Calculate the energy density of pumped hydro electrical storage
(PHES) with Δh = 300m (its urgent pls help)

In architecture, rough stone blocks are referred to as "ashlar." Ashlar masonry is a method of construction that involves using precisely cut and fitted stones to create a flat, smooth surface.

The stones are typically cut to a uniform size and shape to ensure a snug fit with minimal gaps. Ashlar masonry can be used for both load-bearing and non-load-bearing walls, as well as for decorative purposes.

Ashlar masonry has been utilized throughout history and is still used in modern-day construction. It has been incorporated into various architectural styles, such as Gothic, Renaissance, and Baroque. Ashlar masonry is renowned for its durability, strength, and aesthetic appeal.

In summary, ashlar refers to the use of rough stone blocks in masonry construction. This technique involves cutting and fitting the stones precisely to create a flat, smooth surface, and it can be used for both practical and decorative purposes. Ashlar masonry has a long history of use in architecture and is still employed today due to its many benefits.

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The diameter, bolt M27 is known to be 3.0mm in pitch and 41mm across flats.

This is a term that is often seen as the Major diameter. The diameter of a bolt is known to be a kind of a Shank diameter which is said to be often shown or expressed in the unit called millimeters in regards to Metric bolts.

This is so due to the  fact that it is almost the same as the Major or Thread diameter.

Therefore, The diameter, bolt M27 is known to be 3.0mm in pitch and 41mm across flats.

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Transformer plays a crucial role in the electrical system. It transfers energy from one circuit to another through electromagnetic induction. It is an essential electrical device that is used to increase or decrease the voltage in a circuit.

A transformer primary winding operates at 240V, the secondary at 120V, and the load is 1,500W. If the transformer is rated 92 percent efficient, what's the current flowing through the primary winding of this transformer.

Formula: The current flowing through the primary winding of a transformer is given by; Primary Current, I1 = Power Input / Voltage Input Where: Power Input = Power Output / Efficiency Voltage Input = V2 = 120VPrimary Current, I1 = Power Input / Voltage InputI1 = (1500/0.92) / 240VI1 = 7.14 A

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Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod \(\overline T = 548 \ K\)

\(\rho = 7900 \ kg/m^3\)

\(K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K\)

\(\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657\)

Here, we can't apply the lumped capacitance method, since Bi > 0.1

\(\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\\)

\(0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81\)

\(t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins\)

However, on a single rod, the energy extracted is:

\(\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J\)

Hence, for centerline temperature at 50 °C;

The surface temperature is:

\(T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}\)

Traditional firewalls faced challenges due to their limited capabilities and inability to effectively handle evolving threats and complex network environments.

Traditional firewalls were designed to provide basic network security by controlling inbound and outbound traffic based on predefined rules. However, as technology advanced and cyber threats became more sophisticated, traditional firewalls struggled to keep up. Here are some key limitations they faced:

1. Inability to inspect encrypted traffic: Traditional firewalls lacked the ability to decrypt and inspect encrypted traffic, which made it difficult to detect and mitigate threats concealed within encrypted data packets.

2. Lack of application-level awareness: They primarily focused on filtering traffic based on IP addresses and ports, often neglecting the application layer. This limitation made it easier for attackers to exploit vulnerabilities in specific applications and bypass firewall restrictions.

3. Limited visibility into user behavior: Traditional firewalls had limited capabilities to monitor and analyze user behavior, making it challenging to identify insider threats or unusual patterns of network usage.

4. Difficulty handling complex network environments: As networks grew in complexity with the adoption of cloud services, mobile devices, and remote work, traditional firewalls struggled to enforce consistent security policies across different network segments and platforms.

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A disk interface that is designed and developed to use parallel data transfer but with high reliability and an advanced command set is: C. PATA.

A hard-disk drive can be defined as an electro-mechanical, non-volatile data storage device that is made up of magnetic disks (platters) that rotates at high speed.

