southbound on v257, at what time should you arrive at dbs vortac if you crossed over cpn vortac at 0850 and over divid intersection at 0854?a.0939b.0943c.0947

A) Select the Search icon and enter the term into the search field. A definition will show up in the Key Terms section of the search results.

Key Terms are words or phrases used to describe a concept or idea. They are often used in the field of academia to help define the scope of a particular study or research project. Key Terms are also used in business and legal documents to provide a concise and precise description of a concept or idea. By providing clear definitions for each key term, it helps to ensure that all parties involved in a discussion or agreement are on the same page.

C) Select the Table of Contents or the Navigation icon and navigate to the Glossary.
D) Select the key term in your eText and a pop-up definition will appear.

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The option that is not a derived units is known to be option  E) kg.

The units that are known to be used for any form of derived quantities are said to be called the derived units.

They are:

Note that this unit are derived because they are known to be derived because they have to be solved for using different ways. The others such as kg m-3 is one that need to be derived to arrive at it.

Therefore, based on the above, one can say that The option that is not a derived units is known to be option  E) kg because it is one that cannot be derived,

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The following are derived units EXCEPT

Options

A) kg m-3

B)N

C)Ns

D)m3

E)kg

As the angle of the ramp is increased, the force parallel increases. Hence, option (a) can be considered as the correct answer.

When the angle of a ramp is increased, the force parallel to the ramp, also known as the parallel component of the gravitational force, does increase. This is because the component of gravity acting parallel to the ramp increases with the angle. However, it's important to note that the total gravitational force acting on an object remains constant regardless of the angle of the ramp.As the angle of the ramp increases, the force required to push or pull an object up the ramp against gravity increases. This is due to the increase in the vertical component of the gravitational force, which opposes the motion up the ramp. The parallel force required to overcome this increased vertical force also increases.

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Answer:

Any exchange of data that transmits the data in a digital form is a digital communication.

Answer:

A) F = 0.011 N

B) ΔP = 5.6 N/m²

Explanation:

We are given;

surface tension coefficient; S = 0.07 N/m

Radius; r = 2.5 mm = 0.025 m

A) Formula to find the surface tension force(F) is given by;

F = SL

Where L is effective length = 2πr

F = 0.07 × 2π × 0.025

F = 0.011 N

B) Formula for difference in pressure droplet is;

ΔP = 2S/r

Thus;

ΔP = (2 × 0.07)/0.025

ΔP = 5.6 N/m²

Answer:

25 mins into hours = 0.416667 hours

25 inches as mm = 635

Explanation:

Answer:

Used to recall the direction a standard screw

GPS stands for Global Positioning System in a car

It is a navigation system that uses satellites to provide location and time information to a GPS receiver, such as the one found in a car. This allows the car to accurately determine its position and provide directions to a desired destination. The GPS receiver calculates the car's position using the information received from the satellites and displays it on a map. With this information, the car's navigation system can provide directions to a destination, and the driver can follow these directions to reach their desired location. GPS is a common feature in many modern cars and is used for a variety of purposes, including navigation, tracking, and emergency services.

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Answer:

yrt a

Explanation:

Answer:

18 gauge

Explanation:

Standard automotive primary wire is 18 gauge. (internet search) Hope this helps

Answer:

Understand how the network works where she is working at

Explanation:

Answer:

The answer is below

Explanation:

The complete question is attached.

a) Bernoulli equation is given as:

\(P+\frac{1}{2}\rho V^2+ \rho gz=constant\\\\\frac{P}{\rho g} +\frac{V^2}{2g} +z=constant\\\)

Where P = pressure, V = velocity, z = height, g = acceleration due to gravity and ρ = density.

