Answers
Answer:
mountain ranges may be
Related Questions
A duck is cooked in the kitchen oven for 4 hours. Knowing that the oven, powered by 220 V, absorbs a current of 20 A and uses energy costing 0.048 € / kWh, how much does it cost to cook the duck?
Answers
Explanation:
cooking of duck will cost 48000
by the help of the method of rate × A + €
Which of the following is an example of a tax
Answers
Answer:
A tax is a monetary payment without the right to individual consideration, which a public law imposes on all taxable persons - including both natural and legal persons - in order to generate income. This means that taxes are public-law levies that everyone must pay to cover general financial needs who meet the criteria of tax liability, whereby the generation of income should at least be an auxiliary purpose. Taxes are usually the main source of income of a modern state. Due to the financial implications for all citizens and the complex tax legislation, taxes and other charges are an ongoing political and social issue.
I dont know I asked this to
Explanation:
Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required, in Btu/lbm, for this compression. The gas constant of air is R.
Answers
Answer:64.10 Btu/lbm
Explanation:
Work done in an isothermally compressed steady flow device is expressed as
Work done = P₁V₁ In { P₁/ P₂}
Work done=RT In { P₁/ P₂}
where P₁=13 psia
P₂= 80 psia
Temperature =°F Temperature is convert to °R
T(°R) = T(°F) + 459.67
T(°R) = 55°F+ 459.67
=514.67T(°R)
According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm
Work = RT In { P₁/ P₂}
0.06855 x 514.67 In { 13/ 80}
=0.06855 x 514.67 In {0.1625}
= 0.06855 x 514.67 x -1.817
=- 64.10Btu/lbm
The required work therefore for this isothermal compression is 64.10 Btu/lbm
An unknown impedance Z is connected across a 380 V, 60 Hz source. This causes a current of 5A to flow and 1500 W is consumed. Determine the following: a. Real Power (kW) b. Reactive Power (kvar) c. Apparent Power (kVA) d. Power Factor e. The impedance Z in polar and rectangular form
Answers
Answer:
a) Real Power (kW) = 1.5 kW
b) Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA) is 1.9 KVA
d) the Power Factor cos∅ is 0.7894
e) the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω
Explanation:
Given that;
V = 380v
i = 5A
P = 1500 W
determine;
a) Real Power (kW)
P = 1500W = 1.5 kW
therefore Real Power (kW) = 1.5 kW
b) Reactive Power (kvar)
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
∅ = cos⁻¹ (0.7894)
∅ = 37.87°
Q = VIsin∅
Q = 380 × 5 × sin( 37.87° )
Q = 1.1663 KVAR
Therefore Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA)
S = P + jQ
= ( 1500 + J 1166.3 ) VA
S = 1900 ∠ 37.87° VA
S = 1.9 KVA
Therefore Apparent Power (kVA) is 1.9 KVA
d) Power Factor
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
Therefore the Power Factor cos∅ is 0.7894
e) The impedance Z in polar and rectangular form
Z = 380 / ( S∠-37.87) = V/I
Z = ( 60 + j 46.647) Ω
Z = 76 ∠ 37.87° Ω
Therefore the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3
/s and at a
velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Figure.
Determine the force acting on the shaft (which is
also the force acting on the bearing of the shaft) in
the axial direction.
Answers
Answer:
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.
Step-by-step solution:
Step 1 of 5
Given data:-
The velocity of water is .
The water flow rate is.
A roadway is to be designed on a level terrain. The roadway id 500 ft. Five cross-sections have been selected at 0 ft, 125 ft, 250 ft, 375 ft, and 500 ft. the cross sections have areas of 130 ft^2, 140 ft^2, 60 ft^2, 110 ft^2, and 120 ft^2. What is the volume needed along this road assuming a 6% shrinkage?
