Computer-controlled instrument panel dimming is being discussed. Technician A says the body computer dims the illumination lamps by varying resistance through a rheostat that is wired in series to the lights. Technician B says the body computer can use inputs from the panel dimming control and photo cell to determine the illumination level of the instrument panel lights on certain systems.Who is correct?A. A onlyB. B only.C. Both A and B.D. Neither A nor B.

The answer explains how to determine if a power cycle is reversible or irreversible based on heat transfer data and how to calculate thermal efficiency and compare it with Carnot efficiency.

Question: A system executes a power cycle while receiving 1050 kJ by heat transfer at a temperature of 525 K and discharging 700 kJ by heat transfer at 350 K. Determine whether the cycle is internally reversible, irreversible, or impossible. Determine the thermal efficiency using the given heat transfer data and compare it with the Carnot efficiency.

a. Using the given data, we can calculate the change in entropy for the system to determine if the cycle is reversible or irreversible. If the total change in entropy is zero, the cycle is reversible. If not, it is irreversible.

b. To calculate thermal efficiency, use the formula η = W/Qh, where W is the work done and Qh is the heat input. Compare this efficiency value with the Carnot efficiency, which gives the maximum possible efficiency for the given temperature range.

Answer:

As many variables as we can coherently communicate in 2 dimensions

Explanation:

Visualization is a descriptive analytical technique that enables people to see trends and dependencies of data with the aid of graphical information tools. Some of the examples of visualization techniques are pie charts, graphs, bar charts, maps, scatter plots, correlation matrices etc.

When we utilize a visualization on paper/screen, that visualization is limited to exploring as many variables as we can coherently communicate in 2-dimensions (2D).

An arithmetic circuit with select variables S1 and S2 generates eight different operations involving n-bit data inputs A and B, with the output being conditional on the values of S1, S2, and carry in Cin. These operations include various forms of addition, complementing, and incrementing. The circuit design integrates basic digital circuits like adders and incrementers.

Arithmetic Circuit Design for Various Operations

The design of an arithmetic circuit for executing multiple operations based on select variables S1 and S2, as well as input carry Cin, encompasses the creation of a multi-functional digital circuit. This circuit needs to handle eight specific functions with two n-bit inputs, A and B, with the results being contingent on the values of S1, S2, and Cin. The operations are as follows:

Each operation can be implemented by combining basic digital circuits like adders, incrementers, and complement generators within the arithmetic circuit design.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer: Material selection

Explanation: As a biological engineer you need to be conversant with the composition, properties and processing of materials. This is important because when you design equipments and machines using your 3D model. During the fabrication process your knowledge of the composition of materials e.g 5% iron,

Their properties e.g ductility, non corrosive e.t.c and the way each and every materials being used in fabrication were processed plays a big role on how the 3D model would appear in real word. It’s durability and performance, so as a biological engineer this knowledge helps you make the right decision in materials selection.

Answer:

a) 23.39

b) 44977.08 N

c) 1922.92N

d) 454.31 MPa

e) 8.32 MPa

f) [tex] 3.47*10^-^3 [/tex]

Explanation:

a) fiber-matrix load ratio:

Let's use the formula :

[tex] \frac{F_f}{F_m} = \frac{V_f E_f}{V_m E_m}[/tex]

[tex] = \frac{0.3 * 131 GPa}{0.7 * 2.4 GPa} = 23.39 [/tex]

b & c)

Total load is given as:

Fc = Ff + Fm

46900 = Fm(23.39) + Fm

46900 = 24.39 Fm

Actual load carried by matrix=

[tex]F_m = \frac{46900}{24.39}[/tex]

= 1922.92N=> answer for option c

Actual load carried by fiber, Ff:

Ff = 46900 - 1922.92

Ff = 44977.08 N => answer option b

d)

Let's find area of fiber, A_f.

[tex] A_f = V_f * A_c[/tex]

Ac = Cross sectional area =300mm²

= 0.3 * 300 = 99 mm²

Area of matrix=

[tex] A_m = V_m * A_c[/tex]

= 0.7 * 300 = 231 mm²

Magnitude of the stress on the fiber phase:

[tex] \sigma _f= \frac{F_f}{A_f} [/tex]

[tex] \sigma _f= \frac{44977.08}{99} = 454.31 MPa [/tex]

e) Magnitude of the stress on the matrix phase.