\(\frac{P}{\rho g} +\frac{V^2}{2g} +z=constant\\\\\frac{P}{\gamma} +\frac{V^2}{2g} +z=constant\\\\\frac{P_1}{\gamma} +\frac{V_1^2}{2g} +z_1=\frac{P_2}{\gamma} +\frac{V_2^2}{2g} +z_2\\\\but \ z_1=z_2,P_1=0,V_1=0,V_2=60\ mph=88\ ft/s. Hence:\\\\\frac{P_2}{\gamma} =-\frac{V_2^2}{2g} \\\\P_2=\gamma*-\frac{V_2^2}{2g} =\rho g*-\frac{V_2^2}{2g} \\\\P_2=-\frac{V_2^2}{2}*\rho=-\frac{(88.8\ ft/s)^2}{2} * 0.00238\ slug/ft^3=-9.22\ lb/ft^2\\\\P_2+\gamma_{H_2O}h-\gamma_{oil}(1/12 \ ft)=0\\\\\)

\(\gamma_{oil}=0.9\gamma_{H_2O}=0.9*62.4\ lb/ft^3=56.2\ lb/ft^3\\\\Therefore:\\\\-9.22\ lb/ft^2+62.4\ lb/ft^3(h)-56.2\ lb/ft^3(1/12\ ft)=0\\\\h=0.223\ ft\)

b)

\(\frac{P}{\gamma} +\frac{V^2}{2g} +z=constant\\\\\frac{P_2}{\gamma} +\frac{V_2^2}{2g} +z_2=\frac{P_3}{\gamma} +\frac{V_3^2}{2g} +z_3\\\\but \ z_3=z_2,V_3=0,V_2=60\ mph=88\ ft/s. \\\\\frac{P_2}{\gamma}+\frac{V_2^2}{2g} = \frac{P_3}{\gamma}\\\\\frac{P_3-P_2}{\gamma}=\frac{V_2^2}{2g} \\\\P_3-P_2=\frac{V_2}{2g}*\gamma=\frac{V_2^2}{2g}*\rho g\\\\P_3-P_2=\frac{V_2}{2}*\rho=\frac{(88\ ft/s^2)^2}{2}*0.00238\ slg\ft^3\\\\P_3-P_2=9.22\ lb/ft^2\)

A student-centered classroom is NOT a place where the teacher lectures while students take notes and then study on their own.

A student-centered classroom is a learning environment where the teacher shifts the focus from themselves to the students, creating a space that is more interactive and collaborative. In this type of classroom, the teacher works as a facilitator, guiding and supporting students as they take control of their learning.

In a student-centered classroom, the teacher addresses different learning styles of his/her students, meaning that the teacher considers the various ways in which students process and retain information and provides multiple methods of instruction to accommodate different learning preferences. This can include visual aids, group work, hands-on activities, and technology.

Another hallmark of a student-centered classroom is that the teacher incorporates students' suggestions for topics for future class discussions. This approach to learning values the ideas and interests of students, making the content of the class more relevant and engaging. This strategy also promotes active participation and student agency in their learning process.

In a student-centered classroom, the teacher encourages students to take responsibility for their own learning. This means that students are empowered to set their own learning goals, monitor their progress, and reflect on their achievements. The teacher serves as a guide and support, helping students develop the skills and strategies necessary for successful learning.

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The format of a folder target in a DFS (Distributed File System) namespace is a UNC (Uniform Naming Convention) path.

In a DFS namespace, a folder target is a location or path where the actual data is stored. It can be a shared folder on a remote server or a local path on the same server. The UNC path format is used to specify the location of the folder target. UNC paths consist of a server name or IP address followed by the share name and the directory path. For example, "\server\share\folder". This format allows clients to access the folder target regardless of the physical location or server hosting the data. DFS provides a unified view of the distributed data by using UNC paths to map the logical namespace to the physical storage.

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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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Answer:

do the wam wam

Explanation:

Answer:

4

Explanation:

Option C. Small service desks are effective if the service desk has a broad scope of responsibilities.

Small service desks can be effective if they have a broad scope of responsibilities. This is because having a wide range of responsibilities allows them to provide excellent customer service. They can be more efficient and responsive to customer needs and can manage a larger volume of calls and inquiries. This is especially true if the service desk is well-trained and equipped with the right tools and resources. Furthermore, having a small service desk can be cost-effective since it requires fewer staff and resources. Additionally, small service desks can provide world-class service if they have satisfied customers and have the right processes and procedures in place. By having a broad scope of responsibilities, small service desks are able to offer customers the best possible service, making them an effective and efficient choice.

