Series aiding is a term sometimes used to describe voltage sources of the same polarity in series. If a 5 V and a 9 V source are connected in this manner, what is the total voltage?

Answers

Answer 1

Answer:Total Voltage = 14V

Explanation: it is possible that a circuit  can contain more than one source of electromotive force which can cause flow of current in the same or opposite direction . When the  connection to  voltage sources  allows for current  from the voltage sources to flow in  same direction,it is termed  Series aiding  Thus, the  Total/effective voltage in a series aiding circuit is  computed as the sum of series aiding voltages .

 Here we have the series aiding voltages to be 5V and 9V ,

therefore,

Total Voltage = 5V + 9V

= 14V


Related Questions

My computer has a mass of 0.031080997078386 slug the Earth's surface.

a. What is its mass in pounds mass (lbm) on Mars where the acceleration of gravity is 5.35 ft/sec^2?
b. What is its weight in pounds force (lbf) on the Mars surface where the acceleration of gravity is 5.35 ft/sec^2 ?

Answers

Answer:

The answer is "0.187 lbm and 1 lbf".

Explanation:

The mass = [tex]0.031080997078386\ slug[/tex]

Calculating mass on Mars:

[tex]\to m=m_g\frac{g}{g_e}[/tex]

        [tex]=0.031080997078386 \times \frac{32.2}{5.35}\\\\=0.187 \ lbm[/tex]

[tex]\to W=mg_e[/tex]

        [tex]=0.187 \times 5.35\\\\=1 \ lbf[/tex]

. Chemical manufacturers must present which Information on the product's label?

A) Product identifier

B) O Contact Information for the manufacturer

C) O Hazard pictograms

D) All of the above

Answers

Chemical manufacturers must present Product Identifier, Contact Information for the manufacturer and Hazard pictograms on the product's label.

What is a product label?

A product label means a display of written, printed or graphic material that is printed on or affixed to a product or its immediate container.

Let's consider which of the following information must be presented on the product's label by chemical manufacturers.

A) Product identifier. Yes. The product identifier can also be found in Material Safety Data Sheet.B) Contact Information for the manufacturer. Yes. It should provide a phone number or mail to contact the manufacturer.C) Hazard pictograms. Yes. Hazard pictograms form part of the international Globally Harmonized System of Classification and Labelling of Chemicals and alert us to the presence of a hazardous chemical.D) All of the above. Yes.

Chemical manufacturers must present Product Identifier, Contact Information for the manufacturer and Hazard pictograms on the product's label.

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A generator has a voltage constant, KE, of 0.01 volts per rpm. Find the voltage when it is driven at 2400 rpm

a. 60 V
b. 24 V
c. 72 V
d. 54 V

Answers

Answer:

Total voltage = 24 V

Explanation:

Given:

Volts per rpm = 0.01

Total rpm = 2400

Find:

Total voltage

Computation:

Total voltage = Volts per rpm x Total rpm

Total voltage = 0.01 x 2400

Total voltage = 24 V

) A Car is moving with a non-uniform velocity towards East.
Its velocity changes at different time intervals. Calculate the instantaneous
velocity at time 3 sec. The distance is given by equation 2t2 – 4t

Answers

Answer:

Instantaneous velocity = 8m/s

Explanation:

Given the following data;

Distance = 2t² - 4t

Time, t = 3 secs.

To find the instantaneous velocity;

[tex] Velocity = \frac {distance}{time} [/tex]

[tex] V(t) = \frac {dd}{dt} [/tex]

We would differentiate the equation for the distance with respect to time, t.

[tex] \frac {dd}{dt} = \frac {d(2t^{2} - 4t)}{dt}[/tex]

[tex] \frac {dd}{dt} = 4t - 4[/tex]

Substituting the value of "t" into the above equation, we have;

[tex] V(3) = 4(3) - 4[/tex]

[tex] V(3) = 12 - 4[/tex]

Instantaneous velocity = 8m/s

A vortex tube receives 0.3 m^3 /min of air at 600 kPa and 300 K. The discharge from the cold end of the tube is 0.6 kg/min at 245 K and 100 kPa. The discharge from the hot end is at 325 K and 100 kPa. Determine the irreversibility.