[tex] \sigma _m = \frac{F_m}{A_m} [/tex]

[tex] \sigma _m = \frac{1922.92}{231} = 8.32 MPa[/tex]

f) Strain in fiber = [tex] \frac{\sigma _f}{E_f} [/tex]

[tex]= \frac{454.31*10^6}{131*10^9} = 3.47*10^-^3[/tex]

Strain in matrix = [tex] \frac{\sigma _m}{E_m} [/tex]

[tex]= \frac{8.32*10^6}{2.4*10^9} = 3.47*10^-^3[/tex]

Composite strain = [tex] (E_f *V_f) + (E_m * V_m) [/tex]

[tex] (3.47*10^-^3 * 0.3) + (3.47*10^-^3 * 0.7) = 3.47*10^-^3 [/tex]

Find the attachments for complete solution

The Otto cycle has a thermal efficiency of approximately 56%. Given the power output, the required rate of heat input to the system is approximately 161 KW.

The first step to determining the thermal efficiency of an Otto cycle is to calculate it using the formula

ηth = 1 - (1/rγ-1)

, where ηth is the thermal efficiency, γ is the heat capacity ratio (1.4 for air), and r is the compression ratio.

In this case, substituting the given values r = 10.5 and γ = 1.4, we obtain ηth ≈ 0.56 or 56%.

The second step is to use the basic principle of power, which states power = energy/time. From the given repetition rate, the cycle time is 1/2500 minute = 0.0004 minute.

The total rate of heat input is therefore power/ ηth = 90kW / 0.56 ≈ 161 kW.

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Answer:

a) πa⁵ Po² / 6E

Vacuum has infinite volume that's why I didn't consider that for total energy calculations

Explanation:

Check attached images for explanation and solutions to b and c

The question pertains to the calculation of final properties of air after an adiabatic process, using principles of thermodynamics applicable to an ideal gas in a closed, adiabatic system. Final pressure, work done, and entropy produced must be calculated, with specific formulas applying to each aspect.

The question is asking to calculate the final properties of air inside a rigid, insulated tank after being stirred by a paddle wheel. This is a thermodynamics problem specifically regarding a closed system where work is done without heat transfer (adiabatic process).

Since the tank is rigid, the volume remains constant, and since it is adiabatic and insulated, we know no heat is exchanged with the surroundings. For an ideal gas undergoing a process in a closed system, the relationship between pressure and temperature is given by Gay-Lussac's law, which states that P1/T1 = P2/T2 where P is pressure and T is temperature in Kelvin. The pressure will increase as the temperature increases.

For a rigid container, there is no change in volume; hence no boundary work is done. However, work is done by the paddle wheel, and this is equivalent to the change in the internal energy of the gas since it is an adiabatic process. Using the first law of thermodynamics, the work can be found. Since the internal energy of an ideal gas depends only on temperature, the work done by the gas would be the difference in internal energy, which is a function of the difference in temperatures and the specific heat at constant volume.

Since the process is adiabatic and there is work done on the gas, entropy is produced within the system. However, without interaction with the surroundings, the entropy exchanged with the environment is zero. The change in entropy of the system will thus be a function of the initial and final states of the gas, considering that entropy is a state function.

Answer:

4.071 veh/h.

Explanation:

Step one: calculate the estimated free flow speed by using the formula below;

= 75.4 - F(L) - F(C) - 3.22T^ 0.84.

The value of F(L) = 1.9 m/h and F(C) = 0.8 m/h.

Hence,

free flow speed = 75.4 - 1.9 - 0.8 - 3.22(3/6)^0.84.

free flow speed= 75.4 - 1.9 - 0.8 - 1.799

free flow speed= 70.901 mi/h = 70 mi/h.

Step two: determine the adjustment factor by using the formula below;

The adjustment factor = 1/ [1 + Pm (Em - 1) + Pn (En - 1)].

The adjustment factor = 1/[ 1+ 10/100 ( 2.5 - 1) + 0(2.0 - 1] = 0.869.

