Answer:
Technician B is correct.
Explanation:
Two technicians are discussing the a series-parallel brake lamp circuit that does not illuminate.
The statement made by Technician A saying that an open in one of the parallel branches could be the cause has no ground because an opening in any branches has not to do with the issue since there is no disconnection The statement made by Technician B saying that a shorted (or stuck closed) brake lamp switch wired in series could be the cause is correct.
What tool should be used to loosen or tighten brake or fuel lines?
The tool you would use are brake line wrenches.
The tool should be used to loosen or tighten brake or fuel the lines the tool you would use are brake line wrenches.
What is crosshead screws?A crosshead screw is simple metal machine that can be used to fasten one object to another. They have an X–shaped slot at the head of the screw where the screwdriver is inserted. Phillips Head Screwdriver can be used to loosen or tighten crosshead screws.
The Philips head screwdriver is a perfect instrument that can be used to loosen or tighten crosshead screws when fastening an object onto another one. The Philips head screwdriver should be the same as the width of slotted screw head in order to fit onto the head of the screw.
Therefore, The tool should be used to loosen or tighten brake or fuel the lines the tool you would use are brake line wrenches.
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when breathing or the heart has stopped, brain damage can occur within___ minutes
Answer:
Answer:4 minutes
Explanation:
Permanent brain damage begins after only 4 minutes without oxygen, and death can occur as soon as 4 to 6 minutes later.
6, Large, centralized power-generating stations are often located near ___. A) areas without natural energy sources B) abundant energy sources C) metropolitan areas D) rural areas
Answer: abundant energy sources
Explanation:
A power generating station which is sometimes referred to as the power plant is simply an industrial facility that is used for the generation of power.
Large, centralized power-generating stations are often located near abundant energy sources. It should be noted that most power stations burn fossil fuels like oil, coal, natural gas, in order to be able to generate power and hence, they have to be located where there's abundance of the energy sources.
The Department of Transportation regulations require which parameter to be recorded on the shipping paper for hazard class Non Flammable Compressed Gases?
Answer:
Number of cylinders of each gas
Explanation:
The Department of Transportation regulations Title 49 of the United States Code of Federal Regulations specifies that the number of cylinders of each gas being transported should be recorded on the shipping paper for gases classified under hazard class 2.2 of which one of them are the Non Flammable compressed gases.
The definition of gases in division 2.2 encompasses compressed gas, liquefied gas, compressed gas in solution, asphyxiant gas and oxidizing gas. They must meet the absolute pressure of 280kPa or greater at 20 degree Celsius.
prescription eye protection suitable for an electrician should always include
Answer:
If prescription lenses are required, they must be covered by protective goggles or the lenses and frames must be ANSI approved.
Explanation:
this includes side shields for angular protection. Goggles and safety glasses are primary protectors. Goggles fit around the area of the face near the eye. hope this helps you :)
A student is studying science and hopes to one day become employed as a research scientist for a chemical company. What can you infer about this student?
Explanation:
A student as given in the context is persuing his study in science and one days he or she aspires to become a research scientist in some chemical company.
From the above information, we can say that the student is a bright student and has a experimentalist mind set and wants to explore and also research more on science and the principles of scientific laws.
Answer: this student has strong analytical and reasoning skills.
Explanation:
what ratio between differential gain and common-mode gain is called
Answer:
CMRR(Common Mode Rejection Ratio) is the ratio of differential gain and the common mode gain.
Explanation:
A commercial aircraft is in steady flight. The overall lift to drag coefficient is 25. The other flight data are: (i)Weight of aircraft (including crew and passengers) = 70,000 kg (ii) Flight speed 400 km/hr (iii) Average wing span = 50 m (iv) Average wing chord = 7 m When in operation 20% of fuel power is lost. Find the power required for this steady flight. (5M)
Explanation:
weight of aircraft including crew and passengers is equal to 70000 kg therefore option is a thecorrect answer
what type of address do computer use to find something on network?
ip address
url address
mac address
isp address
Answer:
The answer would be ISP address
Explanation:
An Internet Protocol address is a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication. An IP address serves two main functions host or network interface identification and location addressing which can help you a lot.
The type of address the computer use to find something on network is ISP address.
What is ISP?ISP is simply Internet service provider. They are companies which provides internet services to people. They charge based on the quantity of their service used at a particular time. ISP companies also provide softwares for its users.
