The DHCP Server is software running on some server OS (like Windows Server, or Linux). O True O False

Answer: True

Explanation:

Answer:

on the façade, entrance, and column capitals

Answer: Increase in minimills

Explanation:

According to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.

They are experts in construction projects, supporting the design of plans, estimating costs, planning work methods, etc.

Research can facilitate the work of building technicians in the following ways:

Therefore, we can conclude that according to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.

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Using the formula of comfort condition it is possible to determine the safe operating speed on the highway.

In physics, we define angular deviation as the angle resulting from the deflection of light when reflected from a prism. A prism has the ability to separate white light into several other colors, which are reflected by the object.

The comfort condition formula is given by: [tex]L=2(\frac{NV^{3} }{C} )^{\frac{1}{2} }[/tex]

Where, L is the length of the vertical curve, N is the grade, V is the safe speed operation and assuming the standart value for C (0,6 m/s3) it is possible to calculate:

[tex]C = 0,6 m/s^{3} = 1,968 ft/s^{3}\\ N= N1 - N2 = 7,5-0 = 7,5% = 0,075[/tex]

Changing the values in the formula:

[tex]L=2(\frac{NV^{3} }{C}){\frac{1}{2} } \\300 = 2(\frac{0,075.V^{3} }{1,968})^{\frac{1}{2} } \\V = 83,89 ft/s[/tex]

So, the safe operating speed on the highway is 83,89 ft/s.

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The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is; 35535 N

We are told that for an alloy, the yield stress is 345-MPa and the elastic modulus (E) is 103-GPa.

Now, we want to find the maximum load that may be applied to a specimen with a cross-sectional area of 130-mm² without plastic deformation. Thus;

We are given the parameters;

Yield Stress; σ = 345 Mpa = 345 * 10⁶ Pa

Elastic Modulus; E = 103 GPa = 103 * 10⁹ Pa

Cross sectional Area; A = 130 mm² = 103 * 10⁻⁶ m²

Formula for stress without Plastic deformation is;

σ = F_max/Area

where;

σ is stress

F_max is maximum force

Area is Area

Thus making maximum force the subject of the formula gives;

F_max = σ * A

Plugging in the relevant values for stress and area gives us;

F_max = 345 * 10⁶ * 103 * 10⁻⁶

F_max = 35535 N

The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is gotten to be 35535 N

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Answer:

acidity, time,oxygen

Explanation:

These are some conditions for the growth of bacteria

Answer:

[tex]u = 260.22m/s[/tex]

[tex]S_{max} = 1141.07ft[/tex]

Explanation:

Given

[tex]S_0 = 89.6ft[/tex] --- Initial altitude

[tex]S_{16.5} = 0ft[/tex] -- Altitude after 16.5 seconds

[tex]a = -g = -32.2ft/s^2[/tex] --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

[tex]S = ut + \frac{1}{2}at^2[/tex]

The final altitude after 16.5 seconds is represented as:

[tex]S_{16.5} = S_0 + ut + \frac{1}{2}at^2[/tex]

Substitute the following values:

[tex]S_0 = 89.6ft[/tex]       [tex]S_{16.5} = 0ft[/tex]     [tex]a = -g = -32.2ft/s^2[/tex]    and [tex]t = 16.5[/tex]

So, we have:

[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2[/tex]

[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45[/tex]

[tex]0 = 89.6 + 16.5u- 4383.225[/tex]

Collect Like Terms

[tex]16.5u = -89.6 +4383.225[/tex]

[tex]16.5u = 4293.625[/tex]

Make u the subject

[tex]u = \frac{4293.625}{16.5}[/tex]

[tex]u = 260.21969697[/tex]

[tex]u = 260.22m/s[/tex]

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

[tex]v=u + at[/tex]

At the maximum height:

[tex]v =0[/tex] --- The final velocity

[tex]u = 260.22m/s[/tex]

[tex]a = -g = -32.2ft/s^2[/tex]

So, we have:

[tex]0 = 260.22 - 32.2t[/tex]

Collect Like Terms

[tex]32.2t = 260.22[/tex]

Make t the subject

[tex]t = \frac{260.22}{ 32.2}[/tex]

[tex]t = 8.08s[/tex]

The maximum height is then calculated as:

[tex]S_{max} = S_0 + ut + \frac{1}{2}at^2[/tex]

This gives:

[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2[/tex]

[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22[/tex]

[tex]S_{max} = 89.6 + 260.22 * 8.08 - 1051.11[/tex]

[tex]S_{max} = 1141.0676[/tex]

[tex]S_{max} = 1141.07ft[/tex]

Hence, the maximum height is 1141.07ft

The frequency bands that are used primarily for short-range distances include the following:

GSM is an abbreviation for global system for mobile communication and it can be defined as an open and digital cellular network technology that is designed and developed to be used for the transmission of both mobile voice and data services at the same time.

