Code the Communication System of a Tank Bot: Recieves
commands/requests from the [Main Controller] to obtain mapping
information.
-Uses the "range interface" to return distance/layout
information
Jus

Answer:

The average current will be "10.12 mA".

Explanation:

In the transformer:

⇒  \(\frac{V1}{V2}=\frac{N1}{N2}\)

where,

V1 = 120

N2 = 1

N1 = 10

So that,

⇒  \(V2=V1\times \frac{N2}{N1}\)

On putting the values,

         \(=120\times \frac{1}{10}\)

         \(=12V \ rms\)

The full wave rectifier would conducts during positive and negative half.

⇒  \(Vmax=12-0.7\)

               \(=11.3V \ rms\)

⇒  \(Vpeak=11.3\sqrt{2}\)

               \(=15.9V\)

⇒  \(Vaverage=2Vmax \ \pi\)

                     \(=10.12 \ V\)

The amount conducted by each diode is 50%.

⇒  \(Iaverage=\frac{Vaverage}{R}\)

                    \(=\frac{10.12}{1}\)

                    \(=10.12 \ mA\)

The purpose is to compare the length of a flat key with that of a square key and suggest reasons for why a flat key may be preferred in some cases.

The paragraph describes a problem of estimating the length of a flat key required to transmit a torque equal to the elastic torque capacity of a round shaft of diameter d.

To solve this problem, the dimensions from figure 17.1 are needed, and the key and shaft are assumed to be made of the same ductile material with the key being tightly fitted at its top and bottom.

The length of a square key is also estimated, and a possible reason why a flat key might be preferred in some cases is suggested.

The length of the flat key is estimated using the elastic torque capacity of the round shaft, which is related to its diameter, material, and length.

The length of the square key is also estimated, and the comparison between the two lengths is used to suggest that a flat key might be

preferred in some cases because it can transmit the same torque as a square key while being shorter and therefore easier to manufacture and install.

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Answer:

Impermeable

Explanation:

Something that is impermeable does not allow water or liquid to pass through it.

Answer: The outer surface temperature is \(377.5^{o}C\).

Explanation:

Given: Heat = 3 kW (1 kW = 1000 W) = 3000 W

Area = 10 \(m^{2}\)

Length = 2.5 cm (1 cm = 0.01 m) = 0.025 m

Thermal conductivity = 0.2 W/m K

Temperature (inner) = \(415^{o}C\)

Formula used is as follows.

\(q = KA \frac{(t_{in} - t_{out})}{L}\)

where,

K = thermal conductivity

A = area

L = length

\(t_{in}\) = inner surface temperature

\(t_{out}\) = outer surface temperature

Substitute the values into above formula as follows.

\(q = KA \frac{(t_{in} - t_{out})}{L}\\3000 W = 0.2 \times 10 \times \frac{415 - t_{out}}{0.025 m}\\t_{out} = 377.5^{o}C\)

Thus, we can conclude that the outer surface temperature is \(377.5^{o}C\).

two functions or characteristics of ROM in a router are:

1. It contains firmware for basic device operation.

2. It stores bootstrap code to initiate the router's operating system.

Read-Only Memory (ROM) in a router is a non-volatile memory that retains its contents even when the router is powered off. It contains firmware that is essential for basic device operation, such as the router's Power-On Self Test (POST) and diagnostics. Additionally, ROM stores the bootstrap code, which is executed when the router boots up and is responsible for initializing the router's hardware and loading the operating system into memory. This process is critical to the router's operation as without it, the router would not be able to function properly.

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What are two functions or characteristics of ROM in a router?

Answer:

To calculate the coolant boiling point with a 15 psi pressure cap, we need to consider the base boiling point of water, the increase in boiling point due to the 50-50 mix of coolant, and the increase in boiling point per psi of pressure.