Disk Management can be defined as a type of utility that is designed and developed to avail end users an ability to convert two or more basic disks on a computer system to dynamic disks.

In Computer technology, PATA is a disk interface that is designed and developed to use parallel data transfer but with high reliability and an advanced command set.

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Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, \($h_0=2.2$\) inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

\($\Delta h = H_0-h_f=\mu^2R$\)

     \($=0.2^2 \times 14 = 0.56$\) inches

\($h_f = 2.2 - 0.56$\)

     = 1.64 inches

Roll strip contact length (L) = \($\sqrt{R(h_0-h_f)}$\)

                                             \($=\sqrt{14 \times 0.56}$\)

                                             = 2.8 inches

Absolute value of true strain, \($\epsilon_T$\)

\($\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$\)

Average true stress, \($\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$\) Psi

Roll force, \($L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$\)

                                 = 788,900 lb

For SI units,

Power = \($\frac{2 \pi FLN}{60}$\)  

           \($=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$\)

           = 10399.81168 W

Horse power = 13.9357

Answer:

The correct answer would be a) Engineering Controls.

Explanation:

If the controls are handled correctly, you can reduce and eliminate hazards so no one gets hurt. Engineering controls are absolutely necessary to prevent hazards.

Hope this helped! :)

Personal  protective equipment (PPE) is appropriate for controlling hazards

PPE are used for exposure to hazards when safe work practices and other forms of administrative controls cannot provide sufficient additional protection, a supplementary method of control is the use of protective clothing or equipment. PPE may also be appropriate for controlling hazards  while engineering and work practice controls are being installed.

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Answer:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

P=10, P=5,  P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10,   θ = 45°  rectangular coordinates

x = r cosθ  ,   y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

  = 7.07

y =P sinθ = 10 sin 45°

  = 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos  30° = 4.33

y = 5 sin  30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos  60° = 12.5

y = 25 sin  60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos  90° = 0

y = 54 sin  90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos  (-34°) = 53.88

y = 65 sin  (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos  (-56)° = 53.12

y = 95 sin  (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos  20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos  (-42)° = 5.94

y = 8 sin  (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos  (-65)° = 14.79

y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

Please refer to the uploaded Manning's roughness modifier PDF file to determine the appropriate roughness coefficient (n) for the given conditions and use it in the Manning's equation to calculate the flow rate (Q).

To determine the flow rate in the unlined open channel, we can use Manning's equation:

Q = (1.49 / n) * A * R^(2/3) * S^(1/2)

where:

Q is the flow rate,

n is the Manning's roughness coefficient,

A is the cross-sectional area of flow,

R is the hydraulic radius, and

S is the slope of the channel.

Given:

Width (B) = 12 ft

Depth (y) = 3 ft

Side slopes (h:v) = 4:1

Slope (S) = 0.001 ft/ft

First, let's calculate the cross-sectional area of flow (A):

A = B * y + (h * y^2) / 2

= 12 ft * 3 ft + (4 * 3 ft^2) / 2

= 36 ft^2 + 18 ft^2

= 54 ft^2

Next, let's calculate the hydraulic radius (R):

R = A / P

= A / (B + 2y)

= 54 ft^2 / (12 ft + 2 * 3 ft)

= 54 ft^2 / 18 ft

= 3 ft

Now, we need to determine the Manning's roughness coefficient (n) using the provided Manning's roughness modifier table (PDF file). Please refer to the uploaded file to find the appropriate roughness coefficient for the given conditions.

Assuming you have the Manning's roughness coefficient (n), substitute all the values into Manning's equation to find the flow rate (Q).

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Answer:

240

Explanation:

The highest voltage typically encountered on the job by a residential electrician is C. 240 volts.

Voltage is the measure of the difference in electrical power between two points in a circuit.

It is like the force that pushes electric charges in a circuit and is measured in volts (V) and affects how strong the electric current flows in a wire.

