A 750-m section of river runs parallel to a channel 1,000 m away. an aquifer connecting the two has hydraulic conductivity equal to 7. 0 m/day and an average thickness of 10 m. the surface of the river is 5. 0 m higher than the surface of the channel. estimate the rate of seepage (qseepage) from the river to the channel

I am so confused, can you explain it?

Answer:

The equilibrium temperature is 109.7°C

The total entropy change for the process is 0.23 kJ/K

Explanation:

The parameters given are;

The initial temperature of aluminium, T₁ₐ = 140°C = 413.15 K

Mass, m₁, of aluminium block = 28 kg

The specific heat of aluminium, c₁, at 400 K = 0.949 kJ/(kg·K)

The initial temperature of iron, T₁ₙ = 60°C = 333.15 K

Mass, m₂, of iron block = 36 kg

The specific heat of iron, c₂, at room temperature = 0.45 kJ/(kg·K)

The heat lost by the aluminium = Heat gained by the iron

The heat lost by the aluminium = m₁ × c₁ × (T₁ₐ - Tₓ)

The heat lost by the aluminium = 28 × 0.949 × (413.15 - Tₓ)

The heat gained by the iron =  m₂ × c₂ × (Tₓ - T₁ₙ)

The heat gained by the iron =  36 × 0.45 × (Tₓ - 333.15)

Hence;

28 × 0.949 × (413.15 - Tₓ) = 36 × 0.45 × (Tₓ - 333.15)

26.572 × (413.15 - Tₓ) = 16.2 × (Tₓ - 333.15)

∴ (26.572 + 16.2) × Tₓ = 26.572 × 413.15 + 16.2 × 333.15

42.772·Tₓ = 16375.2518

Tₓ = 382.85 K = 109.7°C

The total entropy change of the process is given by the following relation;

\(\Delta S = \Delta S_{Fe} + \Delta S_{Al}\)

\(=m_{1}c_{1}ln\dfrac{T_{x}}{T_{1n}}+m_{2}c_{2}ln\dfrac{T_{x}}{T_{1a}}\)

\(=28 \times 0.949 \times ln\dfrac{382.85}{413.15}+36 \times 0.45 \times ln\dfrac{382.85}{333.15} = 0.23 \, \dfrac{kJ}{K}\)

Assuming a data analyst prepares to communicate to an audience about an analysis project and considers what the audience members hope to do with the data insights, this describes establishing the setting: False.

In Computer technology, data is any representation of factual instructions or information in a formalized and structured manner, especially as a series of binary digits (bits), symbols, characters, quantities, or strings that are used on computer systems in a company.

A data analyst can be defined as an expert or professional who is saddled with the responsibility of inspecting, transforming, analyzing, and modelling data with the sole aim of discovering useful information, providing insights, and creating informed conclusions, in order to support decision-making and create a completed report.

In Computer technology, audience engagement in data storytelling simply refers to a process that is typically focused on considering all of the issues that audience members hope to do and achieve with the data insights provided by a data analyst.

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Answer:

THE thing is 435

Explanation:

Some of the trade-offs that must be made while transitioning from an individually tailored solution to such an enterprise-level solution are about as continues to follow.

As a result, the aforementioned response seems to be correct.

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Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation (\(\dot Q\)), measured in BTU per hour, is represented by this formula:

\(\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})\) (1)

Where:

\(\epsilon\) - Emissivity, dimensionless.

\(A\) - Surface area of the student, measured in square feet.

\(\sigma\) - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

\(T_{s}\) - Temperature of the student, measured in Rankine.

\(T_{b}\) - Temperature of the bus, measured in Rankine.

If we know that \(\epsilon = 0.90\), \(A = 16.188\,ft^{2}\), \(\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}\), \(T_{s} = 554.07\,R\) and \(T_{b} = 527.67\,R\), then the heat transfer rate due to electromagnetic radiation is:

\(\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]\)

\(\dot Q = 417.492\,\frac{BTU}{h}\)

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

\(Q = \dot Q \cdot \Delta t\) (2)

Where \(\Delta t\) is the heat transfer time, measured in hours.

If we know that \(\dot Q = 417.492\,\frac{BTU}{h}\) and \(\Delta t = \frac{1}{3}\,h\), then the net energy transfer is:

\(Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)\)

\(Q = 139.164\,BTU\)

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Answer:

Motorcyclist Fatality and Injury Rates per Vehicle Miles Traveled, 1998-2008. 10. 9. Fatalities in School Transportation Related Crashes,. 1998-2008.

