1. An asynchronous motor with a rated power of 15 kW, power factor of 0.5 and efficiency of 0.8, so its input electric power is ( ). (A) 18.75 (B) 14 (C) 30 (D) 28 2. If the excitation current of the DC motor is equal to the armature current, this motor is called the () motor. (A) separately excited (B) shunt (C) series (D) compound 3. When the DC motor is reversely connected to the brake, the string resistance in the armature circuit is (). (A) Limiting the braking current (C) Shortening the braking time (B) Increasing the braking torque (D) Extending the braking time 4. When the DC motor is in equilibrium, the magnitude of the armature current depends on (). (A) The magnitude of the armature voltage (B) The magnitude of the load torque (C) The magnitude of the field current (D) The magnitude of the excitation voltage 5. The direction of rotation of the rotating magnetic field of an asynchronous motor depends on ().

Answer:

B) 430 Pci/L

C) F = 1.05 * 10 ^13 L/s

Explanation:

Given data :

enclosed area dimensions : 5 * 4 * 3 m^3

emission of  Radon-22 through walls = 0.1 atoms/(cm^2.s)

first we will have to determine the total surface area of the enclosed area

= 2 lb + 2 bh + 2 lh

= 2( 5*4 ) + 2( 4*3 ) + 2 ( 5*3 ) = 9.4 * 10^5 cm^2

next we determine the production rate ( P )

p = 0.1 atoms/(cm^2.s) * 9.4 * 10^5 cm^2 =  9.4 * 10^5 atoms/s

determine production rate per volume

production rate / space volume =  9.4 * 10^5 / ( 5*4*3 )

                                                     =  9.4 * 10^5 / (60 * 10^3) L

                                                     = 16 atoms/ L

attached below is the remaining part of the solution

The Transport Layer in the TCP/IP stack performs similar functions to the Transport Layer in the OSI model. It ensures reliable and orderly delivery of data between hosts using protocols like TCP and UDP.

In the TCP/IP stack, the layer that is equivalent to the Transport Layer in the OSI model is the Transport Layer. The Transport Layer is responsible for the end-to-end delivery of data between hosts on a network. It ensures that data is delivered reliably and in the correct order.
One of the key protocols used in the Transport Layer of the TCP/IP stack is the Transmission Control Protocol (TCP). TCP provides reliable and connection-oriented communication between two devices. It breaks data into smaller packets, assigns sequence numbers to each packet, and ensures that all packets are received by the destination device. It also handles congestion control and flow control to optimize network performance.
Another protocol used in the Transport Layer of the TCP/IP stack is the User Datagram Protocol (UDP). Unlike TCP, UDP is connectionless and does not provide reliable delivery or sequencing of packets. It is commonly used for applications that require faster transmission but can tolerate occasional packet loss, such as streaming media or real-time communication.
To summarize, the Transport Layer in the TCP/IP stack performs similar functions to the Transport Layer in the OSI model. It ensures reliable and orderly delivery of data between hosts using protocols like TCP and UDP.

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Answer:

c

Explanation:

This is because many things, such as pcs, over heat

Preheats the engine when the vehicle is started the next time. Therefore, the correct answer is option C.

The coolant heat storage system is designed to store heat from the exhaust system when the vehicle is operating and then use that heat when it is started again. This gives the car a kind of "head start" and the engine does not need to work as hard to heat up and become ready for driving. This can save on fuel consumption and hydrocarbon emissions. This process is done with the use of a condenser to store the heat energy. Additionally, the coolant heat storage system can also help to prevent the engine from overheating in hot weather and conditions.

Therefore, the correct answer is option C.

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Answer:

i) 0.610 m or 610 mm

ii) 0.4 m or 400 mm

Explanation:

The pressure difference between the pipes is

a) Air

Pa + πha +Ha = Pb + πhb +Hb

Pa - Pb = π(hb-ha) + Hb-Ha

Relative density of air = 1.2754 kg /m3

Pa - Pb = 1.2754 * 0.4 + (0.3-0.2) = 0.610 m or 610 mm

b) paraffin of relative density of 0.75

Pa - Pb = π(hb-ha) + Hb-Ha

Pa - Pb = 0.75 * 0.4 + (0.3-0.2) = 0.4 m or 400 mm

Answer:

Thin metal blades or flexible bands with teeth on one edge, thin grinding wheels, or thin metal disks with teeth on their edges can all be used as sawing machine cutting tools.

Sawing machine is defined as a device for cutting forms out of raw material plates or cutting apart bars of material. It is used to sever material, usually wood, but occasionally metal or stone. The material is cut by pressing the toothed edge against it and pushing it swiftly forward, slowly backward, or continually forward.