In many countries, the United States included, residential electrical systems often use a split-phase setup with a voltage of 120/240 volts.

This voltage is widely used for things like household appliances, lights, and other electrical needs in homes.

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Answer:

Explanation:

Considering the flow of mercury in a tube:

When it comes to laminar flow of mercury, the thermal entry length is quite smaller than the hydrodynamic entry length.

Also, the hydrodynamic and thermal entry lengths which is given as DLhRe05.0= for the case of laminar flow. It should be noted however, that Pr << 1 for liquid metals, and thus making the thermal entry length is smaller than the hydrodynamic entry length in laminar flow, like I'd stated in the previous paragraph

Answer:

9 1/2

Explanation:

3 4/5 is 3.8

2/5 is 0.4

divide 3.8 by 0.4 you get 9.5 which is 9 1/2 in a fraction

It is very important to know where online information comes from in order to validate, authenticate and be sure it's the right information

Online informations are information which are available on the internet such as search engines, social handles and other websites

In conclusion, it is very important to know where online information comes from in order to validate, authenticate and be sure it's the right information

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how am I going to do this, I have a friend that might be able to help I will check

When a function has a basic period of the type f(x+k)=f(x) f(x+k)=f(x), k is referred to as the period of the function, and f is referred to as a periodic function.

A function is a kind of rule that, for one input, it gives you one output. Image source: by Alex Federspiel. An example of this would be y=x2. If you put in anything for x, you get one output for y. We would say that y is a function of x since x is the input value.

i) For periodic signal x1(t+T1)=x1(t)

==> cos(t+T1)=cos(t)

The above equation will hold for T1=2nπ

Thus, the fundamental period will be T1=2π

ii) For periodic signal x1(t+T1)=x1(t)

==> sin(π(t+T2))=sin(πt)

==> sin(πt+πT2))=sin(πt)

The above equation will hold for πT2=2nπ

Thus, the fundamental period will be T2=2

b) Now, for x3(t)

to be periodic k_2 \over k_1=T_1 \over T_2

==> k_2 \over k_1=2\pi \over 2=\pi

Since, π

is not an integer Hence, x3(t) is not periodic c) i) P(x1(t))=12π∫20\picos2(t)dt

==> P(x1(t))=14π∫20π(1+cos(2t))dt

==> P(x1(t))=14π[1+0.5sin(2t)]20π

==> P(x1(t))=12

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patience

problem-solving skills

ability to handle complex calculations

Explanation:

Engineering's fundamental building block is arithmetic, and biomedical engineers unquestionably require solid maths abilities. Geometry and calculus are frequently used by biomedical engineers while assessing and developing medical solutions. Thus, option A,C,D is correct.

To better the lives of patients, biomedical engineers analyse challenges in biology and medicine and create solutions. A good aptitude for engineering, an imaginative spirit, and a love for working in healthcare are all required for this position. The human anatomy fascinates biomedical engineers, and they enjoy addressing problems.

No more difficult than any other type of engineer, at least. For those with a talent for engineering, this field is rigorous but rewarding, even though it is generally difficult to break into. No engineering discipline is more demanding than biomedical engineering.

Therefore, patience, problem-solving skills, ability to handle complex calculations talents do Biomedical Engineers need.

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Question:

All of the following are advantages of using a pressure transducer rather than a vacuum gauge EXCEPT:

A. greater accuracy.  

B. easier identification of the cylinder.

C. measuring higher pressures.

D. ability to see the levels graphically

Answer:

The correct answer is D) ability to see the levels graphically

Explanation:

The above question derives from the Rudiments of Automotive Technology.

The function of the pressure transducer is to enable the diagnostic who is testing the engine vacuum to detect the cylinder with a faulty vacuum.

The pressure transducer does allow its user to see the vacuum graphically NOT  the levels.

Cheers.

