The tailstock
a. is used to mount the lathe center
b. cannot be used for taper turning
C. is removed when long work must be turned
D. holds the inner end of the work being turned

Answer:

Without the indispensable advice of the business community, TVET Institutions will be unable to cover the gap in career knowledge required by the business community.  To develop workers who possess the knowledge and skills required by today's business entities, there is always the continual need for the educational institutions (gown) to regularly meet the business community (town).  This meeting provides the necessary ground for the institutions to develop programs that groom the workforce with skills that are needed in the current workplace.  Educational institutions that do not seek this important advice from the business community risk developing workers with outdated skills.

Explanation:

TVET Institutions mean Technical and Vocational Education and Training Institutions.  They play an important role in equipping young people to enter the world of work.  They also continue to develop programs that will improve the employability of workers throughout their careers.  They regularly respond to the changing labor market needs, adopt new training strategies and technologies, and expand the outreach of their training to current workers while grooming the young people for work.

The order of magnitude of the drag force from lowest to highest is:

b. air-N-H < Od. H-air-N < a. air-H-N < Oe. N-air-H < Of. N-H-air < c. H-N-air

The drag force on a flat plate can be estimated using the formula:

F = 0.5 x rho x v² x Cd x A

where F is the drag force, rho is the density of the fluid, v is the velocity of the fluid, Cd is the drag coefficient, and A is the area of the plate.

Supposing that the plate is 1 m wide (in the direction perpendicular to the flow), the area of the plate is 2 m².

The drag coefficient for a flat plate depends on the Reynolds number of the flow, which is given by:

Re = rho x v x L / mu

where L is the length of the plate and mu is the dynamic viscosity of the fluid.

For air at 350K and atmospheric pressure, the density is approximately 1.16 kg/m³ and the dynamic viscosity is approximately 2.97e-5 Pa x s. Using these values, we can calculate the Reynolds number for air:

Re = 1.16 x 5 x 2 / 2.97e-5 = 390,582

The drag coefficient for a flat plate at this Reynolds number is approximately 0.664. Applying this value and the other values calculated, estimate the drag force on the plate:

F_air = 0.5 x 1.16 x 5² x 0.664 x 2 = 19.4 N

For hydrogen at 350K and atmospheric pressure, the density is approximately 0.084 kg/m³ and the dynamic viscosity is approximately 8.46e-6 Pa x s. Using these values, calculate the Reynolds number for hydrogen:

Re = 0.084 x 5 x 2 / 8.46e-6 = 99,409

The drag coefficient for a flat plate at this Reynolds number is approximately 1.24. Using this value and the other values we have calculated, estimate the drag force on the plate:

F_H = 0.5 x 0.084 x 5² x 1.24 x 2 = 4.2 N

For nitrogen at 350K and atmospheric pressure, the density is approximately 1.02 kg/m³ and the dynamic viscosity is approximately 1.86e-5 Pa x s. Using these values, we can calculate the Reynolds number for nitrogen:

Re = 1.02 x 5 x 2 / 1.86e-5 = 548,387

The drag coefficient for a flat plate at this Reynolds number is approximately 0.696. Using this value and the other values we have calculated, estimate the drag force on the plate:

F_N = 0.5 x 1.02 x 5² x 0.696 x 2 = 18.0 N

Therefore, the order of magnitude of the drag force from lowest to highest is:

b. air-N-H < Od. H-air-N < a. air-H-N < Oe. N-air-H < Of. N-H-air < c. H-N-air

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Answer: Plants and animals can be agents of mechanical weathering. The seed of a tree may sprout in soil that has collected in a cracked rock. As the roots grow, they widen the cracks, eventually breaking the rock into pieces. Over time, trees can break apart even large rocks.

I searched it, hope it helps! Have a great day!

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= \(35000\times (\frac{F}{P}, 4 \ percent,5 )\)

= \(35000\times 1.216 7\)

= \(42584.50\)

Now,

time (t) will be:

= \(5\times 12\)

= \(60 \ monthly \ payments\)

So, monthly payment will be:

= \(42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )\)

= \(42584.50\times 0.0193\)

= \(821.88\)

Answer: the attached picture is the answer.