What is IP address?An Internet Protocol address refers to a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication.
Therefore, the correct answer is option D
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A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 200 kPa. Determine the mole number and the mass of the helium in the balloon.
Answer:
number of mole is 31342.36 moles
mass is 125.369 kg
Explanation:
Diameter of the spherical balloon d = 9 m
radius r = d/2 = 9/2 = 4.5 m
The volume pf the sphere balloon ca be calculated from
V = [tex]\frac{4}{3} \pi r^3[/tex]
V = [tex]\frac{4}{3}* 3.142* 4.5^3[/tex] = 381.75 m^3
Temperature of the gas T = 20 °C = 20 + 273 = 293 K
Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa
number of moles n = ?
Using
PV = nRT
where
P is the pressure of the gas
V is the volume of the gas
n is the mole number of the gas
R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1
T is the temperature of the gas (must be converted to kelvin K)
substituting values, we have
200 x 10^3 x 381.75 = n x 8.314 x 293
number of moles n = 76350000/2436 = 31342.36 moles
We recall that n = m/MM
or m = n x MM
where
n is the number of moles
m is the mass of the gas
MM is the molar mass of the gas
For helium, the molar mass = 4 g/mol
substituting values, we have
m = 31342.36 x 4
m = 125369.44 g
m = 125.369 kg
The mole number and the mass of the helium in the balloon are 250.801 kilomoles and 1003.706 kilograms, respectively.
Let suppose that helium contained in the spherical balloon behaves ideally, the mole number ([tex]n[/tex]), in kilomoles, is determined by the following expression:
[tex]n = \frac{P\cdot V}{R_{u}\cdot T}[/tex] (1)
Where:
[tex]P[/tex] - Pressure, in kilopascals.[tex]V[/tex] - Volume, in cubic meters.[tex]R_{u}[/tex] - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.[tex]T[/tex] - Temperature, in Kelvin.The volume and mass of helium ([tex]m[/tex]), in kilograms, is described by these two formulas:
[tex]V = \frac{4\pi}{3}\cdot R^{3}[/tex] (2)
[tex]m = n\cdot M[/tex] (3)
Where:
[tex]R[/tex] - Radius of the sphere, in meters. [tex]M[/tex] - Molar mass, in kilograms per kilomole.If we know that [tex]P = 200\,kPa[/tex], [tex]R = 9\,m[/tex], [tex]R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex], [tex]T = 293.15\,K[/tex] and [tex]M = 4.002\,\frac{kg}{kmol}[/tex], then the mole number and the mass of the helium in the balloon are:
[tex]V = \frac{4\pi}{3}\cdot (9\,m)^{3}[/tex]
[tex]V \approx 3053.628\,m^{3}[/tex]
[tex]n = \frac{(200\,kPa)\cdot (3053.628\,m^{3})}{\left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot(293.15\,K)}[/tex]
[tex]n = 250.580\,kmol[/tex]
[tex]m = (250.801\,kmol)\cdot \left(4.002\,\frac{kg}{kmol} \right)[/tex]
[tex]m = 1003.706\,kg[/tex]
The mole number and the mass of the helium in the balloon are 250.801 kilomoles and 1003.706 kilograms, respectively.
We kindly invite to check this question on ideal gases: https://brainly.com/question/16211117
The cross-section of a rough, rectangular, concrete() channel measures . The channel slope is 0.02ft/ft. Using the Darcy-Weisbach friction method, determine the maximum allowable flow rate through the channel to maintain one foot of free board(freeboard is the vertical distance form the water surface to the overtopping level of the channel). For these conditions, find the following characteristics(note that FlowMaster may not directly report all of these):
a) Flow area
b) Wetted perimeter
c) Hydraulic radius(A/P) :
d) Velocity
e) Froude number
Answer:
The following are the answer to this question:
Explanation:
In point a, Calculating the are of flow:
[tex]\bold{Area =B \times D_f}[/tex]
[tex]=6\times 5\\\\=30 \ ft^2[/tex]
In point b, Calculating the wetter perimeter.
[tex]\bold{P_w =B+2\times D_f}[/tex]
[tex]= 6 +2\times (5)\\\\= 6 +10 \\\\=16 \ ft[/tex]
In point c, Calculating the hydraulic radius:
[tex]\bold{R=\frac{A}{P_w}}[/tex]
[tex]=\frac{30}{16}\\\\= 1.875 \ ft[/tex]
In point d, Calculating the value of Reynolds's number.
