all of the following statements are true except: a. turning the ac generator rapidly on and off to achieve a precise output is called voltage width modulation. b. the voltage regulator can be located on either the ground side or the positive side of the field circuit. c. the powertrain control module (pcm), or voltage regulator controls the current passing through the windings of the electromagnetic field in the rotor. d. electronic voltage regulation requires no moving parts, so it can cycle electronically between 10 and 7,000 times per second.

The equivalent force-couple system at O is the force and couple experienced when at point O due to the applied force at point A

The equivalent force couple system at O due to force F are;

Force, F =  (8.65·i - 4.6·j) KN

Couple, M₀ ≈ 40.9 k kN·m

The reason the above values are correct is as follows:

The known values for the cantilever are;

The height of the beam = 0.65 m

The magnitude of the applied force, F = 9.8 kN

The length of the beam = 4.9 m

The angle away from the vertical the force is applied = 26°

The required parameter:

The equivalent force-couple system at the centroid of the beam cross-section of the cantilever

Solution:

The equivalent force-couple system is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;

The equivalent force \(\overset \longrightarrow F\) = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j

Which gives;

The equivalent force \(\overset \longrightarrow F\) ≈ (8.65·i - 4.6·j) KN

The couple acting at point O due to the force F is given as follows;

The clockwise moment = 9.8 kN × cos(28°) × 4.9

The anticlockwise moment = 9.8 kN × sin(28°) 0.65/2

The sum of the moments = Anticlockwise moment - Clockwise moments

∴ The sum of the moments, ∑M, gives the moment acting at point O as follows;

M₀ = 9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9  ≈ 40.9 kN·m

The couple acting at O, due to F,  M₀ ≈ 40.9 kN·m

The equivalent force couple system acting at point O due the force, F, is as follows

F =  (8.65·i - 4.6·j) KN

M₀ ≈ 40.9 k kN·m

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Answer:

Named Binary Tag ( NBT)

Explanation:

The format is designed to store data in a tree structure made up of various tags

The meaning of the given word NBT is called; National Benchmark Test.

The meaning of the given word NBT is called National Benchmark Test.

NBT ( national benchmark test ) is defined as a national test which must be written and passed by High school students that intend to go to the university ( i.e. passport to the university ).

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At 1.2mpa pressure and 50c

What is pressure?
By pressing a knife against some fruit, one can see a straightforward illustration of pressure. The surface won't be cut if you press the flat part of the knife against the fruit. The force is dispersed over a wide area (low pressure).

a)Mass flow rate of the refrigerant
Therefore h1= condenser inlet enthalpy =278.28KJ/Kg
saturation temperature at 1.2mpa is 46.29C
Therefore the temperature of the condenser

T2 = 46.29C - 5
T2 = 41.29C

Now,
d)power consumed by compressor W = 3.3KW
Q4 = QL + w = Q4
QL = mR(h1-h2)-W
= 0.0498 x (278.26 - 110.19)-3.3
=5.074KW
Hence refrigerator load is 5.74Kg

(COP)r = 238/53
(Cop) = 4.490

Therefore the above values are the (a) mass flow rate of the refrigerant, b) the refrigerant load, c) the cop, and d) the minimum power input to the compressor for the same refrigeration load.

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Answer:

\(w_{out}=319.1\frac{BTU}{lbm}\)

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

\(h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}\)

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

\(s_2=s_1\)

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

\(x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965\)

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

\(h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}\)

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

\(h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}\)

Best regards.

The thickness of the brick layer required to ensure an outer surface temperature no greater than 40 degrees Celsius is 5.36 cm. The rate of conduction through the wall is 0.65 kW per m^2 of wall surface.

