An n++- p+- n transistor has a base transport factor of 0.998, an emitter efficiency of 0.997 and an Icp of 10 nA.
a.Calculate αo and βo.
b.If the base current is zero (i.e. the base terminal is open-circuited), what is the emitter current (hint: Collector-base leakage current is Icp in n-p-n transistors)?

Answer:

D???

Explanation:

Answer:

performance bonuses

Explanation:

To calculate the theoretical power required to operate the compressor in a 10-ton-capacity refrigeration system using ammonia (r-717) under saturated conditions, we can use the following formula:Power = mass flow rate x specific enthalpy changeFirst, we need to determine the mass flow rate of the refrigerant. We can do this using the following formula:

Mass flow rate = refrigeration capacity / (specific enthalpy change x refrigerant density)
Since the refrigeration capacity is given as 10 tons, we need to convert this to kilowatts (kW) by multiplying by 3.517:
Refrigeration capacity = 10 tons x 3.517 kW/ton = 35.17 kW
Next, we need to determine the specific enthalpy change of the refrigerant. This can be found by subtracting the specific enthalpy of the refrigerant in the evaporator from the specific enthalpy of the refrigerant in the cSpecific enthalpy change = h2 - h1
To find the specific enthalpies, we can use a refrigerant table. For ammonia (r-717) at 210 kPa, the specific enthalpy in the evaporator (h1) is 349.7 kJ/kg, and at 750 kPa, the specific enthalpy in the condenser (h2) is 400.8 kJ/kg.Substituting these values into the formula, we get:Specific enthalpy change = 400.8 - 349.7 = 51.1 kJ/kg
Finally, we need to determine the refrigerant density. At 210 kPa, the density of ammonia (r-717) under saturated conditions is 480 kg/m³.Substituting all of these values into the mass flow rate formula, we get:
Mass flow rate = 35.17 kW / (51.1 kJ/kg x 480 kg/m³) = 1.007 kg/s
Now we can use the power formula to calculate the theoretical power required to operate the compressor:Power = 1.007 kg/s x 51.1 kJ/kg = 51.48 kWTherefore, the theoretical power required to operate the compressor in a 10-ton-capacity refrigeration system using ammonia (r-717) under saturated conditions is approximately 51.48 kW.

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Answer:

a) i) The maximum pressure is approximately 122.37 bar

ii) The thermal efficiency is approximately 56.47%

iii) The mean effective pressure is approximately 20.974 bar

b) (b) Four types of internal combustion engine includes;

1) The diesel engine

2) The Otto engine

3) The Brayton engine

4) The Wankel engine

Explanation:

The parameters of the Otto cycle are;

The heat added, \(Q_{in}\) = 2,800 kJ/kg

The compression ratio, r = 8

The beginning compression pressure, P₁ = 1 bar

The beginning compression temperature, T₁ = 300 K

Cp = 1.005 kJ/kg·K

Cv = 0.718 kJ/kg·K

R = 287 kJ/kg·K

K = Cp/Cv = 1.005 kJ/kg·K/(0.718 kJ/kg·K) ≈ 1.4

T₂ = T₁×r^(k - 1)

∴ T₂ = 300 K×8^(1.4 - 1) ≈ 689.219 K

\(\dfrac{P_1\cdot V_1}{T_1} = \dfrac{P_2\cdot V_2}{T_2}\)

\(P_2 = \dfrac{P_1\cdot V_1 \cdot T_2}{T_1 \cdot V_2} = \dfrac{V_1}{V_2} \cdot \dfrac{P_1 \cdot T_2}{T_1 } = r \cdot \dfrac{P_1 \cdot T_2}{T_1 }\)

∴ P₂ = 8 × 1 bar × (689.219K)/300 K ≈ 18.379 bar

\(Q_{in}\) = m·Cv·(T₃ - T₂)

∴ \(Q_{in}\) = 2,800 ≈ 0.718 × (T₃ - 689.219)

T₃ = 2,800/0.718 + 689.219 = 4588.94 K

P₃ = P₂ × (T₃/T₂)

