electrical engineering​

Electrical Engineering

Answers

Answer 1

Answer:

Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.

Explanation:


Related Questions

QUESTION 6
Which of the following is NOT a resume format?
01. Chronological
O2. Portfolio
3. Functional
04. Combination

Answers

It would be 2 Portfolio

Which one of the following answer options are your employers responsibility

Answers

Where are your answer options?

Answer:

Implement a hazard communication program

Explanation: i took the quiz

Poly(cis-1,4-isoprene), or natural rubber (NR), has a tendency crystallize. The Tm of this polymer is slightly below room temperature, although lightly-crosslinked NR can partially crystallize at room temperature when stretched. Apparently, Tm is elevated upon stretching which allows for crystallization above the Tm of the unstretched polymer. Explain.

Answers

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.

How does distribution add value to goods and services being sold,
including intellectual property?

Answers

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Explanation:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.

Explanation:

hope it helps <33

The two structural members, one of which is in tension and the other in compression, exert the indicated forces on joint o. determine the magnitude r of the resultant r of the two forces and the angle θ which r makes with the positive x axis (measured counterclockwise from the x axis).

Answers

X axis is y axis and x
We're going to remedy it with the parallelogram law.
\alpha = one hundred eighty - 30 - 70 = eighty degrees
R = sqrt(2^2 + 3^2 - 2(2)(three) cos(80)
R = three.30 kN, we can conclude now that the value of the ensuing of R is 3.30 kN

sin \beta /3 = sin(eighty)/three.304 = 63.4 stages

3.3 kN
180 + 33.4 = 213.4 degrees
63.4 - 30 = 33.4


Forces always act in equal and opposite pairs

Answers

You are correct forces always act in the equal of opposite pairs

4. What element is missing from construction drawings?
A. Physical arrangement of specific electrical equipment
B. Electrical layout
C. Electrical connections
D. Side elevation views

Answers

Answer:

C Electrical Connections

Explanation:

In reading says . However, electrical

connections aren’t shown in construction drawings.

Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?

a bachelor’s degree then a master’s degree
vocational school certificate or master’s degree
on-the-job training or vocational school certificate
associate’s degree then a bachelor’s degree

Answers

Answer:

it is indeed C

Explanation:

Answer:

c

Explanation:

what's nested piezometer? ​

Answers

Explanation:

Nested piezometers indicate an upward flow if the elevation of the top of the water in the piezometer tube that penetrates the aquifer to the deeper point is greater than the elevation of the water in the shallower tube

A 4 stroke over-square single cylinder engine with an over square ratio of 1.1,the displacement volume of the engine is 245cc .The clearance volume is 27.2cc the bore of this engine is ?

Answers

Answer:

10.007

Explanation:

Assuming we have to find out the compression ratio of the engine

Given information

Cubic capacity of the engine, V = 245 cc

Clearance volume, V_c = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

Volume of the engine V =[tex]\frac{\pi}{4} D^2L[/tex]

plugging values we get

245 = [tex]\frac{\pi}{4} D^3/1.1[/tex]

Solving we get D =7 cm

therefore,  L= 7/1.1 =6.36 cm

Now,

the compression ratio is given as:

r =(V+V_c)/V_c

on substituting the values, we get

r = (245+27.2)/27.2 =10.007

Hence, Compression ratio = 10.007

what is the relationship between turtuosity and diffusion?​

Answers

As noted earlier, in sediments, or in any porous system, the presence of solid particles causes the diffusion paths of species to deviate from straight lines. Consequently, the diffusion coefficients of species must be corrected for the tortuosity.

What two units of measurement are used to classify engine sizes?

Answers

liters or cubic inches
metric units of cubic centimetres or litres

A power winch is designed to raise a 4,961 N load 10 meter in 2 minutes. The winch is designed with a 283 mm diameter drum taking up the wire lifting the load. The drum will be connected to a 9 rpm motor through a gearbox. What is the minimum torque (Nm) that the motor shaft coupling should be designed to transmit

Answers

Answer:

T = 438.87 N.m

Explanation:

The power required to raise the 4961 N load in 10 meters for 2 minutes is:

[tex]P = \dfrac{4961*10}{2*60}\\ \\ P = 413.42 Nm/sec[/tex]

P = Torque  × W

[tex]413.42 = T \times \dfrac{2 * \pi*9}{60}[/tex]

[tex]413.42 = T \times0.942 \\ \\ T = \dfrac{413.42}{0.942}[/tex]

[tex]T = \dfrac{413.42}{0.942}[/tex]

T = 438.87 N.m

Air at 25 m/s and 15°C is used to cool a square hot molded plastic plate 0.5 m to a side having a surface temperature of 140°C. To increase the throughput of the production process, it is proposed to cool the plate using an array of slotted nozzles with width and pitch of 4 mm and 56 mm, respectively, and a nozzle-to-plate separation of 40 mm. The air exits the nozzle at a temperature of 15°C and a velocity of 10 m/s.

Required:
a. Determine the improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate.
b. Would the heat rates for both arrangements change significantly if the air velocities were increased by a factor of 2?
c. What is the air mass rate requirement for the slotted nozzle arrangement?

