A letter "C" inset in a blue circle following a CT frequency means that the pilot should - 1. use the _________________ frequency as the common Traffic Advisory Frequency.2. Call in the _________ your intentions at that airport when tower is closed.

Answer:

\(-6111.11\ \text{kJ}\)

\(863.28\ \text{kJ}\)

Explanation:

m = Mass of automobile = 1100 kg

v = Velocity of car = 120 km/h = \(\dfrac{120}{3.6}\ \text{m/s}\)

h = Height of hill = 80 m

g = Acceleration due to gravity = \(9.81\ \text{m/s}^2\)

Change in kinetic energy

\(KE=\dfrac{1}{2}m(u^2-v^2)\\\Rightarrow KE=\dfrac{1}{2}\times 1100\times (0-(\dfrac{120}{3.6})^2)\\\Rightarrow KE=-611111.11\ \text{J}\)

Change in kinetic energy is \(-6111.11\ \text{kJ}\)

Change in potential energy is given by

\(PE=mgh\\\Rightarrow PE=1100\times 9.81\times 80\\\Rightarrow PE=863280\ \text{J}\)

The change in potential energy is \(863.28\ \text{kJ}\).

Answer:

F=200kN

Explanation:

When working with multiple ground glass joints in a set-up, the time that you should apply grease to the joints is if the reaction is said to to be at a heated point that is above 150 oC, or, when the joints required is  still rotated if in use.

Despite being made to fit together relatively nicely, ground glass joints are not completely airtight. In certain circumstances, grease must be applied to each joint to make sure of a good seal (such as when employing lower pressure inside an equipment).

Note that  In ambient lighting, ground or frosted glass is frequently used as a heat- and weather-resistant light diffuser, particularly on the glass covers or enclosures for lamp fixtures and occasionally on incandescent bulbs.

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The reversed Carnot cycle is a theoretical cycle that is the most efficient refrigeration cycle possible for a given heat source and sink temperature.

The coefficient of performance (COP) of a reversed Carnot cycle is given by the ratio of the heat removed from the refrigeration space to the work input to the system.The work input to the refrigerator is 200 kW, and the heat rejection is 2000 kW to a heat reservoir at 27°C. Let's assume that the cooling load supplied to the refrigerator is Q_c.  The COP of a reversed Carnot cycle is given by the following equation: COP = Q_c / W where W is the work input to the system. The COP of a reversed Carnot cycle is given by: COP = T_L / (T_H - T_L)  where T_H is the temperature of the heat source and T_L is the temperature of the heat sink. Solving the equation for T_H, we get: T_H = T_L / (1 - COP/T_L)  Substituting the values, we get: COP = Q_c / W  COP = Q_c / 200 kW  COP = (27 + 273) / (T_H - (27 + 273))  T_H = (27 + 273) / (1 - COP/(27 + 273))  T_H = 300 / (1 - COP/300)  Using the values, we get: COP = Q_c / 200 kW   Q_c = COP x 200 kW Q_c = (27 + 273) / (T_H - (27 + 273)) x 200 kW  Plugging in the values, we get:  COP = (27 + 273) / (T_H - (27 + 273))  COP = (27 + 273) / (T_H - 300)  200 kW = (27 + 273) / (T_H - 300) x COP  200 kW = (27 + 273) x COP / (T_H - 300)  (T_H - 300) x 200 kW = (27 + 273) x COP   T_H = ((27 + 273) x COP) / 200 kW + 300   Calculating the COP:  COP = (27 + 273) / (T_H - (27 + 273))   COP = (27 + 273) / (T_H - 300)  Assuming a COP of 4, we can calculate the temperature of the heat source: T_H = ((27 + 273) x 4) / 200 kW + 300  T_H = 300.8 K or 27.65°C  Therefore, the cooling load supplied to the refrigerator is 800 kW, and the temperature of the heat source is 27.65°C.

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Ohm's law, description of the relationship between current, voltage, and resistance.