Explanation:

Assuming:

the switch position connect to 1, hence 4.5V exist at across lamp1

the switch position connects to 2 hence 4.5 V exist across lamp 1 and lamp 2

the switch position connects to 3, hence, 4.5 V exist across lamp 1, lamp 2 and lamp 3.

A mechanical load consisting of an inertia and viscous damping can be described using the following parameters:

1. **Inertia**: Inertia is a measure of an object's resistance to changes in its velocity. It represents the mass or moment of inertia of the load. The inertia determines how much force is required to accelerate or decelerate the load.

2. **Viscous Damping**: Viscous damping is a type of damping that occurs due to the resistance provided by a fluid or material to the motion of the load. It is proportional to the velocity of the load and opposes its motion. The damping coefficient or damping factor determines the strength of the damping effect.

When subjected to external forces or displacements, the load will exhibit a response that depends on the combination of inertia and viscous damping. The inertia component will cause delays and resist changes in motion, while the viscous damping component will dissipate energy and provide resistance to the load's movement.

Understanding the properties of inertia and viscous damping helps in analyzing and predicting the behavior of mechanical systems, such as in vibration analysis or control system design.

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The Reynolds number for the crude oil flowing through the 4-meter pipe is approximately 235,675.

Given:
Density (ρ) = 977 kg/m³
Viscosity (μ) = 0.004 Pa•s
Flow rate (Q) = 3 m³/s
Pipe diameter (D) = 4 m

First, we need to calculate the velocity (v) of the crude oil using the flow rate (Q) and the pipe's cross-sectional area (A). The area of a pipe can be calculated using the formula A = (πD²)/4.

1. Calculate the area (A) of the pipe:
A = (π(4 m)²)/4 = (π(16 m²))/4 = 4π m²

2. Calculate the velocity (v) of the crude oil:
v = Q/A = (3 m³/s)/(4π m²) = 3/(4π) m/s

Now we can use the Reynolds number formula, which is Re = (ρvD)/μ.

3. Calculate the Reynolds number (Re):
Re = (977 kg/m³)(3/(4π) m/s)(4 m)/(0.004 Pa•s) = (977 × 3 × 4)/(4π × 0.004) = (11724)/(4π × 0.004)

Re ≈ 235,675

The Reynolds number for the crude oil flowing through the 4-meter pipe is approximately 235,675.

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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 sliding window technique is utilized to identify the crash frequency of a certain region. A performance measure called "Excess Predicted Average Crash Frequency Using SPFs" will be used to screen Segment A in the urban four-lane divided arterial reference population.

The segment is 0.60 mi long. Let's determine the number of times the performance measure will be applied to Segment A using the sliding window method.In 0.30-mi windows, the section is analyzed. The increment is 0.10 miles long. As a result,0.30 mi long window = 0.60 / 0.30 = 2 windows.0.10 miles long increment = 0.60 / 0.10 = 6 increments.So, the total number of applications = number of windows × number of increments in each window= 2 × 6= 12.The performance measure will be used 12 times on Segment A. Answer: In 200 words. The sliding window technique is utilized to identify the crash frequency of a certain region. A performance measure called "Excess Predicted Average Crash Frequency Using SPFs" will be used to screen Segment A in the urban four-lane divided arterial reference population. The segment is 0.60 mi long. Let's determine the number of times the performance measure will be applied to Segment A using the sliding window method.In 0.30-mi windows, the section is analyzed. The increment is 0.10 miles long. As a result,0.30 mi long window = 0.60 / 0.30 = 2 windows.0.10 miles long increment = 0.60 / 0.10 = 6 increments.So, the total number of applications = number of windows × number of increments in each window= 2 × 6= 12.The performance measure will be used 12 times on Segment A.

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The following commands can be used to display a list of all files and their sizes from Wayne's home directory:

1. `ls -l`

2. `du -h`

1. The command `ls -l` displays a detailed listing of files and directories in the current directory (in this case, Wayne's home directory). The `-l` option stands for "long format" and provides additional information about the files, including their sizes.

