Answer: the mole fraction of D in the mixture is 0.2222
Explanation:
Given that;
mixture analysis shows 2 kmol A, 1 kmol B, 4 kmol C and some unknown kmol of D was present.
4A+3B→2C+D
As from reaction stoichiometry, for every 2 kmol of C produced, kmol of D produced = 1 kmol
so, for 4 kmol C, kmol of D produced = 4/2 × 1 kmol = 2 kmol
Now our mixture has 2 kmol A, 1 kmol B, 4 kmol C and also 2 kmol of D
so, total moles in mixture, we have (2 + 1 + 4 + 2) kmol = 9 kmol
mole fraction of D in mixture will be;
( Kmol of D) / (total moles in mixture) = 2 / 9 = 0.2222
Therefore the mole fraction of D in the mixture is 0.2222
In a CS amplifier, the resistance of the signal source Rsig = 100 kQ, amplifier input resistance (which is due to the biasing network) Rin = 100kQ, Cgs = 1 pF, Cgd = 0.2 pF, gm = 5 mA/V, ro = 25 kΩ, and RL = 20 kΩ. Determine the expected 3-dB cutoff frequency.
Answer:
406.140 KHz
Explanation:
Given data:
Rsig = 100 kΩ
Rin = 100kΩ
Cgs = 1 pF,
Cgd = 0.2 pF, and etc.
Determine the expected 3-dB cutoff frequency
first find the CM miller capacitance
CM = ( 1 + gm*ro || RL )( Cgd )
= ( 1 + 5*10^-3 * 25 || 20 ) ( 0.2 )
= ( 11.311 ) pF
now we apply open time constant method to determine the cutoff frequency
Th = 1 / Fh
hence : Fh = 1 / Th = [tex]\frac{1}{(Rsig +Rin) (Cm + Cgs )}[/tex]
= [tex]\frac{1}{( 200*10^3 ) ( 12.311 * 10^{-12} )}[/tex] = 406.140 KHz
A single phase inductive load draws 10 MW at 0.6 power factor lagging. Draw the power triangle and determine the reactive power of a capacitor to be connected in parallel with the load to raise the power factor to 0.85.
Answer: attached below is the power triangles
7.13589 MVAR
Explanation:
Power ( P1 ) = 10 MW
power factor ( cos ∅ ) = 0.6 lagging
New power factor = 0.85
Calculate the reactive power of a capacitor to be connected in parallel
Cos ∅ = 0.6
therefore ∅ = 53.13°
S = P1 / cos ∅ = 16.67 MVA
Q1 = S ( sin ∅ ) = 13.33 MVAR ( reactive power before capacitor was connected in parallel )
note : the connection of a capacitor in parallel will cause a change in power factor and reactive power while the active power will be unchanged i.e. p1 = p2
cos ∅2 = 0.85 ( new power factor )
hence ∅2 = 31.78°
Qsh ( reactive power when power factor is raised to 0.85 )
= P1 ( tan∅1 - tan∅2 )
= 10 ( 1.333 - 0.6197 )
= 7.13589 MVAR
please help me make a lesson plan. the topic is Zigzag line. and heres the format.
A. Objective
B. Subject matter
C. Learning activities.
D. Assessment.
E. Reinforcement
Explanation:
D. B. C. A. E. Is this a good idea
Indicate similarities between a nucleus and a liquid droplet; why small droplets are stable and very big droplets are not?
Answer:
There are several similarities between the nucleus and a liquid droplet.
Explanation:
A droplet of liquid simply is is very small or tiny drop of liquid. It is also considered as a tiny column of liquid that is surrounded by surfaces that have zero shear stress.
A nucleus on the other hand is an assembly between protons and neutrons. The latter is electrically charged whilst the former is positively charged. The number of protons present in an element is very crucial to the qualities of an element.
The main similarities between a nucleus and a liquid droplet are:
1. a nucleus consists of a large amount of neutrons and protons in the same volume as would a liquid which contains large numbers of molecules in the same volume;
2. both the nucleus and the droplet are similar for their homogeneity in electric charge and density;
3. the molecules exert the same amount for forces towards one another as would the nuclear forces in the nucleons.
4. both of them cannot be compressed
5. both molecules and nucleus are can be subject to nuclear fission which simply mean the breaking apart into smaller units (in the case of the nucleus) or the breaking apart into smaller droplets in the case of the liquid molecule.
6. There are two types of phenomena which occurs in both the liquid droplet and the nucleus which are similar to one another. They are:
Evaporation (in the case of the liquid molecule) and reaction emission (in the case of the nucleus). In evaporation, particles are lost, in Atomic transmutation, particles are lost as well.
B) the forces which determine the stability of droplets are surface tension and gravitation. The smaller the area, the stronger the surface tension available to keep the drops from going out of shape.
Cheers
Which of following are coding languages used in controlling a robot? *
A. Scratch
B. B/B--
C. C/C++
D. Robot Z
Answer:
C/C++
Explanation:
C/C++
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
People tend to self-disclose to others that are in age, social status, religion, and personality.
Answer:people tend to do this when they are in a different environment they lose something or just have something going on in their life
Explanation:
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?
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
Calculate the LER for the rectangular wing from the previous question if the weight of the glider is 0.0500 Newton’s.
Answer:
0.2
Explanation:
Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.
