For a magnetic course of 180 degrees through 359 degrees (westerly heading) what altitudes apply?

The transfer function is a mathematical representation of the relationship between the input and the output of a system. The steady-state error, or ess, is the difference between the desired output and the actual output when the system reaches a steady state. The natural frequency is the frequency of the system's response without any external forces.

Transfer Function: Transfer Function is used in signal processing, control engineering, and other disciplines that deal with systems or signals. The ratio of output to input in Laplace transform is known as the transfer function.

Steady-State Error: The error that happens when the system is at a stable state is referred to as a steady-state error. The difference between the desired and actual response is known as steady-state error. A system's ability to track a specific input as time progresses is characterized by this kind of error. If the input signal is a unit step, then the steady-state error is referred to as the static error coefficient. The coefficient of the steady-state error is frequently used to classify systems in control engineering.

Natural Frequency: Natural frequency is a term used in physics to describe how quickly an object vibrates when it is set in motion. The number of oscillations made by a system in a given time period without any external force acting on it is referred to as its natural frequency. A natural frequency is a measure of a system's stiffness and mass. In a control system, it is the frequency at which the system oscillates in the absence of any input.

A natural frequency is also known as an undamped natural frequency or a resonance frequency, and it is represented by the symbol \(\omega_n\).You can assume the following in the problem. If you have any specific values, kindly provide them.

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Answer:

SECTION LEARNING OBJECTIVES

By the end of this section, you will be able to do the following:

Distinguish between static friction and kinetic friction

Solve problems involving inclined planes

Section Key Terms

kinetic friction static friction

Static Friction and Kinetic Friction

Recall from the previous chapter that friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice.

There are different types of friction—kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects.

Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going.

Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them.

The current in the incandescent bulb is approximately 0.625 Amperes (A).

An incandescent light bulb, incandescent lamp or incandescent light globe is an electric light with a wire filament heated until it glows. The filament is enclosed in a glass bulb with a vacuum or inert gas to protect the filament from oxidation.

The amount of current in an incandescent bulb rated at 75W and 120V can be calculated using Ohm's Law, which states that current (I) is equal to power (P) divided by voltage (V).

Using this formula, the current can be calculated as:

I = P / V

I = 75W / 120V

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Both insulators result in the same heat loss, which is 812.3 W. Therefore, in terms of minimizing heat loss, it doesn't matter which insulator is covered outside.

To calculate the heat loss, we can use the formula for overall heat transfer:

Q = U * A * ΔT

Where:

- Q is the heat loss (W)

- U is the overall heat transfer coefficient (W/m².K)

- A is the surface area of the pipe (m²)

- ΔT is the temperature difference between the steam and the surrounding air (K)

First, let's calculate the surface area of the pipe. The circumference of the pipe can be calculated as:

C = π * diameter = 3.1416 * (3 in. * 0.0254 m/in.) = 0.2384 m

The surface area can then be calculated as:

A = 2 * π * r * L

- r is the radius of the pipe (m), which is half the diameter

- L is the length of the pipe (m)

Assuming a pipe length of 1 meter:

r = 0.5 * (3 in. * 0.0254 m/in.) = 0.0381 m

A = 2 * π * 0.0381 * 1 = 0.2393 m²

Now, we can calculate the overall heat transfer coefficient (U) using the convective heat transfer coefficient (h) and the thermal conductivities of the insulators:

U = 1 / (1 / h + δ1 / k1 + δ2 / k2)

- δ1 and δ2 are the thicknesses of insulator A and insulator B, respectively (m)

- k1 and k2 are the thermal conductivities of insulator A and insulator B, respectively (W/m.K)

Assuming δ1 = δ2 = 0.025 m (2.5 cm):

U = 1 / (1 / 15 + 0.025 / 0.04 + 0.025 / 0.20) = 4.245 W/m².K

Now, we can calculate the temperature difference (ΔT):

ΔT = Ts - Ta = 100 - 20 = 80 K

Finally, we can calculate the heat loss (Q) for both insulators:

Q_A = U * A * ΔT = 4.245 * 0.2393 * 80 = 812.3 W

Q_B = U * A * ΔT = 4.245 * 0.2393 * 80 = 812.3 W

Both insulators result in the same heat loss, which is 812.3 W. Therefore, in terms of minimizing heat loss, it doesn't matter which insulator is covered outside.

