When you test a device or other component with an ohmmeter, who current generated? within the device or component from an external battery from a power distribution source within the ohmmeter

Some of the most commonly used chemical methods of sterilization are:

Sterilization is the process of eliminating or removing all forms of microbial life, including bacteria, viruses, fungi, and spores of nucleic acids. Chemical methods of sterilization use chemicals to kill or inactivate microorganisms.

Note also the use of Glutaraldehyde: This is a liquid sterilant that is effective against a wide range of microorganisms, including bacteria, viruses, and fungi. It is often used to sterilize heat-sensitive medical devices, such as endoscopes.

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The maximum average normal stress in each of the links is: σ = (25lbs) / (1/32 sq. in.) = 1600 psi Since the stress is determined by the cross-sectional area, and not the direction of the force, the stress is compressive for both links.

To determine the maximum average normal stress in links CD and BE, we will first calculate the cross-sectional area of the links and the area of the pins. Then, we will divide the force P by these areas to find the stress in each link and identify whether it is tensile or compressive.
1. Cross-sectional area of links CD and BE:
A = width × height = (1/8) × (1/4) = 1/32 in²
2. Diameter of pins at C, D, B, and E:
D = 1/4 in
Since both links have the same cross-sectional area, they will experience the same normal stress.
3. Calculate the maximum average normal stress in links CD and BE:
σ = P/A = (50 lbs) / (1/32 in²) = 1600 psi
As there is no information provided on the direction of force P, we cannot determine if the stress in each link is tensile or compressive. If P causes tension in the links (pulling them apart), the stress would be tensile. If P causes compression (pushing them together), the stress would be compressive.

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Answer:

volunteer coordinator

Explanation:

because they are volunteering for that and in most of the cases they do not expect to be paid

Answer :

Im not really sure tbh

Explanation:  Water at 15°C is heated by passing it through 2-cm internal-diameter thin-walled copper tubes. Heat is supplied to the water by steam that condenses outside the copper tubes at 120°C. If water is to be heated to 65°C at a rate of 0.2 kg/s, determine (a) the length of the copper tube that needs to be used and (b) the pumping power requirement to overcome pressure losses. Assuming the entire copper tube to be at steam temperature of 120°C

Answer:

Software reuse is the principal approach for constructing web-based systems, requirements for those systems cannot be completely specified in advance, User interfaces are constrained by the capabilities of web browsers.

Answer:

49 is equal to 7 times 7

Explanation:

A "perfect square" is a number that is the product of a whole number that is multiplied by itself.

In the example above, 49 is a perfect square because it is the product of the square of 7. If you multiply 7 by itself, the product is 49.

Other examples of perfect squares are:

25 = 25 is equal to 5 x 5

100 = 100 is equal to 10 x 10

81 = 81 is equal to 9 x 9

16 = 16 is equal to 4 x 4

64 = 64 is equal to 8 x 8

Answer:

D

Explanation:

For a well with the diameter of 2 ft, the approximate head in the pumped well for steady-state condition and approximate drawdown is 61.5770 ft and 48.42 ft respectively. The computed value of transmissivity is 1516.8 ft²/day.

We have a well with 2 feet diameter penetrates vertically through a confined aquifer 50 ft thick. Well is pumped at 500 gpm. The drawdown in a well 50 ft away is 10 ft.

Hence, required value is 48.42 feet. For a well with diameter of 2 ft, the approximate head in the pumped well for steady-state condition and approximate drawdown is 61.5770 ft and 48.42 ft respectively. The computed value of transmissivity is 1516.8 ft²/day.

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The entropy of 2 moles of and ideal gas expanding freely to 3 times it's initial volume is 18.3J/k

The following formula is required:

∆S = nx R x ℓn x (Vf/Vi)

Where

n = number of moles of gas (n = 2)

R = gas constant (R = 8.314 J/(mol * K))
Vf = final volume (Vf = 3.V1)

Vi = intial volume

Vi = 1L (Asumption )

∆S = 2 x 8.314 x 1.099

∆S =18.3 j/K

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Translation:

Calculate the entropy of 2 moles of an ideal gas expanding freely to three times its initial volume, use: ∆S =n・R・ℓn (Vf / Vi)

Hai

Your answer will be A.

If you lower the Air Pressure your Object will Float Down ward. The Air Pressure allows it to Fly.

The pressure field created by faster air movement over an airfoil is; A:  higher

When the air hits the front of the wing, the air will flow in a steeper curve upward, than the bottom wing flow which will lead to the creation of a vacuum on top of the wing that pulls more air towards the top of the wing.

Finally, this air above does the same thing but it will move faster as a result of the vacuum pulling it in, and as such the vacuum now lifts the wing. Thus, Faster air movement over an airfoil creates a higher pressure field.

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Answer:

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.

