The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
[tex]$J=\frac{\pi}{32}d^4$[/tex]
[tex]$J=\frac{\pi}{32}\times (46)^4$[/tex]
J = 207.6 [tex]mm^4[/tex]
So the shear stress at point A is :
[tex]$\tau_A =\frac{Tc_A}{J}$[/tex]
[tex]$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$[/tex]
[tex]$\tau_A = 4913.29 \ MPa$[/tex]
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
Technician A says that all electric motors are DC motors. Technician B says that two types of brushless motors use AC current. Which technician is correct
Answer:
Technician B only.
Explanation:
It is not necessary that all electric motors will use Direct current, some may also use Alternative current. Some electric motors that use DC may use brushes. There are two types of brushless motors that use alternative current.
A network has three independent file servers, each with 90 percent reliability. The probability that the network will be functioning correctly (at least one server is working) at a given time is:
Answer:
The correct answer is "99.9%".
Explanation:
According to the information given in the question,
[tex]P(1 \ fail) = 0.1[/tex]
The probability of all fail will be:
[tex]P(all \ fail) = (0.1)^3[/tex]
[tex]=0.001[/tex]
hence,
[tex]P(not \ all \ fail)= 1-P(all \ fail)[/tex]
[tex]=0.999[/tex]
[tex]=99.9[/tex] (%)
Thus the above is the right answer.
the integral of In3x²/×⁵
If the integral is
[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx[/tex]
substitute u = ln(3x ²) and du = 6x/(3x ²) dx = 2/x dx.
Then x ² = exp(u)/3 and x ⁴ = exp(2u)/9.
The integral is transformed to
[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx = \int \frac{\ln(3x^2)}{2x^4} \times \dfrac2x \,\mathrm dx \\\\ = \int \frac{u}{2\times\dfrac{e^{2u}}9}\,\mathrm du \\\\ = \frac92 \int ue^{-2u}\,\mathrm du[/tex]
Integrate by parts:
[tex]f = u \implies \mathrm df = \mathrm du \\\\ \mathrm dg = e^{-2u}\,\mathrm du \implies g = -\dfrac12 e^{-2u}[/tex]
[tex]\displaystyle \int ue^{-2u}\,\mathrm du = fg - \int g\,\mathrm df \\\\ = -\dfrac12 ue^{-2u} + \displaystyle \frac12 \int e^{-2u}\,\mathrm du \\\\ = -\frac12 ue^{-2u} - \frac14 e^{-2u} + C[/tex]
Then
[tex]\displaystyle \frac92 \int ue^{-2u}\,\mathrm du = -\frac94 ue^{-2u} - \frac98 e^{-2u} + C[/tex]
which in terms of x would be
[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx = -\frac94\times\frac{\ln(3x^2)}{9x^4} - \frac98 \times \frac1{9x^4} + C \\\\ = \boxed{-\frac{\ln(3x^2)}{4x^4}-\frac1{8x^4}+C}[/tex]
A moving-coil instrument, which gives full-scale deflection with 0.015 A has a copper coil having resistance of 1.5 Ohm at 15°C and a temperature coefficient of 1/234.5 at 0 degree C in series with a swamp resistance 3.5 Ohm having a negligible temperature coefficient. Determine the resistance of shunt required for a full-scale deflection of 20 A and the resistant required for a full-scale deflection of 250 V. If the instrument reads correctly at 15°C, determine the percentage error in each case when the temperature is 25°C.
Answer: check answers in pictures (2 p)
The resistance of the shunt required for a full-scale deflection of 20 A is 2.528 Ohm.
What is resistance?Resistance is a measure of an electrical circuit's resistance to current flow. Resistance is measured in ohms, which is represented by the Greek letter omega.
Full-scale deflection current = 0.015 A
Full-scale deflection current for 20 A = 20/0.015 = 1333.33 times full-scale deflection current
The current through the meter coil is given by:
Ic = Im * (Rm / (Rm + Rs))
At full scale, Ic = 1333.33 * 0.015 A = 20 A, Rm = 1.5 Ohm, and Rs = 3.5 Ohm.
Therefore:
20 A = Im * (1.5 Ohm / (1.5 Ohm + 3.5 Ohm))
Im = 20 A * (3.5 Ohm / 5 Ohm)
Im = 14 A
The current through the shunt resistor is therefore:
Ish = Im - Ic
Ish = 14 A - 20 A
Ish = -6 A
Since the shunt resistor is in parallel with the meter, its resistance can be calculated using the following formula:
Rs = Rm * (Im / Ish - 1)
Substituting the values, we get:
3.5 Ohm = 1.5 Ohm * (14 A / (-6 A) - 1)
3.5 Ohm = 1.5 Ohm * (-1.333 - 1)
3.5 Ohm = 1.5 Ohm * (-2.333)
Rs = 1.5 Ohm * (14 A / (-6 A) - 1) / (-2.333)
Rs = 2.528 Ohm
Therefore, the resistance of the shunt required for a full-scale deflection of 20 A is 2.528 Ohm.