Answers
Answer:
51112.5 ft^3
Explanation:
Determine the volume needed along the road when we assume a 6% shrinkage
shrinkage factor = 1 - shrinkage = 1 - 0.06 = 0.94
first we have to calculate the volume between the cross sectional areas (i.e. A1 ---- A5 ) using average end area method
Volume between A1 - A2
= (125 ft - 0 ft) * [(130 ft^2 + 140 ft^2) / 2]
= 125 ft * 135 ft^2
= 16875 ft^3
Volume between A2 - A3
= (250 ft - 125 ft) * [(140 ft^2 + 60 ft^2) / 2]
= 125 ft * (200 ft^2 / 2)
= 12500 ft^3
Volume between A3 - A4
= (375 ft - 250 ft) * [(60 ft^2 + 110 ft^2) / 2]
= 125 ft * (170 ft^2 / 2)
= 10625 ft^3
Volume between A4 - A5
(500 ft - 375 ft) * [(110 ft^2 + 120 ft^2) / 2]
= 125 ft * 115 ft^2
= 14375 ft3
Hence the total volume along the 500 ft road
= ∑ volumes between cross sectional areas
= 16875 ft^3 + 12500 ft^3 + 10625 ft^3 + 14375 ft^3 = 54375 ft^3
Finally the volume needed along this road is calculated as
Total volume * shrinkage factor
= 54375 * 0.94 = 51112.5 ft^3
A rear wheel drive car has an engine running at 3296 revolutions/minute. It is known that at this engine speed the engine produces 80 hp. The car has an overall gear reduction ratio of 10, a wheel radius of 16 inches, and a 95% drivetrain mechanical efficiency. The weight of the car is 2600 lb, the wheelbase is 95 inches, and the center of gravity is 22 inches above the roadway surface. What is the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement?
Answers
Answer:
the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement is 47.8 in
Explanation:
Given that;
Weight of car W = 2600 lb
power = 80 hp = 44000 lb ft/s
Engine rpm = 3296
gear reduction ratio e = 10
drivetrain efficiency n = 95% = 0.95
wheel radius R = 16 in = 1.3333 ft
Length of wheel base L = 95 in =
coefficient of road adhesion u = 0.60
height of center of gravity above pavement h = 22 in
we know that;
Coefficient of rolling resistance frl = 0.01 for good wet pavement
distance of center of gravity behind the front axle lf = ?
Maximum tractive effort (Fmax) = (uW / L) (lf - frl h) / (1 - uh / L)
First we calculate our Fmax to help us find lf
Power = Torque × 2π × Engine rpm / 60 )
44000 = Torque ( 2π×3296 / 60)
Torque = 127.5 lb ft
so
Fmax = Torque × e × n / R
so we substitute in our values
Fmax = 127.5 × 10 × 0.95 / 1.333
Fmax = 908.66 lb
Now we input all our values into the initial formula
(Fmax) = (uW / L) (lf - frl h) / (1 - uh / L)
908.66 = [(0.6×2600/95) (lf - 0.01×22)] / [1 - 0.6×22) / 95]
908.66 = (16.42( lf - 0.22)) / 0.86
781.4476 = (16.42( lf - 0.22))
47.59 = lf - 0.22
lf = 47.59 + 0.22
lf = 47.8 in
Therefore the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement is 47.8 in
What overall material composition would be required to give a material made up of 50wt% mullite and 50wt% alumina at 1400°C?
Answers
Answer: overall composition ⇒ 87 wt% { AL₂O₃] + 13% wt { SiO₂}
Explanation:
Given that;
from the phase diagram SiO₂ - Al₂O₃
alumina at 1400°C
mullite + alumina ranges from 74 - 100% wt
so for 50% mullite and 50wt% alumina
we have;
50/100 = 100 - x / 100 - 74
0.5 = 100 - x / 26
0.5 × 26 = 100 - x
13 = 100 - x
x = 100 - 13
x = 87 wt% { AL₂O₃]
[ 100% - 87% = 13%] 13% wt SiO₂
So overall composition ⇒ 87 wt% { AL₂O₃] + 13% wt { SiO₂}
In beams, why is the strain energy from bending moments much bigger than the strain energy from transverse shear forces? Choose one or more of the following options.
a) The stresses due to bending moments is much more than the stresses from transverse shear.
b) The strains due to bending moments is much more than the strains from transverse shear.
c) The deformations due to bending moments is much more than the deformations from transverse shear.
Answers
Answer:
a) The stresses due to bending moments is much more than the stresses from transverse shear.
c) The deformations due to bending moments is much more than the deformations from transverse shear.
Explanation:
Strain in an object suspended is a function of the stress which the suspended body passed through. The stress which is the function of the force experienced by the body over a given area helps is straining the moment. This lead to the strain energy from bending moment being greater than the strain energy from a transverse shear force.
What is computer programming
Answers
Answer:
Computer programming is where you learn and see how computers work. People do this for a living as a job, if you get really good at it you will soon be able to program/ create a computer.
