A tank with a volume of 40 cuft is filled with a carbon dioxide and air mixture. The pressure within the tank is 30 psia at 70oF. It is known that 2 lb of carbon dioxide was placed in the tank. Assume that air is 80% nitrogen and 20% oxygen and use the ideal gas laws. Calculate ,
Answers
The correct responses are;
(i) Weight percent of nitrogen: 58.6%
Weight percent of oxygen: 14.65%
Weight percent of carbon dioxide: 29.59%
(ii) Volume percent of nitrogen: 64.38%
Weight percent of oxygen: 14.1%
Weight percent of carbon dioxide: 21.53%
(iii) Partial pressure of nitrogen: 19.314 psia
Partial pressure of oxygen: 4.23 psia
Partial pressure of carbon dioxide: 6.459 psia
(iv) Partial pressure of nitrogen: 19.314 psia
Partial pressure of oxygen: 4.23 psia
Partial pressure of carbon dioxide: 6.459 psia
(v) The average molecular weight is approximately 32.02 g/mole
(vi) At 20 psia, 60 °F, the density is 0.11275 lb/ft.³
At 14.7 psia, 60 °F, the density is 0.0844 lb/ft.³
At 14.7 psia, 32 °F, the density is 0.0892 lb/ft.³
(vii) The specific gravity of the mixture 0.169
Reasons:
The volume of the tank = 40 ft.³ = 1.132675 m³
Content of the tank = Carbon dioxide and air
The pressure inside the tank = 30 psia = 206843 Pa
The temperature of the tank = 70 °F ≈ 294.2611 K
Mass of carbon dioxide placed in the tank = 2 lb.
Percent of nitrogen in the tank = 80%
Percent of oxygen in the tank = 20%
(i) 2 lb ≈ 907.1847 g
Molar mass of carbon dioxide = 44.01 g/mol
Number of moles of carbon dioxide = [tex]\displaystyle \mathbf{ \frac{907.1847 \, g}{44.01 \, g/mol}} \approx 20.613 \, moles[/tex]
Assuming the gas is an ideal gas, we have;
[tex]\displaystyle n = \mathbf{\frac{206843\times 1.132674}{8.314 \times 294.2611} }\approx 95.7588[/tex]
The number of moles of nitrogen and oxygen = 95.7588 - 20.613 = 75.1458
Let x represent the mass of air in the mixture, we have;
[tex]\displaystyle \mathbf{ \frac{0.8 \cdot x}{28.014} + \frac{0.2 \cdot x}{32}}= 75.1458[/tex]
Solving gives;
x ≈ 2158.92 grams
Mass of the mixture = 2158.92 g + 907.1847 g ≈ 3066.1047 g
[tex]\displaystyle Weight \ percent \ of \ nitrogen= \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{58.6 \%}[/tex]
[tex]\displaystyle Weight \ percent \ of \ oxygen = \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{14.65\%}[/tex]
[tex]\displaystyle Weight \ percent \ of \ carbon \ dioxide = \frac{907.184}{3066.105} \times 100 \approx \underline{29.59\%}[/tex]
ii.