However, it's worth noting that insulator A has a lower thermal conductivity, which means it is a better insulator and can potentially provide better thermal performance in other aspects such as temperature gradients across the insulation.

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Both insulators result in the same heat loss, which is 812.3 W. Therefore, in terms of minimizing heat loss, it doesn't matter which insulator is covered outside.

To calculate the heat loss, we can use the formula for overall heat transfer:

Q = U * A * ΔT

- Q is the heat loss (W)

- U is the overall heat transfer coefficient (W/m².K)

- A is the surface area of the pipe (m²)

- ΔT is the temperature difference between the steam and the surrounding air (K)

First, let's calculate the surface area of the pipe. The circumference of the pipe can be calculated as:

C = π * diameter = 3.1416 * (3 in. * 0.0254 m/in.) = 0.2384 m

The surface area can then be calculated as:

A = 2 * π * r * L

- r is the radius of the pipe (m), which is half the diameter

- L is the length of the pipe (m)

Assuming a pipe length of 1 meter:

r = 0.5 * (3 in. * 0.0254 m/in.) = 0.0381 m

A = 2 * π * 0.0381 * 1 = 0.2393 m²

Now, we can calculate the overall heat transfer coefficient (U) using the convective heat transfer coefficient (h) and the thermal conductivities of the insulators:

U = 1 / (1 / h + δ1 / k1 + δ2 / k2)

- δ1 and δ2 are the thicknesses of insulator A and insulator B, respectively (m)

- k1 and k2 are the thermal conductivities of insulator A and insulator B, respectively (W/m.K)

Assuming δ1 = δ2 = 0.025 m (2.5 cm):

U = 1 / (1 / 15 + 0.025 / 0.04 + 0.025 / 0.20) = 4.245 W/m².K

Now, we can calculate the temperature difference (ΔT):

ΔT = Ts - Ta = 100 - 20 = 80 K

Finally, we can calculate the heat loss (Q) for both insulators:

Q_A = U * A * ΔT = 4.245 * 0.2393 * 80 = 812.3 W

Q_B = U * A * ΔT = 4.245 * 0.2393 * 80 = 812.3 W

Both insulators result in the same heat loss, which is 812.3 W. Therefore, in terms of minimizing heat loss, it doesn't matter which insulator is covered outside.

However, it's worth noting that insulator A has a lower thermal conductivity, which means it is a better insulator and can potentially provide better thermal performance in other aspects such as temperature gradients across the insulation.

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Answer:

A. Juna's answer is correct because both the number of feet and the number of inches are correct

Explanation:

The drag per unit span on the model can be calculated using a control volume approach as follows:

Where ρ is the density of the fluid, V is the velocity of the fluid, and z is the distance from the top of the test section to the bottom. The integral can be calculated by summing up the velocity of the fluid at each point from the top to the bottom of the test section. This can be expressed as:

Where Vz is the velocity of the fluid at each point in the test section.

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Answer:

Copper and Aluminum

Explanation:

Complete Question:

Problem 8 Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 °F and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by

C_p=10.03+0.0184T  C_p[=]Btu/lbmole- "F ; T[=] °F C,

Answer:

\(P'=0.377hp\)

Explanation:

From the question we are told that:

Initial Temperature T_1=70 F

Final Temperature \(T_2=250pisa =114.94F\)

Efficiency \(E=75\%=0.75\)

Generally the equation for Work-done  is mathematically given by

 \(W=\int C_pT\)

 \(W=10.03(114.94-70 )+0.0184((114.94)^2-70^2 )\)

 \(W=527.21btu/ibmole\)

 \(W=11.982btu/ibm\)

Generally the equation for Efficiency  is mathematically given by

 \(E=\frac{isotropic Power}{Actual P'}\)

 \(E=\frac{P}{P'}\)

Since

Isotropic Power

  \(P=0.0167*11.982btu/ibm\)

 \(P=0.2btu/s\)

Therefore

 \(P'=\frac{0.2}{0.75}\)

 \(P'=0266btu/s\)

Since

 \(1btu/s=1.4148hp\)

Therefore

 \(P'=0.377hp\)

Answer:

\(\triangle h=4.935m\)

Explanation:

From the question we are told that:

Liquid density \(\rho=800\)

Diameter of pipe \(d=4cm \approx 0.004m\)

