent Power = 1.94kVA Reactive Power = 0.62kVAR Resistance R = 15.36 ohms Reactance X = -7.692 ohms

Therefore, the characteristics of the fifth component are:

Resistance = 15.36 ohms

Reactance = -7.692 ohms

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Its True the range before you enter data saves time because it confines the movement of the active cell to the selected range .

What is data
Data is information that has been collected and organized in a meaningful way. Data can be quantitative (numbers) or qualitative (descriptive and subjective information). Data can come from surveys, experiments, observations and other sources. Data is a valuable asset and is used to inform decisions, create insights, and uncover trends and patterns. Data can be structured (structured in a way that makes it easy to search, sort, and analyze) or unstructured (not organized in a conventional way). Data can be used to identify problems, generate and test hypotheses, and measure the success of different strategies. Data is a key component of any organization and is used to help inform decisions, gain insights, and improve operations.

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Answer:

30 lbs

Explanation:

Answer:

The answer above is correct.  30lbs

Explanation:

Answer:

The meole fraction of acetone in the outlet liquid is \(x_1 = 0.0072\)

Explanation:

1.

The schematic diagram to represent this process is shown in the diagram attached below:

2.

the mole fraction of acetone in the outlet liquid is determined as follows:

solute from Basis Gas flow rate \(G_s = 10(1-0.012) =9.88 kmol/hr\)

Let the entering mole be :\(y_1 = 1.2\) % = 0.012

\(y_1 =(\dfrac{y_1}{1-y_1})\)

\(y_1 =(\dfrac{0.012}{1-0.012})\)

\(y_1 =0.012\)

Let the outlet gas concentration be \(y_2\) = 0.1% = 0.001

\(y_2 = 0.001\)

Thus; the mole fraction of acetone in the outlet liquid is:

\(G_s y_1 + L_s x_2 = y_2 L_y + L_s x_1\)

\(9.88(0.012-0.001)=15*x_1\)

\(9.88(0.011) = 15x_1\)

\(x_1 = \dfrac{0.10868}{15}\)

\(x_1 = 0.0072\)

The mole fraction of acetone in the outlet liquid is \(x_1 = 0.0072\)

Moral creativity and innovation are based on original discoveries,  whereas immoral innovation is based on unscrupulous actions.

Innovation refers to the practices aimed at developing new products and services for the well-being of society.

Moral innovation is an expression generally used to describe technological advancements based on intellectual property rights.

In conclusion, moral creativity and innovation are based on original discoveries,  whereas immoral innovation is based on unscrupulous actions.

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Answer: hello the eye bolt diagram is missing attached below is the missing diagram

i) angle = 90°

ii) magnitude ( F2 ) = 692.82

ii) Fr = 400 N

Explanation:

i) Determine the angle ∅

The ∅ = 90° since F2 to be minimum , it is directed perpendicular to the resultant force

ii) Determine the magnitude

By applying the parallelogram law of addition and triangular rule to the sketches attached below

The resultant force and the Magnitude  can be calculated :

F2 = F1sin60°

     = 800 * sin60° = 692.82 N

iii) Fr ( resultant force ) = F1cos60°

    = 800*cos 60° = 400 N  

In the example, we have three bulbs with of r: A, B, and C. Let's examine the conditions of the circuit step by step:

1. When bulb C is installed, bulbs B and C are connected in parallel, creating an equivalent resistance of r/2.2. Bulb A is connected in series with the parallel arrangement of bulbs B and C, and the circuit's overall resistance is 3.3. The total resistance changes to 5r when bulb C is removed, putting bulbs A and B in series.4. Removing bulb C reduces the circuit's current flow since it raises the overall resistance.5. When bulb C is taken out, the battery's current drops to 7/12 of what it was before.

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Answer:

- Ultrasound scanning system

- Magnetic Resonance Imaging (MRI)

- Computed tomography (CT)

- X - Ray scan

Explanation:

\(.\)

Answer:

- USS Ultarsound

- Medical sonography

The power of radio signals reduces with increasing distance from the transmitter because they are electromagnetic waves that are conveyed through the air.

Yet, unlike AM waves, FM waves don't reflect off the ionosphere due to their relatively short wavelengths. FM waves, on the other hand, go past the ionosphere and out into space. The Picture below illustrates this as well. FM waves can't therefore travel very far to reach receivers.

AM radio has substantially longer wavelengths since it runs at a lower frequency. Bigger wavelength waves can pass through solid objects more easily, allowing them to travel further.

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Answer:

Rate of Heat Loss = 336 W

Explanation:

First, we will find the surface area of the cylinder that is modelled as the man:

\(Area = A = (2\pi r)(l)\)

where,

r = radius of cylinder = 30 cm/2 = 15 cm = 0.15 m

l = length of cylinder = 170 cm = 1.7 m

Therefore,

\(A = 2\pi(0.15\ m)(1.7\ m)\\A = 1.6\ m^2\)

Now, we will calculate the rate of heat loss:

\(Rate\ of\ Heat\ Loss = hA\Delta T\)

where,

h = convective heat tranfer  coefficient = 15 W/m²K

ΔT = Temperature difference = 34°C - 20°C = 14°C

Therefore,

\(Rate\ of\ Heat\ Loss = (15\ W/m^2K)(1.6\ m^2)(14\ K)\\\)

Rate of Heat Loss = 336 W

The three main principles in engineering design are: strategic balance, top management approach and team work.