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Answer:

the magnitude of F_A is 752 N

the direction theta of F_A is 57.9°

Explanations:

Given that,

Resultant force = 1330 N in x direction

∑Fx = R

from the diagram of the question which i uploaded along with this answer

FB = 800 N

FAsin∅ + FBcos30 = 1330 N

FAsin∅ = 1330 - (800 × cos30)

FA = 637.18 / sin∅

Now ∑Fx = 0

FAcos∅ - FBsin30 = 0

we substitute for FA

(637.18 / sin∅)cos∅ = 800 × sin30

637.18 / 800 × sin30 = sin∅/cos∅

and we know that { sin∅/cos∅ = tan∅)

so tan∅ = 1.59295

∅ = 57.88° ≈ 57.9°

THEREFORE FROM THE EQUATION

FA = 637.18 / sin∅

we substitute ∅

so FA = 637.18 / sin57.88

FA = 752 N

Answer:

the restraining forces in the x and y-directions are :

\(\mathbf{ F_x =1510.63 \ lbf}\)

\(\mathbf{F_y =6242.17 \ lbf}\)

Explanation:

From the given information;

Let us first calculate the outlet velocity by using continuity equation.

\(A_1v_1= A_2v_2\)

\(\dfrac{\pi}{4}D_1^2 *v_1 = \dfrac{\pi}{4}D_2^2 *v_2\)

where;

\(D_1\) = inlet diameter = 6 - In

\(D_2 =\) outlet diameter = 4 -In

inlet velocity \(v_1 = 5.5 \ ft/s\)

outlet velocity = ???

\(\dfrac{\pi}{4}*6^2 *5.5 = \dfrac{\pi}{4}*4^2 *v_2\)

\(36*5.5= 16 v_2\)

\(198 = 16 v_2\)

\(v_2 = \dfrac{198}{16}\)

\(v_2 = 12 .375 \ ft/s\)  to In; we have

Since 1 ft = 12 inches; thus

12.375 ft/s = (12.375 × 12 ) inches = 148.5 In/s

\(v_1 =\) 5.5 ft/s =  66 In/s

By using the linear momentum  in x-direction for the  volume ; we have the relation:

\(\sum F_x = \sum \dfrac{mdv}{dt}x \\ \\ P_1A_1 - P_2A_2 * \ cos 45 + F_x = (m_2)V_2* \ cos 45 - ( m_1)v\)

where;

\(m = \rho AV\)

\(P_1A_1 - P_2A_2 * \ cos 45 + F_x = (\rho A_2 V_2)V_2* \ cos 45 - ( \rho A_1 V_1)v\)

\(35*\dfrac{\pi}{4}*6^2 - 32.5*\dfrac{\pi}{4}*4^2 * \ cos 45 + F_x = \dfrac{52.5}{12^3}( \dfrac{\pi}{4}*4^2*(148.5)^2 * cos 45 - \dfrac{\pi}{4}*6^2*66^2)\)

\(700.81 + F_x = 2211.44\)

\(F_x = 2211.44-700.81\)

\(\mathbf{ F_x =1510.63 \ lbf}\)

\(\sum F_y =\dfrac{m dvy}{dt}\)

\(- P_2A_2 * \ Sin \ 45 + F_y = (\rho A_2 V_2 )V_2* \ sin 45 - 0\)

\(F_y = (\rho A_2 V_2 )V_2* \ sin 45 + P_2A_2 * \ Sin \ 45 +\)

\(F_ y = 32.5 * \dfrac{\pi}{4}*4^2* sin 45 + \dfrac{52.5}{12^3} *\dfrac{\pi}{4}*4^2*148.5^2*sin45\)

\(F_y = 288.79 +5953.38\)

\(\mathbf{F_y =6242.17 \ lbf}\)

Therefore; the restraining forces in the x and y-directions are :

\(\mathbf{ F_x =1510.63 \ lbf}\)

\(\mathbf{F_y =6242.17 \ lbf}\)

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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Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

\(z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}\)

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

\(50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

90.8777353 m = \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

Answer:

i think its called leggings thats wut my shop teacher told me

Explanation:

30 years

According to research it is estimated that the coal to liquid technology can extend the petroleum use upto 30 years.

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