Now, let's find the resistance required for a full-scale deflection of 250 V:
Full-scale deflection current = 0.015 A
Resistance required for full-scale deflection of 250 V = 250 V / 0.015 A
Resistance required for full-scale deflection of 250 V = 16666.67 Ohm
This resistance is in parallel with the meter coil, so the total resistance in the circuit will be:
Rtotal = Rm || Rext
Rtotal = (Rm * Rext) / (Rm + Rext)
Substituting the values, we get:
Rtotal = (1.5 Ohm * 16666.67 Ohm) / (1.5 Ohm + 16666.67 Ohm)
Rtotal = 1.43 Ohm
This resistance is in series with the meter coil, so the total resistance in the circuit will be:
Rtotal = Rm + Rext + Rs
Substituting the values we get:
1.43 Ohm = 1.5 Ohm + 16666.67 Ohm + Rs
Rs = 1.43 Ohm - 1.5 Ohm - 16666.67 Ohm
Rs = -16666.74 Ohm
Thus, the calculated value of Rs is negative, it means that the shunt resistor must be used instead of a series resistor to achieve a full-scale deflection of 250 V.
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Which of the following is not a part sympathetic activation during the fight or flight response?
Answer:
Digestion functions become more active
Explanation:
I just took the text!
Given : x² + 200x = 166400 The current park is a square, and the addition will increase the width by 200 meters to give the expanded park a total area of 166,400 square meters To Find : the side length of the current square park. Solution: x² + 200x = 166400 => x(x + 200) = 166400 166400 = 320 * 520 => (320)(320 + 200) = 166400 => x = 320 side length of the current square park. = 320 m Learn More: Which expression is a possible leading term for the polynomial ... brainly.In/question/13233517
Answer:
320 m
Explanation:
To find the side length of the current park, x, we solve the quadratic equation for the area of the park
x² + 200x = 166400
x² + 200x - 166400 = 0
We multiply -166400 by x² to get -166400x². We now find the factors of 166400x² that will add up to 200x. These factors are -320x and 520x
So, we re-write the expression as
x² + 200x - 166400 = 0
x² + 520x - 320x - 166400 = 0
We write out the factors of the expression,
x² + 520x - 320x - 320 × 520 = 0
Factorizing the expression, we have
x(x + 520) - 320(x + 520) = 0
(x + 520)(x - 320) = 0
x + 520 = 0 or x - 320 = 0
x = -520 or x = 320
Since x is not negative, we take the positive answer.
So, x = 320 m
Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of 60 ft/s. The ratio of the exit area to the inlet area is 8. Assume that The air is the ideal gas model for the air and ignoring heat transfer, determine
Answer: Hello the question is incomplete below is the missing part
Question: determine the temperature, in °R, at the exit
answer:
T2= 569.62°R
Explanation:
T1 = 540°R
V2 = 600 ft/s
V1 = 60 ft/s
h1 = 129.0613 ( value gotten from Ideal gas property-air table )
first step : calculate the value of h2 using the equation below
assuming no work is done ( potential energy is ignored )
h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778
∴ h2 = 136.17 Btu/Ibm
From Table A-17
we will apply interpolation
attached below is the remaining part of the solution
Please help me:
Use the Node analysis to find the power of all resistors
Explain ROLAP, and list the reasons you would recommend its use in the relational database environment.
Answer:
ROLAP is a branch of OLAP that is used to contain Relational database ( RDB ). which is a very fast database ( quick process of queries )
Very fast to access and also fast in processing queries provides multidimensional view of data / supports multidimensional database schema with RDBMssupports large databasesExplanation:
ROLAP ( Relational On-line Analytical processing ) is a branch of OLAP that is used to contain Relational database ( RDB ).
Advantages of ROLAP ( reasons for the use of ROLAP )
Very fast to access ( fast in processing queries )provides multidimensional view of data / supports multidimensional database schema with RDBMssupports large databasesA rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
122 cm
182 cm
82 cm
22 cm
108.2 cm
Answer:
The new length of the rod is 182 cm.
Explanation:
Given that a rod that was originally 100-cm-long experiences a strain of 82%, to determine what is the new length of the rod, the following calculation must be performed:
100 x 1.82 = X
182 = X
Therefore, the new length of the rod is 182 cm.
