Question 15 In the figure below, if the baud rate is 2400 symbols per second, what is the bit rate (bits per second)? Voltage 000 001 000 011 110 111 Data Transmissions 101 100 000 5 pts

Conductors in Parallel is covered in 310.10(H). Often when working with higher ampacities, we find it to be cumbersome and expensive to keep increasing the size of the wire and conduit. The solution is often to use multiple runs that are connected to a common location on each end. Notice in the 2011 edition of the code that much of this section is highlighted gray, indicating new or revised text. The change here was that the previous code stated that you were permitted to parallel conductors 1/0 AWG and larger; however, it didn’t specifically prohibit you from paralleling smaller conductors, which was the intent and the way it was enforced in all my years of enforcement. However, that’s not what the actual language said, and ambiguous language cancause enforcement issues; therefore, in the 2011 code it was made clear that you are only allowed to parallel conductors 1/0 AWG and larger.

According to Erikson, the major personality attainment of adolescence is identity. The correct option is B).

Erik Erikson was the first to identify identity as a major personality achievement of adolescence and a critical step toward being a productive, content adult.

Creating an identity entails establishing who you are, what you value, and the life paths you pick. There are eight stages of Erickson's theory. Mistrust is the first stage. During this stage, the infant is unsure of their surroundings and looks to their primary caregiver for stability and consistency of care.

Therefore, the correct option is B) identity.

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When it comes to crack growth at a lower applied stress, the situation that would result in such an occurrence is a crack tip with a large radius of curvature. This is the correct option.

Fatigue crack growth refers to the process by which tiny cracks grow and grow in materials subjected to repeated loading. A crack is a slit, a fissure, or a weakness in a substance that extends through it. Cracks are dangerous, particularly in structures, because they have the potential to grow and propagate until they reach a catastrophic level.

A small radius of curvature on a crack tip is less probable to cause crack growth at lower applied stresses. A small radius of curvature has a tendency to magnify the stresses and make them much more serious. A large radius of curvature, on the other hand, makes it easier for stresses to distribute, making it less likely for a crack to grow at lower applied stresses.

Thus, the situation that will result in crack growth at a lower applied stress is a crack tip with a large radius of curvature.

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Answer:

The Coast Guard is the nation's oldest continuous seagoing service with responsibilities including search and rescue (SAR), maritime law enforcement (MLE), aids to navigation (ATON), ice breaking, environmental protection, port security and military readiness.

Explanation:

there

Answer:

Question 1 A 1020 Cold-Drawn steel shaft is to transmit 20 hp while rotating at 1750 rpm. Calculate the transmitted torque in lbs. in. Ignore the effect of friction. Answer with three decimal points. 60.024 Question 2 Based on the maximum-shear-stress theory, determine the minimum diameter in inches for the shaft in Q1 to provide a safety factor of 3. Assume Sy = 57 Kpsi. Answer with three decimal points. 0.728 Question 3 If the shaft in Q2 was made of ASTM 30 cast iron, what would be the factor of safety? Assume Sut = 31 Kpsi, Suc = 109 Kpsi 0 2.1 O 2.0 O 2.5 0 2.4 2.3 O 2.2

Explanation:

hope it helps

i] The density of the pellet is 0.977 g/cm^{3}. ii] The macropore volume in cm^{3}/g is 0.205 cm^{3}/g. iii] The macropore void fraction in the pellet is 25.1%.iv] The micropore void fraction in the pellet is 49.0%. v] The solid fraction of the pellet is 25.9%. vi] The density of the particles is 1.222 g/cm^{3}.

i] To determine the density of the pellet, we can use the formula:

Density = Mass / Volume

Given that the mass of the pellet is 3.15 g and the volume is 3.22cm^{3}, we can calculate the density as follows:

Density = 3.15 g / 3.22 cm^{3}≈ 0.977 \(g/cm^{3\)

ii] The macropore volume in cm3/g can be calculated by dividing the macropore volume of the pellet (0.645 cm3) by the mass of the pellet (3.15 g):

Macropore volume = 0.645 cm^{3} / 3.15 g ≈ 0.205 \(cm^{3} /g\)

iii] The macropore void fraction in the pellet can be calculated using the formula:

Macropore void fraction = Macropore volume / Total volume of the pellet

Total volume of the pellet = Volume - Macropore volume = 3.22 cm^{3}- 0.645 cm^{3} = 2.575 cm^{3}

