a 4-pole induction motor drives a load at 2100 rpm. this is to be accomplished by using an electronic converter to convert a 380v380v dc source into a set of 3-phase ac voltages.
Assuming the slip is 4.5%, the frequency required for the ac voltage is .... Hz

Answer:

true

Explanation:

i took the test and got it right

Answer:

Shallow foundations, often called footings, are usually embedded about a metre or so into soil. ... Another common type of shallow foundation is the slab-on-grade foundation where the weight of the structure is transferred to the soil through a concrete slab placed at the surface.

Explanation:

Because I said so.

Answer:

See explanation below

Explanation:

Hypo-eutectoid steel has less than 0,8% of C in its composition.

It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.

Ferrite has a higher tensile strength than cementite but cementite is harder.

Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:

Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel

Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel

Hyper-eutectoid steel is harder than Hyper-eutectoid steel

Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.

When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because

1. It is harder

2. It has low cost

3. It is lighter

When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because

1. It is ductile

2. It has high tensile strength

3. It is durable

Answer:

(a)

Steels having carbon within 0.02% – 0.8% which consist of ferrite and pearlite are known as hypoeutectoid steel.

Steels having greater than 0.8% carbon but less than 2.0% are known as hypereutectoid steel.

(b)

The proeutectoid ferrite formed at a range of temperatures from austenite in the austenite+ferrite region above 726°C. The eutectoid ferrite formed during the eutectoid transformation as it cools below 726°C. It is a part of the pearlite microconstiutents . Note that both hypereutectoid and hypoeutectoid steels have proeutectoid phases, while in eutectoid steel, no proeutectoid phase is present.

Proeutectoid signifies is a phase that forms (on cooling) before the eutectoid austenite decomposes. It has a parallel with primary solids in that it is the first phase to crystallize out of the austenite phase. If the steel is hypoeutectoid it will produce proeutectoid ferrite and if it is hypereutectoid it will produce proeutectoid cementite.  The carbon concentration for both ferrites is 0.022 wt% C.

(c)

(i) Yield strength: The hypoeutectoid steel have good yield strength and hypereutectoid steels have little higher yield strengh.

(ii) Ductility: The hypoeutectoid steel is more ductile and the ductility has decreased by a factor of three for the eutectoid alloy. In hypereutectoid alloys the additional, brittle cementite on the pearlite grain boundaries further decreases the ductility of the alloy. The proeutectoid cementite restricts plastic deformation to the ferrite lamellae in the pearlite.

(iii) Hardness:  hypoeutectoid steels are softer and hypereutectoid steel contains low strength cementite at grain boundary region which makes it harder than hypoeutectoids.

(iv) Tensile strength: Grain boundary regions of hypereutectoid steel are high energy regions prone to cracking because of cementite in the grain boundaries, its tensile strength decreases drastically even though pearlite is present. Hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains, so grain boundaries being the high energy state region, it has a higher tensile strength.

(d)

I would recommend hypereutectoid steel alloy to make a knife or ax blade

1- Hardness is required at the surface of the blades.

2- Ductility is not needed for such application.

3- Due to constant impact, the material will not easily yield to stress.

(e)

I would choose a hypoeutectoid steel alloy to make a steel that was easy to machine.

1- hypoeutectoid steel alloys have high machinability, hence better productivity

2- It will be used on softer metals, hence its fitness for the application

3- Certain amount of ductility is required which hypoeutectoid steel alloys possess.

Explanation:

See all together above

A device that can connect many computers and sends packets out of every port is known as Hub.

The name of the device that usually connects many computers is known as a hub. It typically connects multiple computer networking devices together. It also acts as a repeater in that it amplifies signals that worsen after traveling long distances over connecting cables.

According to the context of this question, a hub usually connects multiple computers together in a Local Area Network (LAN). All types of information are generally processed or sent to the hub is then sent through each port to every device in the network. It is noted that all types of devices are definitely connected to a network hub and share all available bandwidth equally.

Therefore, a device that can connect many computers and sends packets out of every port is known as Hub.