[tex]\displaystyle Number \ of \ moles \ nitrogen= \frac{0.8 \times 2158.92 \, g}{28.014 \, g/mol} \approx 61.65\, moles[/tex]
[tex]\displaystyle Number \ of \ moles \ oxygen= \frac{0.2 \times 2158.92 \, g}{32 \, g/mol} \approx 13.5\, moles[/tex]
Number of moles of carbon dioxide = 20.613 moles
Sum = 61.65 moles + 13.5 moles + 20.613 moles ≈ 95.763 moles
In an ideal gas, the volume is equal to the mole fraction
[tex]\displaystyle Volume \ percent \ of \ nitrogen= \frac{61.65}{95.763} \times 100 \approx \underline{64.38 \%}[/tex]
[tex]\displaystyle Volume \ percent \ of \ oxygen = \frac{13.5}{95.763} \times 100 \approx \underline{14.1\%}[/tex]
[tex]\displaystyle Volume \ percent \ of \ carbon \ dioxide = \frac{20.613 }{95.763} \times 100 \approx \underline{21.53\%}[/tex]
(iv) The partial pressure of a gas in a mixture, [tex]P_A = \mathbf{X_A \cdot P_{total}}[/tex]
Partial pressure of nitrogen, [tex]P_{N_2}[/tex] = 0.6438 × 30 psia ≈ 19.314 psia
Partial pressure of oxygen, [tex]P_{O_2}[/tex] = 0.141 × 30 psia ≈ 4.23 psia
Partial pressure of carbon dioxide, [tex]P_{CO_2}[/tex] = 0.2153 × 30 psia ≈ 6.459 psia
(v) The average molecular weight is given as follows;
[tex]\displaystyle Average \ molecular \ weight = \frac{3066.105\, g}{95.7588 \, moles} = 32.02 \, g/mole[/tex]
(vi) At 20 psia 70 °F, we have;
Converting to SI units, we have;
[tex]\displaystyle n = \frac{137895.1456\times 1.132674}{8.314 \times 294.2611} \approx 63.84[/tex]
The number of moles, n ≈ 63.84 moles
The mass = 63.84 moles × 32.02 g/mol ≈ 2044.16 grams ≈ 4.51 lb
[tex]\displaystyle Density = \frac{4.51\ lb}{40 \ ft.^3} \approx \underline{0.11275\, lb/ft.^3}[/tex]
When the pressure is 14.7 psia = 101352.93 Pa, and 60 °F
[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 288.7056} \approx 42.8247[/tex]
The mass = 47.8247 moles × 32.02 g/mol ≈ 1531.35 grams = 3.376 lb.
[tex]\displaystyle Density = \frac{3.376 \ lb}{40 \ ft.^3} \approx \underline{ 0.0844\, lb/ft.^3}[/tex]
When the pressure is 14.7 psia = 101352.93 and 32 °F
[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 273.15} \approx 50.5482[/tex]
The mass = 50.5482 moles × 32.02 g/mol ≈ 1618.55 grams = 3.568 lb.
[tex]\displaystyle Density = \frac{3.568 \ lb}{40 \ ft.^3} \approx \underline{ 0.0892 \, lb/ft.^3}[/tex]
(vii) [tex]\displaystyle The \ specific \ gravity \ of \ the \ mixture \ = \frac{\frac{3066.1047\, g}{40 \, ft.^3} }{1.00} = \frac{\frac{6.7596 \, lbs}{40 \, ft.^3} }{1.00} \approx \underline{0.169}[/tex]
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Related Questions
Suppose that you have a block of copper which is 300 g. Relevant material parameters follow.
Material atomic weight Density
Copper 63.5 g/mol 8.96 g/cm^3
Using equipartition, compute the molar heat capacity in J/mol K.
Answers
Answer:
[tex]C_v= 24.942[/tex] J / mol K
Explanation:
Molar Heat Capacity
using the equipartition, the equation :
[tex]$C_v=\frac{d}{2} R$[/tex]
Here, [tex]C_v[/tex] = molar heat capacity
d = degree of freedom
R = Universal gas constant
Degree of freedom, d is 6
Universal gas constant, R = 8.317 J/ mol K
Therefore, the molar heat capacity is :
[tex]$C_v=\frac{d}{2} R$[/tex]
[tex]$C_v=\frac{6}{2} \times 8.314$[/tex]
[tex]C_v= 24.942[/tex] J / mol K
if a person is injured at the hospital during a natural disaster a correct action to take is
Answers
That’s what I learnt sooo..
three forces equal to 3p,5p,7p act simultaneously along the three side AB, BC,and CA of equilateral triangle ABC of side a. find the magnitude, direction and position of the resultant.
Answers
The given parameters include;
side AB = 3pside BC = 5p side CA = 7pFrom the image uploaded;
The resolution of the vectors along the sides of the triangle is as follows;
the y-component = force x sin(θ)the x-component = force x cos (θ)forces -------angle (θ) --------y-component-------x-component
3p ------------- 60⁰ --------- ----2.598 p ------------------- 1.5 p
5p -------------- 60⁰ ------------ (-4.33p) ------------------- 2.5 p
7p -------------- 60⁰ ------------- 0 -------------------------- (-7p)
sum: ( -1.732 p) y (-3P)x
The resultant of the vectors:
R² = (-1.732 p)² + (-3P)²
R² = 11.998 p²
R = √(11.998 p²)
R = 3.46 p
The direction of the vector:
[tex]\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{-1.732}{-3} )\\\\\theta = tan^{-1} (0.577)\\\\\theta = 29.98 \ ^0[/tex]
The position of the vector = 180 - θ = 180 - 29.98 = 150⁰ (second quadrant)
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Technician A says that a continuously variable transmission is an automatic transmission that does not shift gears. Technician B says that a continuously variable transmission has a maximum of six distinct gear ranges. Which technician is correct?