In Computer technology and networking, the global system for mobile communication (GSM) is typically designed and developed to operates at the following bands of frequencies:

In this context, we can infer and logically deduce that the frequency bands that are used primarily for short-range distances such as PAN, include the following:

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Answer:

Industrial users can select from numerous pump technologies for water supply and the handling of other liquid media. A particular favorite is the centrifugal pump. How do centrifugal pumps work? What different delivery principles are there? For which applications are centrifugal pumps especially well-suited?

Here we provide you an overview of the different types of centrifugal pumps. In addition, the operating principles are explained, and typical applications are listed also.

Explanation:

The new slip rings is been  attached to the rotor core during alternator rebuilding by b. Soldered on.

Soldering serves as the way of joining  different types of metals together by melting solder.

This is been used on new slip rings to get it  attached to the rotor core during alternator rebuilding .

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CHECK THE COMPLETE QUESTION BELOW:

How are new slip rings attached to the rotor core during alternator rebuilding?

a. Machine clamped

b. Soldered on

c. Crimped on

d. epoxy glue

Answer:

Camshaft

Explanation:

Answer:

5.118 m^3/hr

Explanation:

Given data:

viscosity of cell broth = 5cP

cake resistance = 1*1011 cm/g

dry basis per volume of filtrate = 20 g/liter

Diameter = 8m ,  Length = 12m

vacuum pressure = 80 kpa

cake formation time = 20 s

cycle time = 60 s

Determine the filtration rate in volumes/hr  expected fir the rotary vacuum filter

attached below is a detailed solution of the question

Hence The filtration rate in volumes/hr expected for the rotary vacuum filter

V' = ( [tex]\frac{60}{20}[/tex] ) * 1706.0670

   = 5118.201 liters  ≈ 5.118 m^3/hr

The demand curve is the graphical representation of the relationship between the price of a good and the quantity demanded for a given period of time.

A demand schedule is a table which shows the quantity demanded of a good or service at different price levels.

A demand schedule can be graphed as a continuous demand curve on a chart where the Y-axis represents the price and the X-axis represents quantity.

Here, a typical representation, the price will appear on the left vertical axis, the quantity demanded on the horizontal axis.

Note that the complete information wasn't found and an overview was given.

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An example of a power consuming device would be a(n) light bulb, a computer device while an example of a non-power consuming device would be a(n) switch or button.

The quantity of energy utilized per unit of time is known as power consumption. Power use is a significant factor in digital systems.

Power consumption is a limiting factor for the battery life of portable devices like laptop computers and cell phones.

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Based on the positive-going zero crossing at 0° of the sine wave, the instantaneous value at 15° is 7.321V.

At 15°, the instantaneous value is:

= 20√2 x Sin(angle)

Solving gives:

=  20√2 x Sin(15°)

= 7.321V

At angel 33°, the instantaneous value is:

=  20√2 x Sin(angle)

=  20√2 x Sin(33)

= 15.405V

At angle 70°:

=  20√2 x Sin(angle)

=  20√2 x Sin(70)

= 26.579V

The full question is:

A certain sine wave has a positive-going zero crossing at 0° and an rms value of 20 V. Calculate its instantaneous value at each of the following angles: a. 15° b. 33° c. 70°

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Airtight plugs are installed in the ends of large sizes of PVC conduit before bending to  A. prevent the conduit from collapsing when heated.

Airtight plugs is necessary in the  PVC conduit so as to be able to avoid the collapse of conduit.

In this case, Airtight plugs are installed in the ends of large sizes of PVC conduit before bending to  A. prevent the conduit from collapsing when heated.