2. The command `du -h` is used to estimate and display the disk usage of files and directories. The `-h` option makes the output human-readable, showing sizes in a more understandable format (e.g., "10K" for kilobytes, "2M" for mega bytes). By running `du -h` in Wayne's home directory, he will obtain a list of files and their respective sizes.

By using either of these commands, Wayne can obtain the desired information about the files and their sizes in his home directory.

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Answer: answer is B

Explanation:

Answer: metamorphic rock

Explanation:

As something of a result, the highest electric field in an unpaid labor p-n junction diode is around 8.93 x 104 V/cm. To calculate the maximum electric field (Emax) in an uncompensated p-n junction diode, we can use the following formula:

Emax = V / W

Where V is the applied bias voltage and W is the depletion width of the diode.

The depletion width can be calculated using the following formula for a double-sided abrupt junction:

W = [(2εε0 / q) * (NA*ND / (NA+ND)) * (Vbi + V)]^0.5

Where ε is the permittivity of silicon, ε0 is the vacuum permittivity, q is the charge of an electron, NA is the acceptor doping concentration, ND is the donor doping concentration, Vbi is the built-in potential, and V is the applied bias voltage.

We can first calculate the built-in potential using the following formula:

Vbi = (kT / q) * ln(NA*ND / ni^2)

Where k is the Boltzmann constant, T is the temperature in Kelvin, and ni is the intrinsic carrier concentration of silicon at 300K, which is approximately 1.5 x 10^10 (cm^-3).

Plugging in the given values, we get:

Vbi = (8.617 x 10^-5 eV/K * 300K / 1.602 x 10^-19 C) * ln(28 x 10^16 cm^-3 * 36 x 10^16 cm^-3 / (1.5 x 10^10 cm^-3)^2)

≈ 0.787 V

Next, we can calculate the depletion width using the formula above:

W = [(2εε0 / q) * (NA*ND / (NA+ND)) * (Vbi + V)]^0.5

= [(2 * 11.7 * 8.854 x 10^-14 F/cm / 1.602 x 10^-19 C) * (28 x 10^16 cm^-3 * 36 x 10^16 cm^-3 / (28 x 10^16 cm^-3 + 36 x 10^16 cm^-3)) * (0.787 V - 1.5 V)]^0.5

≈ 0.168 µm

Finally, we can calculate the maximum electric field using the formula at the beginning:

Emax = V / W

= -1.5 V / (0.168 µm * 10^-4 cm/µm)

≈ -8.93 x 10^4 V/cm

Therefore, the maximum electric field in AP the uncompensated p-n junction diode is approximately 8.93 x 10^4 V/cm.

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The mass of air in the tank calculated by using the  ideal gas law is  79.1 kg.

We can calculate the mass of the air in the tank using the ideal gas law. The ideal gas law states that PV = nRT, where P is the pressure of the gas, V is the volume, n is the amount of substance, R is the universal gas constant, and T is the temperature.

Rearranging the equation to solve for n, we get n = PV/RT. We can then substitute the given values to solve for n:

n = (1.4 MPa)(2 m^3)/[(8.3145 J/molK)(-93°C + 273.15 K)]

n = 79.1 kg

Therefore, the mass of air in the tank is 79.1 kg.

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Complete question:

A tank contains 2m^3 of air at -93 degree celcius and a gage pressure of 1.4MPa. Determine the mass of air in kg. The local atmospheric pressure is 1atm.

To measure the diameters of teflon and steel spheres, a micrometer is typically used.

The bias errors for the micrometer are fixed, systematic errors that are assumed to be zero.

This means that the errors are caused by the instrument itself, rather than from random fluctuations of the measured value. These errors can be caused by a variety of factors, such as misaligned parts, inaccurate measurements, and incorrect calibration.

In order to identify and correct these errors, it is important to understand the source of the errors. The most common sources of bias errors are the instruments themselves, as they may contain inaccurate scales, or may not be properly calibrated.

