how do you test the pressing comb before beginning a service

Answer:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}\)

Explanation:

Given

\(0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,\)

\(0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,\)

\(0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,\)

\(0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.\)

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

\(0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.\)

\(0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\)

\(0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.\)

\(0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.\)

\(0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.\)

\(0.78.\)

The 0.3's is will be plotted as thus:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}\)

The 0.4's is as follows:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}\)

The 0.5's is as follows:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}\)

The 0.6's is as thus:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}\)

Lastly, the 0.7's is as thus:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}\)

The combined steam and leaf plot is:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}\)

Answer:

The required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

Explanation:

We are given current density of 459 A/cm²  and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.

Recall that current density is given by

j = I/A

where I is the current flowing through the wire and A is the area of the wire

A = πr²

but r = d/2  so

A = π(d/2)²

A = πd²/4

so the equation of current density becomes

j = I/πd²/4

j = 4I/πd²

Re-arrange the equation for d

d² = 4I/jπ  

d = √4I/jπ

d = √(4*0.53)/(459π)

d = 0.0383 cm

Therefore, the required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

Answer:

diameter is 1 cm

Explanation:

diameter is 1 cm

Respiratory distress syndrome (RDS) of the newborn is a common breathing disorder in premature infants caused by the lack of surfactant, a substance that helps keep the lungs open. The primary problem resulting from RDS is atelectasis. So the coreect answer is c)atelectasis

Atelectasis is the collapse or incomplete inflation of the lung tissue. Without surfactant, the lungs cannot expand properly, making it difficult for the infant to breathe and causing a range of symptoms such as rapid breathing, grunting, and flaring of the nostrils. The lack of oxygen in the body can also cause pulmonary edema, or fluid buildup in the lungs, and bronchiolar plugging, where mucus blocks the small airways. These complications can be life-threatening for the newborn. In conclusion, the primary problem resulting from RDS in the newborn is atelectasis, which is caused by the lack of surfactant in the lungs.

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ShopKey Pro presents repair information in a customized email template as well as auto repair software format or layout.

In regards to auto repair information tools, the solution that is known to be made available is the use of ShopKey® Pro with 1Search.

This is know  to be a kind of an exclusive auto repair software that helps to streamlines the search acts to bring closer a special  combination of OEM data as well as experience-based Real Fixes.

Therefore, based on the above, one can say that ShopKey Pro presents repair information in a customized email template as well as auto repair software format or layout.

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Answer:

6.564 is the answers i learned that in the american school i live there

Answer:

so??

Explanation:

what's the question??

Answer:

Your computer must have different ports for speakers and microphones. Get a headset that has two cables, one for microphone and the other for a speaker

Explanation:

The old systems were having two ports whereas the modern headsets comes with only one lead. So, if you want to hear and speak from the headset, you must get an old one for it.

Answer:

14.5° ; THD % = 3.873 × 100 = 387.3%.

Explanation:

Okay, in this question we are given the following parameters or data or information which is going to assist us in solving the question efficiently and they are;

(1). "500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor".

(2). Consideration of up to 12th harmonic.

So, let us delve right into the solution to the question above;

Step one: Calculate the Irms and Irms(12th) by using the formula for the equation below;

Irms = reactive power /Vrms = 500/230 = 2.174 A.

Irms(12th) = 1.5 × 10^3/ 12 × 230 = 0.543 A.

Step two: Calculate the THD.

Before the Calculation of the THD, there is the need to determine the value for the dissociation factor, h.

h = Irms(12th)/Irms = 0.543/ 2.174 = 0.25.

Thus, THD = [1/ (h)^2 - 1 ] ^1/2. = 3.873.

THD % = 3.873 × 100 = 387.3%.

Step four: angle AC - Ac converter

theta = sin^-1 (1.5 × 10^3/ 12 × 500) = 14.5°.