Explanation:

The answer to this question is no. This is because all major PC operating systems do not feature a graphical user interface (GUI).

An operating system may be defined as a program that significantly after being initially loaded into the computer by a boot program, manages all of the other application programs in a computer.

The application programs construct the utilization of the operating system by making requests for services through a defined application program interface (API).

The three most common operating systems for personal computers are Microsoft Windows, macOS, and Linux. All modern operating systems use a graphical user interface or GUI.

Therefore, the answer to this question is no. This is because all major PC operating systems do not feature a graphical user interface (GUI).

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Since we cannot have a fraction of a circuit, we round up to the nearest whole number. The minimum number of 20-amp, 277-volt lighting circuits required for the department store is 73.



where 1.73 is the square root of 3, which represents the three-phase power factor. Substituting the given values, we get:
Amps = 400,000 / (1.73 x 277) = 815.2 amps
Maximum load per circuit = 20 amps x 0.8 = 16 amps
Number of circuits = 815.2 / 16 = 50.95 or approximately 51 circuits
Extra circuits = 51 x 0.2 = 10.2 or approximately 10 circuits
Total circuits = 51 + 10 = 61 circuits
In long answer, the minimum number of 20-amp, 277-volt, lighting circuits required for a 150,000 ft2 department store with an actual connected lighting load of 400 kVA and assuming breakers are not rated for continuous use would be 61 circuits. This is calculated based on the total connected lighting load and the maximum load per circuit, while taking into account the safety factor of not exceeding 80% of the circuit capacity and adding extra circuits to avoid
First, let's convert the lighting load from kVA to VA:
400 kVA = 400,000 VA
Number of circuits = (400,000 VA) / (20 amps * 277 volts)
Number of circuits = 400,000 / 5,540
Number of circuits ≈ 72.2

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Inspecting the transmission differential is an important part of maintaining a vehicle's overall health and ensuring proper performance. The differential is responsible for transferring power from the transmission to the wheels, allowing them to turn at different speeds while maintaining traction on the road. Over time, the differential can become worn or damaged, leading to issues like poor acceleration, reduced fuel efficiency, and even damage to other parts of the vehicle.

By inspecting the transmission differential, mechanics can identify any signs of wear or damage before they become more serious problems. This might involve checking the fluid level and quality, inspecting the gears and bearings for signs of wear or damage, and ensuring that everything is properly lubricated and functioning as it should. In some cases, repairs or replacements may be necessary to keep the differential functioning properly.

Regular inspections and maintenance of the transmission differential can also help prevent more serious problems from developing in the future. This can save vehicle owners both time and money, as catching issues early on can prevent more extensive repairs or replacements down the line. Overall, inspecting the transmission differential is an essential part of maintaining a vehicle's performance and ensuring its long-term health and longevity.

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Answer:

How do I calculate voltage drop?

To calculate voltage drop, E, across a component, you need to know the resistance of the component and the current thru it. Ohm's Law is E=I⋅R , which tells us to then multiply I by R . E is the voltage across the component also known as voltage drop

Explanation:

The overall gain of the signal differential from the voltage gain is 1.71V

Voltage gain is the ratio of the output voltage of a circuit to the input voltage. It is usually expressed as a ratio or a percentage. The voltage gain of a circuit is a measure of its amplification capability.

The small signal differential mode voltage gain is calculated as follows:

Gain = (Vout3 - Vout2) / (Vin+ - Vin-)

Gain = (Vout3 - Vout2) / (VCC - VEE)

Gain = (Vout3 - Vout2) / (10V- (-10V))

Gain = (Vout3 - Vout2) / 20V

Gain = (Vout3 - Vout2) / 20

Vout3 = (Rc2 / (Rc2 + R6)) * VCC

Vout3 = (20kΩ / (20kΩ + 16.5kΩ)) * 10V

Vout3 = 0.546 * 10V

Vout3 = 5.46V

Vout2 = (Rc1 / (Rc1 + R7)) * VEE

Vout2 = (20kΩ / (20kΩ + 5kΩ)) * (-10V)

Vout2 = (-3.75V)

Gain = (5.46 - 3.75) = 1.71V

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Electrical inspector

Answer:

D

Explanation:

All of the above

The concentration ratio for the PTC is approximately 1.48, and the length of the parabolic surface is approximately 5.2 meters.