Answers
Answer:
Only Technician A
Explanation:
Continuously variable transmission (CVT) does not shift gears because it simply doesn't have gears. It instead uses 2 pulleys which change constantly and continuously in size, and they are linked by a belt. Continuously variable transmission (CVT) is automatic transmission as mentioned in the question above, largely due to the power the belt continuously transmits, making the vehicle's engine work in the optimum power range.
While Technician B saying, that a continuously variable transmission has a maximum of six distinct gear ranges, which is not correct because continuously variable transmission (CVT) works with an unlimited number of gear ratios within a fixed range.
State three types of maintenance.
Answers
Answer:
Tradicionalmente, se han distinguido 5 tipos de mantenimiento, que se diferencian entre sí por el carácter de las tareas que incluyen:
Explanation:
Mantenimiento Correctivo: Es el conjunto de tareas destinadas a corregir los defectos que se van presentando en los distintos equipos y que son comunicados al departamento de mantenimiento por los usuarios de los mismos.
Mantenimiento Preventivo: Es el mantenimiento que tiene por misión mantener un nivel de servicio determinado en los equipos, programando las intervencions de sus puntos vulnerables en el momento más oportuno. Suele tener un carácter sistemático, es decir, se interviene aunque el equipo no haya dado ningún síntoma de tener un problema.
Mantenimiento Predictivo: Es el que persigue conocer e informar permanentemente del estado y operatividad de las instalaciones mediante el conocimiento de los valores de determinadas variables, representativas de tal estado y operatividad. Para aplicar este mantenimiento, es necesario identificar variables físicas (temperatura, vibración, consumo de energía, etc.) cuya variación sea indicativa de problemas que puedan estar apareciendo en el equipo. Es el tipo de mantenimiento más tecnológico, pues requiere de medios técnicos avanzados, y en ocasiones, de fuertes conocimientos matemáticos, físicos y/o técnicos.
Mantenimiento Cero Horas (Overhaul): Es el conjunto de tareas cuyo objetivo es revisar los equipos a intervalos programados bien antes de que aparezca ningún fallo, bien cuando la fiabilidad del equipo ha disminuido apreciablemente de manera que resulta arriesgado hacer previsiones sobre su capacidad productiva. Dicha revisión consiste en dejar el equipo a Cero horas de funcionamiento, es decir, como si el equipo fuera nuevo. En estas revisiones se sustituyen o se reparan todos los elementos sometidos a desgaste. Se pretende asegurar, con gran probabilidad un tiempo de buen funcionamiento fijado de antemano.
Mantenimiento En Uso: es el mantenimiento básico de un equipo realizado por los usuarios del mismo. Consiste en una serie de tareas elementales (tomas de datos, inspecciones visuales, limpieza, lubricación, reapriete de tornillos) para las que no es necesario una gran formación, sino tal solo un entrenamiento breve. Este tipo de mantenimiento es la base del TPM (Total Productive Maintenance, Mantenimiento Productivo Total).
A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold reservoir at TC = 400 K. For each of the following cases, determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
a.QH = 900 kJ, Wcycle= 450 kJ
b. QH = 900 kJ, Qc = 300 kJ
c. Weycle = 600 kJ, Qc= 400 kJ
d. η = 75%
Answers
Answer:
a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.
Explanation:
Maximum theoretical efficiency for a power cycle ([tex]\eta_{r}[/tex]), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:
[tex]\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\%[/tex] (1)
Where:
[tex]T_{C}[/tex] - Temperature of the cold reservoir, in Kelvin.
[tex]T_{H}[/tex] - Temperature of the hot reservoir, in Kelvin.
The maximum theoretical efficiency associated with this power cycle is: ([tex]T_{C} = 400\,K[/tex], [tex]T_{H} = 1200\,K[/tex])
[tex]\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{r} = 66.667\,\%[/tex]
In exchange, real efficiency for a power cycle ([tex]\eta[/tex]), no unit, is defined by this expression:
[tex]\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\%[/tex] (2)
Where:
[tex]Q_{C}[/tex] - Heat released to cold reservoir, in kilojoules.