Macropore void fraction = 0.645 cm^{3} / 2.575 \(cm^{3}\)≈ 0.251 or 25.1%

iv] The micropore void fraction in the pellet can be calculated using the given micropore volume of the particles (0.40 cm^{3} /g) and the mass of the pellet (3.15 g):

Micropore volume in the pellet = Micropore volume/g x Mass

Micropore volume in the pellet = 0.40 \(cm^{3} /g\) x 3.15 g = 1.26 cm3

Micropore void fraction = Micropore volume in the pellet / Total volume of the pellet

Micropore void fraction = 1.26 \(cm^{3}\) / 2.575 \(cm^{3}\) ≈ 0.490 or 49.0%

v] The solid fraction of the pellet can be calculated by subtracting the sum of macropore and micropore void fractions from 1:

Solid fraction = 1 - (Macropore void fraction + Micropore void fraction)

Solid fraction = 1 - (0.251 + 0.490) ≈ 0.259 or 25.9%

vi] The density of the particles can be determined using the mass of the pellet (3.15 g) and the total volume of the particles:

Total volume of the particles = Volume - Macropore volume = 3.22 \(cm^{3}\)- 0.645 \(cm^{3}\) = 2.575\(cm^{3}\)

Density of the particles = Mass / Total volume of the particles

Density of the particles = 3.15 g / 2.575\(cm^{3}\) ≈ 1.222 \(g/cm^{3}\)

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The lag time is the time it takes for the bacteria to start to break down the organic matter in the sewage. The lag time is longer for the untreated sewage sample because it contains more organic matter.

Carbonaceous BOD is the amount of oxygen that is required to break down the organic matter in sewage. The organic matter in sewage is primarily made up of carbon, so carbonaceous BOD is also known as biochemical oxygen demand (BOD5).

Nitrogenous BOD is the amount of oxygen that is required to break down the nitrogenous compounds in sewage. The nitrogenous compounds in sewage are primarily made up of ammonia, so nitrogenous BOD is also known as ammoniacal oxygen demand (AOD).

The treated sewage sample has a lower carbonaceous BOD and a lower nitrogenous BOD than the untreated sewage sample. This is because the treatment process removes some of the organic matter and nitrogenous compounds from the sewage.

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It is subjective to determine the "most happy" and "most sad" song as it varies from person to person. Different songs can evoke different emotions in different people.

However, some songs that are commonly considered to be happy include "Happy" by Pharrell Williams, "Don't Stop Believin'" by Journey, and "I Will Always Love You" by Whitney Houston. Some songs that are commonly considered to be sad include "Everybody Hurts" by R.E.M., "Tears in Heaven" by Eric Clapton, and "My Heart Will Go On" by Celine Dion.

The emotions that a song evokes in a listener are subjective, meaning that they can vary from person to person. For example, a song that one person finds to be happy and uplifting, another person may find to be sad or melancholy. This is because emotions are personal experiences that are shaped by a person's individual life experiences, beliefs, and attitudes.

Therefore, Regarding the songs mentioned, "Happy" by Pharrell Williams, "Don't Stop Believin'" by Journey, and "I Will Always Love You" by Whitney Houston are considered happy songs because they have upbeat tempos, positive lyrics, and a cheerful sound that can make people feel good.

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I am so confused, can you explain it?

Answer:

a

Explanation:

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out /  W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

Answer:

A. $28

Explanation:

We have the load to be 240V

6000W will be converted to kilowatts

6000/1000 = 6kW

Time heater runs for 36 hours

Cost of electricity = $0.13/Kwh

The question needs us to find cost of running this electric heater

Total kilowatts

= 6kw x 36 hours

= 216kwh

Then the total cost = 216 x $0.13

= 28.08

= $28

Answer:

Agricultural engineer

The common characteristic of all material removal mechanisms in non-traditional machining processes is that they use a tool or electrode to remove material from the workpiece.

This can include processes such as laser cutting, electrochemical machining, electrical discharge machining, and waterjet cutting. These processes are often used when conventional machining methods are inadequate or inefficient, and they are able to achieve high levels of precision and accuracy.

The primary mechanism of material removal in Electron Beam Machining (EBM) is through thermal sputtering. This is a process in which incident electrons, accelerated by a high voltage, strike a surface and cause atoms or molecules to be ejected from the surface.