Answers

Answer:

Irreversibility = 5.361 kW

Explanation:

From the given information:

By applying ideal gas equation at entry:

PV =  mRT

600 × 0.3 = m × 0.287 × 300      (where R = 0.287 kJ/kg)

180 = m × 86.1

m = 180/86.1

m = 2.0905 kg/min

At the hot end, using the same ideal gas equation:

PV = mRT

100 × V = 1.4905 × 0.287 × 325

V = 139.026/100

V = 1.3903 m³/ min

This implies that: The total entropy change = Entropy of the universe

So,

[tex]m\bigg [ c_p \ In \dfrac{T_2}{T_o}-R \ In \dfrac{P_2}{P_o} \bigg] + m_2 \bigg [ c_p \ In \dfrac{T_2}{T_o}- R \ In\dfrac{P_2}{P_o} \bigg][/tex]

[tex]= 0.6\bigg [ 1.004 \ In \dfrac{245}{300}-0.287 \ In \dfrac{100}{600} \bigg] +1.4905\bigg [1.004 \ In \dfrac{325}{300}- 0.287\ In\dfrac{100}{600} \bigg][/tex]

= 0.6[-0.2033 + 0.5142] + 1.4905 [0.08036 + 0.5142]

= 1.0727 kJ/min.K

= 0.01787 kw/K

Irreversibility = [tex]T_o [ \Delta S][/tex]

Irreversibility = 300 × 0.01787

Irreversibility = 5.361 kW

It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less common contexts are listed toward the bottom. According to O*NET, what are common work contexts for Construction Carpenters? Check all that apply. Spend time sitting wear common protective or safety equipment face-to-face discussions exposed to hazardous equipment spend time using hands to handle, control, or feel objects, tools, or controls public speaking

Answers

Answer:

bcde!!

Explanation:

They also tend to be traditional, which means that they enjoy working in structured environments and are typically organized and detail-oriented. Thus option B,C,D, E is correct.

What are the characteristics of Construction Carpenters?

Carpenters continue to have a bright future in their profession. According to Job Outlook, the carpentry industry is expanding quickly. The advantages of carpentry lead to employment security and a long-term career with this kind of industrial growth.

Carpentry is a physically demanding line of work that calls for endurance. You frequently spend the most of your shift standing, moving slowly, and crouching.

Therefore, As well as using hand tools to shape and cut wood, lifting big objects, moving heavy beams, furniture, or equipment are all possible.

Learn more about Construction Carpenters here:

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Electronic dimmers of the type sold for residential use _______ intended for speed control of small motors.

Answers

Solid state switches (incorporating triacs and diacs) speed setpoint adjusted by operator controlled potentiometer... varying the firing rate of the triac circuit.

Depending how old this question is, a plain
rheostat was used to create a voltage divider, reducing voltage to a universal motor.

Maybe they are just looking for solid state switch ??

A light dimmer is usually not interchangeable for motor control.

Air flow is measured in a Venturi meter that has a large diameter of 1.5 m and a small diameter of 0.9 m. A 1:12 scale model with water as the fluid is used to calibrate the meter. For the model, it is determined that when the volume flowrate is 0.07 m3 /s, the pressure drop from the large diameter portion (Section 1) to the small diameter portion (Section 2) is 172 kPa. Calculate:

Answers

This question is incomplete, the complete question is;

Air flow is measured in a Venturi meter that has a large diameter of 1.5 m and a small diameter of 0.9 m. A 1:12 scale model with water as the fluid is used to calibrate the meter. For the model, it is determined that when the volume flowrate is 0.07 m3 /s, the pressure drop from the large diameter portion (Section 1) to the small diameter portion (Section 2) is 172 kPa.

Calculate The corresponding flowrate in the prototype.

Assume a water temperature of 15°C and standard properties of air

Answer:  The corresponding flowrate in the prototype is 10.21 m³/s

Explanation:

Given that;

Lm = Lp/12 and lp = 12Lm,   Qm = 0.07 m³/s,   ΔPm = 172 Kpa

properties of water at 15°C ---- Vm = 1.2015 × 10⁻⁶ m²/s,   Sm = 1001.2 kg/m₂

Also for Air---- Vp = 1.46041 × 10⁻⁵,  Sp = 1.225 kg/m³

Now by Using Reynold's model law; (Vm × Lm)/Vm = (Vp × Lp)/Vp

(Vm × Lm) / 1.2015 × 10⁻⁶ = (Vp ×12 × Lm) / 1.46041 × 10⁻⁵

Vm/Vp = 0.9872

we know that

Discharge = Area × Velocity

Qm/Qp = Lm²/Lpl × Vm/Vp

= (1/12)² × 0.9872

= 6.856 × 10⁻³

so Qp = (0.07 / 6.856 × 10⁻³) = 10.21 m³/s

Therefore The corresponding flowrate in the prototype is 10.21 m³/s

A settling tank has an influent rate of 0.6 mgd. It is 12 ft deep and has a surface area of 8000 ft². What is the hydraulic retention time?