Base Boiling Point of Water: 100°C

Increase in Boiling Point due to Coolant Mix: 6°C

Increase in Boiling Point per psi: 17°C

Calculation for a 15 psi pressure cap:

Base Boiling Point + Increase due to Coolant + (Pressure Increase × Increase per psi)

= 100°C + 6°C + (15 psi × 17°C/psi)

= 100°C + 6°C + 255°C

= 361°C

Therefore, with a 15 psi pressure cap, the cooling system boiling point would be 361°C.

Now let's calculate the boiling point with a 12 psi pressure cap:

Base Boiling Point + Increase due to Coolant + (Pressure Increase × Increase per psi)

= 100°C + 6°C + (12 psi × 17°C/psi)

= 100°C + 6°C + 204°C

= 310°C

With a 12 psi pressure cap, the cooling system boiling point would be 310°C.

If pure water is used instead of a 50-50 mix of coolant, the boiling point would remain at 100°C since there would be no additional increase in boiling point due to the coolant mix.

As for the freezing point, the presence of the coolant mix or the pressure cap does not directly affect the freezing point of water. The freezing point of water is generally 0°C, regardless of the coolant mix or pressure conditions.

Explanation:

Performing the operations A + B, A - C, and C + B - A using the given values of A, B, and C yields the desired results.

In the first operation, A + B, we add the values of A and B together. This operation calculates the sum of the two values. The result of this addition will depend on the specific values assigned to A and B.

In the second operation, A - C, we subtract the value of C from A. This operation calculates the difference between the two values. Again, the result will vary based on the actual values assigned to A and C.

Lastly, in the third operation, C + B - A, we add the value of C to B and then subtract the value of A from the sum. This operation combines addition and subtraction. The specific result will depend on the numerical values of A, B, and C.

Overall, by performing these three operations, we can obtain the desired results and sketch them in the given space.

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The incident light should come from the medium with n2 to undergo total internal reflection.

(a) The incident light should come from the medium with n2 to undergo total internal reflection. Total internal reflection occurs when light travels from a medium with a higher refractive index to a medium with a lower refractive index and the angle of incidence exceeds the critical angle. In this case, since n2 is smaller than n1, the light should originate from the medium with n2 to experience total internal reflection at the interface.

(b) The evanescent field is present in the medium with n1. After total internal reflection, the incident light is completely reflected back into the medium with n1, and it does not propagate into the medium with n2. As a result, the evanescent field is confined to the medium with n1.

(c) To calculate the values of kx and kz in the medium where the field is evanescent, we can use the relationship between the wave vector (k) and the refractive index (n) in each medium. The wave vector in the x-direction (kx) and the wave vector in the z-direction (kz) can be expressed as follows:

kx = (w/c) * n * sin(0₁)

kz = [(w/c)^2 * \(n^2\) - \(kx^2\)]

where:

w = angular frequency of the evanescent field

c = speed of light in vacuum

n = refractive index of the medium

0₁ = angle of incidence

Given:

w = 10¹⁵ rad/s

n1 = 4

n2 = 2

0₁ = 80°

Using the above equations, we can calculate the values of kx and kz in the medium with n1:

kx = (10¹⁵ rad/s / 3x10^8 m/s) * 4 * sin(80°)

kz = [(10¹⁵ rad/s / 3x10^8 m/s) * 4^2 - kx]

After substituting the values and performing the calculations, we can determine the specific values of kx and kz in the medium with n1.

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Answer:160

Explanation:

multiply them all together divide by 1/2

The material engineer is in charge of evaluating the wear and tear of manufacturing process components.

Since the beginning of time, new materials have been essential to human development, prosperity, safety, and quality of life. They rank among the greatest achievements of every era. When it comes to civil, chemical, construction, nuclear, aeronautical, agricultural, mechanical, biomedical, or electrical engineering, new materials always open the door to new technologies.

Engineers and scientists who study materials are still at the forefront of numerous scientific fields. Every time we buy or use a new device, machine, or structure, materials science and engineering has an impact on our lives. Our list of recommended readings contains additional information about the impact of this exciting field.) A realization about every application of materials is what gives rise to the academic field of Materials Science and Engineering: The value of the material is determined by its characteristics. A material may be selected for its strength, electrical properties, heat or corrosion resistance, or a variety of other qualities; However, they all involve properties.