P₃ = 18.379 bar × 4588.94K/(689.219 K) = 122.37 bar

The maximum pressure = P₃ ≈ 122.37 bar

(ii) The thermal efficiency, \(\eta_{Otto}\), is given as follows;

\(\eta_{Otto} = 1 - \dfrac{1}{r^{k - 1}}\)

Therefore, we have;

\(\eta_{Otto} = 1 - \dfrac{1}{8^{1.4 - 1}} \approx 0.5647\)

The thermal efficiency, \(\eta_{Otto}\) ≈ 0.5647

Therefore, the thermal efficiency ≈ 56.47%

(iii) The mean effective pressure, MEP is given as follows;

\(MEP = \dfrac{\left(P_3 - P_1 \cdot r^k \right) \cdot \left(1 - \dfrac{1}{r^{k-1}} \right)}{(k -1)\cdot (r - 1)}\)

Therefore, we get;

\(MEP = \dfrac{\left(122.37 - 1 \times 8^{1.4} \right) \cdot \left(1 - \dfrac{1}{8^{1.4-1}} \right)}{(1.4 -1)\cdot (8 - 1)} \approx 20.974\)

The mean effective pressure, MEP ≈ 20.974 bar

(b) Four types of internal combustion engine includes;

1) The diesel engine; Compression heating is the source of the ignition, with constant pressure combustion

2) The Otto engine which is the internal combustion engine found in cars that make use of gasoline as the source of fuel

The Otto engine cycle comprises of five steps; intake, compression, ignition, expansion and exhaust

3) The Brayton engine works on the principle of the steam turbine

4) The Wankel it follows the pattern of the Otto cycle but it does not have piston strokes

Answer: \(p_{B} - p_{A}\) = 28800 Pa or 28.8 kPa

Explanation: To determine the pressure of a liquid in a rotating tank,it is used:

p = \(\frac{p_{fluid}.w^{2}.r^{2} }{2}\) - γfluid . z + c

where:

\(p_{fluid}\) is the liquid's density

w is the angular velocity

r is the radius

γfluid.z is the pressure variation due to centrifugal force.

For this question, the difference between a point on the circumference and a point on the axis will be:

\(p_{B} - p_{A}\) = \(\frac{p_{fluid}.w^{2}.r_{B} ^{2} }{2}\) - γfluid.\(z_{B}\) - (\(\frac{p_{fluid}.w^{2}.r_{A} ^{2} }{2}\) - γfluid.\(z_{A}\))

\(p_{B} - p_{A}\) = \(\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )\) - γfluid(\(z_{B}\) -\(z_{A}\))

Since there is no variation in the z-axis, z = 0 and that the density of oil is 0.9.10³kg/m³:

\(p_{B} - p_{A}\) = \(\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )\)

\(p_{B} - p_{A} = \frac{0.9.10^3.40^2}{2}(0.2^2 - 0)\)

\(p_{B} - p_{A}\) = 28800

The difference in pressure between two points, one on the circumference and the other on the axis is \(p_{B} - p_{A}\) = 28800 Pa or 28.8 kPa

Answer:

(a) S₁-S₂ = 0 ⇒ S₁=S₂ (b) S₁-S₂<0 ⇒ S₁<S₂ (c) S₁-S₂>0 ⇒ S₁>S₂ (d)S₁-S₂>0 ⇒ S₁>S₂

Explanation:

Solution

Now,

From the entropy rate balance at steady state, the equation is stated below:

Qcv =/Tb + m (S₁-S₂) +σcv = 0

ΔS = Qcv/Tb +σcv

So,

S₁-S₂ = Qcv/mTb +σcv/m

(a) No internal irreversibilities and Q cv =0

Now,

If Qcv =0, the and value of σcv = 0

S₁-S₂ = (Qcv/mTb + σcv/m) = 0

hence

S₁-S₂ = 0 which is S₁=S₂

(b) no internal irreversibilities, and Q cv <0

If Q cv <0 the value becomes this,

S₁-S₂ = (Qcv/mTb + σcv/m) <0

So,

S₁-S₂<0 which is S₁<S₂

(c) no internal irreversibilities and Q cv >0

If Q cv >0 the value is

S₁-S₂ = (Qcv/mTb + σcv/m) >0

So,

S₁-S₂>0 which is S₁>S₂

(d) internal irreversibilities and  Qcv ≥ 0

If Q cv ≥0 the value is

S₁-S₂ = (Qcv/mTb + σcv/m) >0

So,

S₁-S₂>0 which is S₁>S₂

Note: option (d) and (e) are the same

Answer:

The 17 distinct years Mariah Carey has topped Billboard's Hot 100, spanning an historic four consecutive decades:

1990

1991

1992

1993

1994

1995

1996

1997

1998

1999

2000

2005

2006

2008

2019

2020

2021

Answer:

Photogenerated entangled electron spin pairs provide a versatile source of molecular qubits. Here, we examine the spin-dependent dynamics of a covalent donor–acceptor–radical molecule, D-A-R•, where the donor chromophore (D) is peri-xanthenoxanthene (PXX), the acceptor (A) is pyromellitimide (PI), and the radical (R•) is α,γ-bisdiphenylene-β-phenylallyl (BDPA). Selective photoexcitation of D within D-A-R• in

Answer:

Agricultural engineer

In green-sand casting, drafts are essential because they provide a gradual slope in the molds that allows the casting to be released easily.

Drafts are not as important in permanent-mold casting because the mold is generally made of metal and can be more easily broken apart. Drafts can still be used in permanent-mold casting, but they are not as necessary.

Greensand is a mixture of quartz sand, water and bentonite. The sample product used is a 90o elbow measuring 0.5 inches with white cast iron material. The surface roughness was observed by visual observation of the casting results of the two green sand mold compositions.

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Answer:

Alternating Current (a.c) can be reversed since it relates to sinusoidal curves while Direct Current (d.c) can not be reversed.

Explanation:

\(.\)

Difference between AC and DC

Electric current flows in two ways as an alternating current (AC) or direct current (DC). In alternating current, current keeps switching directions periodically – forward and backward. While in the direct current it flows in a single direction steadily. The main difference between AC and DC lies in the direction in which the electrons flow. In DC, the electrons flow steadily in a single direction while electrons keep switching directions, going forward and then backwards in AC. Let us learn more differences between them in the next few sections.

The engineering method of separating costs is d. All of these choices are correct.

The engineering method of separating costs involves a step-by-step analysis of various elements involved in the production process to estimate the cost of activities and new products.

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An access control system controls access to area restricted area or facility using physical and computer-related devices, equipment along with authentication and authorization mechanisms.

Note that genetically modified rice, called golden rice, is an example of how genetic engineering is used to: increase the nutrient quality of food.

Genetically modified foods, also known as genetically engineered foods or bioengineered foods, are foods made from organisms that have had alterations made to their DNA using genetic engineering techniques.

GMO foods are just as healthy and safe to eat as non-GMO ones. Some GMO plants have been genetically engineered to increase their nutritional value. GMO soybeans with better oils, for example, can be used to replace oils containing trans fats.

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Option A is correct.

Option A is correct. A 4096 code transponder with automatic pressure altitude reporting equipment is required for operating an aircraft within Class B airspace.

Class B airspace is generally airspace around the busiest airports, with the highest volume of commercial and private aircraft operations. The requirements for operating an aircraft in Class B airspace are more stringent than in other classes of airspace.

One of the requirements for operating in Class B airspace is the use of a transponder with automatic pressure altitude reporting equipment. This equipment allows air traffic control to track the altitude of the aircraft and maintain separation between other aircraft in the area. Additionally, pilots must receive specific clearance from air traffic control to enter Class B airspace.

While a VOR receiver with DME can be helpful for navigating and communicating with air traffic control, it is not a required piece of equipment for operating within Class B airspace.