Answers

Answer: hydraulic retention time,τ=28.67 hours

Explanation:

The hydraulic retention time  τ (tau),  is given as  The volume of the settling tank(V) divided by the influent flowrate(Q)

τ =V/Q

But Volume is not known  and is given as

Volume =  surface area  x depth of the tank

= 8000 ft² X 12 ft

= 96,000 ft³

Also, the influent flow rate is in mgd ( million gallons per day), we change  it to  ft³/sec so as to be in same unit with the volume in ft³

1 million gallons/day = 1.5472286365101 cubic feet/second

0.6mgd =  1.5472286365101 cubic feet/second  x 0.6

=0.93cubic feet/second

τ =V/Q

96,000 ft³/0.93 ft³/sec

τ=103,225.8 secs

changing to hours

103,225.8 /3600 =28.67 hours

The hydraulic retention time =28.67 hours

At 800K, a plot of ln[cyclobutane] vs t gives a straight line with a slope of -1.6 s-1. Calculate the time needed for the concentration of cyclobutane to fall to 1/16 of its initial value.

Answers

Answer:

hmmm.........

Explanation:

Rear defrosters generally have a relay with a timer. This allows ___.

Answers

This allows. the defogger to shut down after a predetermined length of time the defogger to function just until the rear window is clear the defogger to be independent of the ignition switch none of the above.

A fuel gas containing 45.00 mole% methane and the balance ethane is burned completely with pure oxygen at 25.00 degree C, and the products are cooled to 25.00 degree C. Suppose the reactor is continuous. Take a basis of calculation of 1.000 mol/s of the fuel gas, assume some value for the percent excess oxygen fed to the reactor (the value you choose will not affect the results), and calculate - Q(kW), the rate at which heat must be transferred from the reactor if the water vapor condenses before leaving the reactor and if the water remains as a vapor. Now suppose the combustion takes place in a constant-volume batch reactor. Take a basis of calculation 1.000 mol of the fuel gas charged into the reactor, assume any percent excess oxygen, and calculate -Q(kJ) for the cases of liquid water and water vapor as products.

Answers

Answer:

A)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

Explanation:

Given data :

45.00 % mole of methane

55.00 % of ethane

attached below is a detailed solution

A) calculate - Q(kw)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B ) calculate  - Q ( KJ )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

since combustion takes place in a constant-volume batch reactor

In the final stages of production, a pharmaceutical is sterilized by heating it from 25 to 75°C as it moves at 0.2 m/s through a straight thin-walled stainless steel tube of 12.7-mm diameter. A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the tube. If the tube is 10 m long, what is the required heat flux?

Answers

Answer:

The required heat flux = 12682.268 W/m²

Explanation:

From the given information:

The initial = 25°C

The final = 75°C

The volume of the fluid = 0.2 m/s

The diameter of the steel tube = 12.7 mm = 0.0127 m

The fluid properties for density [tex]\rho[/tex] = 1000 kg/m³

The mass flow rate of the fluid can be calculated as:

[tex]m = pAV[/tex]

[tex]m = \rho \dfrac{\pi}{4}D^2V[/tex]

[tex]m = 1000 \times \dfrac{\pi}{4} \times ( 0.0127)^2 \times 0.2[/tex]

[tex]m = 0.0253 \ kg/s[/tex]

To estimate the amount of the heat by using the expression:

[tex]q = mc_p(T_{final}-T_{initial})[/tex]

q = 0.0253 × 4000(75-25)

q = 101.2 (50)

q = 5060 W

Finally, the required heat of the flux is determined by using the formula:

[tex]q" = \dfrac{q}{A_s}[/tex]

[tex]q" = \dfrac{q}{\pi D L}[/tex]

[tex]q" = \dfrac{5060}{\pi \times 0.0127 \times 10}[/tex]

q" =  12682.268 W/m²

The required heat flux = 12682.268 W/m²

A stream leaving a sewage pond (containing 80 mg/L of sewage) moves as a plug with a velocity of 40 m/hr. A concentration of 50 mg/L is measured 5,000 m downstream. What is the 1st order decay rate constant in the stream?

