As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time t. Know that the rotor starts from rest at t = 0 and that after 10 s the rotor has completed 20 revolutions.Choose the correct equations of motion for the rotor. (You must provide an answer before moving on to the next part.)
a) a = 2kt, w = 3krº, and 8 = 4kr
b) a = {kt, w = ke?, and 0 = }ke?
c) a = kr?, w = jke', and 0 = krº
d) a = kt, w = jke?, and 0 kr

Both Technician A and Technician B are correct regarding the light film of oil observed on the upper area of a shock absorber.

Technician A says that the condition of the light film of oil observed on the upper area of a shock absorber should be considered normal. Technician B says that during cold weather, a rod seal may bleed fluid, causing the oil film. Both of them are correct because a light film of oil on the upper area of a shock absorber may result from any of the following: Heat generated during shock operation and component temperature differences may cause oil to expand and migrate up the piston rod, which may appear as a light film on the upper part of the shock absorber's housing.

During cold weather, a rod seal may bleed fluid, resulting in an oil film on the shock absorber's upper area. During cold weather, fluid seepage past a seal due to lower viscosity is typical. Leakage is occasionally observed when the shock absorber has been in operation for a longer period. A certain amount of leakage past the rod seal and out the weep hole is typical in most high-pressure hydraulic systems. This leakage might be misting or a trickle of fluid.

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Answer:

Vehicle manufacturers specify various reference points on the frame, body, fender, bumper or suspension. On some applications, ride height may even be specified as a distance between two points on the vehicle's suspension, such as the distance from the frame rail to the axle or a suspension control arm.

Answer:

4c4u4civ4j4

Explanation:

x24uu37z3x3uzu2

Answer: A hand-held sledgehammer

Answer:

The answer to your question is a sledgehammer

Explanation:

A hand-held sledgehammer can be used when a great deal of driving power is required.

I hope this helps and have a wonderful day!

Answer: i believe it’s acoustics

Explanation:

Answer:people tend to do this when they are in a different environment they lose something or just have something going on in their life

Explanation:

Decomposition is the process of breaking the Work Breakdown Structure (WBS) into smaller and smaller deliverables. This process is also sometimes referred to as value engineering or detailed specifications. By decomposing the WBS into smaller pieces, it becomes easier to assign tasks, assign costs, and plan out timelines.

The decomposition process begins by taking the major deliverables of the project and breaking them down into smaller tasks. From there, each task is further broken down into even more specific tasks. This process is repeated until all tasks have been broken down into their smallest components.

The purpose of decomposition is to create a well-defined scope of the project so that it can be managed in an efficient manner. It allows managers to easily identify the resources, cost, and timeline of each task, as well as provide a way to evaluate the progress of each task. It also allows for better control of the overall project.

Decomposition is a critical part of the project management process, as it ensures the project is organized and defined. This ultimately leads to an overall better result for the customer.

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9514 1404 393

Answer:

  A.  15 cm²

  B.  15 cm²

Explanation:

The diagram shows the values of a, b, h. Put these in the original formula to find the area.

  A = 0.5(a+b)h

  A = 0.5(3 +7 cm)(3 cm) = 0.5(10 cm)(3 cm) = 15 cm²

__

The base of each triangle is half the difference between the two base lengths.

  triangle base = (7 -3)/2 = 2 . . . cm

The area of a triangle is given by the formula ...

  A = 0.5bh

  A = 0.5(2 cm)(3 cm) = 3 cm²

The area of the central rectangle is given by ...

  A = bh

  A = (3 cm)(3 cm) = 9 cm²

Then the area of the trapezium is ...

  A(triangle) +A(triangle) +A(rectangle) = A(trapezium)

  (3 cm²) +(3 cm²) +(9 cm²) = A(trapezium) = 15 cm²

Answer:

The building that is erected on land that has a ground lease is typically owned by the tenant who has leased the land from the owner of the property. However, the terms of the ground lease may specify that the building and other improvements on the land revert to the owner of the property at the end of the lease term.

