Suppose you connect your laptop into a university network (either via wired ethernet or 802.11 wifi). How does your laptop get assigned an IP address with which it can send datagrams across the internet?
a. IP addresses are unique to each NIC, and therefore, a device does not need to take any action to obtain an IP address. b. Every student is assigned a unique and static IP address for every laptop or device they register with IT.
c. The laptop sends out a special ethernet (or 802.11) frame asking all hosts within the subnet to return their IP addresses. The laptop is free to select any IP address that is not in the returned IP address list d. The laptop sends out a DHCP request over UDP to the local DHCP server to obtain an available IP address.

Answer:

The p-value of the hypothsis 0.00135

Explanation:

Answer:

The energy in pneumatic power systems is easily transmitted to gases, which absorb the energy of the system and lower its efficiency.

Explanation:

I did it on edge and got it right.

Answer:

The energy in pneumatic power systems is easily transmitted to gases, which absorb the energy of the system and lower its efficiency

Thanks for points!!! :D

As the temperature of the material increases, the potential energy of the molecules increases. Thermal expansion occurs due to changes in temperature, and interatomic distances increase as potential energy increases.

Thermal expansion is used in a variety of applications such as rail buckling, engine coolant, mercury thermometers, joint expansion, and others.

It is to be noted that an application of the concept of liquid expansion in everyday life concerns liquid thermometers. As the heat rises, the mercury or alcohol in the thermometer tube moves in only one direction. As the heat decreases, the liquid moves back smoothly.

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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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Answer:

do u speak english

Explanation:

Answer:

Each wattmeter measures a line-to-line voltage between two of the three power supply lines. In this configuration, the total power, watts is accurately measured by the algebraic sum of the two wattmeter values. Pt = P1 + P2. This holds true if the system is balanced or unbalanced.

Loading error can cause the measured value of the signal to be different from the true value of the signal, leading to an error in the measurement.

To plot the signal ratio against the impedance ratio, you could use a software such as M!crosoft Excel or G0ogle Sheets. You would create a column for the impedance ratio and a column for the corresponding signal ratio, and then plot these values using a scatter plot or line graph. You can then observe the trend of the signal ratio as the impedance ratio changes.

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Answer:

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

Explanation:

We require as many variables as there are columns of data in the CSV, and we must first build a reader class instance. These two claims on the unpacking of values into variables are accurate.

To read a CSV file line by line, we can use the readlines() method rather than the readline() method. When used on a file object, the readlines() method produces a list of every line in the file.

The csv. writer() function returns a writer object that creates a delimited string from the user's data.

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Answer:

I am in Eight Grade

Explanation:

Answer:

The answer is 2.25 kΩ

Explanation:

Solution

Given that:

The resistance reading on a DMM'S meter face = 22.5 ohms

The range selector switch = R * 100 range,

We now have to find the actual measure resistance of the circuit which is given below:

The actual measured resistance of the circuit is=R * 100

= 22.5 * 100

=2.25 kΩ

Hence the measured resistance of the circuit is 2.25 kΩ

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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Answer:

The state you live in

Children begin to acquire fine motor skills at the age of three. They are better able to move their fingers on their own and use them for increasingly difficult tasks including holding writing instruments like an adult, using scissors, and creating intricate and detailed drawings.

Children as young as three can successfully perform the fine motor skill of drawing a circle. A five-year-old child's fine motor skills include tying shoelaces, cutting with scissors or a pencil extremely precisely, and drawing a human with several pieces. At this age, kids begin learning about numbers and counting. Speak to your youngster in lengthier than his own phrases that contain actual words to aid in the development of his language abilities. When he says something, repeat it, for instance, "need nana," and then demonstrate how to use more "grown-up" language by saying, "I need nana."

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Project management and development methods for a mobile application for a library are presented using the Systems Development Life Cycle (SDLC) and Scrum, respectively.

The SDLC contains several phases including requirements gathering, design, development, testing, and implementation that are all followed in a systematic and structured manner. Scrum, on the other hand, is an incremental and iterative agile development methodology that places a strong emphasis on adaptability and teamwork.I would advise employing Scrum rather than the SDLC for the library's mobile application development project. Scrum has a number of benefits in this situation. First off, Scrum's iterative and incremental structure promotes regular feedback and adaptation, both of which are essential when creating a customer-focused application. The library can solicit ongoing feedback from users and other stakeholders and make

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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

It is true that the total current in a series circuit is equal to the total current flowing through any resistance in the circuit (IT = I1 = I2 = I3).

Every element of the parallel circuit has the same voltage. The voltage decreases across a series resistor, as you may recall from the previous section. A parallel circuit is an exception. Anywhere in the circuit, there will be a constant voltage.

As electrical current passes through resistors and other series circuit components, the potential in the circuit decreases with each one. As a result, in a series circuit, the voltage fluctuates.

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Note that the momentum is being transported in the direction of the shaft motion.

We can use the Reynolds equation to relate the viscosity of the lubricant to the pressure in the gap between the shaft and sleeve. The Reynolds equation for a slider bearing is given by:

∂(h^3p)/∂x = -12μV

where h is the gap thickness, p is the pressure, x is the axial direction, μ is the viscosity of the lubricant, and V is the velocity of the shaft.