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It can be noted that technician B is correct as amount of energy loss from the traditional drive belt and electric clutch A/C system is too high for hybrid and electric vehicles.

Energy simply means the property that is transferred to a body.

In this case, Technician A says that reversing the direction of refrigerant as with a heat pump system could be done to provide cabin heat. This is incorrect because this is done to remove heat.

Therefore, the correct person is technician B.

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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.

Z≤ -4.852 ft, Maximum efficiency is η≅ 0.88 ≅ 88% is the maximum discharge possible

Solution

Given Data:-

D = 14.62in, N = 2134 rc/min, T=20°C. At T= 20°C ɣ=ρg= 62.35 lb/ft³, vapor pressure. Pv = 49.2 lb/ft².

The efficienies at each flow rate is computal by using formula

η = ρgθH / (550) (bhp)

→ As we can See the maximum efficiency point is at θ = 6ft³/s (close to 6ft³/s)

Maximum efficiency is η≅ 0.88 ≅ 88%

b) Given NPSHR = 16 ft,hg=22ft. Zactual. = 9ft  (below the sea surface)

To avoid cavitation NPSH < Pa - Pv/ρg - Z - hf

Z < Pa - Pv/ρg - hf

Z < 2116 - 49.2/62.35 - 16 - 22 [1 atm = 2116 lb/ft2]

Z≤ -4.852 ft

-> Keeping the pump 9 ft below the surface gives 4.148 ft of marign against cavitation.

Hence it is Sufficient to avoid cavitation.

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Answer:

you said 20 points

Explanation:

The process of die forming that uses a die and a press to form parts from the given processes is called; C: Hydroforming

There are different types of die forming in sheet metal operations. Now, Hydroforming is a specialized type of die forming that uses a high pressure hydraulic fluid to press room temperature working material into a die.

Tailor Welding is a process of making welded blanks made from individual sheets of steel of different thickness, strength and even coating that are joined together by the use of laser welding.

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False. if the level of significance is set at .05 then the experimentwise error will always be .05

The level of significance, typically denoted as alpha (α), represents the threshold for rejecting the null hypothesis in hypothesis testing. If the level of significance is set at 0.05 (or 5%), it means that the researcher is willing to accept a 5% chance of making a Type I error (rejecting the null hypothesis when it is actually true).

However, the experimentwise error rate is not necessarily always equal to the level of significance. The experimentwise error rate refers to the overall probability of making at least one Type I error in a series of hypothesis tests or comparisons within an experiment. It takes into account the possibility of multiple comparisons or tests, which can increase the overall chance of making a Type I error.

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Answer:

a. Compromise the fire rating by one hour.

Explanation:

One hour fire rating is given to materials that can resist the fire exposure. The Acoustical panels controls reverberation and they are used for echo controls. The Fire rating is the passive fire protection which can resist standard fire. The test for fire rating also consider normal functioning of the material.

Answer:

\(\mathbf{\sigma_{max} =638.308 \ MPa}\)

Explanation:

For any large steel plate with an infinite length and width and a center through the crack under tension, the stress intensity factor can be expressed as:

\(K_I = \sigma \sqrt{\pi a}\)

where;

a = 2.5 mm/2

a = 1.25 mm

a = 1.25 × 10⁻³ m

Therefore; the maximum stress in tension capacity can be computed as;

\(\sigma_{max} = \dfrac{K_{I_C}}{\sqrt{\pi a_c}}\)

\(\sigma_{max} = \dfrac{40}{\sqrt{\pi \times 1.25 \times 10^{-3}}}\)

\(\mathbf{\sigma_{max} =638.308 \ MPa}\)

The unknown capacitor in arm CD must have a value of 330 microfarads, and the unknown resistor must have a value of 100 kilo-ohms.

To solve the problem, use the following formula:
Cseries = C1 x C2 / (C1 + C2)

Where C1 is the value of the capacitor in arm AB (47 microfarads) and C2 is the value of the capacitor in arm BC (330 microfarads).