Answer:

LOS = A

Explanation:

Given all the parameters the level of service as seen from the attached graph

is LOS =  A

To determine the LOS from the attached graph

calculate the trial value of Vp

Vp = V / PHF

     = (100 + 150) / 0.95  =  263 pc/h

since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1

next we will calculate the flow rate

flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]

             Fhr  = 1 / 1.035 = 0.966 ≈ 1

next calculate the real value of Vp

Vp = V / ( PHF * N * Fhr * Fp )

     = ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )

Vp ≈ 126 pc/h/In

Next calculate the density

D = Vp /  S  =  126 / ( 45 * 1.61 )  = 1.74 pc/km/In

To determine the various parameters for the given Laval nozzle and the effects of heat addition, we need to consider the equations and principles of compressible flow.

a) The design pressure of the nozzle:

The design pressure of the nozzle is the reservoir pressure, which is given as 2 MPa.

b) The critical back pressure:

The critical back pressure (Pc) is the pressure at which the flow reaches sonic velocity (Mach number = 1) at the narrowest point of the nozzle. It can be calculated using the isentropic flow equations.

Pc = P1 * (2 / (γ + 1))^(γ / (γ - 1))

Here, P1 is the reservoir pressure (2 MPa), and γ is the specific heat ratio of air (assumed to be 1.4 for this calculation).

Pc = 2 MPa * (2 / (1.4 + 1))^(1.4 / (1.4 - 1))

≈ 1.58 MPa

Therefore, the critical back pressure is approximately 1.58 MPa.

c) The exit Mach number:

The exit Mach number (Me) can be calculated using the isentropic flow equations and the ratio of specific heats (γ).

Me = sqrt(((Pc / P2)^((γ - 1) / γ) - 1) * (2 / (γ - 1)))

Here, P2 is the back pressure (1.2 MPa) for this calculation.

Me = sqrt((((1.58 MPa) / (1.2 MPa))^((1.4 - 1) / 1.4) - 1) * (2 / (1.4 - 1)))

≈ 1.36

Therefore, the exit Mach number is approximately 1.36.

d) The new exit Mach number after heat addition:

To calculate the new exit Mach number after heat addition, we need to consider the effects of heat transfer. The added heat increases the energy of the flow, which affects the Mach number.

To determine the new exit Mach number, we need additional information such as the specific heat capacity at constant pressure (Cp) for the air and the mass flow rate of the flow. Without these values, we cannot calculate the new exit Mach number precisely.

Based on the given information and using the principles of compressible flow, we calculated the design pressure of the nozzle as 2 MPa, the critical back pressure as approximately 1.58 MPa, and the exit Mach number as approximately 1.36. However, to determine the new exit Mach number after heat addition, additional information such as the specific heat capacity at constant pressure and the mass flow rate is required.

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Phase Field Dislocation Dynamics (PFDD) is a computational modeling technique used to simulate the behavior of dislocations in crystalline materials.

It combines the phase field method, which describes the evolution of a scalar order parameter, with the dislocation dynamics theory. PFDD allows for the study of dislocation interactions, evolution, and their effect on the material's mechanical behavior.

In the case of modeling non-Schmid behavior in body-centered cubic (BCC) metals, PFDD can be informed by atomistic simulations to incorporate more accurate and realistic dislocation properties.

Atomistic simulations, such as molecular dynamics (MD) or density function theory (DFT), provide detailed information about the atomic-scale behavior of dislocations, including dislocation core structures and their interactions with other defects.

Here is a general outline of how atomistic simulations can inform PFDD modeling for non-Schmid behavior in BCC metals:

1. Atomistic simulations: Perform atomistic simulations, such as MD or DFT, to investigate the dislocation properties and behaviors specific to the BCC metal of interest. This includes studying dislocation core structures, dislocation mobility, and dislocation interactions with other defects.

2. Extracting parameters: Extract relevant parameters from the atomistic simulations that describe the non-Schmid behavior. These parameters can include dislocation line tension, line mobility, and the Peierls barrier, which represents the energy barrier for dislocation motion.

3. Parameterization: Incorporate the extracted parameters into the PFDD model. The non-Schmid behavior can be incorporated by modifying the governing equations or adding additional terms to account for the specific dislocation properties observed in the atomistic simulations.

4. PFDD simulations: Conduct PFDD simulations using the modified model. The PFDD simulations will provide a larger-scale description of dislocation dynamics, interactions, and their effects on material behavior, accounting for the non-Schmid behavior observed in the atomistic simulations.

