For an open system, the is used to consider and analyse all energy changes between states." Steady flow energy equation Continuity equation Momentum equation entropy equation A Moving to another question will save this response.

Answer:

It's equalization of the system

Explanation: If there was a leak it would likely leak all the time even if the car was turned off. Plus, a system leak bad enough to hear would drain the system of refrigerant very quickly and would no longer cool.

Answer: i got you its d

Explanation:had the smae question as you

The Vogel theory is an important tool used in the field of production engineering, especially in petroleum and natural gas engineering.

This theory is named after Dr. Harold F. Vogel, who developed it in the 1950s to optimize the production of crude oil and natural gas from a reservoir. The Vogel theory is based on the concept of maximizing the net present value of the project by optimizing the production rate. It takes into account the production costs, the prices of crude oil and natural gas, and the decline in the production rate over time.

To apply the Vogel theory, one needs to estimate the production costs, the prices of crude oil and natural gas, and the decline in the production rate. The production costs include the costs of drilling, completing, and operating the wells, as well as the costs of transporting and processing the crude oil and natural gas. The optimal production rate is the production rate that maximizes the net present value of the project.

In conclusion, the Vogel theory is an important tool used in production engineering, especially in petroleum and natural gas engineering. This theory helps to optimize the production of crude oil and natural gas from a reservoir by finding the optimal production rate that maximizes the net present value of the project.

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The term used to describe the process of removing refrigerant from a system and filtering and/or cleaning it to meet AHRI 740 standards is reclaiming.

The term used to describe the process of removing refrigerant from a system and filtering and/or cleaning it to meet AHRI 740 standards is "reclaiming." This process involves the removal of refrigerant from a system and the purification of the refrigerant through filtering and/or cleaning to meet the AHRI 740 standards for purity and quality. Once the refrigerant has been reclaimed, it can be reused or sold to other users who require refrigerant for their systems.

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Integrated circuits have revolutionized modern electronics by combining multiple electronic components into a single chip, enhancing performance, compactness, and efficiency.

Integrated circuits, also known as microchips or ICs, have played a pivotal role in revolutionizing modern electronics. These tiny wonders have fundamentally transformed the way electronic devices are designed and manufactured, enabling unprecedented advancements in various industries.

By integrating multiple electronic components onto a single chip, integrated circuits have brought about remarkable improvements in performance, compactness, and efficiency.

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Using the Newton's second law of motion we will see that the acceleration is  0.0875 m/s^2.

The acceleration of a vehicle is determined by the force acting on it and its mass, as described by Newton's second law of motion:

F = m * a

where F is the force acting on the object, m is the mass of the object, and a is the resulting acceleration.

In this case, the force acting on the vehicle is 0.07 N, and the mass of the vehicle is 0.8 kg. Substituting these values into the equation above, we get:

0.07 N = 0.8 kg * a

Solving for a, we get:

a = 0.07 N / 0.8 kg

a = 0.0875 m/s^2

Therefore, the acceleration of the 0.8 kg vehicle powered by 0.07 N of force is 0.0875 m/s^2.

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The average daily flow of water in 10 years would be 21,249 m³/day.

To estimate the average daily flow of water in 10 years,

we have to use the current population and water consumption as a baseline.

We have to find the daily water consumption for the current population,

⇒ 6,083,270 m³ / 365 days = 16,676 m³/day

Now we have to calculate the expected water consumption for the future population,

= 16,676 m³/day x (73,728 / 58,000) = 21,249 m³/day

Therefore,

We can estimate that the average daily flow of water in 10 years would be 21,249 m³/day.

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The question is incomplete. Here is the complete question.

Two kg of a two phase liquid vapor mixture of carbon dioxide (CO₂) exists at -40°C in a 0.05m³ tank. Determine the quality of the mixture, if the values of specific volume for saturated liquid and saturated vapor CO₂ at -40°C are \(v_{f}\) = 0.896 x 10⁻³m³/kg and \(v_{g}=\) 3.824 x 10⁻²m³/kg, respectively.

Answer: x = 1

Explanation: In a phase change of a pure substance, at determined pressure and temperature, the substance exists in two different phases: saturated liquid and saturated vapor.

Quality (x) is the ratio of saturated vapor in the mixture and can be written as:

\(x=\frac{m_{vapor}}{m_{liquid}+m_{vapor}}\)

It has value between 0 and 1: x = 0 for saturated liquid and x = 1 for saturated vapor.