Explanation:
Hope dis helps! :)
The seers were of the opinion that_____ . *
a healthy mind guides a healthy body.
the healthy body needs no exercise.
a healthy mind resides in a healthy body.
the healthy mind resides in every body.
Answers
Answer:
✔️a healthy mind resides in a healthy body.
Explanation:
The seers were of the opinion that "a healthy mind resides in a healthy body."
Just like the English translation of a famous quotation from Thales, pre-Socratic Greek philosopher puts it "a sound mind in a sound body"; which tries to demonstrate the close connections that exists in bodily well-being and one's ability to enjoy life.
The seers were actually of the opinion that a healthy mind resides in a healthy body. It implies that there is connection between the body and the mind. When the body catches an illness, the mind and other parts of the body are affected. When our minds are not healthy, it affects the effective functioning of the body.
So, a healthy mind will definitely be found in a healthy body.
✔️a healthy mind resides in a healthy body.
Explanation:
The seers were of the opinion that "a healthy mind resides in a healthy body."
Just like the English translation of a famous quotation from Thales, pre-Socratic Greek philosopher puts it "a sound mind in a sound body"; which tries to demonstrate the close connections that exists in bodily well-being and one's ability to enjoy life.
The seers were actually of the opinion that a healthy mind resides in a healthy body. It implies that there is connection between the body and the mind. When the body catches an illness, the mind and other parts of the body are affected. When our minds are not healthy, it affects the effective functioning of the body.
So, a healthy mind will definitely be found in a healthy body.
Compute the discharge observed at a v-notch weir. The weir has an angle of 90-degrees. The height above the weir is 3 inches.
Answers
Answer: the discharge observed at a v-notch weir is 66.7 in³/s
Explanation:
Given that;
Notch angle ∅ = 90°
height above the weir is 3 inches { head + head correction factor) h + k = 3 in
Discharge Q = ?
To determine the discharge observed, we us the following expression
Q = 4.28Ctan(∅/2) ( h + k )^5/2
where Q is discharge, C is discharge coefficient, ∅ is notch angle, h is head and k is head correction factor
now we substitute
Q = 4.28 × 1 × tan(90/2) ( 3 )^5/2
Q = 4.28 × 1 × 1 × 15.5884
Q = 66.7 in³/s
Therefore the discharge observed at a v-notch weir is 66.7 in³/s
In an axial flow compressor air enters the compressor at stagnation conditions of 1 bar and 290 K. Air enters with an absolute velocity of 145 m/s axially into the first stage of the compressor and axial velocity remains constant through the stage. The rotational speed is 5500 rpm and stagnation temperature rise is 22 K. The radius of rotor-blade has a hub to tip ratio of 0.5. The stage work done factor is 0.92, and the isentropic efficiency of the stage is 0.90. Assume for air Cp=1005 kJ/(kg·K) and γ= 1.4
Determine the followings. List your assumptions.
i. The tip radius and corresponding rotor angles at the tip, if the inlet Mach number for the relative velocity at the tip is limited to 0.96.
ii. The mass flow at compressor inlet.
iii. The stagnation pressure ratio of the stage and power required by the first stage.
iv. The rotor angles at the root section.
Answers
Answer:
i) r_t = 0.5101 m
ii) m' = 106.73 kg/s
iii) R_s = 1.26
P = 2359.8 kW
iv) β2 = 55.63°
Explanation:
We are given;
Stagnation pressure; T_01 = 290 K
Inlet velocity; C1 = 145 m/s
Cp for air = 1005 kJ/(kg·K)
Mach number; M = 0.96
Ratio of specific heats; γ = 1.4
Stagnation pressure; P_01 = 1 bar
rotational speed; N = 5500 rpm
Work done factor; τ = 0.92
Isentropic effjciency; η = 0.9
Stagnation temperature rise; ΔT_s = 22 K
i) Formula for Stagnation temperature is given as;
T_01 = T1 + C1/(2Cp)
Thus,making T1 the subject, we havw;
T1 = T_01 - C1/(2Cp)
Plugging in the relevant values, we have;
T1 = 290 - (145/(2 × 1005))
T1 = 289.93 K
Formula for the mach number relative to the tip is given by;
M = V1/√(γRT1)
Where V1 is relative velocity at the tip and R is a gas constant with a value of 287 J/Kg.K
Thus;
V1 = M√(γRT1)
V1 = 0.96√(1.4 × 287 × 289.93)
V1 = 0.96 × 341.312
V1 = 327.66 m/s
Now, tip speed is gotten from the velocity triangle in the image attached by the formula;
U_t = √(V1² - C1²)
U_t = √(327.66² - 145²)
U_t = √86336.0756
U_t = 293.83 m/s
Now relationship between tip speed and tip radius is given by;
U_t = (2πN/60)r_t
Where r_t is tip radius.