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Answer:

It's False

Explanation:

I did the assignment

The magnitude of σ1 that will cause yielding according to the maximum-shear-stress theory is 9.94 ksi.

Yield stress for the zirconium-magnesium alloy, σY = 15.3 ksi

In-plane principal stresses are σ1 and σ2 = −0.54σ1

To find the maximum shear stress theory, the equation used is τ_max=1/2(σ1-σ2)

The maximum shear stress theory states that yielding begins when the maximum shear stress in a part equals or exceeds the shear strength of the material. It is represented as τ_max = τ_yield

Where τ_max is the maximum shear stress in a part and τ_yield is the shear strength of the material. In-plane principal stresses are σ1 and σ2 = −0.54σ1

Let us replace the value of σ2 in terms of σ1σ2 = −0.54σ1,σ1 = 1.85σ2

Substitute the values in the τ_max=1/2(σ1-σ2)

τ_max=1/2(σ1-(-0.54σ1))

τ_max=0.77σ1

Now, τ_yield= σY/2 = 7.65 ksi

Therefore, 0.77σ1 = 7.65

σ1 = 9.94 ksi

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(a) Newtonian approach to derive the equation of motion:

Using Newton's second law of motion, we can write the equation of motion for the spring-mass-damper system as:

m * d^2x/dt^2 + b * dx/dt + k * x = F(t)

Where:

m = mass of the system (1 kg)

b = damping coefficient (3 Ns/m)

k = spring constant (5 N/m)

x = displacement of the mass from its equilibrium position

F(t) = applied force (step input)

Taking Laplace transform on both sides, we have:

m * s^2 * X(s) + b * s * X(s) + k * X(s) = F(s)

Rearranging the equation, we get:

X(s) = F(s) / (m * s^2 + b * s + k)

(b) Changing the spring constant (k) to 5*k:

If the spring constant is increased to 5 times its original value, the system becomes stiffer. This change in the spring constant will result in a higher natural frequency of the system. The natural frequency (wn) is given by:

wn = sqrt(k / m)

By increasing k, the natural frequency of the system will increase, resulting in a faster oscillation and shorter period of the system's response.

(c) Keeping the spring constant (k) the same but increasing the damping constant (b) to 5*b:

Increasing the damping constant while keeping the spring constant the same will result in a higher damping ratio (ζ) for the system. The damping ratio is given by:

ζ = b / (2 * sqrt(k * m))

By increasing the damping constant, the system's response will be more overdamped. It will take longer for the system to reach its equilibrium position, resulting in a slower and smoother response.

(d) Model and Simulation in Engineering:

In summary, modeling is the creation of a simplified representation of a system, while simulation involves running the model to analyze system behavior. Simulation is a valuable tool in engineering as it allows for analysis, prediction, and optimization of complex systems, leading to improved designs and decision-making.

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These five sources of authority were named as coercive, rewarding, legitimate, referent, and expert.

French and Raven (1959) proposed five power dynamics (or bases of power): referent, expert, legitimate, reward, and coercive. Coercive power, expert power, legitimate power, referent power, and reward power are the five types of power that can be seen in organizational behavior. The authority that stems from an organization and gives the leader the ability to exercise power are the bases of power. Power originates from five different system: referent, expert, legitimate, coercive, and reward. Each partnership has its own unique power dynamics at any given time. Because each party has enough negotiating power to ensure mutually beneficial outcomes, mutually dependent relationships frequently produce exceptional value addition.

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According to the scenario mentioned in the question, technician B is correct.

Amperage draw may be defined as a kind of measurement of the power that is significantly being consumed by a blower motor in order to migrate the air through your HVAC system.

When a v-6 motor is being checked for starter amperage draw, it is found that the initial surge current was about 210 amperes and about 160 amperes during cranking. Technician B is correct because he reveals that it is the normal current that draws for a starter motor on a V-6.

Therefore, through this statement, it may assume that technician B is absolutely correct.

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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 statement which is not true is z is available to code that is written outside the circle class.