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Two categories of laser printers are personal and shared. Personal laser printers are designed for individual use, typically in a home or small office setting.

Shared laser printers are designed for use by multiple people in a larger office setting. It is important to note that the terms "active-matrix" and "passive-matrix" do not apply to laser printers, but rather to different types of LCD screens. "Thermal" and "ink-jet" are also not categories of laser printers, but rather different types of printers altogether. Laser printers use toner, rather than ink, to produce high-quality prints.

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The best statement that sums up the relationship between microevolution and macroevolution is: Macroevolution occurs through or as a result of microevolution; it is the outcome of microevolution operating over long timescales at high levels of organization.

Microevolution and macroevolution are interconnected processes in evolutionary biology. Microevolution refers to small-scale genetic changes that occur within populations over relatively short periods of time. These changes include variations in allele frequencies, genetic drift, and natural selection. Macroevolution, on the other hand, refers to the large-scale patterns and processes that result in the formation of new species and the diversification of life over long periods of time.

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Criteria define project goals, while constraints set boundaries and limitations, and both are important in the design process.

Define the term boundaries.

Boundaries refer to the limits or constraints that define the extent or scope of something, such as a system, process, or project. Boundaries can be physical, conceptual, or organizational and may be defined by external factors like laws, regulations, or standards, or by internal factors like resources, capabilities, or objectives. Boundaries can help to clarify responsibilities, prevent misunderstandings, and ensure that activities or outcomes are consistent with expectations and requirements.

The statement that is true about criteria and constraints is that they are both important factors to consider in the design process of a project.

Criteria are the desired features, functions, and performance requirements that a project must meet to be considered successful. They define the goals and objectives of the project and help to evaluate whether or not the project has achieved its intended purpose.

Constraints are the limitations and restrictions that affect the design and development of a project. They can be related to resources, such as time, budget, materials, and manpower, as well as technical, legal, ethical, and environmental considerations. Constraints set boundaries and parameters for the project and may require trade-offs or compromises to be made.

In the design process, criteria and constraints are used to guide decision-making and problem-solving. Designers must balance the desired criteria with the constraints that exist, to arrive at a solution that meets the project goals while also being feasible and practical to implement.

Therefore, both criteria and constraints are important and interrelated factors to consider in the design process of a project.

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This question is not complete, the complete question is;

The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005

Answer:

the length of the pipe is 11583 in or 965.25 ft

Explanation:

Given the data in the question;

Static pressure ratio; p1/p2 = 10

friction coefficient f = 0.005

diameter of pipe, D =4 inch

first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so

4fL\(_{max}\) / D = 57.915

we substitute

(4×0.005×L\(_{max}\)) / 4  = 57.915

0.005L\(_{max}\) = 57.915

L\(_{max}\) = 57.915 / 0.005

L\(_{max}\)  = 11583 in

Therefore, the length of the pipe is 11583 in or 965.25 ft

According to the question of temperature rises to boiling level, the given statement is false.

What is temperature?

Temperature is a measure of how hot or cold something is. It is a physical property of matter and is expressed in degrees on a scale such as Celsius, Fahrenheit, or Kelvin. Temperature is related to the average kinetic energy of molecules in a system, with higher temperatures corresponding to higher average kinetic energy and lower temperatures corresponding to lower average kinetic energy. Temperature can affect the properties of matter and the rate at which chemical reactions occur.

Reheating the food may make the lid easier to remove, but it will not create a vacuum. The vacuum is created when the food cools, and it is not released until the pressure inside the pan drops to atmospheric value.

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We can approximate 1 atm by 100 kpa. if a tire's gage pressure is 200 kpa. The absolute pressure in the tire would be 300 kPa.

The pressure in a tire is expressed in terms of gauge pressure and absolute pressure. Gauge pressure is the pressure measured relative to the atmospheric pressure, whereas absolute pressure is the pressure measured relative to a perfect vacuum.