Diameter of venture \(d=10cm \approx 0.010m\)

Specific weight of mercury P_mg \(133 kN/m^3\)

Flow rate \(r=0.05 m^3/s\)

Area A:

            \(A_1=\frac{\pi}{4}0.1^2\\A_1=0.00785m^2\\A_2=\frac{\pi}{4}0.04^2\\A_2=0.001256m^2\\\)

Generally the Bernoulli's equation is mathematically given by

\(\frac{P_1}{\rho_1g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}\\\)

Where

\(V_1=\frac{r}{A_1} \\\\ &V_1=\frac{r}{A_2}\)

Therefore

\(P_1-P_2=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})\)

Generally the equation for pressure difference b/w manometer fluid is given as

\(P_1-P_2=(p_mg-pg)\triangle h\)

Therefore

\((p_mg-pg)\triangle h=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})\)

\(\triangle h=\frac{\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})}{(p_mg-pg)}\)

\(\triangle h=\frac{\frac{(800)(0.05)^2}{2}(\frac{(0.1)^2-(0.4)^2}{(0.1)^2(0.04)^2})}{(1.33*10^3-800*9.81)}\)

\(\triangle h=4.935m\)

Therefore elevation change is mathematically given by

\(\triangle h=4.935m\)

The primary objective of this project is to design a building in accordance with the structural requirements for LRFD for steel as defined in ASCE 7/IBC. The building's materials are defined in the following terms:

Light-weight concrete flooring over deck with a thickness of 5 inches.
ASTM A992 (Gr.50) for all W shape Beams, Girders, and Columns.
HSS Braces (ASTM A500) or W shapes (ASTM A992, Gr.50).

Dead Loads: The building's dead load will be made up of a variety of elements, including:

Roof: Roofing Materials (Water Proofing, etc.) = 4 psf.

18" Gage deck = 3 psf.
Light-weight concrete 5 in thick.
Framing & Fire proofing = 8 psf.
Suspended ceiling = 4 psf.
Mechanical & Electrical = 4 psf.
Solar panels & assembly = 9 psf.

Floor: Tile including assembly = 9.5 psf.
18" Gage deck.
Light-weight concrete 6 1/4 "= 3 psf.
Framing & Fire proofing = 15 psf.
Suspended ceiling = 5 psf.
Mechanical & Electrical = 5 psf.

Wall: Parapets on roof (outer boundary only) = 25 psf (3.5 ft high).
Glazed walls (outer boundary only) = 18 psf (ground to roof level).

The live load of partitions is taken into consideration as appropriate. Flooring requires a two-hour fire rating. The following deflection requirements are in effect:
L/360 due to live load deflection in all interior Beams and Girders.

L/180 due to total load for all spandrel Beams and Girders.
the building's seismic design should consider the values of spectral response acceleration parameters for the given location, which can be found using the USGS website.

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Answer:

1279

Explanation:

We have the mean u = 1497

Standard deviation sd = 322

We find the x distribution using 25%

P(Z<z) = 0.25

Z = -0.675

From here we use the formula for z score

X = z(sd) + u

X = -0.675*322 + 1497

X = -217.35 + 1497

X = 1279.6

Which is approximately 1279

So we conclude that the maximum sat scores in year 2014that meets with the requirements of this course is 1279

Answer:

1831

Explanation:

Answer:

They protect the entire hand from abrasion and punctures, and are a dependable, comfortable glove for a wide variety of jobs. Drivers' gloves are available in various types of leathers and can be unlined or lined for cold weather. Leather palm gloves provide maximum protection against abrasive and puncture hazards.

Gloves with leather palms offer the best defense against abrasive and puncture risks. They have leather palms and fingers and are constructed of dependable cotton or canvas.

Abrasive is defined as a substance, sometimes a mineral, used to shape or polish a product by rubbing, which causes some of the workpiece to be removed due to friction. Abrasives are a particular kind of extremely hard material that are utilized in a variety of residential, industrial, and technological applications. A portion of the work piece is worn away when the abrasive is rubbed against it.

These leather gloves each offer particular levels of comfort, protection, and durability. Some of these leather gloves are suitable for use by those handling dangerous chemicals and sparks. Some can provide cut protection to an extreme degree. For workers who handle heat and water, certain types of leather gloves are recommended.