Step three;

Calculate the hourly volume;

The value corresponding to the LOS C conditions and free flow speed at 70 mi/h is 1735.

Therefore,

hourly volume, V = 1735 × 0.9 × 0.869 × 3.

Hourly volume, V = 4.071 veh/h.

The calculated hourly volume under maximum Level of Service C conditions on a six-lane freeway with a peak hour factor of 0.90 and 10% heavy vehicles is approximately 11,220 vehicles. This considers factors like the number of lanes, peak hour factor, and percentage of heavy vehicles.

The calculation of the hourly traffic volume on a highway is a complex process that involves the consideration of many factors. For this scenario, you have a six-lane freeway (three lanes each way) that's operating at maximum Level of Service (LOS) C conditions, which implies moderate traffic density where the operation speeds are about 50% of free flow speed with restricted lane changing.

Using the Highway Capacity Manual (HCM) 2010 specifications for multilane highways, the maximum service flow rate for LOS C on rolling terrain is approximately 1,700 passenger cars per hour per lane (pc/hpl). The total volume would then be found by multiplying this number by the number of lanes and then adjusting for the peak hour factor and percentage of heavy vehicles.

Therefore, Hourly Volume = (1700 pc/hpl x 6 lanes) x (0.90 peak hour factor) x (1 + 0.1 heavy vehicles) which totals to approximately 11,220 vehicles. Thus, under these conditions, the hourly volume on this segment of the freeway would be around 11,220 vehicles.

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Answer:

a) Total mass of air in the engine (in kg) = 0.00223 kg

b) power developed by the engine (in hp) = 180.61 hp

Explanation:

Initial temperature, [tex]T_{1} = 280 K\\[/tex]

Initial pressure, [tex]P_{1} = 70 kPa[/tex]

Compression ratio, r = 10

The initial total engine volume, [tex]V_{1} = 2555.6 cm^{3} = 2555.6 * 10^{-6} m^{3}[/tex]

a) Total mass of air in the engine

Using the gas equation, [tex]P_{1} V_{1} = mRT_{1}[/tex]

Where R = 0.287 kJ/kg-K

70 * 2555.6 * 10⁻⁶ = m * 0.287 * 280

[tex]m = \frac{70 * 2555.5 * 10^{-6} }{0.287 * 280}[/tex]

m = 0.00223 kg

b) Power developed by the engine

Heat generated due to combustion, [tex]Q_{in} = 1080 kJ/kg[/tex]

[tex]\frac{T_{2} }{T_{1} }= (\frac{V_{1} }{V_{2} })^{\gamma -1}[/tex]

Compression ratio, [tex]r = \frac{V_{2} }{V_{1} } = 10[/tex]

[tex]\frac{T_{2} }{T_{1} }=10^{0.4}\\\frac{T_{2} }{280 }=10^{0.4}\\T_{2} = 280 * 10^{0.4}\\T_{2} = 703.328 K[/tex]

[tex]Q_{in} = c_{v} (T_{3} -T_{2} )[/tex]

Where Specific capacity of air,  [tex]c_{v} = 0.718 kJ/kg-K[/tex]

[tex]1080 = 0.718 (T_{3} -703.328 )\\1504.18 + 703.328 = T_{3}\\ T_{3} = 2207.51 K[/tex]

[tex]\frac{T_{4} }{T_{3} } = (\frac{V_{2} }{V_{1} } )^{\gamma -1} \\\frac{T_{4} }{T_{3} } = (1/10 )^{0.4} \\\frac{T_{4} }{2207.51 } = (1/10 )^{0.4} \\T_{4} = 2207.51 * (1/10 )^{0.4} \\T_{4} = 878.82 K[/tex]

[tex]Q_{out} = c_{v} (T_{4} - T_{1} )\\Q_{out} = 0.718 (878.82 - 280 )\\Q_{out} = 430 kJ/kg[/tex]

[tex]w_{net} = Q_{in} - Q_{out}\\w_{net} = 1080 - 430\\w_{net} = 650 kJ/kg[/tex]

There are 4 cylinders, k = 4

N = 2800/2

N = 1400

Power developed by the engine,

[tex]P =\frac{ mw_{net} Nk}{60} \\P =\frac{ 0.00223*650* 1400*4}{60} \\P = 134.68 kW[/tex]

1 kW = 1.34102 hp

P = 134.68 * 1.34102

P = 180.61 hp

Answer:

Explanation:

The solutions to these questions can be seen in the following screenshots from a solution manual;

Answer:

788°C

Explanation:

Metal palte details

Thickness = t=10 mm =0.01 m.