5. Validation and analysis: Compare the results of the PFDD simulations with experimental data or other atomistic simulations to validate the model's accuracy. Analyze the simulation results to gain insights into the dislocation behavior, material deformation mechanisms, and the effect of non-Schmid behavior on the overall material response.

By combining the strengths of atomistic simulations and PFDD, this approach allows for a multiscale understanding of dislocation behavior, capturing both the atomistic details and the larger-scale mechanical response of BCC metals. It provides a powerful tool for studying dislocation-mediated plasticity and material deformation in complex systems.

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However, assuming it pertains to the National Electrical Code (NEC), four possible combinations of service equipment that meet the requirements of 110.9 and 110.10 of the NEC could.

Based on the references to "110.9" and "110.10 of the Code," it appears that this question pertains to the National Electrical Code (NEC). Section 110.9 of the NEC states that "equipment intended to interrupt current at fault levels shall have an interrupting rating sufficient for the nominal circuit voltage and the current that is available at the line terminals of the equipment." Section 110.10 requires that "all electrical equipment and materials shall be listed or labeled."

Given these requirements, four possible combinations of service equipment that meet the NEC's requirements could include:

It's important to note that these are just a few possible combinations that could meet the NEC's requirements, and the specific combination required will depend on the specific installation and its requirements. A qualified electrical professional should always be consulted to ensure compliance with all applicable codes and standards.

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Chances of death or injury at low speeds when driver is unbelted are the same as those who are belted drivers. (False)

Beginning with the "high speed road" state of the "road type" variable, it can be seen that the higher risk of fatality occurs when a driver is distracted, is not wearing a seat belt, and is operating a vehicle at a high speed, such as on a motorway or a comparable road. This would be one of the most dangerous scenarios in a traffic accident because 48.52% is the highest value in this study.

However, when a driver is not wearing a seatbelt, is distracted, and is travelling at a low speed, there is only a 25% chance that they will cause a fatality or serious injury. This suggests that a lower driving speed will result in a slower impact speed, which can lower the risk of suffering a serious injury.

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Hai

Your answer will be A.

If you lower the Air Pressure your Object will Float Down ward. The Air Pressure allows it to Fly.

The pressure field created by faster air movement over an airfoil is; A:  higher

When the air hits the front of the wing, the air will flow in a steeper curve upward, than the bottom wing flow which will lead to the creation of a vacuum on top of the wing that pulls more air towards the top of the wing.

Finally, this air above does the same thing but it will move faster as a result of the vacuum pulling it in, and as such the vacuum now lifts the wing. Thus, Faster air movement over an airfoil creates a higher pressure field.

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The minimum required kW size of the synchronous motor to achieve a combined power factor of 0.8 lagging is approximately 1.068 kW.

To find the minimum required kW size of the synchronous motor, we need to calculate the reactive power (Q) of the combined load and then determine the additional real power (Psyn) required to achieve the desired power factor.

Real power consumed by the inductive load (Pind) = 10 kW

Power factor of the inductive load (pf_ind) = 0.75 lagging

Power factor desired for the combined load (pf_comb) = 0.8 lagging

First, we calculate the reactive power (Q) of the inductive load:

Q = Pind * tan(acos(pf_ind))

Q = 10 kW * tan(acos(0.75))

Q = 6.708 kVAR (kilo Volt-Amp Reactive)

Next, we calculate the total apparent power (S_comb) of the combined load:

S_comb = Pind / pf_comb

S_comb = 10 kW / 0.8

S_comb = 12.5 kVA (kilo Volt-Amp)

Now, we calculate the reactive power (Q_comb) required for the combined load to have a power factor of 0.8 lagging:

Q_comb = S_comb * tan(acos(pf_comb))

Q_comb = 12.5 kVA * tan(acos(0.8))

Q_comb = 8.664 kVAR

The synchronous motor needs to supply the additional reactive power (Q_diff) to achieve the desired power factor:

Q_diff = Q_comb - Q

Q_diff = 8.664 kVAR - 6.708 kVAR

Q_diff = 1.956 kVAR

Finally, we calculate the additional real power (Psyn) required for the synchronous motor:

Psyn = sqrt((S_comb)² - (Q_diff)²)

Psyn = sqrt((12.5 kVA)² - (1.956 kVAR)²)

Psyn = 1.068 kW (approximately)

Therefore, the minimum required kW size of the synchronous motor is approximately 1.068 kW.