When related with volumes, quality is rearranged as:

\(x=\frac{v-v_{f}}{v_{g}-v_{f}}\)

Solving for x:

\(x=\frac{0.05-0.896.10^{-3}}{3.824.10^{-2}-0.896.10^{-3}}\)

\(x=\frac{0.049104}{0.037344}\)

x = 1.3

Quality of mixture of carbon dioxide is x = 1, which means it's for saturated vapor.

This question is incomplete, the complete question is;

Find the magnitude of the steady-state response of the system whose system model is given by

dx(t)/dt + x(t) = f(t)

where f(t) = 2cos8t.  Keep 3 significant figures

Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )

Explanation:

Given that;

dx(t)/dt + x(t) = f(t)  where f(t) = 2cos8t

dx(t)/dt + x(t) = f(t)

we apply Laplace transformation on both sides

SX(s) + x(s) = f(s)

(S + 1)x(s) = f(s)

f(s) / x(s) = S + 1

x(s) / f(s) = 1 / (S + 1)

Therefore

transfer function = H(s) = x(s)/f(s) = 1/(S+1)

f(t) = 2cos8t →   [ 1 / ( S + 1 ) ]   →  x(t) = Acos(8t - ∅ )

A = Magnitude of steady state output

S = jw

S = j8

so

A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )

A = 2/√65 = 0.2481

∅ = tan⁻¹( 1/1) = 45°

therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )

Answer:

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

Explanation:

We must estimate the monthly storage change of the lake by considering inflows, outflows, seepage and evaporation losses and precipitation. That is:

\(\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}\)

Where \(\Delta V_{storage}\) is the monthly storage change of the lake, measured in cubic feet.

Monthly inflow

\(V_{inflow} = \left(30\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{inflow} = 77.76\times 10^{6}\,ft^{3}\)

Monthly outflow

\(V_{outflow} = \left(27\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{outflow} = 66.98\times 10^{6}\,ft^{3}\)

Seepage losses

\(V_{seepage} = s_{seepage}\cdot A_{lake}\)

Where:

\(s_{seepage}\) - Seepage length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{seepage} = 1.5\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{seepage} = (1.5\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{seepage} = 2.86\times 10^{6}\,ft^{3}\)

Evaporation losses

\(V_{evaporation} = s_{evaporation}\cdot A_{lake}\)

Where:

\(s_{evaporation}\) - Evaporation length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{evaporation} = 6\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{evaporation} = 11.44\times 10^{6}\,ft^{3}\)

Precipitation

\(V_{precipitation} = s_{precipitation}\cdot A_{lake}\)

Where:

\(s_{precipitation}\) - Precipitation length gain, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{precipitation} = 4.25\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{precipitation} = (4.25\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{precipitation} = 8.10\times 10^{6}\,ft^{3}\)

Finally, we estimate the storage change of the lake during the month:

\(\Delta V_{storage} = 77.76\times 10^{6}\,ft^{3}-66.98\times 10^{6}\,ft^{3}-2.86\times 10^{6}\,ft^{3}-11.44\times 10^{6}\,ft^{3}+8.10\times 10^{6}\,ft^{3}\)

\(\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}\)

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

The volume of water gained and the loss of water through flow,

seepage, precipitation and evaporation gives the storage change.

Response:

Given parameters:

Area of the lake = 525 acres

Inflow = 30 ft.³/s

Outflow = 27 ft.³/s

Seepage loss = 1.5 in. = 0.125 ft.

Total precipitation = 4.25 inches

Evaporator loss = 6 inches

Number of seconds in a month is found as follows;

\(30 \ days/month \times \dfrac{24 \ hours }{day} \times \dfrac{60 \, minutes}{Hour} \times \dfrac{60 \, seconds}{Minute} = 2592000 \, seconds\)

Number of seconds in a month = 2592000 s.