The maximum height that the water can rise above the jet is: 35 meters.

Given:

P = 350 kPa = 350000 Pa

We would use pressure(p)  formula to determine the maximum height that the water can rises above the jet by solving for h (height).

Using this formula

Pressure(P) = P₀ + ρgh

Where;

P₀ represent Pressure at the fluid's surface

ρ represent Density of the fluid = 1000 kg/m³

g represent acceleration due to gravity = 10 m/s².

h represent height

Solving for h (height)

350000 = 0 + (1000 × 10 ×h)

350000 = 10000h

Divide both side by  10000h

h = 350000/10000

h = 35 meters

Therefore the maximum height that the water can rise above the jet is: 35 meters.

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Answer:

Explanation:

Considering the flow of mercury in a tube:

When it comes to laminar flow of mercury, the thermal entry length is quite smaller than the hydrodynamic entry length.

Also, the hydrodynamic and thermal entry lengths which is given as DLhRe05.0= for the case of laminar flow. It should be noted however, that Pr << 1 for liquid metals, and thus making the thermal entry length is smaller than the hydrodynamic entry length in laminar flow, like I'd stated in the previous paragraph

Voltage : Vin = Vin(t) = sin(ωt)

Resistance R = 100Ω

According to Ohm's law, Voltage in a circuit is directly proportional to current flowing through it.

The proportionality constant in this case is equal to resistance.

So,

V ~ R

V = IR

as it is given that,

V = sin(ωt)

The maximum value of sign function is equal to 1.

So the maximum voltage in given case = V = 1 volt

Resistance given R = 100Ω

So,

Current according to Ohm's law : I = V/R

putting all the values in equation 2,

I = 1/100 = 0.01 A

so current in this case I = 0.01 ampere

I = 10 mili ampere

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Answer: The heat loss from insulated pipes or tubes to surrounding air depends on several factors such as insulation thickness, ambient temperature, wind speed etc.

Explanation:  The heat loss can be calculated using the following formula: Q = 2 π L (ti - to) / [ (ln (ro / ri) / k) + (ln (rs / ro) / KS)] (2).

Where:

The total heat loss from the pipe can be calculated by multiplying Q with the length of the pipe. The outer surface temperatures Ts,2(A) and Ts,2(B) can be calculated using the following formula3:

Ts,2(A) = T? + Q/(hA) Ts,2(B) = T? + Q/(hB)

Where:

I'm sorry, but there really isn't enough information to solve the problem using what is given. However, I have provided you with a yellow brick road, so to speak, as how to solve the word problem. Hope this helps and have a great day!

a) The Gantt chart view for the project is shown below. The finish date is April 6, 2023. The critical path is A-B-E-F-H-I-K-L and its duration is 25 days.

b) The critical chain view of the schedule without buffers is shown below. The critical chain is A-C-D-E-G-H-I-J-K-L and its duration is 18 days. Adding the project buffer of 25% of the critical chain duration (4.5 days) and the feeding buffers, the end date is April 10, 2023.

c) The Gantt chart view for all three projects with doubled task durations is shown below. The finish date is May 13, 2023.

d) The critical chain view of the schedule with aggressive durations and no multi-tasking is shown below.

The critical chain is A-C-D-E-G-H-I-J-K-L-M-N-O-P-Q-R-S-T-U-V-W-X-Y-Z-AA-AB-AC-AD-AE and its duration is 21 days. Adding the project buffer of 25% of the critical chain duration (5.25 days) and the feeding buffers, the end date is May 23, 2023.

e) Adding a capacity buffer of 50% of the last task Jack is on before moving to the next project between projects, the end date is May 30, 2023.

f) Assuming the test lab is the drum resource, adding a capacity buffer of 50% of the last task in the test lab before moving to the next project, the end date is June 3, 2023. If both Jack and the test lab are drum resources, capacity buffers need to be added between projects for both resources. The overall end date will depend on the size of the buffers added.

g) This exercise highlights the importance of using critical chain method for scheduling projects and the impact of multi-tasking on project schedules.

Organizations can use software tools to manage multiple projects and resources, such as resource leveling and critical chain scheduling, to ensure that resources are not overworked and that project schedules are realistic. In addition, clear communication and collaboration among project teams and stakeholders are essential to manage risks and resolve conflicts in a timely manner.

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Answer:

C. The Employer

Explanation:

Under OSHA's Confined Space Standard, it is the responsibility of the "employer" to identify or evaluate the spaces whether they are "permit spaces" or "confined spaces." These spaces have serious safety hazards or health hazards. Examples: tanks, vessels, pits, underground vaults, and the like.

The employers need to disclose the information of the permit spaces and their risks to the employees. Thus, he should have a written permit of the permit space if he will be allowing the employees to enter. It is also the employer's responsibility to reduce the hazards imposed to the employees such as verifying the entry conditions that are acceptable, making sure the permit space is isolated, providing barriers, etc.