Answers

Answer:Decay rate constant,k  = 0.00376/hr

Explanation:

IsT Order  Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time  t  and  [A]o  is the inital concentration at time 0, and  k  is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

              5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k  = 0.00376/hr

In wet mill ethanol plants, a total energy of 74,488 Btu (British thermal units, a common energy unit) is used to produce 1 gallon of ethanol. If ethanol can provide 76,330 Btu per gallon, what is the net energy yield for one gallon of ethanol?

Answers

Answer:

The energy yield for one gallon of ethanol is 2.473 %.

Explanation:

The net energy yield ([tex]\% e[/tex]), expressed in percentage for one gallon of ethanol is the percentage of the ratio of the difference of the provided energy ([tex]E_{g}[/tex]), measured in Btu, and the energy needed to produce the ethanol ([tex]E_{p}[/tex]), measured in Btu, divided by the energy needed to produce the ethanol. That is:

[tex]\% e =\frac{E_{g}-E_{p}}{E_{p}} \times 100\,\%[/tex] (1)

If we know that [tex]E_{g} = 76330\,Btu[/tex] and [tex]E_{p} = 74488\,Btu[/tex], then the net energy yield of 1 gallon of ethanol:

[tex]\%e = \frac{76330\,Btu-74488\,Btu}{74488\,Btu}\times 100\,\%[/tex]

[tex]\%e = 2.473\,\%[/tex]

The energy yield for one gallon of ethanol is 2.473 %.

Estimating is important in construction industry because
Select one:
a. Contractors need to know the amount of mark-up
b. Projects are awarded on serious competition
c. Construction industry has tendency for complex constructions
d. The regulations of government for construction industry changes
Your answer is correct,​

Answers

Answer:

a( contractors need to know the amount of markup)

because the contractor should have a long term vision for a proper satisfaction to the people.

Technician A says vehicles with electronic throttle control do not need a separate cruise control module, stepper motor, or cable to control engine speed. Technician B says a faulty brake light switch may cause the cruise control to not operate. Who is correct?

Answers

Answer: its A

Explanation:

The voltage v= 12cos(60t + 45o) is applied to a 0.1 H inductor. Calculate the inductor's Impedance Z = j XL in ohms.

Answers

Answer:

6 Ω

Explanation:

given data :

Voltage ( v ) = 12cos ( 60t + 45° )

L = 0.1 H

calculate the inductor's impedance Z  

Z = jXL

  = jx = 60

  = L = 0.1

hence Z = 60 * 0.1 =  6 Ω

A sinusoidal voltage source has a peak voltage of 12 V and a frequency of 50 Hz. What is the voltage at 10 ms?

Answers

Answer:

0

Explanation:

Given that:

The peak of the voltage [tex]V_{peak}[/tex] = 12 V

The frequency f = 50 Hz

At the time t  = 10 ms

[tex]V = V_p \ sin \ \omega t[/tex]

where;

[tex]\omega = 2 \pi f[/tex]

[tex]V = 1 2\ V \times sin \ ( 2 \pi \times 50 \times 10 \ )[/tex]

V = 12V sin (180)

V = 12 V × 0

V = 0

What is the amount of pearlite formed during the equilibrium cooling of a 1055 steel from 1000°C to room temperature?

Answers

Answer: 98.5% of pearlite was formed during the equilibrium cooling

Explanation:

First we calculate the fraction of pro-eutectoid phase which forms for equilibrium cooling of the 1085 steel from 1000°C at room temperature;

we know that in 1085 steel, last two digits denotes the carbon percentage

so 1085 steel contains 0.85% carbon.

Now from the diagram, carbon percentage is greater than the eutectoid com[psition

i.e 0.85 > 0.76

it is a hyper eutectoid steel

so

fraction of pro eutectoid phase W_Fe₃C = (0.85 - 0.76) / ( 6.7 - 0.76)

= 0.09 / 5.94 = 0.015 = 1.5%

Now, the amount of pearlite formed during the equilibrium cooling of the 1055 steel from 1000°C to room temperature will be;

pearlite (C') = (1 - W_Fe₃C)

= 1 - 0.015

= 0.985 = 98.5%

Therefore 98.5% of pearlite was formed during the equilibrium cooling

What pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water?

Answers

According to the scenario, the pounds per square inch required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water is found to be approximately 2.0 Psig.

What do you mean by the Bubbler system?

The bubbler system may be defined as a type of system that significantly measures water level based on the amount of pressure it takes to push an air bubble out of an orifice line and into the water body. This pressure often referred to as the “line pressure”, requires changes in the elevation of the water.

