The primary driving force for natural convection is a. pressure forces
b. shear stress forces c. None of them d. buoyancy forces e, surface tension forces

Answer:

A

Explanation:

Answer:

A

Explanation:

The fuse breaks the circuit if a fault in an appliance causes too much current to flow. The fuse contains a piece of wire that melts easily. If the current going through the fuse is too great,the wire heats up until it melts and breaks the circuit

Answer:

The answer is below

Explanation:

Case hardening is a form of steel hardening that is applied on mild steel with a high temperature of heat.

It results in material forming a hard surface membrane, while the inner layer is soft.

It is mostly used for universal joints, construction cranes, machine tools, etc.

On the other hand, Through hardening is a form of steel hardening in engineering that involves heat treatment of carbon steel.

It increases the hardness and brittleness of the material.

It is often used for axles, blades, nuts and bolts, nails, etc.

The CIA triad and the Parkerian hex are the fundamental principles of information security.

Parkeriano, or Parkerian hexad: is a set of six elements of information security proposed by Donn B. Parker.

1. Confidentiality.

2. Ownership or Control.

3. Integrity.

4. Authenticity.

5. Availability.

6. Utility

The Parkerian hexagram adds three more attributes to the three classic security attributes of the CIA triangle

these are the fundamental principles of information security.

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Answer:

Option D. is correct

Explanation:

Joe uses one of his CAT5 patch cables to connect two hubs to add a new segment to his local network. As he can only connect to it from a workstation within that segment,  he is not able to reach the new network segment from his workstation.

The most problem is that the technician used a straight-through cable.

Option D. is correct.

Organisational change can be difficult to implement, but by using a structured strategy, it can go more smoothly. The three steps listed below can assist reduce opposition and win the support and buy-in of those who are impacted:

1. Getting Ready for Change: Clearly express the necessity for and advantages of change.

- Involve workers early on, getting their feedback and addressing any issues they may have.   - Offer resources and training to develop the abilities required for the transformation.

  - Create a sense of urgency and a common vision.

2. Managing the Change: - Divide the change into doable steps and establish attainable objectives.   - Establish a change management team to oversee the procedure and handle problems.   - Encourage open channels of communication for discussion and feedback.   - Thank and honour staff members for their contributions and achievements.

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The nonlinear state-space equation for the bicycle motion can be derived using the definitions of input, state, and output variables as given below:

Input variables:
- u1: Steering torque
- u2: Torque at the rear wheel

State variables:
- x1: Steering angle (δ)
- x2: Tilt angle (φ)
- x3: Angular velocity of steering (δ')
- x4: Angular velocity of tilt (φ')

Output variables:
- y1: Steering angle (δ)
- y2: Tilt angle (φ)

Using these variables, the nonlinear state-space equation for the bicycle motion can be derived as follows:

x1' = x3
x2' = x4
x3' = (u1 - Dm*a*h*x4^2*sin(x2) - Dm*b*h*x3*x4*cos(x2))/(J + Dm*h^2)
x4' = (u2 - Dm*a*h*x3^2*sin(x2) + Dm*g*h*sin(x2)*cos(x2))/(Dm*(a^2 + b^2) - Dm*h^2*cos^2(x2))
y1 = x1
y2 = x2

Thus, the nonlinear state-space equation for the bicycle motion is given by the above set of equations. These equations describe the dynamics of the bicycle motion with respect to the input, state, and output variables.

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The volume of the given region in the first octant, bounded by the coordinate planes and the surfaces y=1-x² and z=1-x², is 1/12.

To find the volume of the given region in the first octant , bounded by the coordinate planes and the surfaces y=1-x² and z=1-x² , we can use the triple integral formula:

V = ∫∫∫dV = ∫∫∫dx dy dz

where the region R is given by:

0 ≤ x ≤ 1, 0 ≤ y ≤ 1-x² and 0 ≤ z ≤ 1-x².