Given that the output voltage varies by 5 mV when loaded from 0.01 A to 1 A

Therefore, the regulators output resistance is given by

\(I_{L} = \frac{V_{L} }{R_{L} }\)

\((1.00 - 0.01) A_{} = \frac{5 mV_{} }{R_{L} }\)

\(0.99 A = \frac{5 mV}{R_{L} }\)

∴ The line and load regulation is 0.01 Ω

The output of a difference amplifier is related to the input by the equation:

When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes

6.4 A₁ + 5.1 A₂ = 64.7______(1)

When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes

5.6 A₁ + 4.9 A₂ = 35.7_______(2)

Multiply equation (1) by 5.6  and (2) by 6.4

35.84 A₁ + 28.56A₂ = 362.32 ________(3)

35.84 A₁ + 31.36 A₂ = 228.48__________(4)

Subtract equation (3) from (4)

2.8 A₂ = -133.84

A₂ = -133.84/2.8

A₂ = -47.8

Put the value of  A₂ into equation (1)

6.4 A₁ + 5.1 (-47.8) = 64.7

6.4 A₁ = 64.7 + 243.78

A₁ = 308.48/6.4

A₁ = 48.2

a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)

Common mode gain = 0.4

b) Differential mode gain = (A₁ -A₂)/2

Differential mode gain = (48.2 - (-47.8))/2

Differential mode gain = 96/2

Differential mode gain = 48

c) Common Mode Rejection Ratio (CMRR)

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Answer:

a building project...

Explanation:

New, derived traits of human evolution over the past five million years include all of the following except: elongation. The correct answer to the given question is option D .

Derived traits are new traits that have been passed down to a given organism by their ancestors. Some of the derived traits of human evolution over the past five million years include:

Encephalization: Encephalization is the evolutionary trend of brain size increasing relative to body size over geological time in a lineage. This evolutionary trend was observed in the human lineage.

Lack of body hair: One of the derived traits of human evolution is the lack of body hair. Humans have a far less dense body hair than other apes and primates. The loss of body hair facilitated an increase in sweat gland density, which helps us regulate our body temperature.

Reduced sexual dimorphism: The reduced differences between the sexes in size and appearance over time in a given population is referred to as reduced sexual dimorphism.

Elongation: Elongation is not a derived trait of human evolution over the past five million years, so it is the correct answer.

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We can see here that the maximum value of rolling resistance over which a loaded scraper can maintain is: 780 lb/ton

Rolling resistance is the force that resists the motion of a rolling object on a surface. Maximum rolling resistance refers to the highest level of resistance that an object can encounter while rolling on a particular surface.

For a fully loaded condition:

The total resistance = (gross + rolling) resistance

52% = 0% - rolling resistance

rolling resistance = 52%

The maximum rolling resistance = rolling resistance percentage × speed

The maximum rolling resistance = 52% (15 lb/ton)/(1% resistance)

The maximum rolling resistance = 780 lb/ton

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Answer:

the magnitude of the velocity when t = 3 s is 10.54 m/s.

Explanation:

To solve this problem, we can use the following kinematic equation that relates velocity, acceleration, and time:

v = vo + at

where:

v = final velocity

vo = initial velocity

a = acceleration

t = time

First, we need to find the velocity of the object at time t = 3 s. To do this, we can substitute the given values into the kinematic equation and solve for v:

v = vo + at

v = 8i + (-0.2i+2j+1.5k) x 3

v = 8i - 0.6i + 6j + 4.5k

v = 7.4i + 6j + 4.5k

The magnitude of the velocity is given by:

|v| = sqrt(vx^2 + vy^2 + vz^2)

where:

vx, vy, vz = the x, y, and z components of the velocity vector

Substituting the values from above, we get:

|v| = sqrt((7.4)^2 + 6^2 + (4.5)^2)

|v| = sqrt(54.81 + 36 + 20.25)

|v| = sqrt(111.06)

|v| = 10.54 m/s (approx)

Texas law requires children of certain ages, weights, and heights to be restrained by a federally approved child passenger restraint system, booster seat, or safety belt when transported in any motor vehicle other than a public carrier for hire or school bus. The correct option is C. All of the above.