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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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Tech A says that unitized hubs have a wheel nut with a higher installation torque than serviceable wheel bearings. Tech B says that unitized hubs have the proper bearing end play designed into the assembly once they are torqued properly is Option C: Both a and b

When compared to manually adjusted, PreSet, or LMS hub assemblies, unitized hub assemblies often require a lot more assembly torque and special spindle nut systems.

Therefore, An essential component of the wheel assembly that connects the wheel to the axle is a wheel bearing. A metal ring is used to hold a group of steel balls (also known as ball bearings) or taper (also known as tapered bearings) together. It permits the wheel to spin easily and with little resistance.

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Answer:

  C.  3325 Ω - 3675 Ω

Explanation:

5% of 3500 Ω is ...

  0.05 × 3500 = 175

The lower limit is this amount less than the nominal value:

  3500 -175 = 3325

The upper limit is the nominal value plus the tolerance:

  3500 +175 = 3675

The lower and upper limits are 3325 Ω and 3675 Ω, respectively.

Answer:

the energy is sgdrh35679

To exert a pressure of 1.4 atm at 430 K and occupy a volume of 15.1 L, approximately 21.1 moles of H2 gas are necessary.

Let's solve the problems using the ideal gas law (PV = nRT) and show the work for each problem:

A. What is the volume of 0.5 moles of gas at STP?

Given:

n = 0.5 moles

R = 0.0821 L·atm/(mol·K) (ideal gas constant)

T = 273 K (standard temperature)

Using the ideal gas law, we can rearrange the formula to solve for volume (V):

V = (nRT) / P

Substituting the given values:

V = (0.5 moles * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm

V = 11.2 L

Therefore, the volume of 0.5 moles of gas at STP is 11.2 L.

B. How much pressure would 0.8 moles of a gas at 370 K exert if it occupied 17.3 L of space?

Given:

n = 0.8 moles

R = 0.0821 L·atm/(mol·K) (ideal gas constant)

T = 370 K

V = 17.3 L

Using the ideal gas law to solve for pressure (P):

P = (nRT) / V

Substituting the given values:

P = (0.8 moles * 0.0821 L·atm/(mol·K) * 370 K) / 17.3 L

P = 1.4 atm

Therefore, the pressure exerted by 0.8 moles of gas at 370 K, occupying 17.3 L of space, is 1.4 atm.

C. How much H2 gas is necessary to exert a pressure of 1.4 atm at 430 K if occupying a volume of 15.1 L?

Given:

R = 0.0821 L·atm/(mol·K) (ideal gas constant)

T = 430 K

P = 1.4 atm

V = 15.1 L

Using the ideal gas law to solve for the number of moles (n):

n = (PV) / (RT)

Substituting the given values:

n = (1.4 atm * 15.1 L) / (0.0821 L·atm/(mol·K) * 430 K)

n = 21.1 moles

Therefore, to exert a pressure of 1.4 atm at 430 K and occupy a volume of 15.1 L, approximately 21.1 moles of H2 gas are necessary.

Please note that the values obtained from the Gizmo may vary slightly due to rounding or the use of slightly different values for the ideal gas constant.

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Friction clutches, tooth clutches, magnetic clutches, hydraulic clutches, and centrifugal clutches are the five types of clutches.

The clutch is the mechanism within a vehicle which connects rotating shafts and transfers power from the engine to the wheels. Power does not transfer effectively without a properly functioning clutch, and the automobile will not shift.

Note that the above clutches  and their classifications are based on the mechanisms used to transfer torque, which include frictional force, tooth engagement, magnetic force, hydraulic pressure, and centrifugal force.

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The method that is not the method of the String class is count().

The valid indexes for the string "New York" are:

O through 7

O through 8-1

O through 6

A string is a data type in programming that is used to represent text. Strings are a set of characters that are enclosed in either single or double quotes in Python. A string is a sequence of characters, each of which has a unique index number. Strings can be combined by concatenation to create new strings or indexed and sliced to access individual characters or substrings. Therefore, the method that is not the method of the String class is count(). Methods of the String class are: isalpha(...), len(...), lower(...).

Now, we need to determine the valid indexes for the string 'New York'. So, the correct options are:

O through 7, O through 8-1, and O through 6.

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Answer:

B- Constructive interference will occur, causing the new wave to have more energy than Wave A or Wave B.