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After an analysis of the costs, the demand, and the types of processes, the best cookie production option depends on the demand. Answer: If the demand is from 1 dozen to 600.06, pre-bake is selected, if it is greater, easy-bake. An image of the calculation procedures is attached

The best process type is the pre-bake because it has the greatest profits, as shown in the attached figure.

For the pre-bake type process, the semester benefit is $817.50.

The semester demand would have to be 600.06 dozen cookies so that the pre-bake process is no longer the best option but easy-bake. Below is the solution algorithm.

if __name__ == '__main__':

fixed_cost_prebake = 150

fixed_cost_easybake = 350

variable_cost_prebake = 2.34

variable_cost_easybake = 2.01

sales_price = 4.49

profit_prebake = 0

profit_easybake = 0

sales = 0

while True:

 sales = sales+0.01

 profit_prebake = fixed_cost_prebake+(variable_cost_prebake*sales)

 profit_easybake = fixed_cost_easybake+(variable_cost_easybake*sales)

 if profit_prebake>=profit_easybake: break

sales = int(sales*100.0)/100.0

print("Break Even Point")

print("The demand would have to be: ",sales," so that pre-bake is no longer the best option but easy-bake")

When the demand is greater than 600.06 dozen, the best option is easy-bake process.

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The task is threat mitigation.

A task is an individual or collaborative activity which is designed to achieve a specific outcome. It usually involves a set of actions, processes, or operations which are performed within a specified period of time to reach a pre-defined goal. Tasking is an important part of many activities and projects, and it is necessary for efficient and productive workflows. It can involve anything from a simple task such as writing an email, to a complex task such as developing a new product or managing a large project. Tasking requires planning, coordination and organization in order to be successful, and is essential for ensuring that the desired outcomes are achieved in a timely and cost-effective manner.

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If The terminal voltage of a 2-H inductor is 10(1 – t)V then the current voltage is 42A and energy stored in it is 1764J.

We know that the voltage across an inductor is given by:

v = L(di/dt)

We can rearrange this equation to solve for the current:

di/dt = v/L

Integrating both sides with respect to time, we get:

i(t) = (1/L) * ∫[0,t] v dt + i(0)

Substituting v = 10(1 – t), L = 2H, and i(0) = 2A, we get:

i(t) = (1/2) * ∫[0,t] 10(1 – t) dt + 2

i(t) = -5t² + 15t + 2

At t = 4s, the current flowing through the inductor is:

i(4) = -5(4)² + 15(4) + 2 = 42A

The energy stored in an inductor is given by:

E = (1/2) * L * i²

Substituting L = 2H and i(4) = 42A, we get:

E(4) = (1/2) * 2 * (42)² = 1764J

Therefore, the current flowing through the inductor at t=4s is 42A and the energy stored in it at t=4s is 1764J.

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A subcritical refrigeration cycle is a type of refrigeration cycle that operates below the critical point of the refrigerant. The critical point is the temperature and pressure at which the refrigerant transitions between the liquid and gas phases without any distinction between them.

In a subcritical cycle, the refrigerant remains in the liquid phase during the entire cycle.

The four processes involved in a subcritical refrigeration cycle are:

Compression: The refrigerant enters the compressor as a low-pressure vapor and is compressed to a higher pressure and temperature.

Condensation: The compressed refrigerant flows into the condenser, where it releases heat to the surroundings and changes from a high-pressure vapor to a high-pressure liquid.

Expansion: The high-pressure liquid refrigerant enters the expansion valve or throttle valve, where its pressure is reduced, causing it to partially vaporize and cool.

Evaporation: The partially vaporized refrigerant flows into the evaporator, where it absorbs heat from the surrounding environment, completing the cooling process. The refrigerant then returns to the compressor to start the cycle again.