Thermal conductivity= k =200 W/mC

Heat capacity C=900 J/kg-k.

Initial temp Ti =22 C.

Time for which it has been kept in oven =t=2 min =120 sec.

Air properties.

Temp of air in oven =Ta =900 C

Heat transfer coefficient =h =190 W/m-k.

Temp of block after 2 min= T =?

In problem, it is not given block dimension. Let us assume that Block is 1 m in legth and 1 m in width.

Volume of block =V= 1*1*0.01 =0.01 m3 .(We know thckness =0.01 m).

Surface area of block = A=2(1*1+1*0.01+0.01*1)=2.04 m2.

Biot number for this problem is 0.004656 which is less than 0.1. So Lumped capacitance method is applicable to this problem, according to formulae,

T - Tair /Ti - Tair =e(-hAt/density* C* V) .

T - 900/(22-900) = e(-190*2.04*120 /2500*900*0.01) .

T - 900/(22-900) =e-2.0672 .

T -900 = - 110.09

Temp of block after 2 min =788 °C.

Answer:

Find the attachments for complete solution

Answer:

11. b. False

12. b. False

13. c. 4.5 pounds

14. a. True

15. b. Mine only

Explanation:

The aforementioned are associated with pollution and resource depletion. Pollution refers to a process of contamination of the environment through undesirable and unintended production and usage of resources.

Resource depletion describes the usage of limited or scarce resources.

Find the complete solution in the given attachments

Answer:

19.64 square inches

Explanation:

Area will be (¶d^2)/4

= (3.142 x 5^2)/4

= 19.64 square inches

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

Final temperature: [tex]\( 336.575 \, \text{K} \)[/tex] . Vessel volume:  [tex]\( 100 \, \text{L} \).[/tex]

To solve this problem, we can use the principle of conservation of mass and energy.

1. **Conservation of Mass:** Since the membrane ruptures and the entire volume gets filled, the mass of steam remains constant.

2. **Conservation of Energy:** We can use the first law of thermodynamics (energy conservation) to solve for the final temperature.

Given:

- Initial state: 1 kg of steam at 400°C and 200 bar.

- Final pressure: 100 bar.

Let's solve:

1. Conservation of Mass:

Since mass is conserved, the final mass of steam will remain 1 kg.

2. Conservation of Energy:

Using the first law of thermodynamics:

[tex]\[ Q = m \cdot c_v \cdot \Delta T \][/tex]

where:

- ( Q ) is the heat transferred,

- ( m )is the mass of the substance (1 kg in this case),

[tex]- \( c_v \)[/tex] is the specific heat capacity at constant volume, and

[tex]- \( \Delta T \)[/tex]  is the change in temperature.

For steam, we can consider it to be an ideal gas at these high temperatures and pressures. The specific heat capacity at constant volume [tex](\( c_v \))[/tex]  can be taken as a constant value.

Now, we need to find the final temperature [tex](\( T_f \))[/tex] . We can use the ideal gas law to relate the initial and final states:

[tex]\[ P_1 \cdot V_1 / T_1 = P_2 \cdot V_2 / T_2 \][/tex]

Where:

[tex]- \( P_1 \) and \( T_1 \)[/tex] are the initial pressure and temperature,

[tex]- \( P_2 \) and \( T_2 \)[/tex] are the final pressure and temperature,

[tex]- \( V_1 \) and \( V_2 \)[/tex] are the initial and final volumes, respectively.

Solving:

Given:

[tex]- \( P_1 = 200 \, \text{bar} = 20,000 \, \text{kPa} \)\\- \( T_1 = 400 \, \text{°C} = 673.15 \, \text{K} \)\\- \( P_2 = 100 \, \text{bar} = 10,000 \, \text{kPa\\} \)\\- \( m = 1 \, \text{kg} \)[/tex]

Let's assume specific heat capacity at constant volume [tex](\( c_v \))[/tex]  for steam to be approximately  [tex]\( 2.0 \, \text{kJ/(kg·K)} \)[/tex]  (This is a rough estimation).