The sketch of the region is as shown below:

The volume of this region is then given by:

V = ∫∫∫dx dy dz = ∫1-x² dx∫1-x² dy ∫1-x² dz

= ∫1-x² dx∫1-x² dy ∫1-x² dx

= ∫1-x² dx∫1-x² dy (x - x³/3)

= (x - x³/3)∫1-x² dx∫1-x² dy

= (x - x³/3)∫1 dx∫1-x² dy

= (x - x³/3)(1 - (1-x²)²/2)

= (x - x³/3)(x - x² + x⁴/2)

= (x³ - x⁴/2 - x⁴/2 + x⁵/6)(x - x² + x⁴/2)

= x⁵/6 - x⁶/4 + x⁷/6

= x⁵/6 - x⁶/12

= 1/6 - 1/12 = 1/12.

Therefore, the volume of the given region in the first octant, bounded by the coordinate planes and the surfaces y=1-x² and z=1-x², is 1/12.

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Solving problems with technology can create other problems in society, as job displacement is a common consequence. Therefore, the correct option is (O A) Solving problems with technology can create other problems in society.

A parallel that can be drawn between the disappearance of craftspeople's jobs with the advent of manufacturing and the current trend of manufacturing jobs being replaced by digital solutions is that both phenomena involve the displacement of workers due to technological advancements.

In both cases, the introduction of new technologies and methods of production has resulted in a shift in the job market and the need for different skill sets.

This can lead to job loss and economic challenges for those affected.

Additionally, both situations highlight the importance of adapting and retraining workers to meet the demands of the evolving economy.

While the specific technologies and industries involved may differ, the underlying theme of technological progress impacting the workforce remains consistent.

Therefore, the correct option is (O A)  Solving problems with technology can create other problems in society.

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Advanced distillation technologies

2.1. Heat integrated distillation column (HIDC) ...

2.2. Membrane distillation (MD) ...

2.3. Cyclic distillation. ...

2.4. Cryogenic distillation. ...

2.5. Reactive distillation column.

Answer:

P1+1/2pv2/1+pgh1=P2+1/2pv2/2+pgh2

Answer:

The amount of shaft work the turbine cam do per second is 3660.29 kJ

Explanation:

The given parameters are;

Pressure at entry p₁ = 300 kPa

The mass flow rate,  \(\dot {m}\) = 5.0 kg/sec

The initial temperature, T₁ = 175°C

Therefore;

The enthalpy at 300 kPa and 175°C, h₁ = 2,804 kJ/kg

At the turbine exit, we have;

The pressure at exit, p₂ = 100 kPa

The quality of the steam at exit, in percentage, x₂ = 60%

Therefore, for the enthalpy, h₂ for saturated steam at 100 kPa and quality 60%, we have;

h₂ = 417.436 + 0.6 × 2257.51 = 1771.942 kJ/kg

Heat loss from the turbine, \(h_l\)= 1,500 kW

By energy conservation principle we have;

dE/dt = \(\dot Q\) - \(\dot W\) + ∑\(m_i \cdot (h_i\)+ \(ke_i\) +\(pe_i\)) - ∑\(m_e \cdot (h_e\) + \(ke_e\) +\(pe_e\) )

0 = -\(h_l\)  -  \(\dot W\) + \(m_i \cdot h_i\) - \(m_e \cdot h_e\)  

\(\dot W\) = \(\dot {m}\) × (h₁ - h₂) - \(h_l\) = 5.0×(2,804 - 1771.942) - 1500 = 3660.29 kJ/s

The rate of work of the shaft = 3660.29 kJ/s

The amount of shaft work the turbine cam do per second = 3660.29 kJ.

Given :

Mass of gasoline , m = 50 kg .

Length of tank , l = 0.9 m.

Breadth of tank , b = 0.5 m .

Relative density of gasoline , R.D = 0.75 .

Density of water , \(\rho_{water}=1000\ kg/m^3\) .

To Find :

The depth of tank .