[tex]\bold{Re =\frac{4VR}{v}}[/tex]
[tex]=\frac{4V \times 1.875}{1 \times 10^{-5} \frac{ft^2}{s}}\\\\[/tex]
[tex]=750,000 V[/tex]
Calculating the velocity:
[tex]V= \sqrt{\frac{8gRS}{f}}[/tex]
[tex]= \sqrt{\frac{8\times 32.2 \times 1.875 \times 0.02}{f}}\\\\=\frac{3.108}{\sqrt{f}}\\\\[/tex]
[tex]\sqrt{f}=\frac{3.108}{V}\\\\[/tex]
calculating the Cole-brook-White value:
[tex]\frac{1}{\sqrt{f}}= -2 \log (\frac{K}{12 R} +\frac{2.51}{R_e \sqrt{f}})\\\\ \frac{1}{\frac{3.108}{V}}= -2 \log (\frac{2 \times 10^{-2}}{12 \times 1.875} +\frac{2.51}{750,000V\sqrt{f}})\\[/tex]
[tex]\frac{V}{3.108} =-2\log(8.88 \times 10^{-5} + \frac{3.346 \times 10^{-6}}{750,000(3.108)})[/tex]
After calculating the value of V it will give:
[tex]V= 25.18 \ \frac{ft}{s^2}\\[/tex]
In point a, Calculating the value of Froude:
[tex]F= \frac{V}{\sqrt{gD}}[/tex]
[tex]= \frac{V}{\sqrt{g\frac{A}{\text{Width flow}}}}\\[/tex]
[tex]= \frac{25.18}{\sqrt{32.2\frac{30}{6}}}\\\\= \frac{25.18}{\sqrt{32.2 \times 5}}\\\\= \frac{25.18}{\sqrt{161}}\\\\= \frac{25.18}{12.68}\\\\= 1.98[/tex]
The flow is supercritical because the amount of Froude is greater than 1.
Calculating the channel flow rate.
[tex]Q= AV[/tex]
[tex]=30x 25.18\\\\= 755.4 \ \frac{ft^3}{s}\\[/tex]
A photon has energy of 20keV. What are its frequency and wavelength?
Answer:
So we know that: E = h*(frequency)
but for this equation, we need energy in joules
so since 1eV is 1.6 * [tex]10^{-19}[/tex]
we will multiply the energy in eV to 1.6 * 10^-19
so energy in joules:
20000 * 1.6 * 10^-19
3.2 * [tex]10^{-15}[/tex] j
Now, finding the frequency:
3.2 * 10^-15 = 6.6 * [tex]10^{-34}[/tex] * frequency
frequency = (3.2 * [tex]10^{-15}[/tex]) / (6.6 * [tex]10^{-34}[/tex])
frequency = 4.8 * [tex]10^{18}[/tex] Hz
Finally, finding the wavelength:
we will use a different formula here,
frequency * wavelength = speed of light
wavelength = speed of light / frequency
wavelength = 3 * [tex]10^{8}[/tex] / 4.8 * [tex]10^{18}[/tex]
wavelength = 3 * [tex]10^{-10}[/tex] / 4.8
wavelength = 0.625 * [tex]10^{-10}[/tex]
wavelength = 6.25 * [tex]10^{-11}[/tex] m
frequency = 4.8 * [tex]10^{18}[/tex] Hz
meggers are used to measure
Tech A says that a faulty vacuum booster can affect engine operation. Tech b says that steal brake Ponce can be replaced with a copper line, as it easier to bend into shape. Who is correct ?
Answer:
Tech A
Explanation:
A faulty vacuum booster can actually affect the operation of an engine. Engine stalls when brakes are applied. And this can happen when the diaphragm that is inside the brake booster fails. The failing thus allows air to bypass the seal. When the brakes are then pressed, the engine will actually feel like it will stall, and the idle will most probably drop. Also, asides a reduction in the break performance quality, a stalling engine is very bad and can result to many negative effects.
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating
Answer:
The answer is below
Explanation:
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.