From the ideal gas law:

[tex]\[ T_2 = \frac{{P_2 \cdot V_1 \cdot T_1}}{{P_1 \cdot V_2}} \][/tex]

Given that initially, the other side of the vessel is evacuated, so [tex]\( V_2 \)[/tex] is negligible compared to [tex]\( V_1 \).[/tex]

[tex]\[ T_2 \approx \frac{{P_2 \cdot T_1}}{{P_1}} \]\[ T_2 \approx \frac{{10,000 \times 673.15}}{{20,000}} \]\[ T_2 \approx 336.575 \, \text{K} \][/tex]

Now, we can calculate the final volume using the ideal gas law:

[tex]\[ P_1 \cdot V_1 / T_1 = P_2 \cdot V_2 / T_2 \]\[ V_2 = \frac{{P_2 \cdot V_1 \cdot T_1}}{{P_1 \cdot T_2}} \]\[ V_2 = \frac{{10,000 \times 1 \times 673.15}}{{20,000 \times 336.575}} \]\[ V_2 = \frac{{673,150}}{{6731.5}} \]\[ V_2 = 100 \, \text{L} \][/tex]

So, the final temperature of the steam is approximately

[tex]\( 336.575 \, \text{K} \)[/tex] and the volume of the vessel is [tex]\( 100 \, \text{L} \).[/tex]

Answer:

Explanation:

1. Air flows into the top of the carburetor from the car's air intake, passing through a filter that cleans it of debris. ... When the throttle is open, more air and fuel flows to the cylinders so the engine produces more power and the car goes faster. The mixture of air and fuel flows down into the cylinders

2.Advantages :

Carburetor parts are not expensive as that of fuel injectors, especially EFI, which would give you large savings.

With the use of carburetor you get more air and fuel mixture.

In terms of road test, carburetors have more power and precision.

Carburetors are not restricted by the amount of gas pumped from the fuel tank which means that cylinders may pull more fuel through the carburetor that would lead to denser mixture in the chamber and greater power as well.

Disadvantages :

At a very low speed, the mixture supplied by a carburetor is so weak that it will not ignite properly and for its enrichment, at such conditions some arrangement in the carburetor is required.

The working of carburetor is affected by changes of atmospheric pressure. 

It gives the proper mixture at only one engine speed and load, therefore, suitable only for engines running at constant speed increase or decrease.

More fuels are consumed since carburetors are heavier than fuel injectors.

More air emissions than fuel injectors.

Maintenance costs of carburetor is higher than with fuel injection system.

3. The electronics used in the system will calculate this information and constantly adjust. This type of controlled fuel injection results in a higher power output, greater fuel efficiency and much lower emissions. One of the main issues is that these systems are sophisticated and will cost much more than a carburetor

Answer:

The answer is explained below for the sampling frequency

Explanation:

Solution

Recall that:

The frequency Fm1 = 1 kHz

The frequency Fm2 = 5 kHz

Fs = n (2Fm2)

so,

[Fs = 8 * ( 2 * 5) = 80 8kHz], This the sampling frequency for 8-bit A/D converter

Now,

By increasing quantization level we can also increase the substantial noise ratio

Answer:

Find the attachment for solution

Note: The resistance is measured in ohms (Ω). In the question Ω will replace V, where ever the resistance unit is required .

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step and very detailed solution of the given problem

Answer:Aerospace History Timeline

Explanation:

Answer: P = 0.416 kW

Explanation:

taken a step by step process to solving this problem.

we have that from the question;

the amount of heat rejected Qn = 4800 kJ/h

the cooling effect is Ql = 3300 kJ/h

Applying the first law of thermodynamics for this system gives us

Шnet = Qn -Ql

Шnet = 4800 - 3300 = 1500 kJ/h

Next we would calculate the coefficient of performance of the refrigerator;

COPr = Desired Effect / work output = Ql / Шnet  = 3300/1500 = 2.2

COPr = 2.2

The Power as required gives;

