We intend to measure the open-loop gain (A open ) of an actual operational amplifier. The magnitude of A_open is in the range of 10^6 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the A_open?

Answer:

minimum number of stages ≈ 9 ( from Image number 1 )

Explanation:

Attached below is a detailed solution to the question

Total reflux = R Tending to infinity

we can determine the operation equation by stripping and rectifying section in total reflux condition

minimum number of stages ≈ 9 ( from Image number 1 )

Explanation:

Dry air doesn't contain water vapor .

Answer:

Programmers count the number of lines of code in an algorithm.

Programmers count the number of lines of code in an algorithm. Thus, shorter lines are faster than longer lines is an example of an algorithm.

An algorithm is a finite sequence of exact instructions that is used in mathematics and computer science to solve a class of particular problems or carry out a computation.

For performing calculations and processing data, algorithms are employed as specifications. Conditionals can be used by more sophisticated algorithms to divert code execution along several paths and draw reliable inferences, ultimately leading to automation.

Alan Turing was the first to use terminology like "memory," "search," and "stimulus" to describe human traits as metaphorical descriptions of machines.

A heuristic, on the other hand, is a method for addressing problems that may not be fully articulated or may not provide accurate or ideal solutions, particularly in problem domains where there isn't a clearly defined proper or ideal conclusion.

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Answer:

procedure attached below

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

Explanation:

Given data:

Minimum tensile strength = 865 MPa

Ductility = 10%EL

Desired Final diameter = 6.0 mm

20% cold worked 7.94 mm diameter 1040 steel stock

Describe the procedure you would follow to obtain this material.

assuming 1040 steel experiences cracking at 40%CW

attached below is a detailed procedure of obtaining the material

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

Answer:

N0

Explanation:

It does not seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm

Given data :

Diameter ( d ) = 10 mm

length ( l ) = 150 mm

elasticity ( ∈ ) = 100 GPa

longitudinal strain ( б ) 200 MPa

Poisson ratio ( μ )  ( assumed ) =0.3

Assumption : deformation totally elastic

attached below is the detailed solution to why it is not reasonable .

The Sd value = 0.08 > the calculated Sd value ( 6*10^-3 ) hence it is not reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen

Answer:

B. schematic design

Explanation:

This correct for Plato/edmentum

Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.

A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.

The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.

The schematic design consists of a rough sketch with markings and measurements.

Therefore, the correct option is B. schematic design.

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Answer:

required flow rate is 75.44 gal/min  

Explanation:

Given the data in the question;

Power developed = 250 psi = 1.724 × 10⁶ Pa

hydraulic power W = 11 hp = 11 × 746 = 8206 Watt

now, Applying the formula for pump power

W = pgQμ

where p is density of fluid, Q is flow rate, μ is heat and W is power developed;

W = pgQμ

W = pgμ × Q  

W = P × Q -------- let this be equ 1

so we substitute in our values;

8.2027 kW = 1.724 × 10⁶ Pa × Q

Q = 8206 / 1.724 × 10⁶

Q = 4.75986 × 10⁻³ m³/sec

We know that, 1 cubic meter per seconds = 15850.3 US liquid gallon per minute, so

Q = 4.75986 × 10⁻³  × 15850.3 gallon/min

Q = 75.44 gal/min    

Therefore, required flow rate is 75.44 gal/min    

Answer: hello some parts of your question is missing attached below is the missing part

max shear stress = 1875 psi

minimum yield strength = 14911.76 psi

Explanation:

Since The maximum shear stress in the Beam is already given as 1875 psi , I will calculate for the minimum yield strength

Determine minimum yield strength

attached below is the detailed solution

minimum yield strength = 14911.76 psi

Answer:

a) 35%

b) yes it can be improved by moving the tray near the top

   Tray should be located ( 1 to 2 meters below surface )

   max removal efficiency ≈ 70%

c) The maximum removal will drop as the particle settling velocity = 0.5 m/h

Explanation:

Given data:

flow rate = 8000 m^3/d

Detention time = 1h

depth = 3m

Full length movable horizontal tray :  1m below surface

a) Determine percent removal of particles having a settling velocity of 1m/h

velocity of critical sized particle to be removed = Depth / Detention time

= 3 / 1 = 3m/h

The percent removal of particles having a settling velocity of 1m/h ≈ 35%

b) Determine if  the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray  and the maximum removal efficiency

The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the

Total Maximum removal efficiency

=  percent removal[tex]_{above}[/tex] + percent removal[tex]_{below}[/tex]

= ( d[tex]_{a}[/tex],v[tex]_{p}[/tex] ) . [tex]\frac{d_{a} }{depth}[/tex]  + ( d[tex]_{a}[/tex],v[tex]_{p}[/tex] ) . [tex]\frac{depth - d_{a} }{depth}[/tex]  = 100

hence max removal efficiency ≈ 70%

c) what is the effect of moving the tray would be if the particle settling velocity were equal to 0.5m/h?