Explanation:

What will occur when the trough of Wave A overlaps the trough of Wave B is; B: Constructive interference will occur, causing the new wave to have more energy than Wave A or Wave B.

Wave interference is simply defined as the phenomenon that transpires when two waves meet each other while traveling along the same medium.

Now, this meeting of the waves in interference could either be constructive interference or destructive interference.

Constructive interference is the type that occurs where the waves cross over each other in such a way that the two waves are in phase while Destructive interference is a wave interference that occurs where two waves are completely out of phase.

Looking at the question it says they overlap each other and this denotes constructive interference and a new wave will be formed with more energy than the parent waves.

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Answer:

when car vroom it go electric

Explanation:

i think then it goes to some spinny thing that spins and make the electric and the electic spins the spinny thing and the smoke comes outside the back

Answer:

The goal of the software is to observe the software behavior to meet its requirement expectation. In software engineering, validating software might be harder since client's expectation may be vague or unclear.

Explanation:

The initial enthalpy and the entropy of the saturated water can be found out from the table of A-12E

i.e.   \($h_1= 101 \ \text{Btu/lbm}$\)

       \($s_1 = 0.22717 \text{ Btu/lbmR}$\)

Since the process mentioned above is an adiabatic compression process, the entropy will remain constant throughout the process. Therefore, we take the value of entropy and the final pressure using the table with few interpolations and also approximations to find the final enthalpy. It is given by :

       \($h_2= 116.09 \text{ Btu/lbm}$\)

So the work input from the energy balance equation :

       \($\dot{W} + \dot{m}h_1 = \dot{m}h_2$\)

          \($w=h_2 - h_1$\)

              = 116.09 - 101

              = 15.09

Therefore, \($w= 15.09 \text{ Btu/lbm}$\)

Answer:

Used to recall the direction a standard screw

An open dot within the Airport Symbol means that fuel services are available at the airport. This symbol helps pilots and airport staff identify the availability of fuel for aircraft at a particular location.

An open dot within the Airport Symbol means that the content loaded is available for that airport. A location where airplanes land and take off, where passengers can wait in buildings, and where aircraft are protected.  A small airport called a heliport can be used by helicopters and other vertical lift aircraft. In addition to having limited facilities like fuel or hangars, designated heliports typically have one or more areas for touchdown and liftoff. Actual offices incorporate runways, runways, covers, and strips, which are utilized for the setting down and departure of airplane, for the moving and situating of airplanes on the ground, and for the stopping of airplanes to load and release travelers and freight.

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The Airport Symbol indicates the availability of fuel services by an open dot. This symbol aids airport workers and pilots in determining whether or not there is aeroplane fuel nearby.

The content loaded is accessible for that airport if there is an open dot within the airport symbol. a place where aircraft are protected, where passengers can wait in buildings, and where aeroplanes land and take off.  Helicopters and other vertical lift aircraft can utilise a tiny airport called a heliport. The majority of certified heliports feature one or more spaces for landing and taking off in addition to having a few restricted facilities like fuel or hangars.

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Answer:

The landscape architect uses knowledge of legal requirements to inform design choices.

Explanation:

The landscape architect doesn’t only draw on knowledge of law and government to design grounds for public buildings, but will draw on that knowledge quite often on many projects. The landscape architect’s knowledge of design is not purely intuitive but is a product of study and might involve research. The landscape architect’s knowledge of law and government, however, does come into play in making design choices that are consistent with the law and government regulations.

Every thread have its own stack, register, and PC, so only address space that is shared by all thread for a single process.

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Based on the Chvorinov's rule, the diference in the solidification times of the two castings is 14.092 times the solidification time of the prism casting.