Answers

Answer:

a. 2.30

b. decreases with increasing velocity.

c. 0.179 kg/s.

Explanation:

Without mincing let's dive straight into the solution to the question above.

                                                         [a].

The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:

The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.

While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.

Array of slot noozle = [10 × (2  × 0.004)]/ 20.92  × 10^-6] = 3824.

where A = 4/56 =0.714.

And Ar = [ 60 + 4 (40/2  × 4) - 2 ]^2 ]-1/2 = 0.1021.

N = 2/3 (0.1021)^3/4 [ 2  ×  3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.

h = 24.3  ×  0.030/0.004 = 91.1 W/m^2k.

Therefore; 659.6  × 0.030/0.5 = 39.0 W/m²k.

The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.

The slot noozle = 91.1  ×  0.5  ×  0.5 [ 140 -15] = 2846.87W.

The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.

                                                     [b].

2.3 [ (2^2/3)/ 2^4/5] = 2.1

Thus, it decreases with increasing velocity

                                                      [c].

The  air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.

Where do greywater pipes generally feed into?
-Vent stack
-Water heater
-Waste stack
-Main supply

Answers

Answer:

c   Waste stack

Explanation:

In Databrawl, what team is most powerful?
A) Virus
B) Malware
C) Firewall security
D) Programs

Answers

Answer:

I would say Firewall security. Malware is pretty corrupting, but the firewall I think is good for protection if you have a good one.

Explanation:

Yea, I would say firewall security

How do you explain the application of regulations in locations containing baths, showers and electric floor heating, including the requirements needed?

Answers

Answer:

The application of regulations in locations are very important.

Explanation:

The application of regulations in locations are very important in order to gain more benefit from it because people choose those places that are well regulated and having more facilities. If the location has baths, showers, electric floor heaters and other necessities so the people prefer the place over another and increase of clients occurs which give more benefits to the place owners.

Clear Lake has a surface area of 70 ha. In April the inflow of the lake was 1.5 m3/sec. A dam regulated the outflow of the lake to be 1.25 m3/sec. If the precipitation recorded for the month of April was 7.62 cm and the storage volume increased by an estimated 650,000 m3. What is the estimated evaporation in cubic meters?

Answers

Answer:

The estimated evaporation is 51340 cubic meters.

Explanation:

Let suppose that precipitation is very small in comparison with depth of the Clear Lake. The monthly change in the volume of the lake ([tex]\Delta V[/tex]), in cubic meters, is estimated by the following formula:

[tex]\Delta V = A\cdot \Delta z + (\dot V_{in}-\dot V_{out})\cdot \Delta t -V_{evap}[/tex] (1)

Where:

[tex]A[/tex] - Surface area of the lake, in square meters.

[tex]\Delta z[/tex] - Water precipitation, in meters.

[tex]\dot V_{in}[/tex] - Average water inflow, in cubic meters per second.

[tex]\dot V_{out}[/tex] - Average water outflow, in cubic meters per second.

[tex]\Delta t[/tex] - Monthly time, in seconds.

[tex]V_{evap}[/tex] - Evaporation, in cubic meters.

If we know that [tex]A = 700000\,m^{2}[/tex], [tex]\Delta z = 0.0762\,m[/tex], [tex]\dot V_{in} = 1.5\,\frac{m^{3}}{s}[/tex], [tex]\dot V_{out} = 1.25\,\frac{m^{3}}{s}[/tex], [tex]\Delta t = 2.592\times 10^{6}\,s[/tex] and [tex]\Delta V = 650000\,m^{3}[/tex], then the estimated evaporation is:

[tex]650000 = 701340-V_{evap}[/tex] (2)

[tex]V_{evap} = 51340\,m^{3}[/tex]

The estimated evaporation is 51340 cubic meters.

A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum gage pressure in the tank. Mark that point at the interior bottom of the tank. Draw the free surface at this acceleration.

Answers

Answer: hello your question lacks the required diagram attached below is the diagram

answer :  29528.1  N/m^2

Explanation:

Given data :

dimensions of tank :

Length = 5-m

Width = 4-m

Depth = 2.5-m

acceleration of tank = 2m/s^2

Determine the maximum gage pressure in the tank

Pa ( pressure at point A )  = s*g*h1

    = 10^3 * 9.81 * 3.01

    = 29528.1  N/m^2

attached below is the remaining part of the solution

A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude is 22,000 N (4950 lbf), compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.0

Answers

Answer:

13.4 mm

Explanation:

Given data :

Load amplitude ( F )  = 22,000 N

factor of safety ( N )= 2.0

Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa

calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur

minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm

attached below is a detailed solution

The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod is d = 25 mm and the outside diameter of the steel tube is D= 45 mm. The length of the composite column is L = 761 mm. A force P = 88 kN is applied at the top surface, distributed across both the rod and tube.


Required:
Determine the normal stress σ in the steel tube.

