Answer:
1)ΔL = 0.616 mm
2)Δd = 0.00194 mm
Explanation:
We are given;
Force; F = 52900 N
Initial length; L_o = 207 mm = 0.207 m
Diameter; d_o = 19.2 mm = 0.0192 m
Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²
Now, from Hooke's law;
E = σ/ε
Where; σ is stress = force/area = F/A
A = πd²/4 = π × 0.0192²/4
A = 0.00009216π
σ = 52900/0.00009216π
ε = ΔL/L_o
ε = ΔL/0.207
Thus,from E = σ/ε, we have;
61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)
Making ΔL the subject, we have;
ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)
ΔL = 0.616 × 10^(-3) m
ΔL = 0.616 mm
B) Poisson's ratio is given as;
υ = ε_x/ε_z
ε_x = Δd/d_o
ε_z = ΔL/L_o
Thus;
υ = (Δd/d_o) ÷ (ΔL/L_o)
Making Δd the subject gives;
Δd = (υ × d_o × ΔL)/L_o
We are given Poisson's ratio to be 0.34.
Thus;
Δd = (0.34 × 19.2 × 0.616)/207
Δd = 0.00194 mm
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
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
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
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
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 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?
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
Find the magnitude of the steady-state response of the system whose system model is given by dx(t)/dt+ x(t)-f(t), where f(t) 2cos8t. Keep 3 significant figures
This question is incomplete, the complete question is;
Find the magnitude of the steady-state response of the system whose system model is given by
dx(t)/dt + x(t) = f(t)
where f(t) = 2cos8t. Keep 3 significant figures
Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )
Explanation:
Given that;
dx(t)/dt + x(t) = f(t) where f(t) = 2cos8t
dx(t)/dt + x(t) = f(t)
we apply Laplace transformation on both sides
SX(s) + x(s) = f(s)
(S + 1)x(s) = f(s)
f(s) / x(s) = S + 1
x(s) / f(s) = 1 / (S + 1)
Therefore
transfer function = H(s) = x(s)/f(s) = 1/(S+1)
f(t) = 2cos8t → [ 1 / ( S + 1 ) ] → x(t) = Acos(8t - ∅ )
A = Magnitude of steady state output
S = jw
S = j8
so
A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )
A = 2/√65 = 0.2481
∅ = tan⁻¹( 1/1) = 45°
therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )
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.
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
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.
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
Which type of forming operation produces a higher quality surface finish, better mechanical properties, and closer dimensional control of the finished piece?A. Hot working.B. Cold working.
Answer:
Option B (Cold working) would be the correct alternative.
Explanation:
Cold working highlights the importance of reinforcing material without any need for heat through modifying its structure or appearance. Metal becomes considered to have been treated in cold whether it is treated economically underneath the material's transition temperature. The bulk of cold operating operations are carried out at room temperature.The other possibility isn't linked to the given scenario. Therefore the alternative above is the right one.
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
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
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?
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 person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric conditions, what is the maximum pressure on her hand? What would be the maximum pressure if the "car" were an Indy 500 racer traveling 200 mph?
Answer:
[tex]10.8\ \text{lb/ft^2}[/tex]
[tex]101.96\ \text{lb/ft}^2[/tex]
Explanation:
[tex]v_1[/tex] = Velocity of car = 65 mph = [tex]65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}[/tex]
[tex]\rho[/tex] = Density of air = [tex]0.00237\ \text{slug/ft}^3[/tex]
[tex]v_2=0[/tex]
[tex]P_1=0[/tex]
[tex]h_1=h_2[/tex]
From Bernoulli's law we have
[tex]P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}[/tex]
The maximum pressure on the girl's hand is [tex]10.8\ \text{lb/ft^2}[/tex]
Now [tex]v_1[/tex] = 200 mph = [tex]200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}[/tex]
[tex]P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2[/tex]
The maximum pressure on the girl's hand is [tex]101.96\ \text{lb/ft}^2[/tex]
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?
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
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.