In a subcritical refrigeration cycle, the cooling effect is achieved through the evaporation of the refrigerant in the evaporator, while the condenser rejects heat to the surroundings.

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The device that causes pressure and temperature drop in the refrigeration cycle is the expansion valve.The refrigeration cycle is a cyclic process used to move heat from one place to another. It can be used for air conditioning or refrigeration purposes.

The refrigeration cycle consists of four major components, including compressor, condenser, expansion valve, and evaporator.The compressor is the first component of the refrigeration cycle that receives the refrigerant in a low-pressure state and compresses it to a high-pressure state. The condenser is the second component of the refrigeration cycle that receives the high-pressure refrigerant from the compressor and removes heat from it to change the refrigerant from a high-pressure vapor to a high-pressure liquid.

The third component is the expansion valve, which reduces the pressure and temperature of the refrigerant. The fourth component is the evaporator, which receives the refrigerant from the expansion valve and removes heat from the surrounding environment to change the refrigerant from a low-pressure liquid to a low-pressure vapor.The function of the expansion valve is to cause pressure and temperature drop in the refrigerant.

The expansion valve receives the high-pressure liquid refrigerant from the condenser and releases it through a small orifice. As a result, the pressure and temperature of the refrigerant decrease. The low-pressure liquid refrigerant then enters the evaporator to absorb heat from the surrounding environment and evaporates into a low-pressure vapor. The cycle is then repeated.

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Answer: true

Explanation:

It's crucial to keep in mind that engineers don't always follow these phases of the engineering design process sequentially. To update the design, they frequently go back to a prior phase in the process.

The Engineering Design Process: The First Seven Steps

Determine first. Students write down their ideas regarding the issue in this stage.

2. Request. Asking questions about the issue is the next phase in the engineering design process.

Step 3: Visualize.

Plan as you go through Step 4.

Prototype is step five.

Step 6: Test.

Step 7: Develop.

Students write down their ideas regarding the issue in this stage. Some schools compel students to bring their lunches to class with them throughout the day. Eric cites this as an actual example of defining a problem. What issues does this modification cause? One explanation for this could be that students are required to leave their meals on the floor near to where they sit on the ground.

Problem-finding is the first step in problem-solving.

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Answer:

The figure below shows the free body diagram of the beam.

Free Body Diagram

Write the compatibility equation,

Δ

A

B

+

Δ

B

C

+

Δ

C

D

=

0

⋯

⋯

(

I

)

Here,

Δ

A

B

is the deflection in AB rod,

Δ

B

C

is the deflection in BC rod and

Δ

C

D

is the deflection in the rod CD.

The deflection

Δ

A

B

is,

Δ

A

B

=

4

(

R

A

)

(

a

)

π

d

2

a

E

a

Here,

R

A

is the reaction at point A.

The deflection ${\Delta _{BC}$ is,

Δ

A

B

=

4

(

R

A

−

F

B

)

(

b

)

π

d

2

a

E

a

The deflection ${\Delta _{CD}$ is,

Δ

A

B

=

4

(

R

A

−

F

B

−

F

C

)

(

b

)

π

d

2

s

E

s

Substitute all the values in the equation (I).

4

(

R

A

)

(

8

i

n

)

π

(

9

8

i

n

)

2

(

10.4

M

s

i

)

+

4

(

R

A

−

18

k

i

p

s

)

(

10

i

n

)

π

(

9

8

i

n

)

2

(

10.4

M

s

i

)

+

4

(

R

A

−

18

k

i

p

s

−

14

k

i

p

s

)

(

10

i

n

)

π

(

13

8

i

n

)

2

29

M

s

i

=

0

(

R

A

)

(

0.8

)

(

9

8

i

n

)

2

+

(

R

A

−

18

k

i

p

s

)

(

9

8

i

n

)

2

=

−

(

R

A

−

18

k

i

p

s

−

14

k

i

p

s

)

(

13

8

i

n

)

2

29

10.4

R

A

=

11.917

k

i

p

s

Equate the horizontal forces,

R

A

+

R

B

=

F

B

+

F

C

Substitute all the values in the above equation.