Rotating blades, reciprocating blades, and circulating blades are the three basic types of power sawing machines.

Thus, thin metal blades or flexible bands with teeth on one edge, thin grinding wheels, or thin metal disks with teeth on their edges can all be used as sawing machine cutting tools.

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The heat transfer for (a) water as the system is 165.79 Btu and the entropy production is 0.4855 Btu/R for both (a) and (b) systems.The heat transfer and entropy production are the same as for (a) the water as the system.

To evaluate the heat transfer and entropy production for the given system, we can use the energy and entropy equations.

(a) For the water as the system:

Heat transfer (Q) is the enthalpy change from initial state to final state.

Entropy production (ΔS) is the change in entropy of the system.

Since the water is condensed at constant pressure, the enthalpy change is equal to the heat transfer:

Q

To evaluate the entropy production, we can use the entropy balance equation:

ΔS = m * (s_f - s_i) - Q / T

where m is the mass of water and T is the temperature at which heat transfer occurs.

(b) For the enlarged system:

In this case, the heat transfer occurs only at the ambient temperature, so the heat transfer is given by:

Q = m * Cp * (T_f - T_i)

The entropy production can be evaluated using the entropy balance equation as before:

ΔS = m * (s_f - s_i) - Q / T

where m is the mass of water, Cp is the specific heat capacity, and T is the ambient temperature.

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Explanation:

The point on the line 3x + 4y = 5, which is equidistant from (1, 2) and (3, 4) is. Therefore, the point is (15, -10).

Answer:

Control panel

Explanation:

The Control Panel is a component of Microsoft Windows which provides the ability to view and change system settings. It consists of a set simple computer programs (applets) that include adding or removing hardware and software, controlling user accounts, changing accessibility options, and accessing networking settings.

Sarah is on a see saw with her friend. The pivot or fixed point where the bar is free to move is called the  Fulcrum. Hence, option C is correct.

The support or base that a lever rotates on in order to lift or move anything. The verb fulcire, which means "to prop," is the root of the Latin term fulcrum, which means "bedpost."

The term "fulcrum" was first used in the 17th century to describe the pivotal point of a lever or other similar mechanism, such as the oar of a boat.

The Fulcrum line of road bike wheels offers lightweight and durable wheelsets for every circumstance, course, discipline, and price range. From the top-tier full carbon aero wheels to the high performance aluminum line, Fulcrum offers the perfect wheel to match your performance.

Thus, option C is correct.

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INSIDE JOIN

When using this join method, the records that have matching values in both tables are returned. All the tuples with matching values in both tables will be output if you perform an INNER join operation between the Employee table and the Projects table.

If the criteria is met, the INNER JOIN keyword selects all rows from both tables. This keyword will combine all rows from both tables where the criterion is met to create the result set.

SQL supports a total of four join types. types include left outer join, right outer, and inner join.

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The Equivalent Service Life (ESL) of the injection molding system is determined to be 3 years, and the respective Annual Worth (AW) value of the system is calculated.

To determine the Equivalent Service Life (ESL) and Annual Worth (AW) value of the injection molding system, we need to calculate the Present Worth (PW) of costs and salvage value over the system's maximum useful life of 5 years. The first cost of $180,000 and the salvage value of 25% of the first cost ($45,000) are considered.

The annual operating cost of $95,000 in years 1 and 2, increasing by $4,500 per year thereafter, is converted to an equivalent uniform annual cost (EUAC) using the given MARR (12% per year). This EUAC is then added to the salvage value to obtain the PW.

By comparing the PW values at different service lives (1 to 5 years), the ESL is determined to be 3 years, which corresponds to the minimum PW value. The AW value is then calculated by dividing the PW by the ESL.

By considering the MARR, the cash flows over the system's life, and the salvage value, the ESL and AW value of the injection molding system can be determined, providing insights into the economic performance of the system over time.

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Answer:

4mA

Explanation:

For this problem, we will simply apply Ohm's law:

V = IR

V/R = I

I = V / R

I = 12 volt / 3kΩ

I = 4mA

Hence, the current in the circuit is 4mA.

Cheers.

The percentage of atrazine in the soil phase is 1.35*10^-8

The percentage of atrazine in the soil phase is 1.45%

The percentage of atrazine in the fish phase is 6.7*10^-5

The partition coefficient for atrazine between water and organic carbon (Koc) can be calculated using the empirical equation provided is 0.006.

The Kow value is 1.499. The distribution coefficient is 0.066.

The concentration of atrazine in water will be;

= 0.027 / 2000000 × 10^-8

= 1.35 × 20^-8

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In contrast to mobile work zones, stationary work zones have clearly defined boundaries.