Thus;
r_t = (60 × U_t)/(2πN)
r_t = (60 × 293.83)/(2π × 5500)
r_t = 0.5101 m
ii) Now mean radius from derivations is; r_m = 1.5h
While relationship between mean radius and tip radius is;
r_m = r_t - h/2
Thus;
1.5h = 0.5101 - 0.5h
1.5h + 0.5h = 0.5101
2h = 0.5101
h = 0.5101/2
h = 0.2551
So, r_m = 1.5 × 0.2551
r_m = 0.3827 m
Formula for the area is;
A = 2πr_m × h
A = 2π × 0.3827 × 0.2551
A = 0.6134 m²
Isentropic relationship between pressure and temperature gives;
P1 = P_01(T1/T_01)^(γ/(γ - 1))
P1 = 1(289.93/290)^(1.4/(1.4 - 1))
P1 = 0.9992 bar = 0.9992 × 10^(5) N/m²
Formula for density is;
ρ1 = P1/(RT1)
ρ1 = 0.9992 × 10^(5)/(287 × 289.93)
ρ1 = 1.2 kg/m³
Mass flow rate at compressor inlet is;
m' = ρ1 × A × C1
m' = 1.2 × 0.6134 × 145
m' = 106.73 kg/s
iii) stagnation pressure ratio is given as;
R_s = (1 + ηΔT_s/T_01)^(γ/(γ - 1))
R_s = (1 + (0.9 × 22/290))^(1.4/(1.4 - 1))
R_s = 1.26
Work is;
W = C_p × ΔT_s
W = 1005 × 22
W = 22110 J/Kg
Power is;
P = W × m'
P = 22110 × 106.73
P = 2359800.3 W
P = 2359.8 kW
iv) We want to find the rotor angle.
now;
Tan β1 = U_t/C1
tan β1 = 293.83/145
tan β1 = 2.0264
β1 = tan^(-1) 2.0264
β1 = 63.73°
Formula for Stagnation pressure rise is given by;
ΔT_s = (τ•U_t•C1/C_p) × tan(β1 - β2)
Plugging in the relevant values;
22 = (0.92 × 293.83 × 145/1005) × (tan 63.73 - tan β2)
(tan 63.73 - tan β2) = 0.5641
2.0264 - 0.5641 = tan β2
tan β2 = 1.4623
β2 = tan^(-1) 1.4623
β2 = 55.63°
Consider the string length equal to 7. This string is distorted by a function f (x) = 2 sin(2x) - 10sin(10x). What is the wave formed in this string? a. In=12cos (nit ) sin(max) b. 2cos(2t)sin (2x) - 10cos(10t ) sin(10x) c. n 2 sin 2x e' – 10sin 10x e
Answers
Answer:
hello your question has a missing part below is the missing part
Consider the string length equal to [tex]\pi[/tex]
answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)
Explanation:
Given string length = [tex]\pi[/tex]
distorted function f(x) = 2sin(2x) - 10sin(10x)
Determine the wave formed in the string
attached below is a detailed solution of the problem
A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it to flow over the resistors, where it is heated. Air enters a 1400-W hair dryer at 100 kPa and 22°C and leaves at 47°C. The cross-sectional area of the hair dryer at the exit is 60 cm2. Neglect the power consumed by the fan and the heat losses through the walls of the hair dryer. The gas constant of air is R = 0.287 kPa·m3/kg·K. Also, cp = 1.007 kJ/kg·K for air at room temperature.
determine
(a) the volume flow rate of air at the inlet and
(b) the velocity of the air at the exit.