[tex]Q_{H}[/tex] - Heat gained from hot reservoir, in kilojoules.
[tex]W_{C}[/tex] - Power generated within power cycle, in kilojoules.
A power cycle operates irreversibly for [tex]\eta < \eta_{r}[/tex], reversibily for [tex]\eta = \eta_{r}[/tex] and it is impossible for [tex]\eta > \eta_{r}[/tex].
Now we proceed to solve for each case:
a) [tex]Q_{H} = 900\,kJ[/tex], [tex]W_{C} = 450\,kJ[/tex]
[tex]\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 50\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
b) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{C} = 300\,kJ[/tex]
[tex]\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 66.667\,\%[/tex]
Since [tex]\eta = \eta_{r}[/tex], the power cycle operates reversibly.
c) [tex]W_{C} = 600\,kJ[/tex], [tex]Q_{C} = 400\,kJ[/tex]
[tex]\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 60\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
d) Since [tex]\eta > \eta_{r}[/tex], the power cycle is impossible.
Uses of P-N junction
Answers
Answer:
Explanation:
Two that come to mind:
a semiconductor diode is essentially a PN junctiona transistor is made of two pn junctions.On a ship the price gift is 24 euros .What is the difference in the price on a day when the exchange rate is £1=2378
Answers
Answer:
The answer is "[tex]\$29.7072[/tex]".
Explanation:
Please find the complete question in the attachment file.
As the original prices are 24 euros, [tex]\$ 1.2378[/tex] must be multiplied by 24.
[tex]\to 1.2378 \times 24 = \$29.7072[/tex]
Therefore the new price value will be [tex]\$ 29.7072[/tex]
Dalton needs to prepare a close-out report for his project. Which part of the close-out report would describe
how he would plan and manage projects in the future?
Select an answer:
project highlights
major changes and risks
summary of schedule and cost performance
summary of project management effectiveness
Answers
Answer:
Dalton
The part of the close-out report that would describe how he would plan and manage projects in the future is:
summary of project management effectiveness
Explanation:
The Project Close-out Report is a project management document, which identifies the variances from the baseline plans. These variances are specified in terms of project performance, project cost, and schedule. The project close-out report records the completion of the project and the subsequent handover of project deliverables to others. The project management effectiveness summary details the project's objectives and the achievements recorded, including the lessons learned.
If the constant is added to every observation of data then arithmatic mean obtained is
Answers
Answer:
Explanation:
Increased by the constant. Take a very simple case.
4 + 5 + 6 = 15
The mean is 5 (obtained by dividing the total (15) by the number of terms (3).
Now add a constant say 6
4 + 6 = 10
5 + 6 = 11
6 + 6 = 12
Total = 33/3 = 11
So the mean 5 is increased by the constant 6.
Now do the same thing more symbolically.
4 + c
5 + c
6 + c
Total = 15 + 3c
Divide by 3 you get 5 + c
If you want a more formal proof involving n terms, leave a note.
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are currently reviewing the scenario involving the failure of a nitrogen regulator that provides inert padding to the vapor space of the reactor. Your calculations show that the maximum discharge rate of nitrogen through the existing relief system of the vessel is 0.5 kgls, However, your calculations also show that the flow of nitrogen through the l-in supply pipe will be much greater than this. Thus under the current configuration a failure of the nitrogen regulator will result in an over pressuring of the reactor. One way to solve the problem is to install an orifice plate in the nitrogen line, thus limiting the flow to the maximum of 0.5 kg/s. Determine the orifice diameter (in cm) required to achieve this flow. Assume a nitrogen source supply pressure of 15 bar absolute. The ambient temperature is 25°C and the ambient pressure is 1 atm. 3.
Answers
Answer:
[tex]D=0.016m[/tex]
Explanation:
From the question we are told that:
Discharge Rate [tex]F_r=0.5kgls[/tex]
Pressure [tex]P=15Kpa[/tex]
Temperature [tex]T=25=>298K[/tex]
Ambient pressure is 1 atm.