This means that children who are younger than 8 years old, shorter than 4 feet 9 inches tall, and under 80 pounds must be secured in a child safety seat or booster seat. Children who are 8 years old or older or taller than 4 feet 9 inches must be secured in a safety belt. The purpose of these laws is to protect children in the event of a car accident and to reduce the risk of serious injury or death. The correct option is C. All of the above.

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The Administrative Tools that apply from the given options are: a. Defragment and Optimize Drives c. iSCSI Initiator d. Computer Management

Administrative Tools in an operating system provide access to various utilities and functions that help manage and control system settings, configurations, and resources. These tools are typically used by system administrators or advanced users to perform administrative tasks on a computer or network.

Defragment and Optimize Drives is an Administrative Tool that allows users to analyze and optimize the fragmentation of hard drives. It helps improve system performance by rearranging fragmented files and data on the disk, resulting in faster access times and smoother operation.

iSCSI Initiator is an Administrative Tool used to establish and manage connections to iSCSI (Internet Small Computer System Interface) devices over a network. It enables a computer to communicate with remote storage devices as if they were locally attached, expanding storage capabilities and facilitating data sharing and backup.

Computer Management is a comprehensive Administrative Tool that provides access to various system management utilities. It includes tools for managing disk partitions, device drivers, event logs, system services, user accounts, and more. Computer Management is a centralized console that offers a wide range of administrative capabilities for managing and configuring different aspects of a Windows operating system.

These Administrative Tools are designed to provide control and oversight over critical system functions, ensuring efficient management and maintenance of the computer or network infrastructure.

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By offering a centralized repository for collecting and managing data, databases can considerably improve an organization's data collection process.

It will be very difficult for organizations to function without the use of data base.

Databases are used to store, organize, and access data of any type. They collect information about people, places, or things. That information is gathered in one place so that it may be studied and analyzed. Databases are simply organized collections of data.

Databases allow us to work efficiently with massive amounts of data. They make data updating straightforward and reliable, and they help with accuracy.

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The actual dimensions for the given nominal dimensions 1x2 stick of lumber, a 2x4 stick of lumber are;

Actual dimension of 1 x 2 stick of lumber = ³/₄'' × 3¹/₂''

Actual dimension of 2 x 4 stick of lumber = 1¹/₂'' × 3¹/₂''

The nominal dimension for a sheet of plywood is; 4' x 8'

Lumber sticks are sticks made from timber in forms used mainly in building construction as formwork support for the sheets of plywood used.

There could also be other uses of lumber sticks like making of some basic home furniture's but they are primarily used in building construction.

Now, Lumber sticks could come in different nominal dimensions such as;

Now, in the question, we are dealing with 1x2 stick of lumber and a 2x4 stick of lumber. From general cutting standards in most workshops, the actual sizes are respectively;

Actual size of 1 x 2 stick of lumber = ³/₄'' × 3¹/₂''

Actual size of 2 x 4 stick of lumber = 1¹/₂'' × 3¹/₂''

Now, for a sheet of plywood, the standard size of a sheet of plywood is 4' × 8'.

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Answer:

True

Explanation:

A forklift will tip over if the center of gravity is outside the stability triangle. The stability triangle refers to the area within the three wheels of a forklift that ensures the stability of the machine and prevents tipping. If the center of gravity of the load being carried by the forklift moves outside of the stability triangle, the machine becomes unstable and is at risk of tipping over. It's important to properly balance the load on a forklift and ensure that the center of gravity stays within the stability triangle to avoid accidents and maintain safety.

I hope this information is helpful. I mostly prefer MyAssignmentHelp.com for homework help. Please let me know if you have any additional questions or refer to MyAssignmentHelp.com for further assistance.

