Sketch the p-channel current-source (current mirror) circuit. Let VDD = 1.3 V, V = 0.4 V, Q₁ and Q₂ be matched, and upCox = 80 μA/V². Find the device's W/L ratios and the value of the resistor that sets the value of IREF SO that a 80-µA output current is obtained. The current source is required to operate for Vo as high as 1.1 V. Neglect channel-length modulation.

6.25×10⁻¹⁸ electrons will flow through a copper wire.

Electrons can be defined as the sum of atomic particles that is the career of negative charge, they are responsible for the chemical property of an element, the electrons are present in the shell.

Use formula ( I = ne/t )

Here,

number of electron

electron =1.6 x 10¯¹9 C

We have a relation between charge and current as

Charge = Current×time i.e

Q = i×t.

Now we have current = 10 amp & time = 10sec from given data.

Therefore charge Q = 10×1 = 100 coulombs.

Again we have a relation between charge and no of electrons

I.e no of electrons= total charge / 1.6×10⁻¹⁹

So, no of electrons = ( 100C/1.6×10⁻¹⁸ )

= 6.25×10⁻¹⁸ electrons.

Therefore, 6.25×10⁻¹⁸ electrons will flow through a copper wire carrying 10 amps current for 10 sec

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The allowable normal strength in the c-section is 48000 pounds per square inch.

The allowable stress depends on both the factor of safety imposed on the object and the yield strength or stress at which an object will be permanently damaged. To calculate the allowable normal strength we can divide the yield strength by the factor of safety.

To determine the allowable normal strength output we can use this equation below:

The allowable normal strength = The number yield strength : A stress-concentration factor

The number yield strength is 84 kpsi.

Then convert kpsi to psi : 84 kpsi = 84000 psi

A stress-concentration factor is 1.75

The formula to find allowable normal strength output is:

84000 psi : 1.75 = 48000

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The diagram for the extraction to represent the information given is attached.

Extraction refers to the process of obtaining or removing something from a particular source or location. It can refer to various fields such as chemistry, physics, medicine, and data analysis.

In chemistry, extraction is the process of separating a particular component or substance from a mixture using various methods such as solvent extraction, distillation, or chromatography. In physics, extraction can refer to the removal of energy or particles from a system. In medicine, extraction refers to the removal of a tooth or a foreign object from the body. In data analysis, extraction refers to the process of retrieving specific data or information from a large set of data.

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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)

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

Answer:

545454cm

Explanation:

The blue print drawing of the house shows 666 centimeters, but the real picture of the house is 333 meters. So let the number of cm on the blueprint that represent the distance of 272727 meters be x. Firstly convert the meters to centimeters 666cm = 333m, x=272727m ; then cross multiply, 666cm=33300cm x=27272700cm ; x =(666cm×272727cm)/33300cm =545454cm.

If an internally piloted DCV does not shift, you should use a gauge to: B. check the inlet pressure to the DCV.

DCV is an acronym for Directional Control Valve and it can be defined as a fundamental part of pneumatic and hydraulic systems, which is designed and developed to enable fluid from one or more sources flow into different paths.

In pneumatic and hydraulic systems, you should use a gauge to check the inlet pressure to the DCV when an internally piloted DCV does not shift.

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False. Fire extinguishing equipment cannot be frozen according to Ref. 213/91.

According to Ref. 213/91, fire extinguishing equipment cannot be frozen. Fire extinguishers are essential safety devices designed to combat fires effectively. They contain pressurized agents that are specifically formulated to extinguish different types of fires. Freezing temperatures can significantly impair the functionality of fire extinguishers and render them ineffective in emergency situations.

When fire extinguishing equipment freezes, several issues can arise. First, the contents of the extinguisher may expand as they freeze, potentially leading to ruptures or leaks in the container. This can cause the extinguisher to malfunction or become hazardous when used. Second, freezing temperatures can affect the performance of the extinguishing agent itself. Certain agents, such as water-based solutions, can solidify or lose their effectiveness when exposed to extreme cold.