Answers

Answer:

Explanation:

From the information given:

[tex]E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN[/tex]

The total load is distributed across both the rod and tube:

[tex]P = P_1+P_2 --- (1)[/tex]

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

[tex]\delta_1=\delta_2[/tex]

[tex]\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}[/tex]

[tex]\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}[/tex]

[tex]P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})[/tex]

[tex]P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}[/tex]

[tex]P_2 = 6.6212 \ P_1[/tex]

Replace [tex]P_2[/tex] into equation (1)

[tex]P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN[/tex]

Finally, to determine the normal stress in aluminum rod:

[tex]\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}[/tex]

[tex]\sigma_1 = - 23.523 \ MPa}[/tex]

Thus, the normal stress = 23.523 MPa in compression.

A shaft is made from a tube, the ratio of the inside diameter to the outside diameter is 0.6. The material must not experience a shear stress greater than 500KPa. The shaft must transmit 1.5MW of mechanical power at 1500 revolution per minute. Calculate the shaft diameter

Answers

Answer:

shaft diameter = [tex]\sqrt[3]{0.3512}[/tex] mm = 0.7055 mm

Explanation:

Ratio of inside diameter to outside diameter ( i.e. d/D )= 0.6

Shear stress of material ( Z ) ≤ 500 KPa

power transmitted by shaft ( P ) = 1.5MW of mechanical power

Revolution ( N ) = 1500 rev/min

Calculate shaft Diameter

Given that: P = [tex]\frac{2\pi NT}{60}[/tex]  ---- 1

therefore; T = ( 1.5 *10^3 * 60 ) / ( 2[tex]\pi[/tex] * 1500 )  = 9.554 KN-M

next

[tex]\frac{T}{I_{p} } = \frac{Z}{R}[/tex]

hence ; T = Z[tex]_{p} *Z[/tex]

attached below is the remaining part of the solution

Technician A says an oscilloscope can be used to check ignition system operating voltages. Technician B says an oscilloscope can be used to check output signals from computer system sensors. Who is right?

Answers

Answer:

Both technicians.

Explanation:

An oscilloscope can be defined as an electronic instrument used for observing, measuring, analyzing, and displaying the waveform of an electric signal.

The output of an oscilloscope is a graph of instantaneous velocity with respect to time.

Generally, an oscilloscope can be used to check ignition system operating voltages. Also, it can be used to check output signals from computer system sensors.

Therefore, both Technician A and Technician B are right.

Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the underground water level. The diameter of the pipe is 7 cm on the intake side and 5 cm on the discharge side. Determine (a) the maximum flow rate of water (5-point) and (b) the pressure difference across the pump (5-point). Assume the elevation difference between the pump inlet and the outlet and the effect of the kinetic energy correction factors to be negligible

Answers

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

[tex]P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}[/tex]

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

[tex]\eta_t[/tex] = The efficiency of the pump = 78%

[tex]\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}[/tex]

[tex]Q_{max}[/tex] = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump [tex]Q_{max}[/tex] ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

are there any companies that you can get a job at as an air craft engeer after university​

Answers

Explanation:

most big airports. my father has the same degree and works for southwest airlines

The yield stress of a steel is 250Mpa. A steel rod used for implant in a femurneeds to withstand 29KN. What should the diameter of the rod be not to deform

Answers

Answer:

r = 1.922 mm

Explanation:

We are given;

Yield stress; σ = 250 MPa = 250 N/mm²

Force; F = 29 KN = 29000 N

Now, formula for yield stress is;

σ = F/A

A = F/σ

Where A is area = πr²

Thus;

r² = 2900/250π

r² = 3.6924

r = √3.6924

r = 1.922 mm

Construct the plane-stress yield envelopes in a principle stress space for both the Tresca and the von Mises yield theories using your calculated value of the yield strength to scale the envelopes. Indicate the two equivalent load paths corresponding to pure shear on the yield envelopes. Calculate the shear yield strength of Al 6061-T6 aluminum predicted by the above theories.

Answers

Answer:

Explanation:

The missing part of the question is attached in the diagram below, the second diagram shows the schematic view of the stress-strain curve and the plane stress.

From the given information:

The elastic modulus is:

[tex]E = \dfrac{\sigma}{\varepsilon} \\ \\ E = \dfrac{150 \ MPa}{0.0217} \\ \\ E = 69.124 \ GPa[/tex]

Hence, suppose 0.2% offset cuts the stress-strain curve at a designated point A from the image attached below, then the yield strength relating to the stress axis from the curve will be [tex]\sigma_y[/tex] = 270 MPa.

The shear yield strength by using von Mises criteria is estimated as;

[tex]\tau_1 = \dfrac{\sqrt{2}}{3}\sigma_y \\ \\ \tau_1 = \dfrac{\sqrt{2}}{3}*270 \\ \\ \tau_1 = 127.28 \ MPa[/tex]

The shear yield strength by using Tresca criteria is:

[tex]\tau_2 = \dfrac{1}{2}\sigma_y \\ \\ \tau_2= \dfrac{1}{2}*270 \\ \\ \tau_2 = 135 \ MPa[/tex]

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