Answers
Answer:
a) volume flow rate of air at the inlet is 0.0471 m³/s
b) the velocity of the air at the exit is 8.517 m/s
Explanation:
Given that;
The electrical power Input W_elec = -1400 W = -1.4 kW
Inlet temperature of air T_in = 22°C
Inlet pressure of air p_in = 100 kPa
Exit temperature T_out = 47°C
Exit area of the dyer is A_out = 60 cm²= 0.006 m²
cp = 1.007 kJ/kg·K
R = 0.287 kPa·m3/kg·K
Using mass balance
m_in = m_out = m_air
W _elec = m_air ( h_in - h_out)
we know that h = CpT
so
W _elec = m_air.Cp ( T_in - T_out)
we substitute
-1.4 = m_air.1.007 ( 22 - 47 )
-1.4 = - m_air.25.175
m_air = -1.4 / - 25.175
m_ air = 0.0556 kg/s
a) volume flow rate of air at the inlet
we know that
m_air = P_in × V_in
now from the ideal gas equation
P_in = p_in / RT_in
we substitute our values
= (100×10³) / ((0.287×10³)(22+273))
= 100000 / 84665
P_in = 1.18 kg/m³
therefore inlet volume flowrate will be;
V_in = m_air / P_in
= 0.0556 / 1.18
= 0.0471 m³/s
the volume flow rate of air at the inlet is 0.0471 m³/s
b) velocity of the air at the exit
the mass flow rate remains unchanged across the duct
m_ air = P_in.A_in.V_in = P_out.A_out.V_out
still from the ideal gas equation
P_out = p_out/ RT_out ( assume p_in = p_out)
P_out = (100×10³) / ((0.287×10³)(47+273))
P_out = 1.088 kg/m³
so the exit velocity will be;
V_out = m_air / P_out.A_out
we substitute our values
V_out = 0.0556 / ( 1.088 × 0.006)
= 0.0556 / 0.006528
= 8.517 m/s
Therefore the velocity of the air at the exit is 8.517 m/s
Oil with a kinematic viscosity of 4 10 6 m2 /s fl ows through a smooth pipe 12 cm in diameter at 2.3 m/s. What velocity should water?
Answers
Answer:
Velocity of 5 cm diameter pipe is 1.38 m/s
Explanation:
Use following equation of Relation between the Reynolds numbers of both pipes
[tex]Re_{5}[/tex] = [tex]Re_{12}[/tex]
[tex]\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}[/tex]= [tex]\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}[/tex]
[tex]Re_{5}[/tex] = Reynold number of water pipe
[tex]Re_{12}[/tex] = Reynold number of oil pipe
[tex]V_{5}[/tex] = Velocity of water 5 diameter pipe = ?
[tex]V_{12}[/tex] = Velocity of oil 12 diameter pipe = 2.30
[tex]v_{5}[/tex] = Kinetic Viscosity of water = 1 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s
[tex]v_{12}[/tex] = Kinetic Viscosity of oil = 4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s
[tex]D_{5}[/tex] = Diameter of pipe used for water = 0.05 m
[tex]D_{12}[/tex] = Diameter of pipe used for oil = 0.12 m
Use the formula
[tex]\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}[/tex]= [tex]\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}[/tex]
By Removing square rots on both sides
[tex]{\frac{V_{5}XD_{5} }{v_{5}}}[/tex]= [tex]{\frac{V_{12}XD_{12} }{v_{12}}}[/tex]
[tex]{V_{5}[/tex]= [tex]{\frac{V_{12}XD_{12} }{v_{12}XD_{5}\\}}[/tex]x[tex]v_{5}[/tex]
[tex]{V_{5}[/tex]= [ (0.23 x 0.12m ) / (4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s) x 0.05 ] 1 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s
[tex]{V_{5}[/tex] = 1.38 m/s
A tube of diameter 3 cm and length 3 m has a water flow of 100 cm3/s. If the pollutant concentration in the water is constant at 2 mg/L, find the mass flux (mg/cm2-s) of pollutant through the tube due to advection.
Answers
Answer: the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s
Explanation:
Given that;
Diameter of tube = 3 cm, radius r = 1.5 cm
water flow is 100 cm³/s
pollutant concentration = 2 mg/L
first we find the rate of flow of pollutant
we know that
1 L = 1000 cm³
xL = 100 cm³
100Lcm³ = xL1000cm³
xL = 100/1000
xL = 1/10 L
so 100cm³ = 1/10 L
now pollutant concentration in 100 cm³ = 1/10L × 2mg/L = 0.2 mg
Rate of flow of pollutant = 0.2 mg/s
Mass flux density is the pollutant mass per unit time per unit area
so Area of tube = πr² = 3.14 × 1.5² = 7.065 cm²
So
Mass flux = 0.2 / 7.065
Mass flux = 0.0283 mg/cm².s
Therefore, the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s
A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is 0.0500. Liquid hexane condensate is recovered at a rate of 1.50 L/min.