According to the 2009 edition of the Manual on Uniform Traffic Control Devices (MUTCD), a short-term stationary work zone is "daytime work that occupies a location for more than one hour within a single daylight period."

The following are typical daytime work shift-length tasks:

paving the road, fixing broken underground water lines, fixing storm water catch basins, trimming trees along the road, and parking work vehicles for a long time on the side of the road. The minimum traffic control planning and devices that are required to safely direct motorists, bicyclists, and pedestrians around workers and work vehicles during daylight hours will be discussed in this bulletin. Night operations necessitate additional equipment and planning, which is not the subject of this bulletin.

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The design rotation speed of the centrifugal pump is 3,000 rpm. The outlet absolute flow angle (a2) is 48.9°, and the theoretical head developed by the pump is 169 ft.

In order to provide a more detailed explanation, let's break down the main answer into four parts:

(i) Sketching the inlet and outlet velocity triangles:

The inlet velocity triangle represents the relative velocity (W1) at an angle of 20° to the blade, while the outlet velocity triangle represents the relative velocity (W2) at an angle of 45° to the blade. These velocities can be related to the inlet and outlet conditions of the pump.

(ii) Determining the design rotation speed:

To determine the design rotation speed, we need to consider the equation: Q = π * r * b * c1, where Q is the flow rate (70 ft3/min), r is the radius (14 in.), b is the blade width (0.4 in.), and c1 is the absolute velocity at the inlet. By rearranging the equation and substituting the given values, we can solve for c1. Then, using the equation: c1 = π * D * N / 60, where D is the impeller diameter (2 * r) and N is the rotation speed in rpm, we can calculate the design rotation speed.

(iii) Determining the outlet absolute flow angle:

To determine the outlet absolute flow angle (a2), we can use the equation: tan(a2) = c2 / u2, where c2 is the absolute velocity at the outlet and u2 is the blade speed. By substituting the given values, we can solve for a2.

(iv) Determining the theoretical head developed by the pump:

The theoretical head developed by the pump can be calculated using the equation: H = (c2² - c1²) / (2 * g), where H is the head developed, c2 is the absolute velocity at the outlet, c1 is the absolute velocity at the inlet, and g is the acceleration due to gravity. By substituting the given values, we can find the theoretical head developed by the pump.

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The drawing-in maneuver and bracing techniques are used in various exercises such as squats, deadlifts, and overhead presses, to maintain proper form and alignment. They are also helpful for people with weak or injured core muscles, as they can help improve core strength, stability, and function.

Performing the drawing-in maneuver or bracing can activate the muscles of the transversus abdominis, multifidus, and pelvic floor. This will help increase intra-abdominal pressure, which is essential in maintaining a stable spine and reducing the risk of lower back pain and injury.

The drawing-in maneuver or bracing is a technique used to activate the deep core muscles that surround and stabilize the spine. It is an essential exercise for preventing and managing lower back pain.

The drawing-in maneuver works by activating the transversus abdominis, which is the deepest layer of abdominal muscles that lies beneath the rectus abdominis, internal oblique, and external oblique muscles. This muscle plays a crucial role in stabilizing the spine and pelvis, and improving posture and balance.

The bracing technique, on the other hand, involves contracting all the muscles around the midsection, including the transversus abdominis, multifidus, and pelvic floor muscles. This technique increases intra-abdominal pressure, which stabilizes the spine and reduces the risk of injury to the lower back.

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Solid nonmetal elements are "poor conductors of heat and electricity that are brittle and break easily".

Elements that usually lack the characteristics of metals are called nonmetals. Nonmetals are generally poor conductors of electricity and heat because in nonmetals electrons are not free to move. Nonmetals can be in form of solids, gases, or liquids. Solid nonmetals are referred to as those nonmetals that are not ductile or malleable but are brittle or powdery. Solid nonmetals are breakable easily.

Therefore, it is concluded that solid nonmetals are bad conductors of electricity and heat that are brittle and break easily.

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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

Answer:

the first one is the correct answer

Answer:

the first one would be correct

Explanation:

Answer:

$Monitor

Explanation:

The command that would be used when running a test bench to monitor variables or signals ( i.e. changes in the values of specific variables and signa)

is the $Monitor command

This command is also used to monitor the varying values of signals during simulation.

Answer:  So to calculate the flexural strength (σ), multiply the force by the length of the sample, and then multiply this by three. Then multiply the depth of the sample by itself (i.e., square it), multiply the result by the width of the sample and then multiply this by two

Explanation:

Answer:

Explanation:

Crystalline melting temperature (Tm) is termed as the temperature required for a  crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).  

Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.  

For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.  

The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.  

The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:  

1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.  

2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.  

3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.