To solve this problem, we can use the one-dimensional steady-state heat conduction equation in cylindrical coordinates, with no internal heat generation and assuming constant thermal properties:

q = -kA (dT/dr)

where q is the rate of heat transfer per unit length, k is the thermal conductivity, A is the cross-sectional area, and dT/dr is the temperature gradient. We can apply this equation separately to each layer of the wall, and use the boundary conditions at the inner and outer surfaces to solve for the unknown temperatures and heat fluxes.

Using this approach, we can first solve for the temperature at the outer surface of the steel layer, which is given by:

T_steel,outer = T_steel,inner - (q_steel * L_steel) / (k_steel * A_steel)

where L_steel is the thickness of the steel layer, A_steel is its cross-sectional area, and q_steel is the rate of heat transfer per unit length in the steel layer. We know that the temperature drop across the steel layer is 0.7 degrees Celsius, so we can set:

q_steel = -k_steel * A_steel * (0.7 / L_steel)

We can then use the boundary condition at the outer surface of the brick layer to solve for the required thickness of the brick layer:

q_brick = -k_brick * A_brick * (T_brick,outer - T_steel,outer) / L_brick

where T_brick,outer is the outer surface temperature of the brick layer, and q_brick is the rate of heat transfer per unit length in the brick layer. We want T_brick,outer to be no greater than 40 degrees Celsius, so we can set:

q_brick = -k_brick * A_brick * (40 - T_steel,outer) / L_brick

Setting q_steel equal to q_brick and rearranging for L_brick, we get:

L_brick = k_brick * A_brick * L_steel / (k_steel * (40 - T_steel,outer) - 0.7 * k_brick)

Plugging in the given values, we get L_brick = 0.0536 m, or 5.36 cm.

Finally, we can calculate the rate of conduction through the wall using:

q_wall = (T_steel,inner - T_brick,outer) / (L_steel + L_brick)

Using the values we have already calculated, we get q_wall = \(0.65 kW/m^2.\)

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Scientists use a variety of methods to learn about the layers deep inside Earth. One way is by studying seismic waves, which are waves of energy that travel through the Earth's interior during earthquakes.

By analyzing how seismic waves behave as they pass through different layers of the Earth, scientists can infer the composition and properties of each layer. Another way is by examining rocks and minerals that have been brought to the surface by volcanic activity or mountain building. By analyzing the composition of these rocks, scientists can learn about the deeper layers from which they originated. Additionally, scientists use computer models and simulations to study the behavior and composition of the Earth's interior.

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Answer:

Explanation: 2 is thy answer

The plunge and bearing of the lineation are 248° and 315.5°NW, respectively

The slip lineation on the fault plane has a rake of 68° NE. Rake is the angle between the strike of the lineation and the fault. The fault is oriented N52°E and 83°SE. To calculate the plunge and bearing of the lineation, first, calculate the fault plane normal vector:
Fault plane normal vector = N52°E + 83°SE = 135.5°SE
Next, calculate the plunge and bearing of the lineation by taking the rake of the lineation and adding 180° to it.
Plunge = (68° + 180°) = 248°
Bearing = (135.5°SE + 180°) = 315.5°NW

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Answer:

Beam of 25" depth and 12" width is sufficient.

I've attached a detailed section of the beam.

Explanation:

We are given;

Beam Span; L = 20 ft

Dead load; DL = 0.50 k/ft

Live load; LL = 0.65 k/ft.

Beam width; b = 12 inches

From ACI code, ultimate load is given as;

W_u = 1.2DL + 1.6LL

Thus;

W_u = 1.2(0.5) + 1.6(0.65)

W_u = 1.64 k/ft

Now, ultimate moment is given by the formula;

M_u = (W_u × L²)/8

M_u = (1.64 × 20²)/8

M_u = 82 k-ft

Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.