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Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V\(_s\)/( 1 - D )

given that; V\(_s\) = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

\(I_L\) = V\(_s\) / ( 1 - D )²R

given that R = 12.5 Ω, V\(_s\) = 20 V, D = 0.6

we substitute

\(I_L\) = 20 / (( 1 - 0.6 )² × 12.5)

\(I_L\) = 20 / (( 0.4)² × 12.5)

\(I_L\) = 20 / ( 0.16 × 12.5 )

\(I_L\) = 20 / 2

\(I_L\) = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

\(I_{Lmax\) = [V\(_s\) / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V\(_s\) = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

\(I_{Lmax\) = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

\(I_{Lmax\) = [20 / 2 ] + [ 60 / 20 ]    

\(I_{Lmax\) = 10 + 3

\(I_{Lmax\) = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

\(I_{Lmax\) = [V\(_s\) / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V\(_s\) = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

\(I_{Lmin\) = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

\(I_{Lmin\) = [20 / 2 ] -[ 60 / 20 ]    

\(I_{Lmin\) = 10 - 3

\(I_{Lmin\)  = 7 A

Therefore, the maximum inductor current is 7 A

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

\(I_D\) = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

\(I_D\) = 50 / 12.5

\(I_D\) = 4 A

Therefore, the average current in the diode under ideal components is 4 A

h = 129.2m

Pressure at depth is equal to denstiy of the fluid times the acceleartion due to gravity times the depth:

P = pgh

density at the surface of the ocean is at .101MPa

Take the pressure subtract the pressure due to atmosphere:

1.4 - .101 = 1.299 MPa

1299000 Pa = 1025kg/m^3 * 9.8m/s^2 * h

h = 129.2m

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Answer: Option B

Explanation:

In order to enhance net radiation exchange between a high temperature surface and a low temperature surface in an enclosure, one should: paint both surfaces so that they are close to a blackbody surface but don't polish both surfaces so that they are highly reflective

Answer:

a) 3.5 m

b) 14 secs

c) 1.4 secs

Explanation:

a)  Determine the distance the particle will travel

given velocity ( final velocity ) = 5 m/s

v^2 = u^2 + 2as

s = ( v^2 - u^2 ) / 2a

  = ( 5^2 - 8^2 ) / 2 ( -0.5 * 5^3/2 )

  = 3.5 m

b) Determine the time when v = 1m/s

V = u + at

1 = 8 + (  -0.5 * 1^3/2 ) * t

∴ t = 14 secs

c) Determine the time required for particle to travel 8 m

we will employ both equations above

V^2 = u^2 + 2as

s = 8 m , V = unknown , u = 8 m/s   back to equation

V^2 = 8^2 + 2 ( - 1/2 * V^3/2 ) * 8

∴ V^2 + 8V^3/2 - 64 = 0

resolving the above equation

V = 3.478 m/s

now using the second equation

V = u + at

3.478 = 8 + ( - 1/2 * 3.478^3/2 ) * t

hence : t = 1.4 secs

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

\(P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}\)

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\(\mu\) is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that \(\mu = 0.8\)

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

\(P(X \leq 4) + P(X > 4) = 1\)

We want P(X > 4). Then

\(P(X > 4) = 1 - P(X \leq 4)\)

In which

\(P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)\)

\(P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}\)

\(P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493\)

\(P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595\)

\(P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438\)

\(P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383\)

\(P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077\)

\(P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986\)

\(P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014\)

0.14% probability of observing more than 4 errors in the carpet

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

class RepairSchedule def initialize(mechanics, bays, jobs)

mechanics = mechanics bays = bays jobs = job end

Initialize is the process of setting a variable, object, or data structure to a predefined value before it is used. This value is usually determined by the programming language or the program itself, and is usually set to ensure proper functioning of the program or data structure. Initializing variables in programming is often used to make sure that a variable is properly set up before being used in any calculations or operations. It can also help to avoid any unexpected errors or data corruption. Initializing objects can also help to make sure that an object has the correct properties and methods when it is being used. This can help to make sure that the code is running as expected and avoid any unexpected errors.

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True, phototransistors are becoming more common than photoresistors because silicon is more resistant to high temperatures and airborne contamination.

Phototransistors are electronic switching and power amplification components that rely on light to operate. When light hits the junction, it causes a return current proportional to its brightness. Phototransistors are often used to capture pulses of light and convert them into digital electrical signals.