Therefore, Cseries = 330 microfarads.

Also, the total resistance of arms CD is the sum of the resistance of the resistor (R) and the reciprocal of the capacitive reactance of the capacitor (1/Xc).

Using the following formula:
Rtotal = R + 1/Xc

Where Xc = 1/2πfC,

f is the frequency and C is the capacitance.

For this problem,

Xc = 1/2π(50)(330 x 10-6)

=> 100 kilo-ohm.

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If the AC unit fan is not spinning, the first thing to check is the circuit breaker. Take a look at the circuit breaker associated with the AC unit and see if it's been tripped. A circuit breaker may trip from overheating due to outside temperatures and the motor's heat.

Answer:

good question but i cant solve

The given graph consists of vertices V1, U2, O3, T, V4, V5, and V6. We need to determine whether the graph has an Euler trail and an Euler circuit.

An Euler trail is a path that traverses each edge exactly once, while an Euler circuit is a trail that starts and ends at the same vertex.  To determine if the graph has an Euler trail or an Euler circuit, we need to analyze its connectivity and the degrees of its vertices. Since it is mentioned that the graph is connected, we can assume that there is a path between any two vertices. In an undirected graph, the degrees of the vertices indicate the number of edges connected to each vertex. For a graph to have an Euler trail, there should be at most two vertices with an odd degree. If there are zero vertices with an odd degree, the graph will have an Euler circuit. By examining the given graph, we observe that all the vertices, except for T, have even degrees. Since T has an odd degree, the graph does not have an Euler circuit. However, having exactly two vertices with odd degrees (V1 and T) allows the graph to have an Euler trail. In conclusion, the given graph does not have an Euler circuit but has an Euler trail.

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In the context of buildings, structures designed to mitigate the effects of seismic activity or strong winds are commonly referred to as seismic or base isolators.

These isolators act as shock absorbers to reduce the transmission of vibrations and energy from the ground to the building. Seismic or base isolators typically consist of various materials or devices that can absorb and dissipate energy. Some commonly used types of isolators include:

1. Rubber or Neoprene Bearings: These are made of rubber or neoprene pads that provide flexibility and damping to absorb and dissipate energy during seismic events.

2. Sliding Bearings: Sliding bearings allow the building to slide horizontally on a low-friction surface, such as PTFE (polytetrafluoroethylene). This sliding motion helps to reduce the transfer of seismic forces to the building.

3. Fluid Dampers: Fluid dampers utilize the properties of viscous fluid to absorb and dissipate energy. They consist of cylinders filled with a viscous fluid that flows through small orifices, generating resistance against the motion caused by seismic forces.

4. Steel Springs: Steel springs can be used as shock absorbers to reduce the impact of seismic or wind loads. They provide flexibility and absorb energy by deforming under the applied forces.

These seismic or base isolators are strategically placed at the base or foundation of the building to reduce the transfer of vibrations and energy. By incorporating these shock-absorbing mechanisms, buildings can better withstand seismic events and minimize structural damage.

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Answer:

I'm not 100%sure but I think it's the first one

Answer:

1021c.

Explanation:

''.''

Answer: A prototype

Explanation:

The scale model that proves the initial concept is called a domain model.

A copy or depiction of something where all parts have the same dimensions as the original. A scale model is an image or copy of an object that is either larger or smaller than the object being represented's actual size.

A domain model is a type of conceptual model that is used to depict the structural elements and conceptual constraints within a domain of interest.

A domain model will include all of the entities, their attributes, and relationships, as well as the constraints that govern the conceptual integrity of the structural model elements that comprise that problem domain.

Therefore, a domain model is the scale model that proves the initial concept.

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Answer:

right !

Explanation:

Answer:

Aerobic biological treatment process

Explanation:

Aerobic biological treatment process in which micro-organisms, in the presence of oxygen, metabolize organic waste matter in the water, thereby producing more micro-organisms and inorganic waste matter like CO₂, NH₃ and H₂O.