Effective depth of a beam is given by the formula;

d_eff = d - clear cover - stirrup diameter - ½Main bar diameter

Now, let's adopt the following;

Clear cover = 1.5"

Stirrup diameter = 0.5"

Main bar diameter = 1"

Thus;

d_eff = 25" - 1.5" - 0.5" - ½(1")

d_eff = 22.5"

Now, let's find steel ratio(ρ) ;

ρ = Total A_s/(b × d_eff)

Now, A_s = ½ × area of main diameter bar

Thus, A_s = ½ × π × 1² = 0.785 in²

Let's use Nominal number of 3 bars as our main diameter bars.

Thus, total A_s = 3 × 0.785

Total A_s = 2.355 in²

Hence;

ρ = 2.355/(22.5 × 12)

ρ = 0.008722

Design moment Capacity is given;

M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12

Φ is 0.9

f’c = 4,000 psi = 4 kpsi

fy = 60,000 psi = 60 kpsi

M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12

M_n = 220.03 k-ft

Thus: M_n > M_u

Thus, the beam of 25" depth and 12" width is sufficient.

The correct answer is RJ11 connector is used to connect an analog modem to a telephone line socket.

RJ11: As previously mentioned, analog telephony made extensive use of RJ11 to link the phone and cable. It is still used in landlines today and is mostly used to connect to modems. It cannot fit into an RJ45 slot since it is a 4-slot connector with six pins. RJ11 connects to telephone equipment while RJ45 is typically used with Ethernet cables. RJ45 has eight wires inside, whereas RJ11 only has four. RJ45 is larger than RJ11 in size. The RJ45 cable connector can be plugged into an RJ11 jack, interface, port, or slot, but not the other way around.

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Potential (or voltage) transformer is a precision two-winding transformer used to step down high voltage to allow safe voltage measurement.

The type of transformer being described in the question is called a potential transformer or a voltage transformer. It is a specialized transformer that is designed to step down high voltage to a lower voltage, typically 120 or 240 volts, for safe measurement or monitoring purposes. Potential transformers are commonly used in power systems to provide accurate voltage measurements for meters, relays, and other protective devices.

They are also used in electrical testing and calibration to accurately measure high voltages without the risk of electrical shock or damage to the measuring instrument.

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Answer:ALL FOUR PAIRS

Explanation: first major difference is the gigabit standards require the use of all four pairs (all eight wires), unlike Fast Ethernet which only utilizes two pairs of wires. As a result, in Gigabit Ethernet, all four pairs must be crossed when building a Crossover cable

When making a Gigabit Ethernet crossover cable, two pairs of wires get switched. Ethernet cables are used to connect computer networks that enable communication between different devices. The two most popular types of Ethernet cables are crossover and straight cables.

Ethernet cables use four pairs of wires to transmit data, out of which two pairs are utilized for transmitting and two pairs are used for receiving. In a straight-through Ethernet cable, the wire arrangement is identical at both ends. For this reason, a straight-through cable is employed to connect devices that belong to the same category. In contrast, a crossover Ethernet cable is employed to connect devices belonging to different categories. A crossover cable reverses the transmitting and receiving pairs, allowing the devices to communicate with one another. In a Gigabit Ethernet crossover cable, two pairs of wires are swapped or switched. These pairs are Pins 1-2 (White and Orange) and Pins 3-6 (White and Green). Therefore, we can conclude that two pairs of wires get switched when making a Gigabit Ethernet crossover cable.

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Yes, there would be a difference in MT or tapping frequency if you did not control for errors (i.e., missed taps 8) in either Experiment 1 or 2. The reason is that errors, including missed taps, can impact the tapping frequency, which can in turn affect MT (movement time).

If missed taps occur, it can lead to an increase in movement time, and as a result, a decrease in tapping frequency. This can happen if a person takes longer to correct a missed tap, or if they are unable to complete the intended sequence of taps. Additionally, missed taps can lead to an increase in the variability of tapping frequency, which can further impact MT.

Another factor that can affect MT or tapping frequency is the type of task being performed. For instance, if the task involves complex movements, such as playing a musical instrument, then the presence of errors can have a greater impact on MT and tapping frequency. In contrast, if the task is relatively simple, such as tapping a pencil on a desk, then the effect of errors may be less pronounced.