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The time required to accelerate the car from rest to a speed of 100 km/h at full power on a level road is approximately 8.33 seconds.

To determine the time required, we can use the equation of motion:

\[v = u + at\]

where:

- \(v\) is the final velocity (100 km/h),

- \(u\) is the initial velocity (0 km/h),

- \(a\) is the acceleration, and

- \(t\) is the time taken.

To find the acceleration, we can use the equation:

\[P = Fv\]

where:

- \(P\) is the power (75 kW),

- \(F\) is the force exerted on the car, and

- \(v\) is the velocity.

The force can be calculated using the equation:

\[F = ma\]

where:

- \(m\) is the mass of the car (1600 kg), and

- \(a\) is the acceleration.

Substituting the values into the equations, we have:

\[75 \, \text{kW} = (1600 \, \text{kg}) \cdot a \cdot \left(\frac{100 \, \text{km/h}}{3.6}\right)\]

Solving for \(a\), we find \(a \approx 4.63 \, \text{m/s}^2\).

Now, using the equation of motion, we can solve for \(t\):

\[100 \, \text{km/h} = 0 + 4.63 \, \text{m/s}^2 \cdot t\]

Converting the velocity to meters per second, we get:

\[27.78 \, \text{m/s} = 4.63 \, \text{m/s}^2 \cdot t\]

Solving for \(t\), we find \(t \approx 6 \, \text{seconds}\).

However, it's worth noting that the answer might not be entirely realistic. Real-world conditions such as friction, air resistance, and other factors can affect the acceleration and the time required to reach the desired speed. This calculation assumes ideal conditions.

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The SQL query to add a record to the RockTrack table with the following information is given below:

We can insert a row into a table using the INSERT INTO statement. The basic syntax of the INSERT INTO statement is as follows:

INSERT INTO table_name (column1, column2, column3,...)

VALUES (value1, value2, value3,...);

The SQL query to add a record to the RockTrack table with the following information is:

INSERT INTO RockTrack (TrackId, Name, AlbumId, MediaTypeId, GenreId, Composer, Milliseconds, Bytes, UnitPrice)

VALUES (3700, 'MySQL Melodies', 249, 1, 1, 'John', 123456, 3333333, .99);

Thus, the given row is inserted into the RockTrack table with the given data.

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Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

so volume=0.02375≈0.02m³

Since human activity will have had the least impact on the water at the river's highest point, it should be the cleanest. The two cities are upstream.

What does the game term "farm" mean?

In order to obtain experience, points, or some kind of in-game cash, a player, or someone hired by a player, engages in repetitive behaviors. This practice is known as farming. Staying in a game region with a spawn point that continually produces new objects or adversaries is the norm for farming.

What is a "farm beat"?

FarmBeats continuously monitors moisture and soil data. In order to capture soil temperature and moisture levels and follow them using cloud-based computing models, Microsoft developed the new agriculture research project FarmBeats. It uses solar-powered white space-based Internet connectivity.

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Answer:

Simple In one way. That is if he left it blank

The proper term for associating a Group Policy to a set of AD DS objects is a. Linking. The correct option is a. Linking.

In Active Directory Domain Services (AD DS), linking refers to the process of associating a Group Policy Object (GPO) to a set of AD DS objects such as organizational units, domains, or sites. This allows the GPO settings to be applied to the associated objects, enabling administrators to manage and enforce specific configurations for users and computers within the network.

Linking is the correct term for associating a Group Policy to a set of AD DS objects, as it enables the application of GPO settings to the specified objects in the network. The correct option is a. Linking.

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A chip without PMOS transistor is easier because pMOS transistor would not avaiable to offer a conductive path for a node all the way to the ground voltage.

Microprocessors are constructed out of transistors. In particular, they're built out of metal-oxide-semiconductor (MOS) transistors. There are kinds of MOS transistors — positive-MOS (pMOS) and negative-MOS (nMOS). Every pMOS and nMOS comes prepared with 3 primary components — the gate, the source, and the drain. NMOS and pMOS are the primary types of MOSFET. The primary difference between NMOS and pMOS is that, in NMOS, the source and the drain terminals are made from n-kind semiconductors meanwhile, in pMOS, the source and the drain are made from p-kind semiconductors.