Phototransistors have several key advantages that set them apart from other optical sensors. Some of them are listed below

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Answer:

The answer is V = √2FD ÷ CD × PA

Explanation:

FD = CD × PV²A ÷ 2

V² = 2FD ÷ CD × PA

V = √2FD ÷ CD × PA

Thus, The value of V is V = √2FD ÷ CD × PA

-TheUnknownScientist 72

When the process is in control but does not meet specification, it is referred to as a special cause error.

In statistical process control, a process is considered to be in control when it operates within the defined limits and shows only random variations. However, when a process is in control but does not meet the desired specifications, it indicates the presence of a special cause error.

Special cause errors are attributed to specific factors or events that cause the process to deviate from the expected outcome. These errors are typically unpredictable and require investigation and corrective action to bring the process back within the desired specifications.

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Answer:

Gc(s) = \(\frac{0.1s + 0.28727}{s}\)

Explanation:

comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

attached is the detailed solution and the plot in Matlab

Answer:

a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.

Explanation:

Maximum theoretical efficiency for a power cycle (\(\eta_{r}\)), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:

\(\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\%\) (1)

Where:

\(T_{C}\) - Temperature of the cold reservoir, in Kelvin.

\(T_{H}\) - Temperature of the hot reservoir, in Kelvin.

The maximum theoretical efficiency associated with this power cycle is: (\(T_{C} = 400\,K\), \(T_{H} = 1200\,K\))

\(\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%\)

\(\eta_{r} = 66.667\,\%\)

In exchange, real efficiency for a power cycle (\(\eta\)), no unit, is defined by this expression:

\(\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\%\) (2)

Where:

\(Q_{C}\) - Heat released to cold reservoir, in kilojoules.

\(Q_{H}\) - Heat gained from hot reservoir, in kilojoules.

\(W_{C}\) - Power generated within power cycle, in kilojoules.

A power cycle operates irreversibly for \(\eta < \eta_{r}\), reversibily for \(\eta = \eta_{r}\) and it is impossible for \(\eta > \eta_{r}\).

Now we proceed to solve for each case:

a) \(Q_{H} = 900\,kJ\), \(W_{C} = 450\,kJ\)

\(\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%\)

\(\eta = 50\,\%\)

Since \(\eta < \eta_{r}\), the power cycle operates irreversibly.

b) \(Q_{H} = 900\,kJ\), \(Q_{C} = 300\,kJ\)

\(\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%\)

\(\eta = 66.667\,\%\)

Since \(\eta = \eta_{r}\), the power cycle operates reversibly.

c) \(W_{C} = 600\,kJ\), \(Q_{C} = 400\,kJ\)

\(\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%\)

\(\eta = 60\,\%\)

Since \(\eta < \eta_{r}\), the power cycle operates irreversibly.

d) Since \(\eta > \eta_{r}\), the power cycle is impossible.

Try rebooting the router by unplugging it from the power source, then plugging it back in.

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Answer:

I think it is false!

Explanation:

Answer: I think it's true

Explanation:

Because if you were part of a play, you would have a part but if you work on props, you don't have a part onstage.

Since Walter is editing the camera in his workspace, if he wants to adjust how much of the scene can be seen all at once, the property which he can edit is: D. field of view.

A camera operator can be defined as an individual who is professionally trained to tweak, operate and use a digital camera, especially for the purpose of taking pictures or images (photographs) of an event, place, person, etc.

A camera lens can be defined as a transparent optical instrument that is used in conjunction with a digital camera, so as to refract the rays of light coming into the lens.

A field of view can be defined as a range of the observable physical world that is visible to a camera operator at any period of time, especially through the human eye, on a display monitor (screen), or a digital camera viewfinder.

In this context, we can reasonably infer and logically deduce that the property which Walter can edit is the field of view of his digital camera.

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Answer:

See explanation

Explanation:

Since no figure was given, I'll explain how to do this problem theoretically. The formula for shear flow is \(q=\frac{VQ}{I}\) where V is the shear force, Q is the moment of area (more on this later), and I is the moment of inertia.

The first step to solve this problem is to find the resultant internal forces of the beam. This can be done in several ways, but the easiest is to solve the beam statically and draw a shear diagram to determine the maximum shear force V.