The principles are fundamental concepts that engineers use to develop effective solutions to complex problems. These principles are based on scientific and mathematical principles as well as practical considerations related to the materials, technologies and resources available to the engineer.

The engineering design process involves several stages, including problem identification, research, concept development, prototyping and testing. Throughout each stage, engineers apply various principles to ensure that their designs meet the needs and requirements of the intended users.

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Answer:

A Document in a flowchart  represents a printout, for instance a document or a report.

Flowcharts are used in planning and documenting basic process or projects. Like different sorts of outlines, they help imagine what is happening and along these lines help comprehend a process, and maybe additionally find more subtle highlights inside the process.

Explanation:

A flowchart is essentially a graphical portrayal of steps. It shows steps in consecutive request and is generally used in introducing the progression of calculations or work process. Regularly, a flowchart shows the means as boxes of different sorts, and their request by interfacing them with bolts.

It is likewise a kind of graph that speaks to a work process or cycle. A flowchart can likewise be characterized as a diagrammatic portrayal of a calculation, a bit by bit way to deal with settling an assignment.

The flowchart shows the means as boxes of different sorts, and their request by associating the containers with bolts. This diagrammatic portrayal outlines an answer model to a given issue. Flowcharts are used in investigating, planning, reporting or dealing with a process or program in different fields.

Answer:

A

Explanation:

Answer:

A, D, F

Explanation:

took on edge.

Answer:

The answer is A,D,F

A:observing and interpreting readings on gauges, meters, and charts

D:testing boiler water quality or arranging for testing

F:operating or tending stationary engines, boilers, and auxiliary equipment

Explanation:

I got all right on edge.

Answer:

a. True

Explanation:

Apply the principle of conservation of mass.

and also the expression for the steady flow energy equation.

kinetic and potential energy effects can be neglected.

The given statement by the inventor who is claiming the development of a device that requires no work or heat transfer input yet is able to produce hot and cold air streams at a steady state is definitely false.

Heat transfer may be characterized as a type of process which involves the migration of heat from one object or component to another by numerous mechanisms like conduction, convection, and/or radiation.

The process of heat transfer may occur where there is a temperature difference between two objects exist. It significantly utilizes the mechanism of exchanging thermal energy between two or more physical systems.

According to the concept of physics, no object or thing has the ability to perform its function without the utilization of any source of heat or energy. Then, how it is possible for that device to produce hot and cold air streams at a steady state.

Therefore, the given statement by the inventor is absolutely false.

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Here's a class diagram that represents the entities and their relationships for the video store scenario described:

+---------------------+        +---------------+       +-----------------+

|      Movie          |        |    Member     |       |    Rental       |

+---------------------+        +---------------+       +-----------------+

|  movie_id : int     |        |   member_id : |       | rental_id : int |

|  title : string     |        |   int        |<-------| movie : Movie    |

|  format : string    |        |   name : str |        | member : Member |

|  category : string  |        |   phone : str|        | rental_date : dt|

+---------------------+        |   type : str |        | is_returned : bool |

                               +---------------+        +-----------------+

                               |  credit_card: |                

                               +---------------+                

                               | rent_movies() |                

                               +---------------+                

                                         ^                      

                                         |                      

                               +---------------+                

                               | GoldenMember  |                

                               +---------------+                

                               | rent_limit:   |                

                               +---------------+                

                               | return_movies()|                

                               +---------------+                

                                         ^                      

                                         |                      

                               +---------------+                

                               | BronzeMember |                

                               +---------------+                

                               | rent_limit:   |                

                               +---------------+                

                               | return_movie()|                

                               +---------------+                

The Movie class has attributes movie_id, title, format and category. Each Member has a member_id, name, phone, and type. The Rental class is the association class that connects Movie and Member through the movie and member attributes, respectively, and adds rental_id, rental_date, and is_returned attributes.

There are two sub-classes of Member: GoldenMember and BronzeMember. The GoldenMember has an additional credit_card attribute and a rent_limit method that allows them to rent multiple movies at once. The BronzeMember has a rent_limit method that allows them to rent only one movie at a time. Both sub-classes have a return_movie or return_movies method, respectively, to allow them to return the rented movies.

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Answer:

i personally think it is false

Explanation:

i think this because gear friction reduces next to no power

An assemblage of beams or other components joined by nodes to form a rigid structure is known as a truss also state if the member is in tension or compression.

A truss is a structure that, according to engineering definition, "consists of just two-force members, where the members are arranged so that the assembly as a whole acts as a single entity."