Solution :

We know ,

\(R.D=\dfrac{\rho_{gasoline}}{\rho_{water}}\\\\\rho_{gasoline}=0.75\times 1000 \ kg/m^3\\\\\rho_{gasoline}=750\ kg/m^3\)

Now , volume of gasoline is :

\(V=\dfrac{m}{\rho_{gasoline}}\\\\V=\dfrac{50}{750}\ m^3\\\\V=\dfrac{1}{15}\ m^3\)

Since , the tank is rectangular , volume is given by :

\(V=l\times b\times h\\\\h=\dfrac{V}{l \times b}\\\\h=\dfrac{1}{15\times 0.9\times 0.5}\ m\\\\h=0.148\ m=14.8\ cm\)

Therefore , the answer in centimetres is 14.8 cm .

Hence , this is the required solution .

The answer is analyze

The maximum transverse shear stress is 25.46 psi.

The maximum transverse shear stress, in psi, of a 1/4-8 lead screw with a shear force of 10 lbs can be calculated using the following equation:

Shear Stress (psi) = Shear Force (lbs) x 0.323 x (Threads per Inch).

The maximum transverse shear stress can also be determined by using the formula of \(\tau_{max} = (F_s * r) / I,\)

where \(F_s\) is the shear force, r is the radius, and I is the moment of inertia of the shaft.

The moment of inertia of a solid circular shaft can be determined by the formula of \(I = (\pi * r^{4} ) / 4\).

Here, the diameter of the lead screw is 1/4-8, which means the radius is 1/8 inches. The shear force is given to be 10 lbs.

\(\tau_{max} = (F_s * r) / I,\)

\(\tau_{max} = (10 * \frac{1}{8} ) / (\pi* \frac{1}{8})\)

\(\tau_{max} = 25.46\ psi\)

Therefore, the maximum transverse shear stress in psi of a 1/4-8 lead screw with a shear force of 10 lbs is 25.46 psi.

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the correct answer is to Maven error will disappear when running on a jre rather than a jdk.

Maven is a popular open-source build tool developed by the Apache Group to build, publish, and deploy several projects at once for better project management. The tool provides allows developers to build and document the lifecycle framework.This error usually indicates a problem with the compiler version compatibility, and you can check few places to fix it. For example, you can edit your POM and configure Maven compiler plugin to compile your Java code.

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The technique that can be used when welding steels with poor weldability is preheat the base material to reduce the weld cooling rates (option c)

Weldability refers to the ability of a material to be welded under certain conditions without producing defects, such as cracks or porosity, in the weld zone. Steels with poor weldability can be difficult to weld without producing defects, and often require special techniques to ensure a successful weld.

Preheating the base material is a common technique used when welding steels with poor weldability. By heating the base material before welding, the cooling rate of the weld zone can be reduced, which can help prevent the formation of cracks and other defects.

Preheating also helps to ensure that the material being welded is at a consistent temperature throughout the welding process, which can improve the quality of the weld.

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In the given code, the output " -18-9-17 -1-8" is printed because of the combination of print statements inside the main() and ii() methods. The output will be : public static void main(String[] args) {int x9 = 3;// x9 is initialized to 3System.out.print(ii(x9*2) + " " + (x9 - 4) + ii(x9));}

The above statement first calls the ii() method with an argument of x9*2 = 6, which means the x parameter inside the ii() method is 6. Now, the ii() method does the following:public static int ii(int x){x = x * -3;System.out.print(x);x++;return x;}x is multiplied by -3, so x becomes -18.

The print statement inside the ii() method prints the value of x, which is -18. The value of x is incremented by 1 and the new value of x (-17) is returned to the calling function. Now, the main() method continues with the evaluation of the expression:System.out.print(ii(x9*2) + " " + (x9 - 4) + ii(x9));The second argument in the print statement is (x9 - 4), which is equal to -1 (since x9 is 3). Now, the ii() method is called again, this time with an argument of x9 = 3.

The ii() method performs the following steps:public static int ii(int x){x = x * -3;System.out.print(x);x++;return x;}x is multiplied by -3, so x becomes -9. The print statement inside the ii() method prints the value of x, which is -9. The value of x is incremented by 1 and the new value of x (-8) is returned to the calling function.