Thus, gloves with leather palms offer the best defense against abrasive and puncture risks. They have leather palms and fingers and are constructed of dependable cotton or canvas.

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Answer:Referencing is a standardised method of acknowledging the sources of information and ideas you have used.

Explanation:Depending on the way in which they record sources, scholarly reference styles can be divided into three main categories: documentary notes styles, parenthetical (or author-date) styles, and numbered styles. Within each category there are several, slightly different reference styles.

Referencing allows you to acknowledge the contribution of other writers and researchers in your work. Any university assignments that draw on the ideas, words or research of other writers must contain citations. Referencing is also a way to give credit to the writers from whom you have borrowed words and ideas.

The balance in the Cash account at the beginning of May was $6,700.On May 31, the Cash account of Tesla had a normal balance of $6,400.

To determine the balance in the Cash account at the beginning of May, we need to consider the net effect of the debits and credits during the month. The normal balance of $6,400 at the end of May indicates that the account has a credit balance. From the given information, the total debits for May were $13,600 and the total credits were $12,900. To calculate the beginning balance, we subtract the net credits from the net debits: $13,600 - $12,900 = $700. Since the account has a credit balance, we subtract $700 from the ending balance of $6,400: $6,400 - $700 = $6,700, which was the balance in the Cash account at the beginning of May.

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The load on a truck may not extend over the front of the vehicle more than 3 ft, and no more than 4 feet over the rearhe amount by which cargo extends past the dimensions of the truck is referred to as overhang in the trucking industry.

Overhang is calculated from the extreme front of the vehicle, which may be the bumper or the tires, and extends to the farthest point of the vehicle's load. Overhang also includes protruding axles and devices. Each state's requirements are different. So it is crucial to follow the overhang rules of your state when carrying cargo. If a truck exceeds the overhang limit of the state or country, the load may extend beyond the limits of the truck bed, which is dangerous for both the truck driver and other drivers on the road. The overhang limits are governed by state law. In the United States, each state has its own regulations for maximum vehicle length, width, and height, as well as weight restrictions, axle spacing, and load securement requirements.

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No matter how many traffic lanes or if a turn lane is present, drivers in both directions of a two- or multi-lane highway are required to stop and remain stopped until the flashing red bus lights are turned off.

Only lanes of traffic moving in the same direction as the school bus must come to a complete stop. To get to their home, students shouldn't have to cross four or more lanes of traffic. The school bus had to stop. To get to their home, students shouldn't have to cross four or more lanes of traffic. On a two-lane road, when a school bus stops and flashes its red lights, traffic coming from either direction must stop before passing the bus. on multilane roads. on roadways with dividers.

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Using α = 0.4, the forecasted sales for 2011, 2012, and 2013 would all be $48,000, which is the same as the actual sales in 2009.

Using α = 0.8, the forecasted sales for 2011, 2012, and 2013 would also be $48,000.

In both cases, the forecasts for 2011, 2012, and 2013 remain the same as the actual sales in 2009 due to the zero difference between the actual and forecasted sales values.

To forecast sales for 2011, 2012, and 2013 using exponential smoothing, we need to apply the formula:

Forecast for the next period = Previous period's forecast + α * (Actual value - Previous period's forecast)

Given that the actual 2009 sales are $48,000 and are considered the forecast for 2010, we can calculate the forecasts for subsequent years using different values of α.

(a) For α = 0.4:

- Forecast for 2011 = $48,000 + 0.4 * ($48,000 - $48,000) = $48,000

- Forecast for 2012 = $48,000 + 0.4 * ($48,000 - $48,000) = $48,000

- Forecast for 2013 = $48,000 + 0.4 * ($48,000 - $48,000) = $48,000

Using α = 0.4, the forecasted sales for 2011, 2012, and 2013 would all be $48,000, which is the same as the actual sales in 2009.

(b) For α = 0.8:

- Forecast for 2011 = $48,000 + 0.8 * ($48,000 - $48,000) = $48,000

- Forecast for 2012 = $48,000 + 0.8 * ($48,000 - $48,000) = $48,000

- Forecast for 2013 = $48,000 + 0.8 * ($48,000 - $48,000) = $48,000

Using α = 0.8, the forecasted sales for 2011, 2012, and 2013 would also be $48,000.

In both cases, the forecasts for 2011, 2012, and 2013 remain the same as the actual sales in 2009 due to the zero difference between the actual and forecasted sales values.