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The heat transferred to and the work produced by the steam during this process  is 13781.618 kJ/kg

The Net Change in Enthalpy will be:

= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)

= 1/2 x ( 75 + 225) x (5 - 2)

W = 450 KJ

From the First Law of Thermodynamics, Q = U + W

So, Heat Transfer = Change in Internal Energy + Work Done

= 13331.618 + 450

Q = 13781.618 kJ/kg

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Answer:

19.81 m/s

Explanation:

The total acceleration of the truck (a) is due to the centripetal acceleration and as a result of the linear acceleration. Therefore the total acceleration (a) is given by:

[tex]a^2=a_n^2+a_t^2\\\\where\ a_n=centripetal\ acceleration=\frac{v^2}{r},a_t=linear \ acceleration\\\\But\ since\ the \ speed\ is \ constant, the \ linear \ acceleration(a_t)\ would\ be\ 0.\\\\a^2=a_n^2+a_t^2\\\\a^2=a_n^2\\\\a=a_n=\frac{v^2}{r} \\\\v^2=ar\\\\v=\sqrt{ar} \\\\a=0.4g=0.4*9.81,r=98\ m+2\ m=100\ m\\\\v=\sqrt{0.4*9.81*100} \\\\v=19.81\ m/s[/tex]

Water pressure principles is known to  state that the amount of Fluid pressure is said to be perpendicular to the surface on which it it exert or works on.

Note that  principle is shown by a vessel that is said to be having flat sides and has water. The pressure exerted by the weight of the water is therefore known to be perpendicular to the walls of the storage.

A water butt is known to be a container that is made and also used to store a lot of rainwater.

Note that Water pressure principles is known to  state that the amount of Fluid pressure is said to be perpendicular to the surface on which it it exert or works on.

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Answer:

d

Explanation:

OR criteria, AND criteria

In an OR criteria, it doesn't need all the records to be true. Just one record is enough and all other criterion becomes true.

In an AND criteria, it's unlike the OR criteria and works in opposite. It needs every member of the record to be true to be able to adjudge the whole record as true.

And as such, we have

With OR criteria, only one criterion must evaluate true in order for a record to be selected and with AND criteria, all criteria must be evaluate true in order for a record to be selected.

Answer: This approach is allocated 32 seconds of effective green time. The cycle length is 100 seconds. Determine the average approach delay (using Eq. 7.27). A) 34.8 s.

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

[tex]\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}[/tex]

Solve the following for the real stress and pressure for the stable.[tex]\sigma_{r1}=K(\varepsilon_{r1})^{n}[/tex]

[tex]K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}[/tex]

Solve the following for the true state stress and stress2.

[tex]\sigma_{r2}=K(\varepsilon_{r2})^n[/tex]

     [tex]=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa[/tex]

Different lever designs can be engineered and developed to alter the brake pedal effort required of the driver by using different levels of mechanical advantage.

Mechanical advantage can be defined as a ratio of the output force of a lever to the force acting on it (input force or effort), assuming no losses due to wear, flexibility, tear or friction.

This ultimately implies that, different lever designs can be suitably engineered and developed to alter the brake pedal effort (input force) that is required of the driver, especially by using different levels of mechanical advantage.

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The combination of temperatures that is expected to give the best result is option A: SCY temperature higher than electrode temp.

The answer is option A because the proton conductivity of SCY is known to be one that often goes up along with increasing temperature, and the favorability of reaction goes down or reduces with total temperature.

Hence, It is said to be good for one to to keep the SCY conductor at a higher temperature and as such, The combination of temperatures that is expected to give the best result is option A: SCY temperature higher than electrode temp.

See options below

. Which combination of temperatures is expected to give the best results?

A.

SCY temperature higher than electrode temperature

B.

SCY temperature lower than electrode temperature

C.

SCY temperature the same as electrode temperature

D.

The temperature of the components does not make a difference.

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Answer:

hi there

Explanation:

The working principle of a DC machine is when electric current flows through a coil within a magnetic field, and then the magnetic force generates a torque that rotates the dc motor. The DC machines are classified into two types such as DC generator as well as DC motor.

Answer:

hello some parts of your question is missing attached below is the missing part

answer :

A) Determine changes in the 50-mm dimensions

The changes are : 0.006mm compression  in y-direction

                               0.002 mm expansion in x and z directions

B) the stress required are evenly distributed

Explanation:

Given data :

50-mm cube of graphite fiber reinforced polymer matrix

subjected to 125-KN force in direction 2,

direction 2 is perpendicular to fiber direction ( direction 1 ) and cube is constrained against expansion in direction 3

A) Determine changes in the 50-mm dimensions

The changes are : 0.006mm compression  in y-direction

                               0.002 mm expansion in x and z directions

B) the stress required are evenly distributed

attached below is the detailed solution