According to the context of this question,

The depth of bubbles produced by the bubbler system = 4 ft 7 inches.

The pounds per square inch = 2.31 Psig.

∴ The pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water = 4 ft 7 inches/2.31 = 2.03 Psig ≅ 2Psig.

Therefore, according to the scenario, the pounds per square inch required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water is found to be approximately 2.0 Psig.

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A MBR treatment plant has a flow of 0.3 mgd with 220 mg/l BOD (the soluble portion is 120 mg/l) and 230 mg/l suspended solids (184 mg/l is volatile). Effluent soluble BOD is 2 mg/l. The design calls for 5000 mg/l MLSS and sludge age of 20 days with a Y = 0.8, kd = 0.4. Calculate the aeration basin volume, detention time, BOD loading, ratio, and waste solids, both VSS and TSS.

Answers

Answer:

attached below is the detailed solution

A) 8288.77 cu.ft

B) 4.96 hours

C) Vss = 131.21 IbVss/day

   Tss = 164 IbTss/day

D) attached below

E ) 0.2

F) 287.23 Ib/day

Explanation:

A) Determine the aeration basin volume

Given

∅c = 20 days

Y = 0.8Ib VSS/Ib BOD

Q = 0.3 mgd

So = 120 mg/l

Se = 2 mg/l

X = 5000 mg/l

Kd = 0.04 per day

attached below is the detailed solution

B) Determine the detention time using this relation

t = ( V / Q )* 24

  = ( 0.062 / 0.3 ) * 24  = 4.96 hours

C ) Determine Vss and Tss

we first calculate the excess biomass Px then assuming Vss ratio to be 80% for sludge age of 20 days

Vss = 131.21 IbVss/day

Tss = 164 IbTss/day

D ) determine BOD loading

  Q = 0.3 mgd , BOD = 220mg/l , V = 8288.77 cu.ft

solution attached below

e) food to microorganism ratio

F/M  = 0.2

solution attached below

f) determine the waste solid

waste solid = Q * SS * % removal of suspended solids

where : Q = 0.3 , SS = 220mgl , % = 50 %

waste solids = 0.3 * 230 * 0.5 * 8.34  = 287.23 Ib/day

the differences between building technology vs architectural technology​

Answers

Architectural technologists are concerned more with the technical and functional elements of the building, including building code, drafting and CAD, conceptual drawings and 3D models They also generally cost less than an architect.

A building wall consists of 12-in clay brick and 1/2-in. Fiberboard on one side. If the wall is 10 ft high, determine the load in pounds per foot that it exerts on the floor.

Answers

Answer:

1.16 k/ft

Explanation:

From the given information;

Using table 1.3 for Minimum design dead loads;

For 12-in clay brick,

the obtained min. design dead load = 115 psf

For Fiberboard 1/2 in. ceilings, the minimum design dead load is 0.75 psf

To start with the load that is being exerted on the floor as a result of the clay brick wall ([tex]L_1[/tex] ), we have:

[tex]L_1 = Load \times h[/tex]

[tex]L_1 = 115 \times 10[/tex]

[tex]L_1 = 1150 \ lb/ft[/tex]

To calculate the load exerted on the floor as a result of the 1/2 fireboard, we have:

[tex]L_2 = Load \times h[/tex]

[tex]L_2 = 0.75 \times 10[/tex]

[tex]L_2 = 7.5 \ lb/ft[/tex]

The total load exerted on the floor = [tex]L_1 + L_2[/tex]

The total load exerted on the floor = 1150 + 7.50

The total load exerted on the floor = 1157.50 lb/ft

To (k/ft), we get:

[tex]= 1157.50 \ lb/ft \times \dfrac{1 \ k}{1000 \ lb}[/tex]

= 1.157 k/ft

≅ 1.16 k/ft

A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as shown. Determine the velocity v of the motorcycle when s=200 m. At this point, also determine the value of the derivative dv/ds.

Answers

Answer:

Follows are the solution to this question:

Explanation:

Calculating the area under the curve:  

A = as

   [tex]=\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}[/tex]

Calculating the kinematics equation:

[tex]\to v^2 = v^2_{o} + 2as\\\\[/tex]

        [tex]=0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}[/tex]

Calculating the value of acceleration:  

[tex]\to a= \frac{dv}{dt}[/tex]

[tex]=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}[/tex]

[tex]\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\[/tex]

         [tex]=\frac{0.092}{s}[/tex]

An oxygen–nitrogen mixture consists of 35 kg of oxygen and 40 kg of nitrogen. This mixture is cooled to 84 K at 0.1 MPa pressure. Determine the mass of the oxygen in the liquid and gaseous phase.