Generally the equation for Density is mathematically given by
[tex]\rho=\frac{PM}{RT}[/tex]
[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]
[tex]\rho=16.958kg/m^2[/tex]
Generally the equation for Flow rate is mathematically given by
[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]
Where
[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]
[tex]Q=1.4[/tex]
[tex]\mu= Discharge\ coefficient[/tex]
[tex]\mu=0.68[/tex]
Therefore
[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]
[tex]A=2.129*10^{-4}[/tex]
Where
[tex]A=\frac{\pi}{4}D^2[/tex]
[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]
[tex]D=0.016m[/tex]
The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Sieve analysis Sieve Size No. 4 (4.75 mm) No. 10 (2.00 mm) No. 40 (0.425 mm) No. 200 (0.075 mm) Percent passing by weight 80 60 30 10 Atterberg limits Liquid limit (LL) Plastic limit (PL 31 25
(a) Classify this soil according to USCS system, providing the group symbol for it. Show how you arrive at the final classification.
(b) According to USCS system, what is a group name for this soil?
(c) Is this a clean sand? If not, explain why.
Answers
Answer: hello the complete question is attached below
answer:
A) Group symbol = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Explanation:
A) Classifying the soil according to USCS system
( using 2nd image attached below )
description of sand :
The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also
The soil is a sand given that more than 50% particles passed from No 4 sieve
The soil can be a clean sand given that fines ≤ 12%
The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time
Group symbol as per the 2nd image attached below = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
In a major human artery with an internal diameter of 5mm, the flow of blood, averaged over the cardiac cycle is 5cm3·s−1. The artery bifurcates (splits) into two identical blood vessels that are each 3mm in diameter. What are the average velocity and the mass flow rate upstream and downstream of the bifurcation? The density of blood is 1.06g·cm−3
Answers
9514 1404 393
Answer:
see attached
Explanation:
Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.
(5 cm³/s)(1.06 g/cm³) = 5.3 g/s
If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:
(5.3 g/s)/2 = 2.65 g/s
__
The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...
(5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s
Downstream, we have half the volumetric flow and a smaller area.
(2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s
which type of clectrical circuit is represented by this diagram?
Answers
Answer:
parallel
Explanation:
All components in this circuit are tied in parallel. Each component experiences the same voltage from one terminal to the other. It is a parallel circuit.
How many numbers multiple of 3 are in the range [2,2000]?
Answers
g A motor driving a 1000-W water pump has a power factor of 0.80 lagging; a second motor driving a 600-W water pump has a power factor of 0.60 lagging assuming the motors are working under 120- Vrms, 60-HZ AC. (a) When both motors working together what is combined power factor
Answers
complete Question
A motor driving a 1000-W water pump has a power factor of 0.80 lagging; a second motor driving a 600-W water pump has a power factor of 0.60 lagging assuming the motors are working under 120-Vrms, 60-HZAC. When both motors working together what is the combined power factor? If a 200-μF capacitor is connected to the above system (two motors) what is the new combined power factor?
Answer:
[tex]p.f'=0.960[/tex]
Explanation:
First motor Power [tex]P=1000W[/tex]
First motor Power factor [tex]P.f=0.80[/tex]
Second motor Power [tex]P=600W[/tex]
Second motor Power factor p.f=0.60
Voltage [tex]V=120Vrms[/tex]
Frequency [tex]F=60Hz[/tex]
Capacitor [tex]C=200\mu F[/tex]
Generally power in Var is given as
For First Motor
[tex]Q=\frac{1000}{0.8}\sqrt{1-0.8^2}[/tex]
[tex]Q=750Var[/tex]
For Second Motor
[tex]Q=\frac{600}{0.6}\sqrt{1-0.6^2}[/tex]
[tex]Q=800Var[/tex]
Generally the equation for The Reactive Power is mathematically given by
[tex]Q_c=\frac{V^2}{X_c}[/tex]
Where
[tex]X_c=\frac{1}{2 \pi fc}[/tex]
[tex]X_c=\frac{1}{2 \pi 60*200*10^{-6} }[/tex]
[tex]X_c=13.3[/tex]
Therefore
[tex]Q_c=-\frac{120^2}{13.3}\\\\Q_c=-1085.97j[/tex]
Giving
Total Power Drawn by Supply
[tex]P_t=(1000+j750)+(600+800)-j1085.97[/tex]
[tex]P_t=1600+464.03j[/tex]
Therefore
[tex]p.f'=\frac{1600}{\sqrt{1600^2+464.03^2}}[/tex]
[tex]p.f'=0.960[/tex]
What are three types of land reform
Answers
Answer:
Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);
Types of Land Reform
Abolition of Intermediaries
The first step taken by the Indian government under land reforms post-independence was passing the Zamindari Abolition Act. The primary reason of a backward agrarian economy was the presence of intermediate entities like, jagirdars and zamindar who primarily focussed on collecting sky-rocketing rents catering to their personal benefits, without paying attention to the disposition of farms and farmers. Abolition of such intermediaries not only improved conditions of farmers by establishing their direct connection with the government but also improved agricultural production.