Answer:

1779 veh/h

Explanation:

Given :

Peak hour factor, PHF = 0.84

Heavy vehicle adjustment factor, \($F_{HV}$\) = 0.847

Analysis flow rate, \(v_p\) = 1250 pc/h/ln

We know that,

Analysis flow rate, \($v_p=\frac{v}{(PHF)(N)(F_g)(F_{HV})}$\)

Here, v = peak hour volume

         \($F_g$\) = grade adjustment factor which is 0.99 for rolling terrain.

               > 1200 pc/h/ln

\($v_p=\frac{v}{(PHF)(N)(F_g)(F_{HV})}$\)

\($1250=\frac{v}{(0.84)(2)(1)(0.847)}$\)

v = 1779 veh/h

Therefore, the peak hour volume, v = 1779  veh/h

a. The equal-frequency partitioning for the given sales price records would be as follows: Bin 1: 5, 10, 11, 13  Bin 2: 15, 35, 50, 55  Bin 3: 72, 92, 204, 215

(a) Equal-Frequency (Equidepth) Partitioning:

Equal-frequency partitioning divides the data into bins of equal frequency. In this case, we have 12 records, so we need to partition them into three bins.

Now, we can assign the records to the bins based on their order. Starting from the lowest value, we assign four records to each bin until all records are assigned. If there are any remaining records, we distribute them evenly across the bins.

Using this method, the equal-frequency partitioning for the given sales price records would be as follows:

Bin 1: 5, 10, 11, 13

Bin 2: 15, 35, 50, 55

Bin 3: 72, 92, 204, 215

(b) Equal-Width Partitioning:

Equal-width partitioning divides the data into bins of equal width or range. In this case, we need to determine the range of the data and then divide it into three equal-width bins.

The range of the data is the difference between the maximum and minimum values, which is 215 - 5 = 210.

Each bin will have a width of 210 / 3 = 70. Starting from the minimum value, we assign the records to the bins based on their value falling within the corresponding width range.

Using this method, the equal-width partitioning for the given sales price records would be as follows:

Bin 1: 5, 10, 11, 13, 15, 35

Bin 2: 50, 55, 72, 92

Bin 3: 204, 215

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Answer:

The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.

Explanation:

Let suppose that vehicle is moving at constant velocity. By Newton's Law of Motion, the force given by engine must be equal to the sum of the rolling resistance and the aerodynamic drag force of the air. And by definition of power, we have the following formula:

\(\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v\) (1)

Where:

\(\dot W\)- Power, in pounds-force-feet per second.

\(f\) - Rolling resistance coefficient, no unit.

\(W\) - Weight of the passanger car, in pounds-force.

\(\rho\) - Density of air, in pounds-mass per cubic feet.

\(C_{D}\) - Drag coefficient, no unit.

\(A\) - Projected frontal area, in square feet.

\(v\) - Vehicle speed, in feet per second.

\(g_{c}\) - Pound-mass to pound-force ratio, in pounds-mass to pound-force.

If we know that \(f = 0.02\), \(W = 3,550\,lbf\), \(\rho = 0.08\,\frac{lbm}{ft^{3}}\), \(C_{D} = 0.34\), \(A = 23.3\,ft^{2}\), \(v = 80.685\,\frac{ft}{s}\) and \(g_{c} = 32.174\,\frac{lbm}{lbf}\), then the power required by the car is:

\(\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v\)

\(\dot W = 10901.941\,\frac{lbf\cdot ft}{s}\)

\(\dot W = 19.623\,h.p.\)