It is crucial to store fire extinguishers in suitable environments that are above freezing temperatures. This ensures that the equipment remains in optimal condition and is ready for immediate use during emergencies. Regular inspections and maintenance are also essential to identify any signs of damage or deterioration that may compromise the functionality of fire extinguishers.

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PN junction diode and bipolar junction diode are two types of diodes with distinct characteristics and functionalities.

In summary, the main difference between PN junction diode and bipolar junction diode lies in their construction and operating principles. PN junction diodes are based on a single PN junction and primarily used for rectification, while bipolar junction diodes are composed of multiple regions and employed for signal amplification and switching.

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Answer:

D). Eesha  pays more than the minimum payment each month.

Explanation:

Eesha would be considered most creditworthy among the given persons as she not only pays on time but also repays more than the minimum amount assigned to pay each month. In order to test the creditworthiness of an individual, his ontime debt paying capability is tested at first followed by the past credit repayment history and the credit score. Except Eesha, all the given candidates have failed to make timely repayment of their debts and hence, they cannot be considered creditworthy.

Answer:

F=200kN

Explanation:

Answer:

0.05%

Explanation:

From the question, we have;

The yield strength of the mild steel, \(\sigma _c\) = 43.5 ksi

Young's modulus of elasticity, ∈ = 29,000 ksi

The total strain, \(\epsilon _c\) = 0.2% = 0.002

The inelatic strain \(\epsilon_c^{in}\) is given as follows;

\(\epsilon_c^{in}\) = \(\epsilon _c\) - \(\sigma _c\)/∈

Therefore, we have;

\(\epsilon_c^{in}\) = 0.002 - 43.5/(29,000) = 0.0005

Therefore, the inelastic strain, \(\epsilon_c^{in}\) = 0.0005 = 0.05%

Taking the inelastic strain as the residual strain, we have;

The residual strain = 0.05%

The control group in this study would be a location at least several kilometers away from the research study area, such as the area marked with a blue circle on the map.

A control group is a group of people or things that are not subjected to any special treatment or manipulation in an experiment or test. It is used as a reference point to compare results from the experimental group, which is the group that is exposed to the particular treatment or manipulation. The control group serves to ensure that any observed changes in the experimental group can be attributed to the experiment and not any other outside factors. Control groups are used in both scientific experiments and marketing tests to ensure that any results are accurate and valid.

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Answer:

2.9237 ft/s

Explanation:

Given the data in the question;

A₁ = 9-inch × 9-inch = 81 in² = 81 / 144 = 0.5625 ft²

V₁ = 6.1 ft/s

A₂ = 13 in × 13 in = 169 in² = 1.17361 ft²

v₂ = ?

using the the equation if continuity

( Rate of volumetric flow is constant )

A₁V₁ = A₂V₂

we substitute

0.5625 ft² × 6.1 ft/s = 1.17361 ft² × V₂

3.43125 ft³/s = 1.17361 ft² × V₂  

V₂  = 3.43125 ft³/s  / 1.17361 ft²

V₂ = 2.9237 ft/s

Therefore, the speed of the air flowing into the room is 2.9237 ft/s

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Answer:

what are the options available?

With a professional, do an automotive diagnostic

Verification and confirmation in

Step 1. This step in the vehicle diagnostic process enables the expert to look for potential issues.

Step 2: Identify the issue.

Step 3: Area isolation.

4. Make repairs.

The final step is a check.

the procedure of determining a diagnosis, disease, or injury based on its indications and symptoms. To aid in the diagnosis, testing like blood tests, imaging tests, and biopsies may be done in addition to a physical examination and health history.

Using a computer scan tool, the repairman will obtain the diagnostic issue code data from the computer. A fully qualified technician can more correctly identify and address the issue by using this information.

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Using a computer scan tool, the repair technician will retrieve the diagnostic trouble code information from the computer. A properly trained technician can use this information to more accurately locate and repair the problem.

As part of the standard operating procedure, many experts recommend that shops perform a pre-scan of all vehicle computer modules, as well as a scan after the vehicle is repaired (SOP). This is not only good business practice, but it also helps to communicate with the customer about potential vehicle flaws that may not have been part of the customer's original concern.

