Answer:
a and c
Explanation:
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 ?
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]
) 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
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 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.
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
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
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.
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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.
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
What pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water?
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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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?
Answer: its A
Explanation:
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.
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]
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,
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.
the differences between building technology vs architectural technology
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?
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
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.
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
Rear defrosters generally have a relay with a timer. This allows ___.
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?
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 %.
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:
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
Electronic dimmers of the type sold for residential use _______ intended for speed control of small motors.
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.
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
What is the amount of pearlite formed during the equilibrium cooling of a 1055 steel from 1000°C to room temperature?
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
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.
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
The voltage v= 12cos(60t + 45o) is applied to a 0.1 H inductor. Calculate the inductor's Impedance Z = j XL in ohms.
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
. 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
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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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.
Answer:
hmmm.........
Explanation:
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
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
What is the best way to collaborate with your team when publishing Instagram Stories from Hootsuite?
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.
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
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?
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
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?
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 sinusoidal voltage source has a peak voltage of 12 V and a frequency of 50 Hz. What is the voltage at 10 ms?
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
Help this is very hard and I don't get it
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
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?
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