Given that:
number of poles (p) = 4, frequency (f) = 60 Hz
1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:
[tex]n_s=\frac{120f}{p}=\frac{120*60}{4}=1800\ rpm[/tex]
2) The slip (s) = 0.05
The speed of the motor (n) is the speed of the rotor, it is given as:
[tex]n=n_s-sn_s\\\\n=1800-0.05(1800)\\\\n=1800-90\\\\n=1710\ rpm[/tex]
3) s = 0.04
The rotor frequency is the product of the supply frequency and slip it is given as:
[tex]f_r=sf\\\\f_r=0.04*60\\\\f_r=2.4\ Hz[/tex]
4) At standstill, the motor speed is zero hence the slip = 1:
[tex]s=\frac{n_s-n}{n_s}\\ \\n=0\\\\s=\frac{n_s-0}{n_s}\\\\s=1[/tex]
The rotor frequency is the product of the supply frequency and slip it is given as:
[tex]f_r=sf\\\\f_r=1*60\\\\f_r=60\ Hz[/tex]
List and describe three classifications of burns to the body.
AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.
Explanation:
Answer:
AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.
Explanation:
Using Raoult’s law, estimate the boiling temperature and mole fractions in the vapor phase that is in equilibrium with a liquid having 0.2608 mol fraction of ethanol, in a mixture ethanol/water at P = 1 atm.
If possible, solve non-numerically.
Answer:
[tex]T=92.16 \°C[/tex]
[tex]y_{et}=0.433\\\\y_w=0.567[/tex]
Explanation:
Hello,
In this case, the Raoult's law for this problem is:
[tex]y_{et}P=x_{et}P_{et}^{sat}\\\\y_{w}P=x_{w}P_{w}^{sat}[/tex]
Which can be written as:
[tex]P=x_{et}P_{et}^{sat}+x_{w}P_{w}^{sat}[/tex]
Thus, by using the Antoine equation, we can symbolically represent the the temperature at which such mixture boil:
[tex]1atm=0.2608*10^{(8.13484-\frac{ 1662.48}{238.131+T})}/760+0.7392*10^{(5.40221-\frac{1838.675}{-31.737+T-273.15})}/1.01325[/tex]
The solution, by numerical iteration process (there is not way to solve it analytically) is 92.16 °C considering the data extracted from NIST database. Next, vapor fractions are:
[tex]y_{et}=x_{et}*10^{(8.13484-\frac{ 1662.48}{238.131+T})}/760/P\\\\y_{et}=0.2608*10^{(8.13484-\frac{ 1662.48}{238.131+92.16})}/760/1atm\\\\y_{et}=0.433\\\\y_w=1-y_{et}=1-0.433\\\\y_w=0.567[/tex]
Regards.
Which of the following terms involves the study of all the systems, devices, knowledge, processes, organizations, items, and people involved in the operation and creation of practical artifacts?
Technology
Mathematics
Science
Engineering
Answer: Engineering
Explanation:
Engineering is a most integral aspect of human development since human settlement first began. It has propelled and will continue to propel our progress as a species in the way it knows how.
Engineering is the study of the science of all things that relate to the operation and creation of practical artifacts and the knowledge gained from this study is what is applied to human development and enables our progress in technology.
The answer is:
D) EngineeringA precision miling machince wighing 1000lb is supported on a rubber mount. The force deflection relationship of the rubber mount I'd given by F=3500x+55x^3. Where the force F and the deflection x are measured in pounds and inches, respectively. Determine the equivalent linerized spring constant of he rubber mount at its static equilibrium position.
Answer:
0.2846 in
Explanation:
The static equilibrium position of the rubber mount ( [tex]x^*[/tex]), under the weight of the milling machine, can be determined from:
[tex]1000=3500(x^*)+ 55(x^*)^3\\\\55(x^*)^3+ 3500x^*-1000=0\\\\Solving\ for\ the\ roots\ using \ a \ calculator\ or\ matlab\ gives:\\ \\x^*_1=0.28535\\x^*_2=-0.14267+7.89107i\\x^*_3=-0.14267-7.89107i\\\\We\ are\ using\ the \ real\ root\ which\ is\ x^*_1=0.28535\\[/tex]
[tex]k=\frac{dF}{dx}|_{x^*}\\ \\k=\frac{d}{dx}(3500x+55x^3)|_{x^*}\\\\k=3500+165x^2|_{x^*}\\\\k=3500+165(x^*)^2\\\\k=3500+165(0.28535)^2=3513.435\ lb/in[/tex]
The static equilibrium position is at:
[tex]x=\frac{F}{k}=\frac{1000}{3513.435} =0.2846\ in[/tex]
What is best for a busy student to do for better results in school?