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Answer:

Let's look at the most common problems and tips for overcoming IT challenges.

1) DISTRIBUTE YOUR BUDGET CORRECTLY. The cost of new technologies often stops small businesses from adopting them. ...

2) HIRE AN INTERNAL TEAM OR THIRD-PARTY IT SPECIALISTS. ...

3) UNDERSTAND THE IT SPHERE. ...

4) MAKE THE SYSTEM SAFE. ...

5) SCALE IT. ...

CONCLUSION.

Answer:

Diameter of pipe is 0.535 ft

Explanation:

see attachment, its works out 1st half

D) Identify the problem step comes first in the decision-making process.

In psychology, decision-making (also spelled decision making and decisionmaking) is regarded as the cognitive process resulting in the selection of a belief or a course of action among several possible alternative options. It could be either rational or irrational. The decision-making process is a reasoning process based on assumptions of values, preferences and beliefs of the decision-maker. Every decision-making process produces a final choice, which may or may not prompt action. Research about decision-making is also published under the label problem solving, particularly in European psychological research. Decision-making can be regarded as a problem-solving activity yielding a solution deemed to be optimal, or at least satisfactory. It is therefore a process which can be more or less rational or irrational and can be based on explicit or tacit knowledge and beliefs. Problem solving and critical thinking refers to the ability to use knowledge, facts, and data to effectively solve problems.

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Energy transfer is factor that affects unit operation this is because each processing of change requires energy transfer.

Unit operations involves reactions that lead to physical change or chemical transformation.

It include separating of mixtures which can be done by filtering, crystallization and polymerization.

Unit operation often results into changes that can be seen or visible changes.

Energy transfer is one the factors that affect unit operation.

Therefore, Energy transfer is factor that affects unit operation this is because each processing of change requires energy transfer.

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Answer: So you are dealing with maximum and minimum weights and you want to know what MINIMUM number of supporting strands for this block and tackle system are needed I believe. If so you are dealing with economic imbalances Though we are not worrying about money Right? Right we need physics which Physics study matter and how it moves You would need 8 STRANDS

Explanation: Step By Step

Explanation:

उह्ह्नमजज्ल्ह्ह्बनुतनकुहक्जो

Answer:

it should be A bc I took the test a while back

Answer: See explanation

Explanation:

69.48 / 7.2 = 965 as written in the question isn't correct. The correct answer should have been:

69.48 / 7.2 = 9.65

It should be noted that when we divide, the answer that one gets cannot be more than the two numbers that were used for the division. Hence, there's no way that 965 would be more than 69.48 and 7.2.

Answer:

Quick Breads

Explanation:

I think this is correct. I am not for sure i just looked it up and it says word for word that this is the answer.

Answer:

Quick Breads

Explanation:

I got it right on my Plato test

Answer:

A) About \(9.273 \times 10^{-9}\) newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) \(F_g=\dfrac{GM_1M_2}{r^2}\) , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

\(F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}\)

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

Answer:

Explanation:

For first isobaric process (expansion with constant pressure):

\(W_{1}=P_{1}(V_{2}-V_{1})=(0.43\times 10^{6})(2.9-0.28)=1.1266 \times 10^{6} J=1.1266 MJ(+) (Expansion)\)

Now, for the first isochoric: \(W_{2}=0\), and \(Q_{2}=\Delta U=U_{3}-U_{2}\)

For the second isobaric process (compression with constant pressure):

\(W_{3}=P_{2}(V_{1}-V_{2})=(4.3\times 10^{6})(0.28-2.9) =-11.266\times 10^{6}J=-11.266 MJ (Compression)\)

For the last isochoric: \(W_{4}=0, Q_{4}=\Delta U=U_{4}-U_{3}\)

So, the total work per cycle: \(W=W_{1}+W_{2}+W_{3}+W_{4}=1.1266+0-11.266+0=-10.1394 MJ\)

As a pilot operating under 14 CFR Part 121, you can be contacted about your next-day duty assignment during the last two hours of the required rest period.