Let the span of the rectangular wing be 0.225 m
Let the chord of the rectangular wing be 0.045 m.
Then, the area of any rectangular chord is
A = chord * span
A = 0.045 * 0.225
A = 0.010 m²
And using the weight of the glider given to us from the question, we can find the LER for the wing.
LER = Area / weight.
LER = 0.010 / 0.05
LER = 0.2.
Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2
Please mark brainliest...
Answer: 0.2025
Explanation: I got it correct
A rectangular channel 3-m-wide carries 12 m^3/s at a depth of 90cm. Is the flow subcritical or supercritical? For the same flowrate, what depth will five critical flow?
Answer:
Super critical
1.2 m
Explanation:
Q = Flow rate = [tex]12\ \text{m}^3/\text{s}[/tex]
w = Width = 3 m
d = Depth = 90 cm = 0.9 m
A = Area = wd
v = Velocity
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
[tex]Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}[/tex]
Froude number is given by
[tex]Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5[/tex]
Since [tex]F_r>1[/tex] the flow is super critical.
Flow is critical when [tex]Fr=1[/tex]
Depth is given by
[tex]d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}[/tex]
The depth of the channel will be 1.2 m for critical flow.
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
A roadway with a rough-asphalt pavement has a cross slope of 2%, a longitudinal slope of 2.5%, a curb height of 8 cm, and a 90-cm-wide concrete gutter. If the flow rate in the gutter is 0.07 m/s, determine the size (W XL, in mm) and interception capacity (m/s) of a reticuline grate that should be used to intercept as much of the flow as possible.
a. Reticuline grate size?
b. Interception capacity?
Answer:
b
Explanation:
Copy bits 3..0 in $s1 to 6..3 in $s2. Bits 6..3 in $s2 are already set to 0. Registers$s0 0..01111$s1 0..0101$s3 0
Answer:
Following are the solution to this question:
Explanation:
To copy 3.0 bits in 50 dollars or run at 50 dollars, it takes just 3.0 bits as well as other bits but masks, and 50 dollars.
Instead of shifting the $ 50 by 3 bits to 6...3 bits of [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex] This procedure instead took place at $53 and $50
AND [tex]\$ \ 50,\$ \ 50,0*0000 000f[/tex], take 3..0 bits
SLL [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex]Shifts the bits to 6..3
O R [tex]\$ \ 53,\$ \ 53,\$ \ 50 ,[/tex] coping to [tex]\$ \ 53[/tex]
How many snaps points does an object have?
Answer:
what do you mean by that ? snap points ?
Products exit a combustor at a rate of 100 kg/sec, and the air-fuel ratio is 9. Determine the air flow rate. a. 9 kg/sec b. 90 kg/sec c. 100 kg/sec d. 10 kg/sec
Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer
Explanation:
Given that;
product of combustor flow rate m = 100 kg/s
air-fuel = 9
Airflow rate = ?
⇒We know that in the combustor, air fuel are mixed and then ignited,
⇒air fuel products are exited at the combustor
let air and fuel be a and b respectively
⇒ a + b = 100 kg/sec ----- let this be equation 1
now
⇒ air / fuel = 9
a / b = 9
a = 9b -----------let this be equation 2
now input a = 9b in equation 1
9b + b = 100 kg/sec
10b = 100 kg/sec
b = 10 kg/sec
we know that
a = 9b
so a = 9 × 10 = 90 kg/sec
Therefore the air flow rate a is 90 kg/sec
A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 318 g/mol
b. 204 g/mol
c. 175 g/mol
d. 298 g/mol
Answer: Option D) 298 g/mol is the correct answer
Explanation:
Given that;
Mass of sample m = 13.7 g
pressure P = 2.01 atm
Volume V = 0.750 L
Temperature T = 399 K
Now taking a look at the ideal gas equation
PV = nRT
we solve for n
n = PV/RT
now we substitute
n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )
= 1.5075 / 32.7579
= 0.04601 mol
we know that
molar mass of the compound = mass / moles
so
Molar Mass = 13.7 g / 0.04601 mol
= 297.7 g/mol ≈ 298 g/mol
Therefore Option D) 298 g/mol is the correct answer
A three-phase motor rated 25 hp, 480 V, operates with a power factor of 0.74 lagging and supplies the rated load. The motor efficiency is 96%. Calculate the motor input power, reactive power and current.
Answer:
the motor input power is 19.42 KW
the Reactive power is 17.65 KVAR
Current is 31.56 A
Explanation:
Given that;
V = 480V
h.p = 25 hp
p.f = 0.74 lagging
n_motor = 96%
so output = 25hp
and we know that;
1hp = 746 watt
watt = hp × 1hp
so output in watt = 25 × 746 = 18650 Watt = 18.65 KW
n_motor = (output / input) × 100
96 = 1865 / Input
96Input = 1865
Input = 1865 / 96
Input = 19.42 KW
Therefore the motor input power is 19.42 KW
P = √( 3 × V × I × cos∅)
19.42 = √( 3 ×480 × I × 0.74)
I = 31.56 A
Therefore Current is 31.56 A
Q = √( 3 × V × I × sin∅)
we know that
cos∅ = 0.74
so ∅ = cos⁻¹(0.74) = 42.26
so we substitute
Q = √( 3 × 480 × 31.56 × sin(42.26))
= 17.65 KVAR
Therefore the Reactive power is 17.65 KVAR
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