Finally, the print statement inside the main() method prints the values returned by the two ii() method calls, which are -17 and -8, respectively. This results in the output " -18-9-17 -1-8".Thus, the output "-18-9-17 -1-8" is generated because of the combination of print statements inside the main() and ii() methods.

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1. Inverse square law states that the intensity of light varies inversely with the square of the distance from the source. It can be represented mathematically as: I = k/d², where I is the intensity of light, d is the distance from the source and k is a constant of proportionality.

This law is illustrated by the fact that as the distance from the source increases, the intensity of light decreases proportionally to the square of the distance.2. Given, diameter of the circular area, d = 200 mRadius of the circular area, r = d/2 = 100 mLamp illuminates a circular area of 200 meters in diameter with the illumination of the disc increasing uniformly from 0.5 meter-candle at the edge to 2 meter-candle at the center. The average illumination can be calculated as follows:Average illumination of the disc, I = (0.5 + 2)/2 = 1.25 meter-candleThe total light received can be calculated as follows:Total light received = (2πr² × I) = (2 × π × 100² × 1.25) = 78,540 lumensAverage candle power of the source can be calculated as follows:Average candle power = Total light received/4π = 78,540/4π = 6250 lumens3. Floodlighting is the use of high-intensity artificial light to illuminate a large area.

The purpose of floodlighting is to provide a bright and uniform light over a large area, typically for outdoor sports fields, stadiums, and other large events. It can be achieved using various types of lighting fixtures, such as floodlights, spotlights, and high-intensity discharge lamps. Suitable diagrams for floodlighting are shown below:

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All answers are mentioned below.

Isentropic pressure refers to the pressure that results from a thermodynamic process that occurs at constant entropy. In other words, the entropy of a system does not change during an isentropic process, and therefore the pressure changes in a specific manner based on the initial and final states of the system. Isentropic processes are typically adiabatic, meaning that no heat is added or removed from the system during the process. These processes can occur in many different forms, such as in a compressor, a nozzle, or a thermodynamic cycle. In each of these cases, the isentropic pressure change can be calculated using the first and second laws of thermodynamics, along with the specific conditions of the process.

a. The inlet to the compressor can be analyzed using the isentropic relationship between pressure and temperature. Given the ambient conditions of P = 11.0 kPa and T = 216.7 K, the stagnation pressure and temperature can be found using the equation:

P0 = P * (1 + (γ - 1) / 2 * M^2)^(γ / (γ - 1))

where γ = 1.4 is the ratio of specific heats for air.

Solving for P0, we find that the stagnation pressure is 22.0 kPa. To find the stagnation temperature, we use the equation:

T0 = T * (1 + (γ - 1) / 2 * M^2)

Solving for T0, we find that the stagnation temperature is 533.3 K.

Next, the pressure ratio of the compressor is given as 10 and the isentropic efficiency is 90%, so the actual pressure rise can be calculated as:

P2/P1 = η * P2/P1,isentropic

where η is the isentropic efficiency. Solving for P2, we find that the pressure at the exit of the compressor is 86.1 kPa.

The temperature rise in the compressor can be calculated using the equation:

T2 = T1 * (P2 / P1)^((γ - 1) / γ)

Solving for T2, we find that the temperature at the exit of the compressor is 793.3 K.

b. In the combustor, the velocities are low, so the stagnation and static pressures are equal, and the stagnation pressure falls by 5%. The fall in stagnation pressure can be expressed as:

P0,comb = P0 * (1 - ΔP / P0)

where ΔP is the fall in stagnation pressure. Solving for P0,comb, we find that the stagnation pressure in the combustor is 20.38 kPa.

Next, the stagnation temperature at turbine entry is given as 1400 K, and the turbine efficiency is 90%. The temperature at turbine exit can be found using the equation:

T3 = T2 * ηturbine

Solving for T3, we find that the temperature at turbine exit is 996.7 K.