Volume change due to flow, \(V_{fl}\) = (30 ft.³/s - 27 ft.³/s) × 2592000 s = 7776000 ft.³

1 acre = 43560 ft.²

Therefore;

525 acres = 525 × 43560 ft.² =  2.2869 × 10⁷ ft.²

Volume of water in seepage loss, \(V_s\) = 0.125 ft. × 2.2869 × 10⁷ ft.² = 2,858,625 ft.³

Volume gained due to precipitation, \(V_p\) = 0.354167 ft. × 2.2869 × 10⁷ ft.² = 8,099,445.123 ft.³

Volume evaporation loss, \(V_e\) = 0.5 ft. × 2.2869 × 10⁷ ft.² = 11,434,500 ft.³

Which gives;

ΔV = 7776000 - 2858625 + 8099445.123 - 11434500 = 1582823.123

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Note that the positive half cycle, the peak current is:  16.27mA

During the negative half cycle, the peak current is 12mA. The above is computed on the assumption that amplitude is 16.97mA

The peak current is the largest amount of current that an output may provide for short periods of time. When a power source or an electrical device is turned on for the first time, a large amount of current flows into the load, beginning at zero and increasing until it reaches a maximum value known as the peak current.

The formula for Load resistor is used to compute the peak current.

I = V/RL

⇒ (|16.97| -0.7) / 1kΩ

= 16.27mA.

When the half cycle is negative:
|  = |-12| / 1kΩ
I = 12mA.

Both positive and negative are computed on the assumption that amplitude is 16.97mA

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Answer:

Creating an app is both an expression of our self and a reflection of what we see is missing in the world. We find ourselves digging deep into who we are, what we would enjoy working on, and what needs still need to be fulfilled. Generating an app idea for the first time can be extremely daunting. Especially with an endless amount of possibilities such as building a church app.

The uncertainty has always spawned a certain fear inside creators. The fear of creating something no one will enjoy. Spending hundreds of dollars and hours building something which might not bring back any real tangible results. The fear of losing our investment to a poor concept is daunting but not random. But simple app ideas are actually pretty easy to come by.

Great app idea generation is not a gift given to a selected few, instead, it is a process by which any of us are able to carefully explore step by step methods to find our own solution to any problem. Whether you are a seasoned creator or a novice, we have provided a few recommendations to challenge and aid you as you create your next masterpiece.

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Answer:

116,090 Btus

Explanation:

A. moisture content around the anodes. Routine monitoring of a cathodic protection system usually does NOT include moisture content around the anodes.

Routine monitoring of a cathodic protection system typically includes measuring the structure-to-electrolyte potentials, monitoring the rectifier voltage and current output, and ensuring interference control bond current. However, measuring the moisture content around the anodes is not typically part of the routine monitoring process. This is because the anodes are designed to operate in a moist environment, and their effectiveness is based on their ability to corrode in the electrolyte. Instead, the focus is on monitoring the performance of the system in terms of its ability to protect the structure from corrosion, which is achieved through the other monitoring methods mentioned above.

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Answer:

Alternating

Explanation:

To condition ECG raw signals, a circuit consisting of an amplifier and a bandpass filter can be used. The circuit diagram for this can be drawn as follows:

Input (RA- right arm, LA- left arm) --> Pin 2 of Instrumentation Amplifier (INA) --> Pin 6 of INA --> 0.1 μF capacitor --> Pin 5 of INA --> 100 kΩ resistor --> Pin 4 of INA --> 0.1 μF capacitor --> Pin 3 of INA --> Ground

The INA is a differential amplifier that amplifies the ECG signals and removes common mode noise. The output of the INA is then passed through a bandpass filter to remove unwanted frequencies and enhance the desired frequencies. The bandpass filter can be designed using passive components such as resistors and capacitors.

The output of the circuit can be taken from the output pin of the bandpass filter. The pin numbers of the amplifier and output of the circuit are marked in the diagram as Pin 2 and Pin 5 of INA and the output pin of the bandpass filter, respectively.

This circuit can be useful for conditioning ECG signals for further analysis and diagnosis of heart conditions.

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Answer: Hello your question is incomplete attached below is the complete question

answer : V(out) (t) = 1 - e^-100t

Explanation:

The equation of the output voltage as a function of time assuming at t = 0 switch closes and capacitor will be discharged when t < 0

V(out) (t) = 1 - e^-100t

attached below is the step by step explanation  

According to  fatigue cracking in flexible pavements occurs due to repeated loading from tire-pavement interaction and aging of asphalt binder. Excessive moisture content in pavements after a rainfall is not mentioned as a cause for fatigue cracking.