The second step to solving this problem is to determine the location of the neutral axis of the cross section if it is not given. The formula for the neutral axis is  \(NA = \frac{\sum y*A}{\sum A}\). The y in this equation represents the middle of the small shapes that the web is divided into. An I-beam can be thought of as 3 rectangles, while a T-beam can be thought of as 2. The A in this formula represents the area of each of the rectangles (an I-beam will have 3 of these and a T-beam will have 2).

The third step for this problem is to find the moment of inertia. There are several formulas for moment of inertia depending on the shape of the cross section. I-beam's and T-beams both can be thought of as multiple rectangles, so they have the same base formula of \(I=\frac{1}{12}bh^3\) where b is the base of the rectangles and h is the height. For I-beams, the easiest way to calculate moment of inertia is to think of the entire cross section as a big rectangle that had two smaller rectangles cut out of it. The formula for this moment of inertia becomes \(I=\frac{1}{12} b_{big}h^{3} _{big}-\frac{1}{6}b_{small}h^{3}_{small}\). Note that this form of moment of inertia already takes into account subtracting 2 small rectangles. For T-beams, this approach will not work, so the parallel axis theorem must be used. The moment of inertia for the T-beam becomes \(I=\frac{1}{12}b_{1} h^{3}_{1} +b_{1}h_{1}dy_{1}^{2} +\frac{1}{12}b_{2} h^{3}_{2} +b_{2}h_{2}dy_{2}^{2}\) where the terms with the subscript 1 represent the first rectangle and the terms with the subscript 2 represent the second rectangle. The dy terms represent the distance from the center of that specific rectangle to the neutral axis.

The fourth step for this problem is to find Q. The formula to find Q is \(Q=\sum y'A'\) where y' represents the distance from the neutral axis to the center of the "wanted" point and A' is the area of the rectangle that has the wanted point at its center. (This would be the area above or below the thickness (t) if you were solving for maximum shear \(\tau=\frac{VQ}{It}\)).

The last step for this problem is to substitute the found values into the formula for shear flow \(q=\frac{VQ}{I}\). V came from step 1, Q came from step 4, and I came from step 3.

Answer:

Sarah is a professional interior designer.

Therefore, Sarah carries the title of a _registered___ interior designer.

Explanation:

Some states require only the licensed designers to carry out interior designs, while some others allow both the licensed and unlicensed interior designers to carry on the work of interior designers.  In the states where only the licensed designers are allowed to use the title of interior designer, they are registered after passing the state-prescribed examinations and meeting some professional experience requirements. The most common exam is the National Council for Interior Design Qualification (NCIDQ) examination, which only bachelor degree holders in interior design are allowed to sit for.

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:

The plotted diagram of line distance versus the transmission speed is uploaded below.

Explanation:

Given the data in the question;

Bit length B = 1000 bits

propagation velocity V = 2 × 10⁸ m/s

now, we know that the bit length of a link is expressed as;

B = R × d/V

where V is propagation velocity

d is the distance

R is the transmission speed

B is bit length

so we substitute

1000 = R × d/(2 × 10⁸)

1000 = Rd/(2 × 10⁸)

2 × 10¹¹ = Rd

R = 2 × 10¹¹ / d

R = 2E+11

Hence, we plot the transmission speed versus line distance; as shown in the image BELOW.

From the plot, if the transmission speed increases, the distances between stations decreases and vise versa.

Hence, both are inversely proportional.

The graph plot of line distance versus the transmission speed for a bit length of 1000 bits is; plotted below with Lν = 2 × 10¹¹ m/s

We are given;

Bit length; B = 1000 bits

Propagation velocity; V = 2 × 10⁸ m/s

Formula for bit length of a link is expressed as;

B = ν × L/V

where;

V is propagation velocity

L is the distance

ν is the transmission speed

B is bit length

Thus, plugging in the relevant values gives;

1000 = ν × L/(2 × 10⁸)

Thus;

Lν = 1000 × 2 × 10⁸

Lν = 2 × 10¹¹ m/s

Thus, find the attached image of a graph showing the line distance versus the transmission speed for a bit length of 1000 bits.

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Answer:

sa china po

Explanation:

sana makatulong ako