To determine the concentration ratio and length of the parabolic surface for a Parabolic Trough Collector (PTC) with the given parameters, we can use the following formulas:

Concentration Ratio (CR) = Rim Angle / Aperture Angle

Length of Parabolic Surface (L) = Aperture^{2} / (16 * Focal Length)

First, let's calculate the concentration ratio:

Given:

Rim Angle (θ) = 80º

Aperture Angle (α) = 5.2 m

Concentration Ratio (CR) = 80º / 5.2 m

Converting the rim angle from degrees to radians:

θ_rad = 80º * (π / 180º)

CR = θ_rad / α

Next, let's calculate the length of the parabolic surface:

Given:

Aperture (A) = 5.2 m

Receiver Diameter (D) = 50 mm = 0.05 m

Focal Length (F) = A^{2} / (16 * D)

L = A^{2} / (16 * F)

Now we can substitute the given values into the formulas:

CR =\((80º * (π / 180º)) / 5.2 m\)

L = \((5.2 m)^2 / (16 * (5.2 m)^2 / (16 * 0.05 m))\)

Simplifying the equations:

CR ≈ 1.48

L ≈ 5.2 m

Therefore, the concentration ratio for the PTC is approximately 1.48, and the length of the parabolic surface is approximately 5.2 meters.

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Answer:

Step One: Know What Your Rocket Needs to Do. ...

Step Two: Establish Mission Parameters. ...

Step Three: Call in Experts. ...

Step Four: Start Drawing. ...

Step Five: Whittle Down the Possibilities. ...

Step Six: Pick the Best Design.

Explanation:

Hope this helps and thx for tha fwee pts

The minimum beta is given as 90 for which the tolerance of the transistor can be used.

We have the following values to solve this question with

gm of amplifier = 30mA/V

input resistance = 3000 ohm

We have transconductance as

\(\frac{I_{c} }{V_{T} } = 30mA/V\)

We have to consider the thermal voltage at Vt = 25mV when i tis occuring at the room temperature.

The collector bias would be solved by

Ic = 0.03Vt

= 0.03 x 0.025

= 0.75mA

Then we have to solve for the minimum value of the dc current as

β = gmγm

= 0.03 * 3000

= 90

Hence the minimum beta is given as 90

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Tech A exists right. Transmission mode refers to the process of moving data from one device to another.

Actually, the act of transmitting involves sending and spreading analog or digital information impulses. Transmission technology primarily refers to responsibilities associated with physical layer protocols, such as line coding, demodulation, and modulation.

The movement of data from one digital device to another is known as data transmission. Through point-to-point data streams or channels, this transmission is carried out.

Transmission mode refers to the process of moving data from one device to another. The communication mode is another name for the transmission mode. Each communication channel has a direction that goes with it, and the direction is provided by the transmission medium.

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Answer: hello your question is incomplete below is the complete question

An ideal Otto Cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression process, air is at 98kPa and 20C. The pressure is doubled during the constant volume heat addition process. Assuming constant specific heats, determine:the amount of heat transferred to the air

answer : 609.804 kj/kg

Explanation:

Given data:

compression ratio (r)= 9.2

pressure given(p1) = 98 kPa

Initial temperature =  20 + 273 = 293 k

pressure during constant volume heat addition process = 2p1

note : specific heat at constant pressure and specific heat at  constant volume varies with temperature

we use T = 300k because it is closest to T1 = 293 k

hence at T = 300 K ( ideal gas properties of air )

             \(u_{1}\)   = 214.07 Kj/kg

            \(v _{r1}\) = 621.2

To get  \(v_{r2}\) = \(v_{r1} * \frac{v_{2} }{r}\) = 621.2 * 1 / 9.2 = 67.52

ALSO    at \(v_{r2}\) = 67.52 ( from ideal gas properties )

                \(u_{2}\) = 518.9 kj/kg

                T2 = 708.32 k

next we apply the gas equation

\(\frac{p1v1}{T1} = \frac{p2v2}{T2}\)

hence  p2 = (9.2) * \(\frac{708.32}{293} * 98\) = 2179.59 kpa

to determine T3  due to the constant volume heat addition

\(\frac{T3}{T2} = \frac{P3}{P2}\)

Hence T3 = p3/p2 * T2 = 2( 708.32 ) = 1416.64 k

At T3 = 1416.64 k ( from ideal gas properties )

\(u_{3}\) = 1128.704 kj/kg

\(v_{r3}\) = 8.592

Determine the amount of heat transferred to the air

\(q_{in} = ( u_{3} - u_{2} )\)

     = ( 1128.704  - 518.9 )

     = 609.804 kj/kg

Answer:

Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's activities.

hope it helps pa brainliest pa follow nadin tysm

Answer:

Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's activities

thx for following me

Answer: c. Centre of pressure​

Explanation:

Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.