Answer:

I can help but I need to know what it looking for

When a neuron sends a signal, they create electrical events called actions potentials and chemical neurotransmitters.

Basically, a neuron is a fundamental units of the brain and nervous system. They are nerve cells responsible for: receiving sensory input from the external world, sending motor commands to the muscles, and for transforming and relaying electrical signals.

Being that neurons also communicates, they do so using electricals and chemicals means. So one neuron can communicate to the other via electrical events which are action potentials and chemical neurotransmitters.

An action potential causes one neuron to release chemical neurotransmitter which will help or hinder the other neuron from firing its own action potential.

In summary, neurons are essentially electrical devices having action potentials and chemical neurotransmitters as the fundamental units of sending signals or communicating with other neurons.

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A commemorative coin is not an example of currency as commemorative coins are coins with a unique design that refers to the occasion on which they were issued and are issued to mark a specific event or issue.

What is a Commemorative coins?

Commemorative coins are coins with a unique design that refers to the occasion on which they were issued and are issued to mark a specific event or issue. Although some nations also produce commemorative coins for everyday use, many coins in this category are only intended for collectors.

Commemorative coins can be seen as being of one of three types:

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Answer:

A PULLY

Explanation:

HAD THIS ONE THAT IS THE CORRECT ANWSER

Answer:

The answer is B. a pulley

Explanation:

I hope I answered your question:)

Focal points are typically arranged in a grid composition at the intersection of horizontal and vertical guidelines to create an interesting and dynamic visual layout.

This type of composition can be used to lead the viewer's eye around the page, highlighting certain elements and creating a sense of balance. Additionally, this type of composition can be used to create a sense of movement and energy in the layout.

Grid compositions are particularly useful for creating a sense of order and balance in a design. The intersecting horizontal and vertical guidelines create a structured composition that allows for the focal points to be arranged in a way that naturally guides the viewer's eye through the design.

This creates a sense of movement and energy as the viewer is taken on a journey through the design.

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The values for low and high on the third call to binary search are low = 3 and high = 4.

The first call to binary search with input list (3, 7, 13, 18, 24, 29, 31, 33, 41), target item x = 14, low = 1, and high = 9, would proceed as follows:

The middle index is (1 + 9) // 2 = 5.

The middle element is 24.

14 is less than 24, so we update high to 4 and repeat the search with the lower half of the list.

The second call to binary search would have low = 1 and high = 4 and would proceed as follows:

The middle index is (1 + 4) // 2 = 2.

The middle element is 13.

14 is greater than 13, so we update low to 3 and repeat the search with the upper half of the lower half of the list.

The third call to binary search would have low = 3 and high = 4, and the search would terminate with an unsuccessful result, since the sublist (13, 18) does not contain the target element 14.

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The transformer has 387 turns on the primary winding and 96 turns on the secondary winding.

To solve this problem, we need to use the transformer equation:
V1/V2 = N1/N2
where V1 and V2 are the voltages on the primary and secondary windings, respectively, and N1 and N2 are the number of turns on the primary and secondary windings, respectively.
Given the transformer rating of 20 KVA, we can calculate the current on the primary and secondary windings as follows:
I1 = 20,000/480 = 41.67 A
I2 = 20,000/120 = 166.67 A

Next, we need to determine the power factor of the load. We are given that the load absorbs 15 KVA at 0.8 pf lagging when the load voltage is 118 V. To calculate the apparent power of the load, we can use the formula:
S = P / pf
where S is the apparent power, P is the real power (15 KVA), and pf is the power factor (0.8 lagging). Thus:
S = 15,000 / 0.8 = 18,750 VA
Now we can calculate the current on the secondary winding:
I2 = S / V2 = 18,750 / 118 = 158.9 A

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Answer:

A) condenser pressure = 9.73 kPa,

B) 10242 kw

C) 36.9%

Explanation:

given data

entrance pressure of steam = 12.5 MPa

temperature of steam = 550⁰c

flow rate of steam = 7.7 kg/s

outer pressure = 2 MPa

reheated steam temperature = 450⁰c

isentropic efficiency of turbine( nt ) = 85% = 0.85

isentropic efficiency of pump = 90% = 0.90

From steam tables

at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg,  S3 = 6.6317 kJ/kgK

also for an Isentropic expansion

S4s = S3 .

therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa

h4s = 2948.1 kJ/kg

nt = 0.85

nt (0.85) = \(\frac{h3-h4}{h3-h4s}\) = \(\frac{3476.5 - h4}{3476.5 - 2948.1}\)

making h4 subject of the equation

h4 = 3476.5 - 0.85 (3476.5 - 2948.1)

h4 = 3027.3 kj/kg

at P5 = 2 MPa and T5 = 450⁰c

h5 = 3358.2 kj/kg,  s5 = 7.2815 kj/kgk

at P6 , x6 = 0.95  and s5 = s6

using nt = 0.85 we can calculate for h6 and h6s

from the chart attached below we can see that

p6 = 9.73 kPa, h6 = 2463.3 kj/kg

B) the net power output

solution is attached below

c) thermal efficiency

thermal efficiency = 1 - \(\frac{qout}{qin}\) = 1 - ( 2273.7/ 3603.8) = 36.9% ≈ 37%

Answer:

d

Explanation:

OR criteria, AND criteria

In an OR criteria, it doesn't need all the records to be true. Just one record is enough and all other criterion becomes true.

In an AND criteria, it's unlike the OR criteria and works in opposite. It needs every member of the record to be true to be able to adjudge the whole record as true.

And as such, we have

With OR criteria, only one criterion must evaluate true in order for a record to be selected and with AND criteria, all criteria must be evaluate true in order for a record to be selected.

Answer:

B. polarity

Explanation:

Of the three types of hazard controls – engineering, administrative, and personal protective equipment (PPE) – PPE is the last line of defense: True.

OSHA is simply an abbreviation for occupational safety and health administration that was created under the Occupational Safety and Health Act, so as to develop work-safety standards for workers and providing precautionary measures against all workplace hazards.

Generally, there are three types of hazard controls and these include the following:

PPE is an acronym for personal protective equipment and it can be defined as a terminology that is used to denote any piece of equipment which offer protection to different parts of the body while working in a potentially hazardous environment.

Also, some examples of personal protective equipment (PPE) used to protect the different parts of the body are:

According to occupational safety and health administration (OSHA), the use of a faulty personal protective equipment (PPE) should be the last line of defense.

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Complete Question:

Of the three types of hazard controls – engineering, administrative, and personal protective equipment (PPE) – PPE is the last line of defense. True or False?

The incidence of nonconformance in Southstar Airlines that is typical of their entire fleet of airplanes is needed. We have been given that in one month, 45 preflight inspections were performed on an airplane at Southstar Airlines, 15 nonconformances were noted and each inspection checks 50 items.

Assuming 2 sigma off-centering, what sigma level does Southstar maintain?

From the given, the nonconformance rate is calculated by taking the ratio of the number of nonconformances to the number of items inspected.

Nonconformance rate = 15/ (45 * 50) = 0.0066667 (approx)

Now, we can calculate the Z-score by using the standard normal distribution table

Z-score = 2sigma off-centering = 2

looking up a Z-score table we obtain that 0.0066667 corresponds to 2.11 standard deviations or 2.11 sigma level approx.

Southstar Airlines maintains a sigma level of about 2.11 if this incidence of nonconformance is typical of their entire fleet of airplanes.

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There are different aspect of drilling operations. What the operator have to do to ensure hole concentricity and a good surface finish is to Increase the spindle speed one and a half times the normal rate.

Countersinking is often used to produce a conical hole linking the angled shape on the bottom side of a flat-head screw.

There is a required spindle speed used for countersinking operation . The lead out rule is of 1/3 the speed of a drill of the same size often work for almost all countersinking and counterboring operations.

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Answer:

1021c.

Explanation:

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