The pressure at turbine exit can be found using the equation:

P3 = P2 * (T3 / T2)^(γ / (γ - 1))

Solving for P3, we find that the pressure at turbine exit is 0.212 MPa.

c. The velocity of the jet can be found by finding the velocity of an isentropic expansion from the stagnation conditions at the exit of the propulsive nozzle to the ambient conditions. The velocity can be found using the equation:

V = (2 * (γ / (γ - 1)) * R * T0 * (1 - (P / P0)^((γ - 1) / γ)))^0.5

where R is the specific gas constant for air. Solving for V, we find that the velocity of the jet is 1069 m/s.

The gross thrust per unit mass flow can be found using the equation:

Fa = V * (1 - (P / P0))

Solving for Fa, we find that the gross thrust per unit mass flow is 1069 N kg

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Solving for Fa, we have find that the gross thrust per unit mass flow is 1069 N·kg.

Isentropic pressure refers to the pressure that results from a thermodynamic process that occurs at constant entropy. In other words, the entropy of a system does not change during an isentropic process, and therefore the pressure changes in a specific manner based on the initial and final states of the system. Isentropic processes are typically adiabatic, meaning that no heat is added or removed from the system during the process. These processes can occur in many different forms, such as in a compressor, a nozzle, or a thermodynamic cycle. In each of these cases, the isentropic pressure change can be calculated using the first and second laws of thermodynamics, along with the specific conditions of the process.

a. The inlet to the compressor can be analyzed using the isentropic relationship between pressure and temperature. Given the ambient conditions of P = 11.0 kPa and T = 216.7 K, the stagnation pressure and temperature can be found using the equation:

P0 = P * (1 + (γ - 1) / 2 * M^2)^(γ / (γ - 1))

where γ = 1.4 is the ratio of specific heats for air.

Solving for P0, we find that the stagnation pressure is 22.0 kPa. To find the stagnation temperature, we use the equation:

T0 = T * (1 + (γ - 1) / 2 * M^2)

Solving for T0, we find that the stagnation temperature is 533.3 K.

Next, the pressure ratio of the compressor is given as 10 and the isentropic efficiency is 90%, so the actual pressure rise can be calculated as:

P2/P1 = η * P2/P1,isentropic

where η is the isentropic efficiency. Solving for P2, we find that the pressure at the exit of the compressor is 86.1 kPa.

The temperature rise in the compressor can be calculated using the equation:

T2 = T1 * (P2 / P1)^((γ - 1) / γ)

Solving for T2, we find that the temperature at the exit of the compressor is 793.3 K.

b. In the combustor, the velocities are low, so the stagnation and static pressures are equal, and the stagnation pressure falls by 5%. The fall in stagnation pressure can be expressed as:

P0,comb = P0 * (1 - ΔP / P0)

where ΔP is the fall in stagnation pressure. Solving for P0,comb, we find that the stagnation pressure in the combustor is 20.38 kPa.

Next, the stagnation temperature at turbine entry is given as 1400 K, and the turbine efficiency is 90%. The temperature at turbine exit can be found using the equation:

T3 = T2 * ηturbine

Solving for T3, we find that the temperature at turbine exit is 996.7 K.

The pressure at turbine exit can be found using the equation:

P3 = P2 * (T3 / T2)^(γ / (γ - 1))

Solving for P3, we find that the pressure at turbine exit is 0.212 MPa.

c. The velocity of the jet can be found by finding the velocity of an isentropic expansion from the stagnation conditions at the exit of the propulsive nozzle to the ambient conditions. The velocity can be found using the equation:

V = (2 * (γ / (γ - 1)) * R * T0 * (1 - (P / P0)^((γ - 1) / γ)))^0.5

where R is the specific gas constant for air. Solving for V, we find that the velocity of the jet is 1069 m/s.

The gross thrust per unit mass flow can be found using the equation:

Fa = V * (1 - (P / P0))

Solving for Fa, we find that the gross thrust per unit mass flow is 1069 N kg

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You get a result immediately from Euler's formula:

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

The statement given "Recognition can be described as perception and long-term memory working together" is true because recognition can be described as perception and long-term memory working together.