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It is important to keep welders properly tuned because it "Reduces air pollution" (Option A).

Welding machinery, welding guns, and welders are among the most important items for a welding expert to have. Welding machines provide heat that melts steel pieces, allowing them to be bonded.

Most welders also use an angle grinder to smooth out joints, wire brushes to clean or abrade steel surfaces before welding etc.

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Full Question:

Why is it important to keep portable welders properly tuned?

a-Reduce air pollution

b-So they run quieter

c-To produce smoother welding current

d-All of the above

The taper in the entire length of the shaft is **0.505 mm**.

A shaft is a mechanical component used to transmit rotational motion and power between different parts of a machine. It is a long, cylindrical rod typically made of metal, and it plays a crucial role in various applications such as engines, pumps, and industrial machinery.

To calculate the taper in the entire length of the shaft, we can use the formula:
Taper = (Taper per unit length) x (Total length).

In this case, the taper per unit length is 0.05 mm in 101.5 mm, which is equivalent to 0.00049 mm per mm.

The total length of the shaft is 406 mm. By substituting these values into the formula,

we get: Taper = 0.00049 mm/mm x 406 mm = 0.19954 mm

. Rounding this value to three decimal places, the taper in the entire length of the shaft is approximately 0.505 mm.

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An airplane flying across the sky experience drag force determined by the factors including the speed of flight, coefficient of skin friction and the reference surface area

The boundary layer thickness is approximately 0.233 cm

The total skin friction drag, is approximately 265 N

Reason:

First part:

Given parameters are;

Chord length, L = 3 m

Velocity of the plane, V = 200 m/s

Density of the air, ρ = 1.225 kg/m³

Viscosity of the air, μ = 1.81 × 10⁻⁵ kg/(m·s)

The Reynolds number is given as follows;

\(R_{eL} = \dfrac{\rho \times V \times L}{\mu}\)

Therefore;

\(R_{eL} = \dfrac{1.255 \times 200 \times 3}{1.81 \times 10^{-5}} = 4.16022099448 \times 10^7 \approx 4.16 \times 10^7\)

Boundary layer thickness, \(\delta_L\), for laminar flow, is given as follows;

\(\dfrac{ \delta_L }{L}=\dfrac{5.0}{\sqrt{R_{eL} } }\)

\({ \delta_L }=\dfrac{5.0 \times L}{\sqrt{R_{eL} } }\)

Which gives;

\({ \delta_L }=\dfrac{5.0 \times 3}{\sqrt{4.16 \times 10^{7}} } \approx 2.33 \times 10^{-3 }\)

The boundary layer thickness, \(\delta_L\) ≈ 2.33 × 10⁻³ m = 0.233 cm

Second Part

The total skin friction is given as follows;

\(Dynamic \ pressure, q = \dfrac{1}{2} \cdot \rho \cdot V^2\)

Therefore;

\(q = \dfrac{1}{2} \times 1.225 \times 200^2 = 24,500\)

The dynamic pressure, q = 24,500 N/m²

Skin friction drag coefficient, \(C_D\), is given as follows;

\(C_D = \dfrac{1.328}{\sqrt{R_{eL} } }\)

Therefore;

\(C_D = \dfrac{1.328}{\sqrt{4.16 \times 10^7 } } \approx 2.06 \times 10^{-4}\)

Skin friction drag, \(D_f\), is given as follows;

\(D_f\) = q × \(C_D\) × A

Where;

A = The reference area

∴ \(D_f\) = 24,500 N/m² × 2.06 × 10⁻⁴ × 3 m × 17.5 m = 264.9675 N ≈ 265 N

The total skin friction drag, \(D_f\) ≈ 265 N

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Answer:

The correct answer is "\(O (n\times Log n)\)". A further explanation is given below.

Explanation:

Answer:

an ability to identify, formulate, and solve engineering problems. an understanding of professional and ethical responsibility. an ability to communicate effectively. the broad education necessary to understand the impact of engineering solutions in a global, economic, environmental, and societal context