The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a reference time. This rule states that cooling time (t) is directly proportional to the square of the volume (V), in cubic meters, divided to the surface area (A), in square meters. Now we proceed to model each casting:

t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²

t = C · [0.25 · D · L/(0.5 · D + L)]²    (1)

t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²

t' = C · [3 · T · L/(8 · L + 6 · T)]²     (2)

3 · T² = 0.25π · D²     (3)

Where:

If we know that D = 0.3 m, then the thickness of the prism casting is:

\(T = \sqrt{\frac{\pi}{12} }\cdot D\)

T ≈ 0.153 m

And (1) and (2) simplified into these forms:

t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²

t = 0.0329 · C     (1b)

t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²

t' = 0.00218 · C     (2b)

Lastly we find the percentual difference in the solidification times of the two castings by using the following expression:

r = (1 - t'/t) × 100 %

r = (1 - 0.00218/0.0329) × 100 %

r = 93.374 %

The cooling time of the prism casting is 6.626 % of the solidification time of the cylindrical casting. The diference in the solidification times of the two castings is 14.092 times the solidification time of the prism casting. \(\blacksquare\)

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A distance of at least 6 inches should separate the siding from a roof surface.

A distance of at least 6 inches should separate the siding from a roof surface.

This is to prevent moisture and water damage that can result from contact between the two surfaces.

When siding is installed too close to the roof surface, rainwater and other moisture can easily seep into the siding and cause rot, decay, and other forms of damage.

This can result in expensive repairs and replacements over time. To avoid this, it is essential to maintain a sufficient gap between the siding and the roof surface.

In addition, proper installation techniques and materials should be used to ensure that the siding is securely attached and able to withstand exposure to the elements over time.

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For a linear time-invariant (LTI) system, BIBO stability refers to bounded-input bounded-output stability. It means that if the input to the system is bounded, the output will also be bounded.

In the context of BIBO stability, the inverse system is not directly related to the condition you mentioned involving absolute summability. The condition you provided, Σ1h1n1, seems to be related to the impulse response of the system.In order to determine the BIBO stability of an LTI system, you need to examine the properties of its impulse response. The impulse response, denoted by h(n), describes the output of the system when an impulse is applied as the input.For a system to be BIBO stable, the impulse response h(n) must be absolutely summable, meaning that the sum of the absolute values of its elements is finite. Mathematically, it can be expressed as:Σ|hn| < ∞If the impulse response satisfies this condition, then the LTI system is BIBO stable.Please note that the inverse system and its absolute summability are not directly involved in determining BIBO stability. BIBO stability is solely based on the properties of the system's impulse response.

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Explanation:

Conservation of velocity equation

Answer:

The light from the first LED will appear as yellow while the light from the second LED won't appear as Yellow light when passed through a diffraction grating

Explanation:

There are two types of LED's

First LED emits only yellow light  with wavelength of 590 nm

Second LED : emits red light with wavelength of 630 nm and emits green light with a 540 nm wavelength

The second LED produces a yellow light because of the mixture of red and green lights

therefore when the light from the SECOND LED is passed through a  diffraction grating, the light is split into its components ( i,e green and red lights ) this is because the lights are not on the same wavelength

The question is incomplete. The complete question is --

An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Given the rod has a diameter of 1.01 cm and the young's modulus of aluminum is 69 GPa, determine the following :

1. the interference value, i

2. the radial pressure at the interference of the collar and the rod

3. the maximum effective stress in the collar

4. if the yield strength of aluminium is 200 MPa and assume a safety factor of 1.5, will the aluminium collar break

Solution:

Given :

Inner radius of the collar = 1 cm

So, inner diameter, \($d_1$\) = 2 cm

Outer radius of the collar = 12 cm

So, outer diameter, \($d_2$\) = 4 cm

The aluminium rod diameter, d = 1.01 cm

Now, from the figure, we can see that there will be no interference and so the rod will easily insert inside the collar.

1. So, the interference , i =0

2. The radial pressure is also 0.

3. There will be no stress developed. So the maximum effective stress is 0

4. The collar will not break

Answer:

Example problem

Explanation:

(a) The source rate can be calculated as the product of the picture sample size and number of pixel per sample size.

Given:

Picture sample size = 480 * 500. The number of pictures sent per second = 30.

Hence sample size = 480 * 500 * 30 = 7.2e6.

Since each pixel (sample) can take one of 32 values (voltage levels), hence it needs

l

o

g

2

32

=

l

o

g

2

2

5

=

5

∗

l

o

g

2

2

=

5

∗

1

=

5

bits/pixel (bits/sample).

The source rate R is (bps) = 7.26 * 5 = 36e6 bps.