Regulation of Rents
This was in direct response to the unimaginably high rents which were charged by intermediaries during British rule, which resulted in a never-ending cycle of poverty and misery for tenants. Indian government implemented these regulations to protect farmers and labourers from exploitation by placing a maximum limit on the rent that could be charged for land.
Tenure Security
Legislations were passed in all states of the country to grant tenants with permanent ownership of lands and protection from unlawful evictions on expiry of the lease. This law protects tenants from having to vacate a property immediately after their tenure is over unless ordered by law. Even in that case, ownership can be regained by tenants with the excuse of personal cultivation.
1) (30 pts ) Oxygen (O2) flows through a pipe, entering at at 4 m/sec at 10000 kPa, 227oC. For a pipe inside diameter of 3.0 cm, find the volumetric flow rate (m3/sec) and the mass flow rate of the gas (kg/sec) assuming you have an ideal gas
Answers
Complete Question
Nitrogen (N2) flows through a pipe, entering at at 4 m/sec at 1000 kPa, 2270C. For a pipe inside diameter of 3 cm, find the volumetric flow rate (m3/sec) and the mass flow rate of the gas (kg/sec) assuming you have an ideal gas Then using your ideal gas mass flow rate find the rate at which enthalpy enters the pipe (kJ/sec) NO Cp, Cv, k permitted
Answer:
[tex]H=9.91kJ/sec[/tex]
Explanation:
From the question we are told that:
Velocity [tex]v=4 m/sec[/tex]
Pressure [tex]P=1000kPa[/tex]
Temperature [tex]T=227 \textdegree C[/tex]
Diameter [tex]d=3cm=>0.03m[/tex]
Generally the equation for volumetric Flow Rate is mathematically given by
[tex]V_r=(\frac{\pi*d^2}{4}v)[/tex]
[tex]V_r=(\frac{\pi*(0.03)^2}{4} *4)[/tex]
[tex]V_r=0.002827m^3/s[/tex]
Generally the equation for mass Flow Rate is mathematically given by
[tex]m_r=\frac{PV_r}{RT}[/tex]
[tex]m_r=\frac{1000*0.002827}{0.297*(227+273)}[/tex]
[tex]m_r=0.019kg/sec[/tex]
Generally the equation for mass Flow Rate is mathematically given by
Using gas Table for enthalpy Value
[tex]T=500K=>h=520.75kg[/tex]
Therefore
[tex]H=mh[/tex]
[tex]H=0.019*520.75[/tex]
[tex]H=9.91kJ/sec[/tex]
Use a truth table to verify the first De Morgan law ¬(p ∧ q) ≡ ¬p ∨ ¬q.
Answers
Answer:
p q output ¬(p ∧ q)
0 0 1
0 1 1
1 0 1
0 0 0
p q output ¬p ∨ ¬q
0 0 1
0 1 1
1 0 1
0 0 0
Explanation:
We'll create two separate truth tables for both sides of the equation, and see if they match.
The expressions in the question use AND, OR and NOT operators.
The AND operation needs both inputs to be 1 to return a 1.The OR operation needs at least 1 of the inputs to be 1 to return a 1. The NOT operation takes a 1 and turns it into a 0, or takes a 0 and turns it into a 1.Let's start with ¬(p ∧ q)
NOT (0 AND 0) = NOT (0) = 1NOT (0 AND 1) = NOT (0) = 1NOT (1 AND 0) = NOT (0) = 1NOT (1 AND 1) = NOT (1) = 0Now let's move on to the second expression ¬p ∨ ¬q
NOT(0) OR NOT(0) = 1 OR 1 = 1NOT(0) OR NOT(1) = 1 OR 0 = 1NOT(1) OR NOT(0) = 0 OR 1 = 1NOT(0) OR NOT(0) = 0 OR 0 = 0Therefore we can say the two expressions are equivalent.