(a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint : First calculate the molar flow rate of the condensate and note that the rates at which C6H14 and N2 enter the unit must equal the total rates at which they leave in the two exit streams.)
(b) What percentage of the hexane entering the condenser is recovered as a liquid?
Answers
Answer:
A. 72.34mol/min
B. 76.0%
Explanation:
A.
We start by converting to molar flow rate. Using density and molecular weight of hexane
= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17
= 988.5/86.17
= 11.47mol/min
n1 = n2+n3
n1 = n2 + 11.47mol/min
We have a balance on hexane
n1y1C6H14 = n2y2C6H14 + n3y3C6H14
n1(0.18) = n2(0.05) + 11.47(1.00)
To get n2
(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)
0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min
0.18n2-0.05n2 = 11.47-2.0646
= 0.13n2 = 9.4054
n2 = 9.4054/0.13
n2 = 72.34 mol/min
This value is the flow rate of gas that is leaving the system.
B.
n1 = n2 + 11.47mol/min
72.34mol/min + 11.47mol/min
= 83.81 mol/min
Amount of hexane entering condenser
0.18(83.81)
= 15.1 mol/min
Then the percentage condensed =
11.47/15.1
= 7.59
~7.6
7.6x100
= 76.0%
Therefore the answers are a.) 72.34mol/min b.) 76.0%
Please refer to the attachment .
If you make a mistake in polarity when measuring the value of DC voltage in a circuit with a digital VOM, what will happen? A. The meter will be damaged. B. The meter will read positive voltage only. C. The meter will display a negative sign. D. The meter will display OL which states an overload condition.
Answers
Answer:
C. The meter will display a negative sign.
Explanation:
If you use an analog voltmeter and you measure voltage with reverse polarity you will damage it. But in this case we are using a digital multimeter. This kind of multimeter is designed to be able to deal with positive and negative voltages
A smooth ceramic sphere (SG 5 2.6) is immersed in a fl ow of water at 208C and 25 cm/s. What is the sphere diameter if it is encountering (a) creeping motion, Red 5 1 or (b) transition to turbulence, Red 5 250,000
Answers
Answer:
a. 4[tex]\mu m[/tex]
b. 1 m
Explanation:
According to the question, the data is as follows
The Density of water at 20 degrees celcius is 1000 kg/m^3
Viscosity is 0.001kg/m/.s
Velocity V = 25 cm/s
V = 0.25 m/s
Now
a. The creeping motion is
As we know that
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
1 = (1,000 × 0.25 × d) ÷ 0.0001
d = (1 × 0.001) ÷ (1,000 × 0.25)
= 4E - 06^m
= 4[tex]\mu m[/tex]
b. Now the sphere diameter is
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
250,000 = (1,000 × 0.25 × d) ÷ 0.0001
d = (250,000 × 0.001) ÷ (1,000 × 0.25)
= 1 m
I need help please thank for the help on the last one <3
Answers
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
Answers
dutile is the correct answer
The structure of PF3(C6H5)2 is trigonal bipyramidal, with one equatorial and two axial F atoms which interchange positions when heated. Describe the low- and high- temperature 31P and 19F NMR spectra.
Answers
Answer:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
Explanation:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
Consider the following ways of handling deadlock: (1) banker’s algorithm, (2) detect
deadlock and kill thread, releasing all resources, (3) reserve all resources in advance,
(4) restart thread and release all resources if thread needs to wait, (5) resource ordering, and (6) detect deadlock and roll back thread’s actions.
a. One criterion to use in evaluating different approaches to deadlock is which
approach permits the greatest concurrency. In other words, which approach allows
the most threads to make progress without waiting when there is no deadlock?
Give a rank order from 1 to 6 for each of the ways of handling deadlock just listed,
where 1 allows the greatest degree of concurrency. Comment on your ordering.
b. Another criterion is efficiency; in other words, which requires the least processor
overhead. Rank order the approaches from 1 to 6, with 1 being the most efficient,
assuming that deadlock is a very rare event. Comment on your ordering. Does
your ordering change if deadlocks occur frequently?
who can answer part B for me?