The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.

Based on the information provided, we can logically deduce that technician A was right. Therefore, the correct answer is: a. A only.

In order for a technician to diagnose a brake system, the following symptoms should be adequately and appropriately checked and used by the technician:

This ultimately implies that, a symptom you should use when diagnosing brake system troubles is improper braking action.

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For joining the attachments on the steel panel both the options such as welding and gluing are prevalent and highly useful.

Welding is considered an excellent procedure to join the thin sheet attachment of a steel panel. The type of welding used for joining the attachments on steel surfaces is typically called tack welding. Tack welding performs very small welds that do not cause any damage to the surface of the steel panel. Hence, technician A is correct who states that on a steel panel, welding may be used on attachments.

Similarly, attachments on the steel materials have the option to be joined with glues. Cyanoacrylate-based adhesives and polyurethane-based glues are considered best to provide the strong joining of attachments as welding. Therefore technician B is also correct who says that glue may be used on steel panels to join attachments. It is because on a steel surface glue cannnot be useable as it is not able to join attachments.

Since both welds and glues can work on steel material in order to join attachments, both technicians are correct.

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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:

Gens or generators are typically used as a backup power source for buildings or facilities in case of a power outage. When it comes to starting a GPU (graphics processing unit) or a computer's graphics card, it is important to turn off the gens to ensure that the power being supplied to the GPU is consistent and stable.

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The cabin pressure of a pressurized aircraft is typically controlled by c) a pressure sensitive valve that regulates the output pressure of the pressurization pump. This valve is responsible for maintaining the desired cabin pressure during flight.

Option C, a pressure sensitive valve that controls the output pressure of the pressurization pump, is the correct answer. The pressurization system in an aircraft consists of a pressurization pump that supplies compressed air to the cabin. The pressure sensitive valve acts as a regulator, continuously monitoring the cabin pressure and adjusting the output pressure of the pump accordingly.

The valve works in conjunction with cabin altitude sensors that measure the cabin pressure and compare it to a pre-set target pressure. If the cabin pressure is below the desired level, the valve will open, allowing more compressed air from the pressurization pump into the cabin. If the pressure exceeds the target level, the valve will close or reduce the airflow, maintaining the desired cabin pressure.

This control mechanism ensures that the cabin pressure remains at a comfortable and safe level for passengers and crew throughout the flight, despite changes in altitude. It helps prevent the negative effects of high altitude, such as hypoxia or discomfort, by providing a controlled and pressurized environment inside the aircraft cabin.

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Answer:A

Explanation:

Those who want to save money and will use the product for only a few years

Answer:

THE ANSWER is A - those who want to save money and will use the product for only a few years

Explanation: Got it right on edg 2021

Answer: D none of these

Explanation:

A man-made incision, trench, hole, or depression in the Earth's surface is referred to as an excavation. A trench is described as a lengthwise thin excavation dug below the surface of the land. Thus, option D is correct.

The sidewalls of the excavation can be benched or slopped, or trench shoring or shielding devices can be used. Observe the rules for benching, slope, and the use of shielding and shoring equipment.

Therefore, In contrast to a wider gully, a trench is a sort of excavation in or in the earth that is often deeper than it is wide. Instead of a straightforward hole or pit, the object was a thin ditch.A trench is defined as a shallow excavation that is drilled along its length.

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Answer:

The answers are in the explanation. The pH is 5.91

Explanation:

The CH3NH2 reacts with HCl as follows:

CH3NH2 + HCl → CH3NH3⁺ + Cl⁻

When 200mL of HCl are added, the moles of CH3NH2 and HCl are reacting completely producing CH3NH3+ and Cl-. That means the species present are:

no H+. All reacted

yes H2O. Because the water is present in the solutions of HCl and CH3NH2

yes Cl-. Is a product of the reaction

Yes CH3NH2. Is consumed in the reaction but comes from the equilibrium of CH3NH3+

yes CH3NH3+. Is the other product of the reaction. MAJOR SPECIES

When 300.00mL of HCl are added, 100mL are in excess:

yes H+. Is in excess: H+ + Cl- = HCl in water. MAJOR SPECIES. Determine the pH of the solution.

yes H2O. Is present because the reactants are diluted.  

yes Cl-. Is a product of reaction and comes from HCl.