Attached the truth table to verify the first De Morgan's law ¬(p ∧ q) ≡ ¬p ∨ ¬q:
What is the explanation of the truth table?As you can see from the attached truth table, the truth values for ¬(p ∧ q) and ¬p ∨ ¬q are the same for all combinations of p and q, confirming the validity of the first De Morgan's law.
De Morgan's law is a fundamental principle in propositional logic.
It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the individual propositions.
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lists at least 6 units of measuring atmospheric pressure
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Answer:
On my console displays for the ISS visiting vehicles, three units are used. The Americans use pounds per square inch (psi). The Russians use kilopascals (kPa). The Japanese use Torr - millimeters of mercury (mmHg). A fourth unit is simply the atmosphere, or multiples of it. So, for example, sea level air pressure (which is what we use onboard ISS) is defined as 1 atmosphere. That is equivalent to 14.7 psi, 101.3 kPa, or 760 mmHg.Here N represents newton which is SI unit of Force which is same as Kg.m/s2." role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">2.2.
m represent metre which is SI unit of length.
Kg represent Kilogram which is SI unit of Mass.
m2SI" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">2SI2SI unit of Area.
Hope it helps.
Thanks.
Answer:
Pounds per square inch (psi)
Kilopascals (kPa)
Millimeters of mercury (mmHg)
Pascal (Pa)
Megapascal (MPa)
Atmospheric pressure (atm)
Hope this helps!
Different metabolic control systems have different characteristic time scales for a control response to be achieved. Match the time scale with the control system.
a. Covalent modification
b. Allosteric control
c. Gene expression
1. Seconds to minutes
2. Milliseconds
3. Hours
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Answer:
a. Covalent modification = Seconds to minutes
b. Allosteric control = Milliseconds
c. Gene expression = Hours
Explanation:
Covalent modifications refer to the addition and/or removal of chemical groups by the action of particular enzymes such as methylases, acetylases, phosphorylases, phosphatases, etc. For example, histones are chromatin-associated proteins covalently modified by enzymes that add methyl groups (histone methylation), acetyl groups (histone acetylation), phosphate groups (histone phosphorylation), etc. Moreover, allosteric control, also known as allosteric regulation, is a type of regulation of the enzyme activity by binding an effector molecule (allosteric modulator) at a different site than the enzyme's active site, thereby triggering a conformational change on the enzyme upon binding of an effector. Finally, gene expression encompasses the cellular processes by which genetic information flows from genes to proteins (i.e., transcription >> translation). In metabolic pathways, enzymes that are able to catalyze irreversible reactions represent sites of control (for example, during glycolysis, pyruvate kinase is an enzyme that catalyzes an irreversible reaction, thereby serving as a control site). In turn, enzymatic activity is modulated by covalent modifications or reversible binding of allosteric effectors. Finally, metabolic pathways are also modulated by gene regulatory mechanisms that control the transcription of specific enzymes required for such pathways. During these processes, the times required for allosteric regulation, covalent modification (e.g., phosphorylation) and transcriptional control can be counted in milliseconds, seconds, and hours, respectively.
1025 steel wire is stretched with a stress of 70 MPa at room temperature 20 C. If th length is held constant, to what temperature in 'C and 'F must the wire be heated to reduce the stres to 17 MPa?
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Check attached pictureCheck attached pictureCheck attached pictureCheck attached picture
. Bơm kiểu piston tác dụng đơn có áp suất p=0,64 Mpa và lưu lượng Q=3,5 l/s. Xác định tốc độ quay của trục bơm và công suất của bơm nếu biết đường kính piston D=150 mm; bán kính tay quay R=60 mm; hiệu suất thể tích của bơm là 0=0,94; hiệu suất chung của bơm b=0,80.
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Answer:
not understand language
A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance it slides in 10seconds if the kinetic coefficient is 0.4.
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Answer:
21 m
Explanation:
Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.
Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N
The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N
The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N
If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.
Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.
For the block to be in contact with the surface, the vertical forces on the block must balance.
Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,
N = mg + F₃" (since both the weight and the resultant vertical force act downwards)
N = mg + F₃"
Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N
So,
N = mg + F₃"
N = 35 kg × 9.8 m/s² + 8.82 N
N = 343 N + 8.82 N
N = 351.82 N
So, the net horizontal force F = F₃' - f.