P = Qn - Ql  = 4800 - 3300 = 1500 kJ/h = 0.416

P = 0.416 kW

cheers i hope this helps!!!!1

Answer:

0.0198 mm/sec

Explanation:

Given:

d = 8 cm

L = 10 m = 1000 cm

v = 200 m/min

E = 100%

density = 3500 A.m^-2

Using Faraday's law of electrolysis, we have:

Q = n(e) * F

Where,

Q = electric charge

F = Faraday's constant=96500Cmol

n(e) = electron charge = 2

Therefore

Q= 96500 * 2 = 193000

Let's find the area of pipe, A = πr²(L)

But radius, r = [tex] \frac{d}{2} = \frac{4}{2} = 4cm [/tex]

Therefore,

A = 3.142 * 4² * 1000

A = 50272

To find the thickness of the coating deposited, we have :

[tex] = \frac{193000 * 7.31 * 118.7 * 200}{96500 * 50272 * 3500} = 1.98 * 10^-^3[/tex]

= 0.0198 mm/sec

Answer:

b) Commonly used to transmit network signals over great distances.

Explanation:

The transmission of information or data by using microwave radio waves is known as microwave transmission. Microwave transmitter is commonly used to transmit network signals over great distances. It is an electronic device that transmits and receives radio frequency signals ranging from 1GHz to 100GHz.

The microwave transmitter has a wide range of applications and these includes, radio stations, television stations, mobile phones, radio astronomy, radar,

Answer:

d) An omnidirectional wireless technology that provides limited-range voice and data transmission over the unlicensed 2.4-GHz frequency band, allowing connections with a wide variety of fixed and portable devices that normally would have to be cabled together.

Explanation:

Microwave transmission is a technology widely used in the 1950s and 1960s for transmitting signals, such as long-distance telephone calls and television programs between two terrestrial points on a narrow beam of microwaves. In microwave radio relay, microwaves are transmitted on a line of sight path between relay stations using directional antennas, forming a fixed radio connection between the two points. The requirement of a line of sight limits the separation between stations to the visual horizon, about 30 to 50 miles (48 to 80 km). Before the widespread use of communications satellites, chains of microwave relay stations were used to transmit telecommunication signals over transcontinental distances.

Answer:

minimum sight distance = 699 ft

Explanation:

given data

road lane = 4 divided road

median width = 12 ft

grade road = 5%

solution

we take here time gap factor for minor road vehicle when enter to major road  from table

time gap = 8.1  sec

and for median width of 12 ft

time gap = 8.2 + 0.7 ( 1 + [tex]\frac{12}{12}[/tex]  )  

time gap = 9.5 second

so minimum sight distance will be

minimum sight distance = 1.47 × design speed × time gap  

minimum sight distance = 1.47 × 50  × 9.5

minimum sight distance = 699 ft

Answer:

0.00692 cm/Sec

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

No.

Explanation:

The Coefficient of Performance of the reversible heat pump is determined by the Carnot's cycle:

[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]

[tex]COP_{HP} = \frac{297.15\,K}{297.15\,K-280.15\,K}[/tex]

[tex]COP_{HP} = 3.339[/tex]

The power required to make the heat pump working is:

[tex]\dot W = \frac{300\,kW}{3.339}[/tex]

[tex]\dot W = 89.847\,kW[/tex]

The heat absorbed from the exterior air is:

[tex]\dot Q_{L} = 300\,kW - 89.847\,kW[/tex]

[tex]\dot Q_{L} = 210.153\,kW[/tex]

According to the Second Law of Thermodynamics, the entropy generation rate in a reversible cycle must be zero. The formula for the heat pump is:

[tex]\frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}} + \dot S_{gen} = 0[/tex]

[tex]\dot S_{gen} = \frac{\dot Q_{H}}{T_{H}} - \frac{\dot Q_{L}}{T_{L}}[/tex]

[tex]\dot S_{gen} = \frac{300\,kW}{297.15\,K}-\frac{210.153\,kW}{280.15\,K}[/tex]

[tex]\dot S_{gen} = 0.259\,\frac{kW}{K}[/tex]

Which contradicts the reversibility criterion according to the Second Law of Thermodynamics.