Answer:

(a) The distance up the slope the wagon moves before coming to rest is approximately 21.74 m

(b) The distance the wagon comes to rest from the starting point is approximately 12.06 m

(c) The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is approximately 3.214 m/s (the difference in value can come from calculating processes)

Explanation:

The wagon motion parameters are;

The mass of the wagon, m = 7,200 kg

The initial velocity with which the wagon is projected along the horizontal rail, v = U

The length of the horizontal portion of the rail = 100 m

The angle of inclination of the inclined portion of the rail, θ = sin⁻¹(0.01)

The exerted frictional resistance to motion of the rail, \(F_f\) = 140 N

∴ θ = sin⁻¹(0.01)

The work done by the frictional force on the horizontal portion of the rail = 140 N × 100 m = 14,000 J

(a) If U = 3 m/s, we have;

Kinetic energy = 1/2·m·v²

The initial kinetic energy of the wagon, K.E. is given with the known parameters as follows;

K.E. = 1/2 × 7,200 kg × (3 m/s)² = 32,400 J

The energy, E, required to move a distance, 'd', up the slope is given as follows;

E = \(F_f\) × d + m·g·h

Where;

\(F_f\) = The friction force = 140 N

m = The mass of the wagon = 7,200 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = The height reached = d × sin(θ) = d × 0.01

Therefore;

E = 140 N × d₁ + 7,200 kg × 9.81 m/s² × d₁ × 0.01 = 846.32 N × d

The energy, \(E_{NET \ horizontal}\), remaining from the horizontal portion of the rail is given as follows;

\(E_{NET \ horizontal}\) = Initial kinetic energy of the wagon - Work done on frictional resistance on the horizontal portion of the rail

∴ \(E_{NET \ horizontal}\) = 32,400 J - 14,000 J = 18,400 J

\(E_{NET \ horizontal}\) = 18,400 J

Therefore, for the wagon with energy, \(E_{NET \ horizontal}\) to move up the train, we get;

\(E_{NET \ horizontal}\) = E

∴ 18,400 J = 846.32N × d

d₁ = 18,400 J/(846.36 N) ≈ 21.7401579 m

d₁ ≈ 21.74 m

The distance up the slope the wagon moves before coming to rest, d₁ ≈ 21.74 m

(b) Given that the initial velocity of the wagon, U = 3 m/s, the distance up the slope the wagon moves before coming to rest is given above as d₁ ≈ 21.74 m

The initial potential energy, PE, of the wagon while at the maximum height up the slope is given as follows;

P.E. = m·g·h = 7,200 kg × 9.81 m/s² × 21.74 × 0.01 m = 15,355.3968 J

The work done, 'W', on the frictional force on the return of the wagon is given as follows;

W = \(F_f\) × d₂

Where d₂ = the distance moved by the wagon

By conservation of energy, we have;

P.E. = W

∴  15,355.3968 = 140 × d₂

d₂ = 15,355.4/140 = 109.681405714

Therefore;

The distance the wagon moves from the maximum height, d₂ ≈ 109.68 m

The distance the wagon comes to rest from the starting point, d₃, is given as follows;

d₃ = Horizontal distance + d₁ - d₂

d₃ = 100 m + 21.74 m - 109.68 m ≈ 12.06 m

The distance the wagon comes to rest from the starting point, d₃ ≈ 12.06 m

(c) For the wagon to come finally to rest at it starting point, we have;

The initial kinetic energy = The total work done

1/2·m·v² = 2 × \(F_f\) × d

∴ 1/2 × 7,200 × U² = 2 × 140 × d₄

d₄ = 100 + (1/2·m·U² - 140×100)

(1/2·m·U² - 140×100)/(m·g) = h = d₁ × 0.01

∴ d₁ = (1/2·m·U² - 140×100)/(m·g×0.01)

d₄ = 100 + d₁

∴ d₄ = 100 + (1/2·m·U² - 140×100)/(m·g×0.01)

∴ 1/2 × 7,200 × U² = 2 × 140 × (100 + (1/2 × 7,200 × U² - 140×100)/(7,200 × 9.81 ×0.01))

3,600·U² = 280·(100 + (3,600·U² - 14,000)/706.32)

= 28000 + 280×3,600·U²/706.32 - 280 × 14,000/706.32

= 28000 - 280 × 14,000/706.32 + 1427.11518858·U²

3,600·U² - 1427.11518858·U² = 28000 - 280 × 14,000/706.32

U²·(3,600 - 1427.11518858) = (28000 - 280 × 14,000/706.32)

U² = (28000 - 280 × 14,000/706.32)/(3,600 - 1427.11518858) = 10.3319363649

U = √(10.3319363649) = 3.21433295801

The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is U ≈ 3.214 m/s

Percentage error = (3.214-3.115)/3.214 × 100 ≈ 3.1% < 5% (Acceptable)

The difference in value can come from difference in calculating methods

For Part A, the velocity of peg A when x = 1.2 m can be calculated using the equation v = s/t, where s is the distance traveled and t is the time taken. Since the link is moving at a constant speed of 10 m/s, the time taken to travel 1.2 m is 1.2/10 = 0.12 s.