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
Consider diodes in a rectifier circuit. Input voltage is sinusoidal with a peak of +/-10 V. Diode drop is 0.7 V. What is the PIV for each type rectifier 1. 0.7 V 2. 1.4 V 3. 10.7 V 4. 11.4 V Bridge rectifier 5. 19.3 V Full-wave rectifier 6. 8.6 V 7. 9.3 V Half-wave rectifier 8. 7.2 V 9. 12.1 V 10. 12.8 V 11. 10 V
Answer is given below:
Explanation:
Peak inverse voltage (PIV) can be defined as the maximum value of the reverse voltage of the diode, which is the maximum value of the input cycle when the diode is on. In reverse bias. Happens. 9.3V for braid rectifiers cut at 0.7The center tapered rectifier has 2 diodes in parallel so the maximum voltage is 2Vm so the answer to cut off the 0.7 voltage is19.3V. For a half wave rectifier it is Vm i.e. 10 V.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
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}$[/tex]
Thus the answer is option A). 10
Which of the following is an example of a tax
Answer:
A tax is a monetary payment without the right to individual consideration, which a public law imposes on all taxable persons - including both natural and legal persons - in order to generate income. This means that taxes are public-law levies that everyone must pay to cover general financial needs who meet the criteria of tax liability, whereby the generation of income should at least be an auxiliary purpose. Taxes are usually the main source of income of a modern state. Due to the financial implications for all citizens and the complex tax legislation, taxes and other charges are an ongoing political and social issue.
I dont know I asked this to
Explanation:
Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0.846 and 0.049, respectively? Why or why not?
Answer:
Yes it is possible.
Explanation:
This problem is about to possibility to have alloy of iron-carbon for which mass fraction of ferrite, [tex]$W_{\alpha} = 0.846$[/tex] and proeutectoid cementite, [tex]$W_{Fe_3C}=0.049$[/tex]
An alloy formation is possible when the composition values of the two alloy are equal.
Now writing the expression for the mass fraction of total ferrite, we have
[tex]$W_{\alpha}=\frac{C_{Fe_3C}-C_0}{C_{Fe_3C}-C_{\alpha}}$[/tex]
[tex]$0.846}=\frac{6.70-C_0}{6.70-0.022}$[/tex]
[tex]$5.649588 = 6.70 - C_0$[/tex]
[tex]$\therefore C_0 = 1.05 $[/tex] wt. % of C
Now write the expression for the mass fraction of the proeutectoid cementite :
[tex]$W_{Fe_3C}=\frac{C_1-0.76}{5.94}$[/tex]
[tex]$0.049=\frac{C_1-0.76}{5.94}$[/tex]
[tex]$C_1 = 1.05$[/tex] % wt. C
Since, [tex]$C_0 =C_1$[/tex], it is possible to have an alloy of iron - carbon.
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.
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.
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.
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
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.
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
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?
Explanation:
cooking of duck will cost 48000
by the help of the method of rate × A + €
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
dutile is the correct answer
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.
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
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?
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 .
Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.
Sieve size Weight retained (g) No. 4 59.5 No. 8 86.5 No. 16 138.0 No. 30 127.8 No. 50 97.0 No. 100 66.8 Pan 6.3
Solution :
Sieve Size (in) Weight retain(g)
3 1.62
2 2.17
[tex]$1\frac{1}{2}$[/tex] 3.62
[tex]$\frac{3}{4}$[/tex] 2.27
[tex]$\frac{3}{8}$[/tex] 1.38
PAN 0.21
Given :
Sieve weight % wt. retain % cumulative % finer
size retained wt. retain
No. 4 59.5 10.225% 10.225% 89.775%
No. 8 86.5 14.865% 25.090% 74.91%
No. 16 138 23.7154% 48.8054% 51.2%
No. 30 127.8 21.91% 70.7154% 29.2850%
No. 50 97 16.6695% 87.3849% 12.62%
No. 100 66.8 11.4796% 98.92% 1.08%
Pan 6.3 1.08% 100% 0%
581.9 gram
Effective size = percentage finer 10% ([tex]$$D_{20}[/tex])
0.149 mm, N 100, % finer 1.08
0.297, N 50 , % finer 12.62%
x , 10%
[tex]$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$[/tex]
[tex]$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$[/tex]
x = 0.2634 mm
Effective size, [tex]$D_{10} = 0.2643 \ mm$[/tex]
Now, N 16 (1.19 mm) , 51.2%
N 8 (2.38 mm) , 74.91%
x, 60%
[tex]$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$[/tex]
x = 1.6317 mm
[tex]$\therefore D_{60} = 1.6317 \ mm$[/tex]
Uniformity co-efficient = [tex]$\frac{D_{60}}{D_{10}}$[/tex]
[tex]$Cu= \frac{1.6317}{0.2643}$[/tex]
Cu = 6.17
Now, fineness modulus = [tex]$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$[/tex]
[tex]$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$[/tex]
= 4.41
which lies between No. 4 and No. 5 sieve [4.76 to 4.00]
So, fineness modulus = 4.38 mm
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.
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 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
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
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?
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
How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?
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
What overall material composition would be required to give a material made up of 50wt% mullite and 50wt% alumina at 1400°C?
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}