Pre-scan: This entails gaining access to all vehicle modules and downloading any or all stored Diagnostic Trouble Codes (DTCs), including pending codes. Any DTCs that have been stored will be recorded on the work order, and if they are related to a customer, it may be necessary to notify the customer and obtain approval before proceeding with the repair.

Post Scan: After the vehicle has been repaired and before it is released to the customer, a full module scan is performed to ensure that another DTC was not set during the repair in addition to the repair. procedure for repair The results of this follow-up scan should also be documented on the repair order so that they can be added to the driving history.

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Answer:

To calculate the coolant boiling point with a 15 psi pressure cap, we need to consider the base boiling point of water, the increase in boiling point due to the 50-50 mix of coolant, and the increase in boiling point per psi of pressure.

Base Boiling Point of Water: 100°C

Increase in Boiling Point due to Coolant Mix: 6°C

Increase in Boiling Point per psi: 17°C

Calculation for a 15 psi pressure cap:

Base Boiling Point + Increase due to Coolant + (Pressure Increase × Increase per psi)

= 100°C + 6°C + (15 psi × 17°C/psi)

= 100°C + 6°C + 255°C

= 361°C

Therefore, with a 15 psi pressure cap, the cooling system boiling point would be 361°C.

Now let's calculate the boiling point with a 12 psi pressure cap:

Base Boiling Point + Increase due to Coolant + (Pressure Increase × Increase per psi)

= 100°C + 6°C + (12 psi × 17°C/psi)

= 100°C + 6°C + 204°C

= 310°C

With a 12 psi pressure cap, the cooling system boiling point would be 310°C.

If pure water is used instead of a 50-50 mix of coolant, the boiling point would remain at 100°C since there would be no additional increase in boiling point due to the coolant mix.

As for the freezing point, the presence of the coolant mix or the pressure cap does not directly affect the freezing point of water. The freezing point of water is generally 0°C, regardless of the coolant mix or pressure conditions.

Explanation:

Answer:

1.887 minutes

Explanation:

We are given k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol = 20000 cal/mol

To solve this, first of all let's calculate the rate constant(k);

For this question, The formula is;

K(t) = k(300K) × exp[(E/R)((1/300) - (1/T2))]

R is gas constant = 1.987 cal/mol.K

For temperature of 70°C which is = 70 + 273K = 343K, we have;

K(343) = 0.07 × exp[(20000/1.987)((1/300) - (1/343))]

K(343) = 4.7 dm³/mol.min

The design equation is;

dX/dt = -(rA/C_Ao) = K•(C_Ao)²•(1 - X)²/(C_Ao) = (KC_Ao)(1 - X)²

Since there is no change in volume by cause of the state at which the reaction is carried out, that is liquid. Thus, integrating and solving for time for a 90% conversion we obtain;

(0.9,0)∫dX/(1 - X)².dX = (KC_Ao)((t, 0)∫dt

So, we'll get;

0.9/(1 - 0.9) = 4.77 × 1 × t

t = 9/4.77

t = 1.887 minutes

Answer:

1. 4/d and 1

2. Engineering tech with 3, Engineer with 1, and Scientist with 2

Explanation:

I'm pretty sure but tell me if im wrong

The critical crack length for a through crack in a thick plate of 7150-T651 aluminum alloy under uniaxial tension can be determined using fracture mechanics principles. For this particular alloy, the fracture toughness (Kic) is given as 25.5 MPa√m, and the ultimate tensile strength (σf) is 400 MPa.

By assuming a yield strength (Y) equal to the square root of pi, the critical crack length can be calculated.

Fracture mechanics is a field of study that deals with the behavior of materials when subjected to cracks or flaws. The critical crack length represents the maximum allowable size of a crack within a material before it fails catastrophically. In this case, we are considering a through crack in a thick plate of 7150-T651 aluminum alloy that is subjected to uniaxial tension.

To determine the critical crack length, we need to use the fracture toughness value (Kic) and the ultimate tensile strength (σf) of the material. The fracture toughness measures a material's resistance to crack propagation, and it is given as 25.5 MPa√m for this particular aluminum alloy. The ultimate tensile strength represents the maximum stress the material can withstand before failure, and it is specified as 400 MPa.