The FAA regulations under 14 CFR Part 121, specifically 121.471, states that a pilot's rest period must be at least 8 consecutive hours in a 24-hour period. During the last two hours of this rest period, the pilot may be contacted by their employer to discuss their next day duty assignment.

The purpose of this regulation is to ensure that the pilot is well-rested and ready to perform their duties safely. It's also a way to ensure that the pilots are not overworked and fatigued which can impact their performance and decision-making ability while flying.

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Answer:

  see attached

Explanation:

As you know, a full adder produces the binary value in (carry, output) that is the the number of true input bits among the (carry in, A, B) inputs.

The annotated diagram attached shows the bit values for a 4-bit full adder/subtractor.

__

Additional comment

If the numbers are considered "signed", then the top diagram will generate an "overflow" based on the difference between the carry C4 and the output O4.

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

1. Given the below input signal, VIN(t), and the phasors of the transfer function at various frequencies, calculate the expression for the output signal, VOUT(t) in the form VOUT(t)=Acos(200πt+θ1)+Bcos(400πt+θ2).

What is the value of A?

VIN(t)=10cos(200πt+30)+20cos(400πt+45)

H(f)=0.1∠10      f=100Hz

H(f)=1∠−20      f=200Hz

H(f)=2∠30      f=300Hz

H(f)=3∠−40     f=400Hz

H(f)=4∠50      f=500Hz

2. In the same question, what is the value of θ1 in degrees? Enter the value in the box below without the units.

3. In the same question, what is the value of B? Enter the value in the box below without the units.

4. In the same question, what is the value of θ2 in degrees? Enter the value in the box below without the units.

Answer:

Vout(t) = 10cos(200πt + 10°) + 60cos(400πt + 5°)

1. What is the value of A?

A = 10

2. In the same question, what is the value of θ1 in degrees?

θ₁ = 10°

3. In the same question, what is the value of B?

B = 60

4. In the same question, what is the value of θ2 in degrees?

θ₂ = 5°

Explanation:

The output signal Vout(t) is in the form

Vout(t)=Acos(200πt + θ₁) + Bcos(400πt + θ₂)

The input signal Vin(t) is given as

Vin(t) = 10cos(200πt + 30°) + 20cos(400πt + 45°)

The output signal Vout(t) is found by

Vout(t) = H(f) × Vin(t)

Where H(f) is the transfer function at various frequencies and Vin(t) is the input signal.

H(200) = 1 ∠−20°

H(400) = 3 ∠−40°

converting the input signal into phasors

10cos(200πt + 30°) = 10 < 30°

20cos(400πt + 45°) = 20 < 45°

Vout(t) = H(200)×10cos(200πt + 30°) + H(400)×20cos(400πt + 45°)

Vout(t) = (1 ∠−20°)×10cos(200πt + 30°) + (3 ∠−40°)×20cos(400πt + 45°)

Vout(t) = (1 ∠−20°)×(10 < 30°) + (3 ∠−40°)×(20 < 45°)

Multipy the magnitude and add phase angles together

Vout(t) = (1×10 ∠−20° + 30°) + (3×20 ∠−40° + 45°)

Vout(t) = 10 ∠ 10° + 60 ∠5°

Vout(t) = 10cos(200πt + 10°) + 60cos(400πt + 5°)

Comparing it with the general form

Vout(t)=Acos(200πt + θ₁) + Bcos(400πt + θ₂)

1. What is the value of A?

A = 10

2. In the same question, what is the value of θ1 in degrees?

θ₁ = 10°

3. In the same question, what is the value of B?

B = 60

4. In the same question, what is the value of θ2 in degrees?

θ₂ = 5°