F = 155.41 N - 0.4 × 351.82 N
F = 155.41 N - 140.728 N
F = 14.682 N
Since F = ma, where a = acceleration of block,
a = F/m = 14.682 N/35 kg = 0.42 m/s²
To find the distance the block moved, x we use the equation
x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²
Substituting the values of the variables into the equation, we have
x = ut + 1/2at²
x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²
x = 0 m + 1/2 × 0.42 m/s² × 100 s²
x = 0.21 m/s² × 100 s²
x = 21 m
So, the distance moved by the block is 21 m.
beacuse thye want them to hav egoood and thye wn thme tto
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Answer:
I don't understand the question
Trình bày sự khác nhau của Dây chuyền đẳng nhịp đồng nhất, dây chuyền đẳng nhịp không đồng nhất, cho ví dụ minh họa
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Plexiglas, polymethylmethacrylate (PMMA), is often used as display windows and cases in art galleries and museums. If its flame imparts 32 kW/m^2 to the surface when it burns estimate the energy release rate of a 3 m^2 square sheet (one side). Assume its vaporization temperature is 350°C. Calculate with units, provide work.
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This question is incomplete, the complete question is;
Plexiglas, polymethylmethacrylate (PMMA), is often used as display windows and cases in art galleries and museums. If its flame imparts 32 kW/m^2 to the surface when it burns estimate the energy release rate of a 3 m^2 square sheet (one side). Assume its vaporization temperature is 350°C. Calculate with units, provide work.
a ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10⁻¹¹ kW/m².K⁴
b ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] - [tex]q_{rad[/tex]
c ⇒ [tex]m"[/tex] = [tex]q"[/tex]/L, where L = 1.6 kJ/g
d ⇒ Q = [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g
Answer:
a) [tex]q_{rad[/tex] = 8.54 kW/m²
b) [tex]q"[/tex] = 23.46 kW/m²
c) [tex]m"[/tex] = 14.6625 g/m²
d) Q = 1095.2888 kJ
Explanation:
Given the data in the question;
[tex]q_{flame[/tex] = 32 kW/m²
Area; A = 3m²
vaporization temperature; T = 350°C = ( 350 + 273 )K = 623 K
Now,
a) ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10¹¹ kW/m².K⁴
we substitute
[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × ( 623 K)⁴
[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × 150644120641 K⁴
[tex]q_{rad[/tex] = 8.54 kW/m²
b) ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] - [tex]q_{rad[/tex]
we substitute
[tex]q"[/tex] = 32 kW/m² - 8.54 kW/m²
[tex]q"[/tex] = 23.46 kW/m²
c) ⇒ [tex]m"[/tex] = [tex]q"[/tex]/L, where L = 1.6 kJ/g
we substitute
[tex]m"[/tex] = 23.46 / 1.6
[tex]m"[/tex] = 14.6625 g/m²
d) ⇒ Q = [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g
we substitute
Q = 14.6625 × 3 × 24.9
Q = 1095.2888 kJ
Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend auaching fins both inside and outside the tubes?
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Answer:
Fins should be attached outside the tube Fins can be attached on both sides when convection coefficient of air inside the tube is equal to the convection coefficient of atmospheric air outside the tubeExplanation:
The main function of the fins that are to be added is to ensure the speedy transfer of heat from the Hot air.
The fins should be attached outside the tube because the convection coefficient of air is higher inside the tube than the convection coefficient of the outside air ( atmospheric air ), BUT
When convection coefficient of air inside the tube is equal to the atmospheric air outside the tube, it is recommended that the fins can be added on both sides of the tube ( i.e. in and outside the tube )
Theo Anh / Chị, để đáp ứng yêu cầu phát triển nền kinh tế thị trường định hướng Xã hội Chủ nghĩa ở Việt Nam trong bối cảnh thời đại hiện nay, cần chú trọng giải quyết những vấn đề gì ?
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tôi nghĩ
A technician wants to implement a dual factor authentication system that will enable the organization to authorize access to sensitive systems on a need-to-know basis. What should be implemented during the authorization stage?
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Answer: Biometrics
Explanation:
Dual factor authentication refers to an electronic authentication method whereby a user will only be granted an access to an application or a website after the user has successfully been able to present two pieces of evidence which then grants access to the application or website.
Since the technician wants to implement a dual factor authentication system, the biometrics should be implemented during the authorization stage.
Biometrics refers to the body measurements and the calculations that are related to the characteristics of humans. Biometric authentication is used as a form of identification.