One atmospheric pressure is commonly defined as 100 kPa, so we can approximate 1 atm to be 100 kPa. If a tire's gauge pressure is 200 kPa, it means that the tire pressure is 200 kPa above atmospheric pressure. To convert this gauge pressure to absolute pressure, we add it to the atmospheric pressure of 100 kPa. Thus, the absolute pressure in the tire would be 200 kPa + 100 kPa = 300 kPa.

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Answer:

Surface Parallelism When a surface is to be parallel to a datum plane, the feature control frame is connected by a leader to the surface, or to an extension line from the surface.

Explanation:

75743366

Explanation:

The graphs 1) Scatter plot, 2) Line graph, 3) Pie chart, and 4) Bar graph can all be used to depict data.

The strength of a linear relationship between two variables is measured using the Pearson correlation coefficient, where r = 1 signifies a perfect positive correlation and r = -1 denotes a perfect negative correlation.

This number is calculated using at least three different formulas, and these formulas serve as a kind of representation of several approaches to the issue. However, each of the various methods yields the same result for r.

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Answer:

thanks for the poiunts

Explanation:

Answer:

a. 89.5 N b. 1.59 N

Explanation:

a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?

Using Boyle's law, we find the final pressure at the compressed state given that the initial pressure is atmospheric pressure

So, P₁V₁ = P₂V₂ where P₁ = initial atmospheric pressure in syringe = 1 atm = 1.013 × 10⁵ N/m²,V₁ = initial volume of syringe = 5 ml, P₂ = final pressure in syringe at compression and V₂ = final volume of syringe = 1 ml

So, making P₂ subject of the formula, we have

P₂ = P₁V₁/V₂

Substituting the values of the variables into the equation, we have

P₂ = P₁V₁/V₂

P₂ = 1.013 × 10⁵ N/m² × 5 ml/1 ml

P₂ = 1.013 × 10⁵ N/m² × 5

P₂ = 5.065 × 10⁵ N/m²

Since pressure, P = F/A where F = force and A = cross-sectional area of syringe = πd²/4 where d = diameter of syringe = 1.5 cm = 1.5 × 10⁻² m.

So, F = PA

F = P₂πd²/4

substituting the values of the variables into the equation, we have

F = P₂πd²/4

F = 5.065 × 10⁵ N/m²π(1.5 × 10⁻² m)²/4

F = 5.065 × 10⁵ N/m²π(2.25 × 10⁻⁴ m²)/4

F = 35.8 × 10/4 N

F = 8.95 × 10

F = 89.5 N

b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?

Since the pressure in the syringe after compression is constant, we have

P₂ = F₁/A₁ where F₁ = force exerted on thumb and A₁ = cross-sectional area of  opening of syringe = πd₁²/4 where d = diameter of opening of syringe = 2 mm = 2 × 10⁻³ m.

So, F₁ = P₂A₁

F = P₂πd₁²/4

substituting the values of the variables into the equation, we have

F = P₂πd²/4

F = 5.065 × 10⁵ N/m²π(2 × 10⁻³ m)²/4

F = 5.065 × 10⁵ N/m²π(4 × 10⁻⁶ m²)/4

F = 15.91 × 10⁻¹

F = 1.591 N

F ≅ 1.59 N

Stresses in a Newtonian fluid due to shear. A fluid cannot resist shearing forces while it is at rest. These forces cause it to continuously distort.

Many non-Newtonian fluids are shear-thinning or pseudoplastic, that is the viscosity h) decreases with increasing shear-rate γ. The product, however, constantly grows with increasing shear-rate since the size of the shear stress always does as well.

A fluid that has constant viscosity and a shear rate that is directly proportional to the shear stress is referred to as a Newtonian fluid.

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Answer:

Problem definition

Explanation:

The engineering design is a series of step that the engineer gradually take in other to create a functioning product or process. There are many different models of the engineering design process , but they can be shrunk into four basic steps which are

The students are within the first step, and the problem here is 'an appropriate means for the students to keep their drinks cold, for 3 hours for the ball game.'

Answer:

I think Microsoft Corporation