Answer:
Set high and clear expectations for quality work
Don't attempt to cram all your studying into one session
Explanation:
Participate in Class
eat a well balanced diet
Find perfect place to Study.
HavE a NIce dAY ;}
When entering a freeway you should always:
A. Slow down and proceed when it is safe.
B. Stop and make sure there is no traffic approaching.
C. Accelerate to the same speed as the freeway traffic and merge smoothly.
D. Go as fast as you can and swing abruptly into traffic.
Explain why the following acts lead to hazardous safety conditions when working with electrical equipmenta. Wearing metal ring or braceletsb. Being barefootc. Working on a damp concrete floord. Touching grounded conductors while working on electrical equipmente. Working on electrical equipment with sweaty hands
Answer:
Please find the answer in the explanation
Explanation:
In Engineering, safety is very essential and very important for all engineers to stick to.
A.) Wearing metal ring or bracelets.
When their is discharg of electrical charges, wearing of metal rings or braceletsb can allow charges to pass through them into the body which can eventually lead to electrical shock.
B.) Being barefoot
Being bare footed is very dangerous because someone can mistakenly step on naked wire which can lead to electrical shock.
C. Working on a damp concrete floor.
It is very hazardous to be working on a damp concrete floor because of the water moisture. It is very unsafe for any electrical job to be done on any wet area because water can conduct electricity which can lead to electrical shock.
d. Touching grounded conductors while working on electrical equipment
Grounded conductor can allow charges to flow through them. So, it is very unsafe to have them have contact with the body because of electrical charges.
e. Working on electrical equipment with sweaty hands
A sweaty hands contain some content of water which can conduct electricity and lead to electrical shock.
A civil engineer designs a wheelchair accessible ramp next to a set of steps leading up to a building. The height from the ground to the top of the stairs is 3ft. Based on ADA codes, the slope must be 1:12 or less. (Slope is equal to the rise of the ramp divided by the run of the ramp.) What is the IMA of this ramp if the engineer uses a slope of 1:12?
Answer: IMA = 12.042
Explanation:
Given that;
Height from the ground h is 3 ft
Slope of ramp s is 1:12
Horizontal length of the ramp x will be 3 × 12 = 36 ft
Now to get the IMA ( ideal mechanical advantage,)
IMA = length of the ramp over / height of the ramp
IMA = (√( 36² + 3²)) / 3
IMA = (√ 1305 ) / 3
IMA = 36.124 / 3
IMA = 12.042
The flow rate on an arterial is 1,800 veh/h, evenly distributed over two lanes. If the average speed in these lanes is 40 mi/h, what is the density in veh/h/ln?
Using
q = k*v,
where q = traffic flow rate, k = density, v = space mean speed.
so, 1800 = k x 40
so, k = 1800/40 = 45 veh/mie on two lanes,
so k = 45/2 = 22.5veh/h/in
List 10 uses for surveying in areas other than land surveying. Select all that apply. a. topographic surveying b. satellite surveying c. aerial surveying d. optical tooling e. marketing surveying f. control surveying g. as-built surveying h. statistical surveying i. telephone surveying j. construction surveying k. alignment surveying l. mine surveying m. solar surveying
Answer:
b. satellite surveying
c. aerial surveying
d. optical tooling
e. marketing surveying
f. control surveying
h. statistical surveying
i. telephone surveying
k. alignment surveying
l. mine surveying
m. solar surveying
Explanation:
A Survey is an act of examination of the features of a subject or material under consideration. Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.
Land surveying finds application in construction where a survey is made on all the structures found in a constructed property. Topographic surveying deals with examining the natural and man-made feature of a piece of land. As-built survey as the name implies examines the features and location of a building during or recently after construction. These three are examples of land surveys.
Some type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying
A Survey means an examination of a features, subject or material under consideration.
Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.
However, some other type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying.