It mentions that higher binder content in asphalt concrete mixture and excessive moisture content can both contribute to fatigue cracking, while heavy traffic volume, bad construction quality control (QC), and dramatic temperature changes are also factors. Based on this information, the least likely cause of fatigue cracking in flexible pavements is "maximum allowable asphalt content with low viscosity asphalt in asphalt layers".

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The region of convergence (ROC) for a stable system with the transfer function H(s) = -2(0.2)(s-2)/(s-1) is the entire complex plane except for the pole at s = 1.

To determine the ROC, we analyze the poles of the transfer function. The system is stable if all the poles are located in the left-half of the complex plane (i.e., have negative real parts) or if there are no poles at all.

In this case, the transfer function has a pole at s = 1. Since the pole is a single point and it has a positive real part, it lies in the right-half plane. Therefore, the ROC excludes the pole at s = 1.

Hence, the region of convergence for this stable system is the entire complex plane excluding the point s = 1.

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Answer:

1/2 semesters before completion

Explanation:

Program completion and review are both a necessary part and parcel of a successful completion of any program. The recommended period for applying for program completion and review is one to two semesters before the program is to be completed. This is mandatory, so that there is different time span for review and any other issues that are needed to be carried out. And then afterwards, finally the certificates are then issued to the qualified person.

When performing a complete governor system adjustment after servicing the carburetor, the technician should adjust the normal primary governor spring first before adjusting the governed idle spring. This will ensure that the engine operates at the correct maximum speed before adjusting the idle speed.

The normal primary governor spring is responsible for controlling the engine speed under normal operating conditions, while the governed idle spring is responsible for controlling the engine speed when the throttle is in the idle position. Adjusting the normal primary governor spring first will ensure that the engine is operating at the correct speed under normal operating conditions. Once the primary governor spring is adjusted, the governed idle spring can then be adjusted to ensure that the engine is operating at the correct speed when the throttle is in the idle position.It is important to follow the manufacturer's instructions and specifications when adjusting the governor system on an engine to ensure proper operation and prevent damage to the engine.

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Answer:

ANSI A sized paper is commonly referred to as Letter and ANSI B as Ledger or Tabloid.

Explanation:

The load choices that are available in the question are:

a. The 6.8 Zener diode

what is required would be 3 zener diodes. These would be able to provide 3 x 6.8 = 20. 4 output

The resistance would be 3 x 10 ohms = 30 Ohms.

The requirement here would be 4 zener diodes. This would have to provide 4 x 5.1 = 20.4 volts of output.

The resistance that it would have to give would be 4 x 25 = 100 ohms of resistance.

Therefore from the calculations that we have above we would have to conclude that the load regulation that is here without R would be

30 ohms for the case of a , while it would have to be 100 ohms for the case of b.

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The mass balance equation for the sewage lagoon can be expressed as:

Qin x Cin = Qout x Cout + V x R

Where:

Qin = influent flow rate = 8,640 m^3/d

Cin = influent concentration of biodegradable contaminant = 100 mg/L

Qout = effluent flow rate = Qin (since there are no losses or gains of water)

Cout = effluent concentration of biodegradable contaminant = 20 mg/L (maximum allowable concentration)

V = volume of lagoon = 10 ha x 10,000 m^2/ha x 1 m = 100,000 m^3

R = biodegradation reaction rate (kg/m^3/d) x density of water = k x Cin

Assuming a first-order reaction, the biodegradation rate can be expressed as:

R = k x Cin

Substituting the values in the mass balance equation and rearranging:

k = (Qin x (Cin - Cout)) / (V x Cin)

k = (8,640 m^3/d x (100 mg/L - 20 mg/L)) / (100,000 m^3 x 100 mg/L)

k = 0.0691 d^-1

Therefore, the biodegradation reaction rate coefficient that must be achieved for a first-order reaction is 0.0691 d^-1.

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The values of sum1 and sum2 if the operands in the expressions are evaluated left to right are; sum1 = 46 and sum2 = 48

In computer programming, the term operand is defined as the object of a mathematical operation or it is the object or quantity that is operated on.

A) If the operands in the expressions are evaluated left to right, then the values of sum1 and sum2 are;

sum1 = (10/2) + 41 = 46

sum2 = 41 + (14/2) = 48

B) If the operands in the expressions are evaluated right to left, then the values of sum1 and sum2 are;

sum1 = (14/2) + 41 = 48

sum2 = 41 + (10/2) = 46

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Answer: Maybe A

Explanation:

A is the correct answer

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