Answers

Answer:

The mass of oxygen in liquid phase = 14.703 kg

The mass of oxygen in the vapor phase = 20.302 kg

Explanation:

Given that:

The mass of the oxygen [tex]m_{O_2}[/tex] = 35 kg

The mass of the nitrogen [tex]m_{N_2}[/tex] = 40 kg

The cooling temperature of the mixture T = 84 K

The cooling pressure of the mixture P = 0.1 MPa

From the equilibrium diagram for te-phase mixture od oxygen-nitrogen as 0.1 MPa graph. The properties of liquid and vapor percentages are obtained.

i.e.

Liquid percentage of [tex]O_2[/tex] = 70% = 0.70

Vapor percentage of [tex]O_2[/tex] = 34% = 0.34

The molar mass (mm) of oxygen and nitrogen are 32 g/mol and 28 g/mol respectively

Thus, the number of moles of each component is:

number of moles of oxygen = 35/32

number of moles of oxygen =  1.0938 kmol

number of moles of nitrogen = 40/28

number of moles of nitrogen = 1.4286  kmol

Hence, the total no. of moles in the mixture is:

[tex]N_{total} = 1.0938+1.4286[/tex]

[tex]N_{total} = 2.5224 \ kmol[/tex]

So, the total no of moles in the whole system is:

[tex]N_f + N_g = 2.5224 --- (1)[/tex]

The total number of moles for oxygen in the system is

[tex]0.7 \ N_f + 0.34 \ N_g = 1.0938 --- (2)[/tex]

From equation (1), let N_f = 2.5224 - N_g, then replace the value of N_f into equation (2)

0.7(2.5224 - N_g) + 0.34 N_g = 1.0938

1.76568 - 0.7 N_g + 0.34 N_g = 1.0938

1.76568 - 0.36 N_g = 1.0938

1.76568 - 1.0938 = 0.36 N_g

0.67188  = 0.36 N_g

N_g = 0.67188/0.36

N_g = 1.866

From equation (1)

[tex]N_f + N_g = 2.5224[/tex]

N_f + 1.866 = 2.5224

N_f = 2.5224 - 1.866

N_f = 0.6564

Thus, the mass of oxygen in the liquid and vapor phases is:

[tex]m_{fO_2} = 0.7 \times 0.6564 \times 32[/tex]

[tex]m_{fO_2} = 14.703 \ kg[/tex]

The mass of oxygen in liquid phase = 14.703 kg

[tex]m_{g_O_2} = 0.34 \times 1.866 \times 32[/tex]

[tex]m_{g_O_2} = 20.302 \ kg[/tex]

The mass of oxygen in the vapor phase = 20.302 kg

Consider a circuit element, with terminals a and b, that has vab= -12V and iab= 3A. Over a period of 2 seconds, how much charge moves through the element? If electrons carry the charge, which terminal do they enter? How much energy is transferred? Is it delivered to the element or taken from it?

Answers

Answer:

a) 6 coulombs

b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current

c) = -72 joules

Energy is taken from element

Explanation:

Given data:

V ab = -12 v

I ab = 3A

period ( t ) = 2 seconds

a) determine how much charge moves through the element

q = I * t

  = 3 * 2 = 6 coulombs

b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current

c) determine how much energy is transferred

= Vab * Iab * t

= -12 * 3 * 2

= -72 joules

Energy is taken from element

What is the best way to collaborate with your team when publishing Instagram Stories from Hootsuite?

Answers

Implement a robust approval process with multiple touch points so every team member has input
Save the Story as a draft, then share it with your team via Hootsuite Inbox to gather feedback before publishing
Schedule Stories in the Planner so they can be reviewed and edited by the approver directly
Add specific publishing notes in the composer to help guide the teammate ultimately publishing the Stories

Help this is very hard and I don't get it

Answers

Answer:

yes it is very hard you should find a reccomended doctor to aid in your situation. But in the meantime how about you give me that lil brainliest thingy :p

Air is compressed by a 30-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as a result of heat transfer to the surrounding medium at 20°C. Determine the rate of entropy change of the air.

Answers

Answer:

-0.1006Kw/K

Explanation:

The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,

ΔS = Q/T

Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

S(air) = - Q/T(air) .......1

Where S.air =

Q = 30-kW

T.air = 298k

Substitute the values into equation 1

S(air) = - 30/298

= -0.1006Kw/K

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