Yes CH3NH2. The reactant is over but comes from the equilibrium of CH3NH3+

yes no CH3NH3+. Yes. Is a product and remains despite HCl is in excess.

To find the pH:

At equivalence point the ion that determines pH is CH3NH3+. Its concentration is:

0.100L * (0.200mol/L) = 0.0200 moles / 0.300L = 0.0667M CH3NH3+

The equilibrium of CH3NH3+ is:

Ka = Kw/kb = 1x10-14/4.4x10-4 = 2.273x10-11 = [H+] [CH3NH2] / [CH3NH3+]

As both [H+] [CH3NH2] comes from the same equilibrium:

[H+] =  [CH3NH2] = X

2.273x10-11 = [X] [X] / [0.0667M]

1.5159x10-12 = X²

X = 1.23x10-6M = [H+]

As pH = -log [H+]

pH = 5.91

The pH at the equivalent point for this titration is "5.91".

\(CH_3NH_2 = 0.200\ M\\\\ \text{volume} = 100.0\ mL = 0.100\ L\\\\HCl = 0.100\ M\\\\\)

We must now quantify the pH well at the equivalence point.

We know that even at the point of equivalence, moles of acid and moles of the base are equivalent. As such, first, we must calculate the number of moles of the given base.

Calculating the Moles in \(CH_3NH_2 = 0.200\ M \times 0.100\ L = 0.0200\ moles\)

Calculating the Moles in \(HCl = 0.0200 \ moles\)

Calculating the volume of \(HCl\):

\(\to \text{Molarity} = \frac{ \text{moles}}{\text{volume \ (L)}} \\\\\to \text{Volume} = \frac{\text{moles}}{\text{molarity}}\\\\\)

                \(= \frac{0.0200 \ moles}{ 0.100\ M}\\\\= 0.200 \ L\\\\= 200 \ mL\\\\\)

Calculating the reaction among the acid and base:

\(CH_3NH_2 + HCl \longrightarrow CH_3NH_3^{+} + Cl^-\)

\(0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200\)

Therefore the conjugate acid of the bases exists at the standard solution.

Then we must calculate the new molar mass of \(CH_3NH_3^+\).

Total volume\(= 100 + 200 = 300\ mL = 0.300\ L\)

\([CH_3NH_3^+] = \frac{0.0200\ mole}{ 0.300\ L}= 0.0667\ M\)

Using the ICE table

\(CH_3NH_3^+ + H_2O \longrightarrow CH_3NH_2 + H_3O^+\)

\(I \ \ \ \ \ \ \ \ \ \ 0.0667 \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ 0\\\\C\ \ \ \ \ \ \ \ -x\ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ +x\\\\E \ \ \ \ \ \ \ \ \ \ \ \ 0.0667-x \ \ \ \ \ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ \ \ \ \+x\\\\\to Ka = \frac{[CH_3NH_2] [H_3O^+] }{[CH_3NH_3^+]}\)

Calculating \(K_a\) from \(K_b\)

\(\to K_a \times K_b = 1\times 10^{-14}\\\\\to K_a = \frac{1\times 10^{-14}}{4.4\times 10^{-4}} = 2.27\times 10^{-11}\\\\\)

                           \(= 2.27\times 10^{-11} \\\\= x\times \frac{x}{(0.0667-x)}\)

The x in the 0.0667-x can be ignored since the Ka value is just too small and it also does not follow the five percent criteria.

\(\to 2.27 \times 10^{-11} \times 0.0667 = x_2\\\\\to x_2 = 1.515\times 10^{-12}\\\\\to x = 1.23\times 10^{-6}\ M\\\\\to [H_3O^+] = x = 1.23\times 10^{-6}\ M\\\\\)

We have the formula to calculate pH.

\(\to pH = - \log [H_3O^+] = - \log 1.23\times 10^{-6}\ M= 5.91\)

The pH at the equivalent point for this titration is "5.91".

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