Recognition is the ability to identify someone or something previously encountered or known. Recognition occurs when perceptual processes (e.g. vision, hearing) work together with long-term memory to match a current stimulus with stored information.

When we recognize something, we are able to recall information about it and apply that information to our current situation. This allows us to quickly and accurately identify familiar objects, people, places, or events.

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To determine the average shearing stress in the pin at b, we need to find the shear force acting on the pin.

(a) First, let's analyze the forces acting on member bd. The pin at b connects member bd to member ab. Assuming member ab is loaded, we can calculate the shear force at b using the equation F = P / (2 * sin(theta)), where P is the load on member ab and theta is the angle between member ab and bd.

Once we have the shear force, we can calculate the average shearing stress using the formula average shearing stress = shear force / area of the pin. The area of the pin can be calculated using the equation area = pi * (d^2) / 4, where d is the diameter of the pin.

(b) To find the average bearing stress at b in member bd, we need to determine the force acting on the pin and the contact area between the pin and the member. The force can be calculated using the equation F = P / (2 * cos(theta)). The average bearing stress is then calculated as average bearing stress = force / area of contact.

(c) For member abc, we are given that it has a 10 x 50-mm uniform rectangular cross-section. To find the average bearing stress at b in member abc, we need to calculate the force acting on the pin and the contact area between the pin and the member. The force can be calculated using the equation F = P / (2 * cos(theta)).

The average bearing stress is then calculated as average bearing stress = force / area of contact. The contact area can be found by multiplying the width and thickness of the rectangular cross-section.

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Answer:

\(X=-\$19318\)

Explanation:

From the question we are told that:

Initial cost \(P= $100,000\)

Annual benefits \(A= $12,500\)

Salvage value \(S= $30,000\)

Interest rate \(I=10\%=>0.10\)

Time \(t=8years\)

Generally the equation for Net Project worth X is mathematically given by

  \(X=-P+A+S\)

Where

Present worth of Annual benefits A is

 \(A'=A(P/a,0.10,8)\)

 \(A'=12500*5.3349\)

 \(A'=\$66686.25\)

Present worth of Salvage Price S is

 \(S'=S(P/a,0.10,8)\)

 \(S'=30000*0.46651\)

 \(S'=\$13995.3\)

Therefore

 \(X=-P+A'+S'\)

 \(X=-100000+66686.25+\$13995.3\)

 \(X=-\$19318\)

Answer:

yes maybe

Explanation:

I have a project and everyone in the world will know about it.

Answer:

both

Explanation:

The correct option to complete the Java ArrayList class's resize() method is arrayData[i] = newArray[i];.

The purpose of the resize() method is to resize the array to the new allocation size provided as a parameter. The method creates a new array of the given size and copies the elements from the old array to the new array. In the for loop, the index i iterates over the old array elements, and each element is copied to the corresponding index of the new array using the assignment operator. Therefore, the correct option to complete the statement is arrayData[i] = newArray[i];.

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Recursion is the process by which a method or function repeatedly calls itself. Recursive function refers to a function that is repeatedly invoked, either directly or indirectly.

To comprehend recursion, different instances will be given. Let's now examine the recursion syntax. We must first supply the base case answer before developing the recursive algorithm. The major problem is then described in terms of lesser difficulties.

methodName (T parameters…)

{  

   if (precondition == true)          

//precondition or base condition

   {

       return result;

   }

   return methodName (T parameters…);

       //recursive call

}

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It means that the gas with the lowest molecular weight will have the highest effusion rate.

The given gases' effusion rates are listed in order from highest to lowest. The effusion rate of a hydrogen molecule is the highest, whereas that of a hydrocarbon is the lowest.

A gas will effuse faster when it is lighter and more slowly when it is heavier. Helium (He) will have the highest rate of effusion since it has the lowest molecular weight (atomic weight, in this example).

The following equation can be used to compare the rate of effusion for two gases: The effusion rates in this case are inversely related to the square root of the gas molecules' masses. A container contains an amalgam of neon and argon gas.

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