11.917

k

i

p

s

+

R

B

=

18

k

i

p

s

+

14

k

i

p

s

R

B

=

20.083

k

i

p

s

Thus the reaction at point A is

11.917

k

i

p

s

and reaction at B is

20.083

k

i

p

s

.

(b)

The deflection at point C is,

Δ

C

=

4

(

R

B

)

(

b

)

π

d

2

s

E

s

Substitute all the values in the above equation.

Δ

C

=

4

(

20.083

k

i

p

s

(

1000

l

b

1

k

i

p

s

)

)

(

10

i

n

)

π

(

13

8

i

n

)

2

29

M

s

i

(

10

6

l

b

/

i

n

2

1

M

s

i

)

=

3.33

×

10

−

3

i

n

Answer:

Fifty (50) percent. [50%]

Explanation:

Water heater is a home appliance that comprises of an electric or gas heating unit as well as a water-tank where water is heated and stored for use.

When using an alternative method of sizing with two vent connectors for draft hood-equipped water heaters, the effective area of the common vent connector or vent manifold and all junction fittings shall not be less than the area of the larger vent connector plus fifty (50) percent of the areas of smaller flue collar outlets.

A water heater is primarily vented with an approved and standardized plastic or metallic pipe such as flue or chimney, which allows gas to flow out of the water heater into the surrounding environment.

For a draft hood-equipped water heater, both the water heater and the barometric draft regulators must be installed in the same room. Also, the technician should ensure that the vent is through a concealed space such as conduit and should be labeled as Type L or Type B.

The minimum capacity of a water heater should be calculated based on the number of bathrooms, bedrooms and its first hour rating.

The Arduino Software (IDE) is a program created to assist in the development of Arduino boards and compatible microcontrollers. It is an open-source platform that includes a C++ compiler and a cross-platform IDE that runs on Windows, Mac, or Linux.

The software is used to write and upload code to the Arduino board. It includes a code editor with syntax highlighting, a serial monitor for debugging, and a library manager to easily add pre-written code libraries. The software is user-friendly, and it is simple for beginners to learn and use.

The Arduino Software (IDE) has a simple interface that is easy to navigate and understand. Users can start by selecting the board type and the port to which it is connected. After this, they can open a new sketch and start writing code. The code editor has several features to help write and debug code, such as auto-completion and error highlighting.

In addition to writing code, the IDE is also used to upload the code to the Arduino board. The software takes care of the compilation, linking, and uploading of the code, making it easy for users to test their projects.

The Arduino Software (IDE) also includes a serial monitor that allows users to view the output of the code as it runs on the board. This feature is useful for debugging and testing code. The software also has a library manager that makes it easy to add pre-written code libraries to the project.

The Arduino Software (IDE) is an essential tool for anyone working with Arduino boards and compatible microcontrollers. It is user-friendly and easy to learn, making it an excellent choice for beginners. The software includes several features to help write and debug code, making it easy to test and refine projects. Overall, the Arduino Software (IDE) is an essential tool for anyone looking to create exciting projects with Arduino boards and microcontrollers.

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Mel's decision is that she will be able to hire some of the most productive salespeople who work for the other two dealerships.

Customers are persuaded to purchase goods or services by salespeople. From the first point of contact all the way through to the point of sale, you direct the customer's experience by using one-on-one communication to match your product with their requirements.

To increase sales in the short and long terms, you employ a variety of strategies. Salespeople can effectively communicate with customers, are open to their needs, and can convince them of the worth of their goods.

The right tool, service, product, or experience can be connected to customers by salespeople. Sales can be closed in a variety of ways by people with a variety of personalities. Regardless of your sales personality, it's useful to know the universal skills for success in sales.

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Answer:

C - Prototype model

Explanation:

I believe it's C but Hope this Helps!

Explanation Operations management is an area of management concerned with designing and controlling the process of production and redesigning business operations in the production of goods or services.[1] It involves the responsibility of ensuring that business operations are efficient in terms of using as few resources as needed and effective in meeting customer requirements.

Answer:

Mass, in physics, quantitative measure of inertia, a fundamental property of all matter.

Explanation:

Mass is the matter that makes up objects