[2] The term "two-force member" refers to a structural element where only two locations are subject to force. Trusses are normally made up of five or more triangular units made of straight members whose ends are joined at joints known as nodes, despite the fact that this strict definition permits the members to be united in any form and stable arrangement.

Glider AE: Uniformly distributed load on

beam CD: 150(12)(4/12) + 490(18.3/144) = 662.3 lb/ft.

Uniformly distributed load=490(32.7/144)=111.3 lb/ft concentrated load at c=8279 lb concentrated load at A and E=[150(6)(4/12)+490(18.3/144)]

(25/2)=4529lb

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Answer:

a) the mass flow rate of the steam is  \(\mathbf{m_1 =6.92 \ kg/s}\)

b) the exit velocity of the steam  is \(\mathbf{V_2 = 562.7 \ m/s}\)

c) the exit area of the nozzle is  \(A_2\) = 0.0015435 m²

Explanation:

Given that:

A steam with 5 MPa and 400° C enters a nozzle steadily

So;

Inlet:

\(P_1 =\) 5 MPa

\(T_1\) = 400° C

Velocity V = 80 m/s

Exit:

\(P_2 =\) 2 MPa

\(T_2\) = 300° C

From the properties of steam tables  at \(P_1 =\) 5 MPa and \(T_1\) = 400° C we obtain the following properties for enthalpy h and the speed v

\(h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg\)

From the properties of steam tables  at \(P_2 =\) 2 MPa and \(T_1\) = 300° C we obtain the following properties for enthalpy h and the speed v

\(h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg\)

Inlet Area of the nozzle = 50 cm²

Heat lost Q = 120 kJ/s

We are to determine the following:

a) the mass flow rate of the steam.

From the system in a steady flow state;

\(m_1=m_2=m_3\)

Thus

\(m_1 =\dfrac{V_1 \times A_1}{v_1}\)

\(m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}\)

\(m_1 =\dfrac{0.4 }{0.057838 }\)

\(\mathbf{m_1 =6.92 \ kg/s}\)

b) the exit velocity of the steam.

Using Energy Balance equation:

\(\Delta E _{system} = E_{in}-E_{out}\)

In a steady flow process;

\(\Delta E _{system} = 0\)

\(E_{in} = E_{out}\)

\(m(h_1 + \dfrac{V_1^2}{2})\) \(= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})\)

\(- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})\)

\(- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})\)

\(- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})\)

\(- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})\)

\(V_2^2 = 316631.29 \ m/s\)

\(V_2 = \sqrt{316631.29 \ m/s\)

\(\mathbf{V_2 = 562.7 \ m/s}\)

c) the exit area of the nozzle.

The exit of the nozzle can be determined by using the expression:

\(m = \dfrac{V_2A_2}{v_2}\)

making \(A_2\) the subject of the formula ; we have:

\(A_2 = \dfrac{ m \times v_2}{V_2}\)

\(A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}\)

\(A_2\) = 0.0015435 m²

Answer:intrusive test

Explanation:

(a) To determine the Schmid factor for all slip systems in a single crystal of iron (bcc) pulled in tension along the [120] direction, we need to consider the valid slip systems and their corresponding Schmid factors.

In bcc crystals, slip occurs on specific slip systems characterized by a combination of crystallographic planes and directions. The valid slip systems in iron (bcc) include {110}<111>, {112}<111>, and {123}<111>.

To calculate the Schmid factor for each slip system, we need to determine the dot product between the slip direction and the applied tensile stress direction, as well as the dot product between the slip plane normal and the tensile stress direction.

For example, for the {110}<111> slip system:

Slip direction: [110]

Slip plane normal: [111]

Tensile stress direction: [120]

Schmid factor = (Dot product of slip direction and tensile stress direction) * (Dot product of slip plane normal and tensile stress direction)

By calculating the dot products for each slip system and applying the formula, we can determine the Schmid factors.

(b) The tensile stress at which the crystal will flow plastically depends on the critical resolved shear stress (CRSS) for the slip system with the highest Schmid factor. The CRSS represents the stress required to initiate slip in a particular slip system.

Once we identify the slip system with the highest Schmid factor, the corresponding CRSS value can be obtained from experimental data or material properties. The tensile stress at which plastic flow will occur is equal to or greater than the CRSS for that slip system.

It's important to note that the exact values for Schmid factors, CRSS, and the tensile stress required for plastic flow can vary depending on the specific crystallographic orientation and material properties of the iron (bcc) single crystal.

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Answer:

Option B

Explanation:

Atoms of ferromagnetic materials have north and south pole but these atoms are oriented in random directions due to which they do no exhibit magnetic properties until unless they are brought into influence of any external temporary or permanent magnetic field.

Under the influence of external magnetic force, the atoms of the ferromagnetic material get oriented in a particular direction.

Hence, option B is correct

Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa

Explanation:

Given that;

patm = 79 kPa, h = 13 in of H₂O,

A sketch of the problem is uploaded along this answer.

Now

pA = patm + 13 in of H₂O ( h × density × g )

pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)

pA = 82.23596 kPa

the absolute static pressure in the gas cylinder is 82.23596 kPa