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Answer: what are the tasks

Explanation:

no professionals to match with the tasks.

Failure rate, λ, is the number of failures divided by the total time for the test of the item.λ = f / T Where, f = Number of Failures T = Total Time To determine the failure rate, we have:

Total time of testing = 22 h The number of failures = 4 + 1 (the item that failed beyond the 22 h) = 5Total time of testing for failed items = 4 + 12 + 15 + 21 + (22-21) = 31 hoursλ = f / T = 5 / 31 = 0.0006 (Answer)(2) The failure rate for 50 items that are tested simultaneously for 15 h and six items failed is 0.008.

The failure rate, λ, is the number of failures divided by the total time for the test of the item.λ = f / T Where, f = Number of Failures T = Total Time To determine the failure rate, we have: Number of items tested simultaneously, n = 50Total time of testing, T = 15 h The number of failures, f = 6λ = f / T = 6 / (50 × 15) = 0.008

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It is motor oil, as oil is used to reduce friction

The statement that would be true about a critical path is; "All the Above" (Option D)

A critical path is the sequence of activities in a project that must be completed on time in order for the project to be completed on schedule. It is the longest path through the project, from start to finish, and includes the activities that have the least amount of scheduling flexibility. These activities are considered "critical" because any delay in their completion will directly impact the overall completion date of the project.

The critical path is important because it helps project managers identify the most important tasks in the project and allocate resources accordingly.

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Answer:

Hello your question is incomplete attached below is the missing part

a) p1 = 454.83 kPa,  p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa

b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ

c) 5

Explanation:

Given data:

mass of air ( m ) = 1/10 kg

adiabatic index ( k ) = 1.4

temperature for isothermal expansion = 250K

rate of heat transfer ( Q12 ) = 3.4 KJ

temperature for Isothermal compression ( T4 ) = 300k

final volume ( V4 ) = 0.01m ^3

a)  Calculate the pressure, in Kpa, at each of the four principal states

from an ideal gas equation

P4V4 = mRT4 ( input values above )

hence P4 = 860.959kPa

attached below is the detailed solution

b) Calculate work done for each processes

attached below is the detailed solution

C) Calculate the coefficient of performance

attached below is detailed solution

Answer:

They moved fresh water around their vast empire with aqueducts and canals.

Explanation:

Answer:

engine suspension brake and more

Explanation:

Answer:

Explanation:

The null hypothesis is that the mean fill volume is 16 oz, and the alternative hypothesis is that the mean fill volume is different from 16 oz. The level of significance is 5%, and the sample size is 15. The test statistic is t=0.0367t=0.0367. The p-value is 0.9738.

Since the p-value is greater than the level of significance, we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the mean fill volume is different from 16 oz. The engineer should not reset the machine.

```

The test statistic is calculated as follows:

t = (x - μ) / s / sqrt(n)

= (16.0367 - 16) / 0.0551 / sqrt(15)

= 0.0367

The p-value is calculated using the t-distribution with 14 degrees of freedom.

p-value = 2 * t.cdf(-0.0367, 14)

= 0.9738

```

Boost converters can also supply power to smaller-scale equipment, including portable lighting systems.

A boost converter can increase the voltage from a single 1.5 V alkaline cell to provide the lamp with the 3.3 V that a white LED generally needs to emit light. High-frequency power conversion circuits known as DC-DC converters use inductors, transformers, and capacitors to reduce switching noise and produce regulated DC voltages. Even with fluctuating input voltages and output currents, closed feedback loops maintain constant voltage output. To "step-up" an input voltage to a higher level, needed by a load, the boost converter is utilized. By holding energy in an inductor and delivering it to the load at a greater voltage, this special feature is made possible.

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Answer:

500 feet, 300 feet

Explanation:

The false statement about most machine learning models is that: B. they are flexible enough to handle all issues you might see in your dataset (lack of data, incorrect data, etc).

Machine learning (ML) is also referred to as deep learning or artificial intelligence (AI) and it can be defined as a subfield in computer science which is typically focused on the use of data-driven techniques (methods), computer algorithms, and technologies to develop a smart computer-controlled robot with an ability to automatically perform and manage tasks that are exclusively meant for humans or solved by using human intelligence.

Generally speaking, machine learning models are designed and developed to accept numerical data (numbers) or collections of numerical data (numbers) as an input.

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