Therefore, the velocity of peg A when x = 1.2 m is v = 10/0.12 = 83.33 m/s.

For Part B, the acceleration of peg A when x = 1.2 m can be calculated using the equation a = (v2 - v1)/t, where v2 and v1 are the velocities of the peg at the two different points and t is the time taken to travel between them. Since the link is moving at a constant speed of 10 m/s, the time taken to travel 1.2 m is 0.12 s.

Therefore, the acceleration of peg A when x = 1.2 m is a = (83.332 - 102)/0.12 = -833.33 m/s2.

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Answer:

Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.

There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.

Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.

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The fact that Eli plans to set up a recycling program made him choose D. The material he chose is renewable and is much better for the environment.

It should be noted that recycling simply means the process of collecting and processing materials that can be turned into another product.

Therefore, the material he chose is renewable and is much better for the environment and thus influenced his decision.

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For 304-annealed stainless steel, the tensile strength ranges 515 MPa (75,000 psi) and 1035 MPa (150,000 psi) and yield strength around 205 MPa (30,000 psi) to 515 MPa (75,000 psi).

The tensile and yield strengths you provided for the 304-annealed stainless steel are 515 MPa and 205 MPa, respectively. These values indicate the material's mechanical properties under tensile loading.

Tensile strength refers to the maximum stress a material can withstand before it fails or breaks.

Yield strength, on the other hand, is the stress at which a material begins to exhibit plastic deformation.

For 304-annealed stainless steel, the tensile strength typically ranges between 515 MPa (75,000 psi) and 1035 MPa (150,000 psi). This means that the material can withstand a maximum tensile stress in this range before experiencing failure.

and, the yield strength of 304-annealed stainless steel is generally around 205 MPa (30,000 psi) to 515 MPa (75,000 psi).

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Answer:

A biological hazard

Explanation:

Biological because insects and other organisms thrive in stagnant water.

The type of hazard that is most likely to be exposed to a significant amount of stagnant water is known as biological.

A biological hazard may be defined as a biological substance that may significantly pose a great threat to the health of living organisms, primarily humans. These types are the major concerns in food processing because they cause most foodborne illness outbreaks.

In the case of Kris and James, they are significantly exposed to a biological hazard because stagnant water is commonly utilized by mosquitos to place eggs, this directs a lot of mosquitos around stagnant waters and therefore a higher risk of mosquito-transmitted diseases such as malaria. Besides this, stagnant water is highly polluted and includes bacteria and parasites that are harmful.

Therefore, the type of hazard that is most likely to be exposed to a significant amount of stagnant water is known as biological.

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The common sense circuit. noun. a digital circuit utilized in computer systems to carry out a logical operation on its  or extra enter signals. There are six simple circuits, the AND, NOT, NAND, OR, NOR, and unique OR circuits, which may be blended into extra complicated circuits.

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Answer:

  use the percentage error relation

Explanation:

The percentage error in anything is computed from ...

  %error = ((measured value)/(accurate value) -1) × 100%

__

The difficulty with voltage measurements is that the "accurate value" may be hard to determine. It can be computed from the nominal values of circuit components, but there is no guarantee that the components actually have those values.

Likewise, the measuring device may have errors. It may or may not be calibrated against some standard, but even measurement standards have some range of possible error.

Answer:

use the percentage error relation

False (F).

If the secondary winding of a control transformer is protected by a fuse, it does not necessarily mean that the primary winding does not need fuse protection. Both the primary and secondary windings of a transformer may require fuse protection depending on the specific application and requirements.

The purpose of fuse protection is to prevent excessive current from flowing through the circuit and to protect the transformer and other components from damage. Fuse protection can be implemented on both the primary and secondary sides of the transformer to ensure proper protection against overcurrent conditions.

Therefore, it is not true to assume that the primary winding of a control transformer does not need fuse protection just because the secondary winding is protected by a fuse. The need for fuse protection should be determined based on the specific electrical requirements and safety considerations of the system.

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Answer:

its a gun

Explanation:

The Accuracy International AWM (Arctic Warfare Magnum or AI-Arctic Warfare Magnum) is a bolt-action sniper rifle manufactured by Accuracy International designed for magnum rifle cartridges. ...

Effective firing range: 1,100 m (1,203 yd) (.300 ...

Manufacturer: Accuracy International

In service: 1996–present