Read more about surveying
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1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = constant heat addition from 100 kPa and 27∘C to 700 kPa 2-3 Isothermal expansion until V3 = 7v23-1 P = constant heat rejection to the initial state2. Assume air has constant properties with cv = 0.718 kJ/kg K, cp = 1.005 kJ/kg K, R = 0.287 kJ/kg K, and k = 1.4.(a) Sketch the P- and T-s diagrams for the cycle.(b) Determine the ratio of the compression work to the expansion work (the back work ratio).(c) Determine the cycle thermal efficiency.
Answer:
Explanation: Here it is: 67 Hope that helps! :)
5. One of the major road blocks for direct DC power in residences is the ___. A) generation of DC B) lack of standardization C) difficulty of installation D) quantity of components required for this type of system
Answer:
D) quantity of components required for this type of system
Explanation:
Electricity can be transmitted using the alternating and direct currents. The alternating current is one in which the flow of the current diverts at certain time intervals whereas, the direct current is one in which there is a constant one-directional flow of current. The DC is used in batteries and solar panels. Residential areas and business places make use of the AC current.
One of the several reasons why the DC is not used in homes is because unlike the AC it is not easy to build and sustain. Moreso, it has more components compared to the AC. For example, its motor consists of brushes and commutators . The components, example, switches are also large compared to the AC components.
A balanced three phase source with vL=240 V rms is supplying 8 kVA at 0.6 powder factor lagging to two wye connected parallel loads. If one load draws 3kW at unity powder factor , calculate impedence per phase of the second load.
Answer:
2.35 + j8.34 Ω
Explanation:
Voltage = V[tex]_{L}[/tex] = 240 V rms
supplying power = S[tex]_{s}[/tex] = 8 kVA
power factor = pf[tex]_{s}[/tex] = 0.6
Let P₁ represents one load draws 3kW at unity powder factor
The power angle is:
θ[tex]_{s}[/tex] = cos⁻¹ pf[tex]_{s}[/tex] = cos⁻¹ 0.6 = 53.13°
Complex power supplied source is:
S[tex]_{s}[/tex] = S[tex]_{s}[/tex] < θ[tex]_{s}[/tex] = 8<53.13° kVA
Complex power for first load:
S₁ = P₁ = 3kVA
Since the power angle of first load is θ₁ = 0°
According to principle of conservation of AC power, the power of second load is:
S₂ = S[tex]_{s}[/tex] - S₁
= 8<53.13° - 3
= 6.65<74.29° kVA
Since the second load is a Y connected load the phase voltage:
V[tex]_{p}[/tex] = V[tex]_{L}[/tex] / [tex]\sqrt{3}[/tex]
= 240/1.732051
= 138.564
= 138.56 V
Complex power of second load:
S₂ = 3 V[tex]_{p}[/tex]² / Z[tex]_{p}[/tex]
impedance per phase of the second load:
Z[tex]_{p}[/tex] = 3 V[tex]_{p}[/tex]² / S₂
= 3 (138.56)² / 6.65<74.29°
= 3(19198.8736) / 6.65<74.29°
= 57596.6208 / 6.65<74.29°
Z[tex]_{p}[/tex] = 2.35 + j8.34Ω
Technician A states that if someone is asking a question of you, then he or she is showing interest. Technician B states that silence is golden-when used properly. Who is correct?
Answer:
They are both correct
Explanation:
When someone is asking a question of you, it means he or she has either been paying attention, and is interested in you, what you're doing, or what you're saying. Also, silence when used properly can be golden in the sense that it can prevent unnecessary problems from arising, and can save one from a lot of unforeseen problem. Whatever is said cannot be taken back again, and some things should never be said at all, especially in a professional setting.
1. Round off 2553 N to three significant figures.
2. Round off 58342 m to three significant figures.
3. Round off 68.534 s to three significant figures.
Answer:
(1) 2553 N = 2550 N
(2) 58342 m = 58300 m
(3) 68.534 s = 68.5 s
Explanation:
To round off a number to any significant number start from the last digit, round it off to 1 if the number is up to 5 and to 0 if the last digit is less than 5. Add this 1 or 0 to the preceding digit and continue the process until you are left with three non zero digits, if you are rounding off to three significant figures.
(1) Round off 2553 N to three significant figures.
= 2550 N
(2) Round off 58342 m to three significant figures.
= 58300 m
(3) Round off 68.534 s to three significant figures.
= 68.500 s (zero after decimal point is insignificant)
= 68.5 s