Answers
Answer:
b
Explanation:
How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?
Answers
Answer:
hello your question is incomplete attached below is the missing part of the question
Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.
answer : Nd ∝ rt
Explanation:
Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases
Pactive ( active power ) = Efs * F
Pactive = [tex]\frac{q^2Nd^2*Xn^2}{6Eo} * f[/tex]
also note that ; Pactive ∝ Nd2 (
tD = K . [tex]\frac{Vdd}{(Vdd - Vt )^2}[/tex] since K = constant
Hence : Nd ∝ rt
The pascal is actually a very small unit of pressure. To show this, convert 1 Pa = 1 N/m² to lb/ft². Atmosphere pressure at sea level is 14.7 lb/in². How many pascals is this?
Answers
Answer:
pascals is this = 101352.972 Pa
Explanation:
given data
Atmosphere pressure at sea level = 14.7 lb/in²
we convert 1 Pa = 1 N/m² to lb/ft²
so we convert here 14.7 lb/in² to pascals
we know that 1 lb/ft² = 47.990172 N/m²
so
1 lb/ft² × ft²/(12in)² = 47.990172 × 144 N/m²
it will be simplyfy
1 lb/ft² = 6894.76 N/m²
so
14.7 lb/in² = 14.7 × 6894.76 N/m²
14.7 lb/in² = 101352.972 Pa
A cylinder 10 mm in diameter is pulled with a stress of 150 MPa. The diameter elastically decreased by 0.007 mm. Determine Poisson's ratio if the material has a elastic modulus of 100 GPa.
Answers
Answer:Poisson's Ratio,μ = 0.46
Explanation:
Poisson's Ratio is calculate as
μ = transverse/ longitudinal strain
μ = - εt / εl
where
μ = Poisson's ratio
εt = transverse strain
εl = longitudinal strain
Transverse strain can be expressed as
εt = change in diameter / initial diameter
where
εt =transverse strain
change in diameter=0.007mm
initial diameter = 10mm
εt =0.007mm/ 10mm= 0.0007
Longitudinal strain can be expressed as
εl=Stress/ elastic modulus = σ/ E
= Stress = 150 MPa , converting to GPa becomes 150/1000 = 0.15 GPa
εl= 0.15 GPa / 100 GPa= 0.0015
Poisson's Ratio,μ = transverse/ longitudinal strain
( 0.0007 /0.0015) = 0.46 =0.46
Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tiny parallel pipes of the same total cross-sectional area, 4.0 mm2. Volume flow is 1000 mm3/s. The pressure drop for fluid passing through the single pipe is lower than that through the 100 vessel array by a factor of:_______.
A. 10
B. 100
C. 1000
Answers
Solution:
Given that :
Volume flow is, [tex]$Q_1 = 1000 \ mm^3/s$[/tex]
So, [tex]$Q_2= \frac{1000}{100}=10 \ mm^3/s$[/tex]
Therefore, the equation of a single straight vessel is given by
[tex]$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$[/tex] ......................(i)
So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is
[tex]$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $[/tex]
or [tex]$d_1=10 \ d_2$[/tex]
Now for parallel pipes
[tex]$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$[/tex] ...........(ii)
Solving the equations (i) and (ii),
[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$[/tex]
[tex]$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$[/tex]
[tex]$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$[/tex]
[tex]$=\frac{10^6}{10^7}$[/tex]
Therefore,
[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$[/tex]
or [tex]$H_{f_2}=10 \ H_{f_1}$[/tex]
Thus the answer is option A). 10
Given a 12-bit A/D converter operating over a voltage range from ????5 V to 5 V, how much does the input voltage have to change, in general, in order to be detectable
Answers
Answer:
2.44 mV
Explanation:
This question has to be one of analog quantization size questions and as such, we use the formula
Q = (V₂ - V₁) / 2^n
Where
n = 12
V₂ = higher voltage, 5 V
V₁ = lower voltage, -5 V
Q = is the change in voltage were looking for
On applying the formula and substitutiting the values we have
Q = (5 - -5) / 2^12
Q = 10 / 4096
Q = 0.00244 V, or we say, 2.44 mV
A system samples a sinusoid of frequency 230 Hz at a rate of 175 Hz and writes the sampled signal to its output without further modification. Determine the frequency that the sampling system will generate in its output.
a. 120
b. 55
c. 175
d. 230