To limit the total force on the sleeve to less than 2.2 N, we can calculate the maximum pressure as:

P = F/A = 2.2 N / (π(2.6/2)^2) = 0.06 MPa

At a velocity of 6.1 m/s, the flux of momentum in the gap is given by:

Φ = h^3p / 12μ

We can rearrange the Reynolds equation to solve for the viscosity:

μ = h^3p / (-12V(∂p/∂x))

Since the sleeve is free to move axially, we can assume that there is no pressure gradient in the axial direction (∂p/∂x = 0). Therefore, the viscosity can be calculated as:

μ = h^3p / (-12V(∂p/∂x)) = h^3p / (-12V(0)) = 0

This means that the viscosity of the grease can be zero, and the pressure in the gap can still be limited to 0.06 MPa to keep the total force on the sleeve below 2.2 N.

The magnitude of the flux of momentum in the gap is given by:

Φ = h^3p / 12μ = (5.08/100)^3 * 0.06 MPa / (12 * 0 Pa s) = 1.56 x 10^-9 kg m/s

The momentum is being transported in the direction of the shaft motion.

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 P2, a counter-clockwise current will be induced in the copper ring.2)The answer is (d) P2. The number of turns in the secondary coil is 200.

When the north pole of the magnet moves towards the copper ring, an induced electric field and a current will flow in the clockwise direction, according to Faraday's law.

A counter-clockwise current will be induced in the copper ring when the north pole moves away from the copper ring, which is in the opposite direction of the magnetic field and the current in the wire.

The copper ring will rotate counterclockwise, which is perpendicular to the plane of the paper if the magnet is moved away from the copper ring. Hence, at P2, a counter-clockwise current will be induced in the copper ring.2)

Given,Primary coil has 200 turns.Input voltage = 2400 V.Output power = 240 W.I = 2.0 A.

We know that the output voltage of a transformer can be calculated using the formula,Ns/Np = Vp/VsWhere,Ns = number of turns in the secondary coilNp = number of turns in the primary coilVp = input voltageVs = output voltage.

The output voltage, Vs, can be calculated as, Vs = P / Iwhere,P = output powerI = currentThe number of turns in the secondary coil, Ns can be calculated by, Ns = (Vs / Vp) x Np.

Now substituting the given values,240W / 2A = VsVs = 120V2400V / Vs = Np / Ns120 = Ns / 200Ns = 24,000/120 = 200The number of turns in the secondary coil is 200.

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1. When the interviewer asked if you have any questions at the end of the interview don't say no. You should always say yes your interviewer is expecting you to ask a few good questions before ending the interview.

2. Always thank the interviewer for their time and effort to interview you. This would look very good for you and its a nice way to help wrap up the interview.

Answer:

u need to plan it out

Explanation:

u need to plan it out

Answer:

use the turn 1 degrees option and put a repeat loop on it

Explanation:

u can add sound in ur loop

Answer:

The solution is presented in explanation

Explanation:

This problem can be solved in following steps:

1) In the first round the peasant will take the goat to the other side.

2) Now, the peasant will come back alone.

3) The peasant will now take the wolf with him to other side.

4) The peasant will return with the goat to riverbank.

5) Now, he will take cabbage to the other side of the river, where the wolf is already present.

6) Peasant will leave cabbage and wolf on other side and come back to riverbank alone. Since, wolf does not eat cabbage.

7) Now, finally the peasant will take goat to the other side of river.

In this way, all three of them shall be transported to the other side of the river without eating each other.

Answer:

I need help also

Answer 1: The correct answer is B: CAD as it is used to describe 3D modeling programs that are used in fields that require precise and exact real-world measurements

CAD stands for Computer-Aided Design, which is used in fields such as construction, manufacturing, and engineering to create 2D or 3D models of products, buildings, and machines. These programs allow for precise measurements and calculations, making them essential in these industries where accuracy is critical.

Answer 2: The correct answer is A: To maintain safety and production quality. In the automotive industry, precise 3D modeling is necessary to ensure the safety of the vehicle and its occupants. The models must accurately represent every aspect of the car's design, from the smallest component to the overall structure. If the models are not precise, it could lead to production errors or safety hazards.

Answer 3: The correct answer is D: Wheel Physics. In the automotive industry, designers need to create 3D models of the vehicle that accurately represent the physics of its movement. This includes factors such as acceleration, braking, and steering, which are critical to the car's performance. Wheel physics is one such feature that would be helpful in this context.

Answer 4: The correct answer is A: CAD. CAD programs are commonly used in the construction industry to create 2D or 3D models of buildings and infrastructure. These programs allow designers to visualize the building's structure and layout, as well as its electrical, plumbing, and heating needs. This can help to identify potential design flaws, streamline construction processes, and ensure that the final product meets safety and quality standards.

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Answer:

Wood is heavy

Explanation:

To determine the circulation for the given vector field around the enclosed half circle of radius 1, we need to calculate the line integral of the vector field along the boundary of the half circle.

The line integral is calculated by integrating the dot product of the vector field and the differential of the boundary. The circulation is then equal to the line integral divided by 2π. To calculate the line integral, we need to know the vector field and the boundary of the half circle. Once we have this information, we can use the equation for the line integral to calculate the circulation.

To calculate the circulation for the given vector field around the enclosed half circle of radius 1, we need to first calculate the line integral of the vector field along the boundary of the half circle. The line integral is calculated by integrating the dot product of the vector field and the differential of the boundary. The differential of the boundary is equal to the vector from the starting point to the end point of the boundary. The vector field can be expressed as a function of the coordinates of the boundary. Once we have the vector field and the differential of the boundary, we can use the equation for the line integral to calculate the circulation. The circulation is then equal to the line integral divided by 2π.

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