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answer

d just too the test

To estimate the hydraulic conductivity of the unconfined aquifer, we can use the equation:

Q = 2πKhdwhere Q is the flow rate, h is the rise in water level, d is the diameter of the well, and K is the hydraulic conductivitySubstituting the given values, we have:

100 gpm = 2πK(15 ft)(2 ft)

K ≈ 1.32 ft/dayb) To calculate the height the water table will be mounded above the original water level, we can use the Theis equation:

S = Q/4πT * W(u)

where S is the drawdown, Q is the flow rate, T is the transmissivity, W(u) is the well function, and u is a dimensionless parameter that depends on the aquifer properties and pumping ratI this case, we want to find the height of the groundwater mound, which is equal to the drawdown. We can assume that the groundwater mound is symmetric and therefore the drawdown at the center of the mound is half of the total drawdown.

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False, Good housekeeping eliminates accident and fire hazards. It also maintains safe, healthy work conditions; saves time, money, materials, space, and effort; improves productivity and quality; boosts morale; and reflects an image of a well-run, successful organization.

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The continuity equation allows finding the results for the speed and flow of the outlet pipe are:

Fluid mechanics studies the movement and fluid dynamics, it is described  by two fundamental relationships:

          \(P_1 + \frac{1}{2} \rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g y_2\)  

Where P is the pressure, v the velocity, y the height, the subscripts are two selected points of interest in the pipe.

            \(A_1v_1 = A_2 v_2\)

Where A₁ and A₂ is the area of ​​the inlet and outlet pipes and v₁ and v₂ are the velocities of the fluid at the inlet and outlet

They indicate the inlet and outlet diameters of the pipe are d₁ = 0.02 m and d₂ = 0.1 m, the inlet velocity v₁ = 10 m / s, with the continuity equation let's find the outlet velocity

         \(v_2 = ( \frac{A_1}{A_2} )^2 \ v_1\)  

the cross section of the tavern is

        A = \(\pi r^2 = \pi \ d^2/4\)  

Where r is the radius and d is the diameter

Let's substitutes

        \(v_2 = (\frac{d_1}{d_2} )^2 \ v_1\)  

        \(v_2 = (\frac{0.02}{0.1} )^2 \ 10\)  

        v₂ = 0.4 m / s

The flow  (Q) is defined as the amount of liquid that passes through the section of the pipe per unit of time, we can see that this is the same conservation of mass, therefore

        Q = A v

        Q = \(\pi \frac{d_2^2}{4} \ v_2\)π d2 ^ 2 v2

        Q = \(\pi \ \frac{0.1^2}{4} \ 0.4\)

        Q = 3.14 10⁻³ m ^ 3 / s

As the transported liquid is water

         1 liter water = 1 kg

Consequently

         1 m³ = 1000 liter = 1000 kg of water

Let's reduce the flow

         Q = 3.14 10⁻³ m³/s ( \(\frac{10^3 kg}{1 m^3}\) )

         Q = 3.14 kg / s

In conclusion using the continuity equation we can find the results for the speed and flow of the outlet pipe are:

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Answer:

I think it is D

Explanation:

I did a couple mins of research on this topic but there is no clear line that separates a physical and chemical reaction, so do with that as you will. i hope I somewhat helped.

Given sentence: ''To change a logic gate to its alternate representation''  is False. Because to replace the gate with its opposite type (e.g. replace an AND gate with an OR gate, or a NAND gate with a NOR gate).

Logic is the study of correct reasoning. It includes both formal and informal logic. Formal logic is the science of deductively valid inferences

To change a logic gate to its alternate representation, a simple two-step process is followed:

Invert the output of the gate (change 1 to 0 or 0 to 1).

Replace the gate with its opposite type (e.g. replace an AND gate with an OR gate, or a NAND gate with a NOR gate).

There is no need for a third step.

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Answer:

t5g5gtttttttttttttttttttttttttttttttttttttttttttt

Explanation:gt555555555555555555555555555555555555555555555555

Answer:

dawbkjbjwwjhjfbfjewfaekfhawkjndwkja

Explanation: dum*as*

The scale of the blueprint tells us the boundary of the drawing to real space. The correct option is a.

It is a photographic print in white on a bright blue background or blue on a white background, particularly useful for copying maps, mechanical drawings, and architect's designs

The plan of the construction is a project prediction that is slightly auto cad. The scale factor provides information on the dimensions and measurements of things represented by smaller copies in maps, blueprints, scale drawings, and scale models. The scale tells us how much actual length some length on the scale graphic represents.

Therefore, the correct option is a, Boundary.

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