In fracture mechanics, the critical crack length can be calculated using the equation:

a = (Kic / (Y * σf))²

Here, Y represents the yield strength of the material. In this case, Y is assumed to be equal to the square root of pi (√π). By substituting the given values into the equation, we can determine the critical crack length for the 7150-T651 aluminum alloy under uniaxial tension.

The calculation of the critical crack length will provide a value in terms of the crack size (a). This value represents the maximum allowable crack length before failure occurs. It is crucial to consider the critical crack length in engineering applications to ensure the structural integrity and safety of components made from the aluminum alloy.

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Using the Newton's second law of motion we will see that the acceleration is  0.0875 m/s^2.

The acceleration of a vehicle is determined by the force acting on it and its mass, as described by Newton's second law of motion:

F = m * a

where F is the force acting on the object, m is the mass of the object, and a is the resulting acceleration.

In this case, the force acting on the vehicle is 0.07 N, and the mass of the vehicle is 0.8 kg. Substituting these values into the equation above, we get:

0.07 N = 0.8 kg * a

Solving for a, we get:

a = 0.07 N / 0.8 kg

a = 0.0875 m/s^2

Therefore, the acceleration of the 0.8 kg vehicle powered by 0.07 N of force is 0.0875 m/s^2.

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When using a radio, it is important to wait for a short duration before speaking to avoid cutting off the first words of your transmission. This waiting time is commonly known as "transmitting time" or "key-up time."

The recommended duration to wait before speaking is usually around 1 to 2 seconds. This allows the radio system to establish a connection and for any signal delays to settle before transmitting your voice.

By waiting for this brief period, you ensure that your entire message is transmitted clearly without any parts being cut off. It is a good practice to give a moment of silence before starting to speak on the radio to ensure effective communication.

Remember, clear and concise transmissions are crucial for effective communication over a radio system.

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Answer:The two-second rule is a rule of thumb by which a driver may maintain a safe trailing distance at any speed. The rule is that a driver should ideally stay at least two seconds behind any vehicle that is directly in front of his or her vehicle.

Explanation:

The pedal does not push the master cylinder farther into the master cylinder bore.

What is the advantage of using a left foot?

The theory used in this question is the principle of using a left foot to apply pressure to the brake pedal while seating the brake pads. By using the left foot on the brake pedal while the right foot applies pressure, the technician is able to prevent the master cylinder from being pushed farther into the master cylinder bore.

The steps for this process are as follows:

1. Place the brake pads into the caliper.

2. Place the left foot on the brake pedal.

3. Use the right foot to apply pressure to the pedal, pushing it down slowly.

4. Allow the pressure from the left foot to prevent the master cylinder from being pushed farther into the master cylinder bore.

5. When the brake pads are seated correctly, release the pressure from the pedal.

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The deformation at the point of B is 1.09 mm

Deformation refers to the change in shape or size of a material due to the application of external forces or stress. It is a common phenomenon in materials science and engineering, and it plays an important role in the behavior and performance of materials under various conditions.

E = 70 Gpa

P = 90 KN

For every section

A and B

125 - 90

= 35

For B and C

75 + 50 = 125

For C and D

50

we have

\(\frac{35*10^3*1.75}{800*10^6*70*10^9}\)

= \(\frac{61250}{800000000 * 5.6*10^1^9}\)

= 1.09 mm

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Hotel Fire Safety Directors are not authorized to order a full building evacuation of guests. Therefore, option (C) is correct.

Hotel Fire Safety Directors are individuals who are responsible for ensuring the safety of guests and employees in the event of a fire emergency in a hotel or other hospitality establishment.

They are typically appointed by the hotel management and must have knowledge of fire safety regulations and protocols, as well as the ability to lead and manage a team during an emergency.

Hotel Fire Safety Directors are responsible for implementing and maintaining the Fire Safety Plan, which includes conducting fire drills, training and supervising the Fire Brigade, and distributing applicable parts of the Fire Safety Plan to all employees. They must also work closely with the local fire department to ensure compliance with local fire safety regulations and protocols.

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