The maximum removal will drop as the particle settling velocity = 0.5 m/h

Answer:

Following are the responses to these question:

Explanation:

They might believe that it was an enhanced layout because the quality is not updated. For instance, its new XS Max iPhone does have a better display than the iPhone X, however, the performance wasn't enhanced. It also has the same processor or graphic cards however a bigger pixel every centimeter ratio. When its output AND is not altered, the layout doesn't change, basically the very same item.

Answer:

If an Improvement creates no significant change in a product’s performance, then it is a(n)  Superficial  design improvement.

Explanation:

DIFFERENT BREED!

Answer:

True

Explanation:

Those are the exact uses of a resistor

Answer:

attached below

Explanation:

Attached below is  a detailed solution to the question above

Step 1 : determine the Reynolds number using the characteristics of Air at 45°c

Step 2 : calculate the Nusselt's number

Step 3 : determine heat transfer coefficient

Step 4 : calculate heat transfer ratio and thermal resistance

Repeat steps 1 - 4 for each value of diameter from 0.05 to 0.5 m

attached below is a detailed solution

Answer:

the entrained fluid flowrate is 150 liters/s

Explanation:

Given the data in the question;

we determine the flow rate of water though the jet by using the following expression;

Q₂ = A₂ × V₂

where Q₂  is the flow rate of water though the jet, A₂ is the cross sectional area of the jet( 0.01 m² ) and V₂ is the jet velocity( 30 m/s )

so we substitute

Q₂ = 0.01 m² × 30 m/s

Q₂ = 0.3 m³/s

Next we determine the flow rate of water through the pump by using the following expression

Q₃ = A₃ × V₃

where Q₃  is the flow rate of water though the pump, A₃ is the cross sectional area of the pump( 0.075 m² ) and V₃ is the average velocity of mixing( 6 m/s )

so we substitute

Q₃ = 0.075 m² × 6 m/s

Q₃ = 0.45 m³/s

so to calculate the flow pumping rate of water into the water jet pump, we use the expression;

Q₁ + Q₂ = Q₃

we substitute

Q₁ + 0.3 m³/s = 0.45 m³/s

Q₁ = 0.45 m³/s - 0.3 m³/s

Q₁ = 0.15 m³/s

we know that 1 m³/s = 1000 Liter/second

so

Q₁ = 0.15 × 1000 Liter/seconds

Q₁ = 150 liters/s

Therefore, the entrained fluid flowrate is 150 liters/s

Answer:

i dont know th answer can u help ?

Explanation:

Answer:

a) 24.07 m

b)  4 m

c)  14 number of drops

d) p = number of passes

e)  Dcd = 2.27

0.69 m

Explanation:

Given data:

Depth  ( D )= 7.6 m below ground surface

dynamic compaction ( w )  = 15-ton , diameter of tamper = 2.0 m , thickness = 1.4 m

Determine :

A) drop height ( H )

  D = n √wH

 therefore H = 361 / 15 = 24.07 m

where : D = 7.6 m ,  n = 0.4 , w = 15

B) Drop spacing

drop spacing = average of ( 1.5 to 2.5 )  * diameter  of tamper

                       = 2 * 2.0m =  4 m

C) number of drops

since the applied energy for fine grained soils and day fills range from 250 - 350 kj/m^2  the number of drops can be calculated using the relation below

AE = [tex]\frac{NWHP}{SPACING ^2}[/tex]

w = 15, H = 24.07 , Np = ?  , AE = 300 kj/m^2

∴ Np = 4800 / 361.05  = 13.3

the number of drops at one pass =  14

D) number of passes

p = number of passes

E) estimated crater depth and settlement

crater depth ( Dcd ) = 0.028 [tex]N_{d} ^{0.55} \sqrt{wtIt}[/tex]

Nd = 14 ,  wt = 15, It = 24.07

therefore : Dcd = 2.27

estimate settlement is within 3 to 5% therefore the improved settlement

= 2.27 * 0.04 * 7.6 = 0.69 m

Answer:

The resulting pressure is 300 kilopascals.

Explanation:

Let consider that gas within the closed vessel behaves ideally. By the equation of state for ideal gases, we construct the following relationship for the isothermal relationship:

[tex]P_{1}\cdot V_{1} = P_{2} \cdot V_{2}[/tex] (1)

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressure, measured in kilopascals.