The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem

Here is the table summarizing the costs and benefits that should be considered in the financial and economic evaluation of the water treatment facility construction:

| Category                | Cost or Benefit Description                                    |

|-------------------------|---------------------------------------------------------------|

| A. Cost of land purchase| Cost of acquiring land for constructing the water treatment facility |

| B. Construction costs   | Expenses incurred during the construction of the facility      |

| C. Salaries of human resources   | Ongoing salaries and wages for the employees working at the facility |

| D. Inspection and maintenance costs | Costs associated with regular inspection and maintenance of the facility |

| E. Opportunity cost of land    | Value of the land if it were used for an alternative purpose    |

| F. Opportunity cost of capital investment | The return that could be earned from investing the capital elsewhere |

| G. Income from selling water to consumers | Revenue generated from selling treated water to the residents |

| H. Cost savings due to eliminating the cost of energy to boil water | Savings achieved by avoiding the need for residents to boil water |

| I. Time saving for the residents due to not having to boil the water | Value of time saved by the residents due to not needing to boil water |

| J. Saving on health-related costs due to accidentally drinking unsafe water | Reduction in healthcare expenses due to improved water quality |

| K. Residual value of the land at the end of the service life | Estimated value of the land at the conclusion of the facility's lifespan |

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Here is the ER diagram for the Cruise Business Case:

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

|   Employee  |          |    Ships    |

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

| EmployeeID  |          |   ShipID    |

| LastName    |          | ShipName    |

| FirstName   |          | Description |

| Type        |          +-------------+

| ParrotID    |<---------|     Crew    |

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

                         |   Captain   |

                         |  ParrotID   |

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

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

|   Parrot     |         |  Dependents  |

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

| ParrotID     |<--------|  DependentID |

| ParrotName   |         |  FirstName   |

| Color        |         |  EmployeeID  |

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

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

|   Missions   |         | MissionShip |

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

| MissionID    |         | MissionID   |

| MissionName  |         | ShipID      |

| ScheduledDate|         +-------------+

| QuotedPrice  |

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

The tables and their relationships are as follows:

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The maximum height that the water can rise above the jet is: 35 meters.

Given:

P = 350 kPa = 350000 Pa

We would use pressure(p)  formula to determine the maximum height that the water can rises above the jet by solving for h (height).

Using this formula

Pressure(P) = P₀ + ρgh

Where;

P₀ represent Pressure at the fluid's surface

ρ represent Density of the fluid = 1000 kg/m³

g represent acceleration due to gravity = 10 m/s².

h represent height

Solving for h (height)

350000 = 0 + (1000 × 10 ×h)

350000 = 10000h

Divide both side by  10000h

h = 350000/10000

h = 35 meters

Therefore the maximum height that the water can rise above the jet is: 35 meters.

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You can read about it here https://www.cornellcollege.edu/information-technology/policies/technology-policies/limitations-of-laptops.shtml

B only. The specifications state that the draw should be between 10 and 30 milliamps, which is equal to 0.01-0.03 amps. The DVOM reading of 0.251 amps is significantly higher than the specified range, indicating that there is a problem with the vehicle's electrical system. Technician A is incorrect in stating that the draw is within the specification range. Technician B is correct in saying that the draw is too high.

Your question is: A vehicle is being tested for a draw against the battery with the ignition switch in the OFF position. The specifications state the draw should be between 10 and 30 milliamps. The DVOM reads 0.251 amps. Who is correct between Technician A and Technician B?

Technician A says the draw is within the specification range, while Technician B says the draw is too high. In this case, Technician B is correct. The draw should be between 10 and 30 milliamps, but the DVOM reads 0.251 amps, which is equivalent to 251 milliamps. This value is higher than the specified range. Therefore, the correct answer is B. B only.

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The proper mounting height for receptacles can vary depending on the application and local building codes. However, a common recommended mounting height for receptacles in a commercial setting is 18 inches above the finished floor to the center of the receptacle.

Therefore, for the six 20-amp 120-volt receptacles shown along column line c, between columns 3 and 4 on sheet e6, the proper mounting height would be 18 inches above the finished floor to the center of each receptacle.

So the possible answers are:

18 inches above the finished floor to the center of each receptacle.

It's not possible to determine the exact mounting height without additional information about the specific building codes and requirements for the application.

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