[tex]V_{1}[/tex], [tex]V_{2}[/tex] - Initial and final volume, measured in litres.

If we know that [tex]\frac{V_{1}}{V_{2}} = 2[/tex] and [tex]P_{1} = 150\,kPa[/tex], then the resulting pressure is:

[tex]P_{2} = P_{1}\times \frac{V_{1}}{V_{2}}[/tex]

[tex]P_{2} = 300\,kPa[/tex]

The resulting pressure is 300 kilopascals.

Answer:

the speed of your poop

Explanation:

Answer:

the carburizing time necessary to achieve a carbon concentration is 31.657 hours

Explanation:

Given the data in the question;

To determine the carburizing time necessary to achieve the given carbon concentration, we will be using the following equation:

(Cs - Cx) / (Cs - C0) = ERF( x / 2√Dt)

where Cs is Concentration of carbon at surface = 0.90

Cx is Concentration of carbon at distance x = 0.30 ; x in this case is 4 mm = ( 0.004 m )

C0 is Initial concentration of carbon = 0.10

ERF() = Error function at the given value

D = Diffusion of Carbon into steel

t = Time necessary to achieve given carbon concentration ,

so

(Cs - Cx) / (Cs - C0) = (0.9 - 0.3) / (0.9 - 0.1)

= 0.6 / 0.8

= 0.75

now, ERF(z) = 0.75; using ERF table, we can say;

Z ~ 0.81; which means ( x / 2√Dt) = 0.81

Now, Using the table of diffusion data

D = 5.35 × 10⁻¹¹ m²/sec at (1100°C) or 1373 K

now we calculate the carbonizing time by using the following equation;

z = (x/2√Dt)

t is carbonizing time

so we we substitute in our values

0.81 = ( 0.004 / 2 × √5.35 × 10⁻¹¹ × √t)

0.81 = 0.004 / 1.4628 × 10⁻⁵ × √t

0.81 × 1.4628 × 10⁻⁵ × √t = 0.004

1.184868 × 10⁻⁵ × √t = 0.004  

√t = 0.004 / 1.184868 × 10⁻⁵

√t = 337.5903

t = ( 337.5903)²  

t = 113967.21 seconds

we convert to hours

t = 113967.21 / 3600

t = 31.657 hours

Therefore, the carburizing time necessary to achieve a carbon concentration is 31.657 hours

Answer:

represnt

Explanation:

Answer:

a. The time required for the tank to empty halfway is presented as follows;

[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]

b. The time it takes for the tank to empty the remaining half is presented as follows;

[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]

The total time 't', is presented as follows;

[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

[tex]\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}[/tex]

[tex]\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}[/tex]

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

[tex]dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh[/tex]

The time for the tank to drop halfway is given as follows;

[tex]\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]

[tex]t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]

[tex]t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]

[tex]t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]The time required for the tank to empty halfway, t₁, is given as follows;

[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

[tex]\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]

[tex]t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)[/tex]

[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]

The total time, t, to empty the tank is given as follows;

[tex]t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}[/tex]

[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]

Answer:

B. The Hall-Petch Relation

Explanation:

The Hall-Petch relation indicates that by reducing the grain size the strength of a material is increased up to the theoretical strength of the material however when the material grain size is reduced below 20 nm the material is more susceptible to creep deformation and displays an "inverse" Hall-Petch Relation as the Hall-Petch relation then has a negative slope (k value)

The Hall-Petch relation can be presented as follows;

[tex]\sigma_y[/tex] = [tex]\sigma_0[/tex] + k·(1/√d)

Where;

[tex]\sigma_y[/tex] = The strength

σ₀ = The friction stress

d = The grain size

k = The strengthening coefficient

The model equation for the reverse Hall-Petch effect is presented here as follows;

[tex]\sigma_y[/tex] = 10.253 - 10.111·(1/√d)

Answer:

a) 570 kWh of electricity will be saved

b) the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

[tex]P_{incandescent}[/tex] = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find  the total power consumption by the fluorescent lamp

[tex]P_{fluorescent}[/tex] = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

[tex]P_{saved}[/tex] = [tex]P_{incandescent}[/tex]  - [tex]P_{fluorescent}[/tex]

[tex]P_{saved}[/tex] = 750 - 180

[tex]P_{saved}[/tex]  = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved =  570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved =  570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ /  million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c)  If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

Answer:

is a mathematical representation of something three-dimensional.

Explanation:The typical base of a the model is a 3D mesh; the structural build consists of polygons.