Elliptic curve cryptography is considered as the latest and probably the one with a future. Having seen RSA in earlier modules, in which ways do YOU think elliptic cryptography is more advanced than RSA. You may read other material or get this information from the internet to answer this question. But make sure to provide necessary references when you do cite others.
Answer:
The answer is below.
Explanation:
Some of the ways, how I think elliptic cryptography is more advanced than RSA are the following:
1. ECC - Elliptic Curve Cryptography uses smaller keys for the same level of security, particularly at greater levels of security.
2. ECC can work well and at a faster rate on a small-capacity device compared to RSA
3. It uses offer speedier SSL handshakes that enhance security
4. It offers fast signatures
5. It allows signatures to be computed in two stages, which enables lower latency than inverse throughput.
6. Relatively quick encryption and decryption
The main water line into a tall building has a pressure of600kPaat 5mbelow ground level. Apump brings the pressure up so the water can be delivered at 200kPaat the top floor 150maboveground level. The volumetric flow rate is 0.01m3/s. Consider the liquid water as an incompressiblesubstance. Assume steady state operation, negligible changes in kinetic energy, and the water temperature remains constant. Determine the required pumpingpower [kW, hp].
Answer:
Required pump power = - 11.205 kW
Explanation:
Pi = 600 kPa
Zi ( initial height ) = - 5m
P2 = 200 kPa
Ze ( top floor above ground level ) = 150m
Flowrate = 0.01 m^3/s = 10 kg/s
Temperature = 10°C ( assumed value ) remains constant
calculate the value of pumping power required ( i.e. work done )
Applying the energy equation
Hi + 1/2(vi)^2 + gZi = He + 1/2(Ve)^2 + gZe + W
given negligible internal/kinetic energy difference in "h's = the pv terms
W = ( Pi Vi - Pe V2 ) + g( Zi - Z2 )
= ( (600 * 0.001) - (200 * 0.001) ) + (9.81/1000) ( - 5 - 150 )
= - 1.1205 Kg/kJ
required pump power = flowrate * -1.1205
= 10 * ( - 1.1205 ) = - 11.205 kW
Transients (surges) on a line can cause spikes or surges of energy that can damage delicate electronic components. A SPD device contains one or more ________________ than bypass and absorb the energy of the transient.
Answer:
I think ( MOV Metal oxide varistors )
Transients (surges) on a line can cause spikes or surges of energy that can damage delicate electronic components. A SPD device contains one or more MOV Metal oxide varistors than bypass and absorb the energy of the transient.
An ideal gas within a piston-cylinder assembly undergoes a Carnot refrigeration cycle. The isothermal compression occurs at 325 K from 2 bar to 4 bar. The isothermal expansion occurs at 250 K. Determine:
a. the coefficient of performance
b. the heat transfer to the gas during the isothermal expansion, in kj per kmol of gas
c. the magintude of the net work input, in kj per kmol of gas.
Answer:
a) [tex]\mu=3.3[/tex]
b) [tex]Q=1440.7KJ/Kmol[/tex]
c) [tex]W=1872.9KJ/Kmol[/tex]
Explanation:
From the question we are told that:
Initial Temperature [tex]T_1=325k[/tex]
initial Pressure [tex]P_1=2 bar[/tex]
Final Pressure [tex]P_2=4 bar[/tex]
iso-thermal expansion [tex]T_2=250k[/tex]
a)
Generally the equation for Coefficient of performance is mathematically given by
[tex]\mu=\frac{T_2}{T_1-T_2}[/tex]
[tex]\mu=\frac{250}{325-250}[/tex]
[tex]\mu=3.3[/tex]
b)
Generally the equation for Heat Expansion is mathematically given by
[tex]Q=RT_2 In(\frac{P_2}{P_1})[/tex]
Where
R=Gas constant
[tex]R=8.314462618[/tex]
Therefore
[tex]Q=8.314462618*250 In(\frac{4}{2})[/tex]
[tex]Q=1440.7KJ/Kmol[/tex]
c)
Generally the equation for work input is mathematically given by
[tex]W=RT_1 In(\frac{P_2}{P_1})[/tex]
[tex]W=8.314462618*250 In(\frac{4}{2})[/tex]
[tex]W=1872.9KJ/Kmol[/tex]
The coefficient of performance is 3.33, the heat transfer in the isothermal expansion is 1440.71kJ/K.mol and the work input is calculated as 1872.92kJ/K.mol
Given Data:
T1 = 325KP1 = 2 barP2 = 4 barT2 = 250KIsothermal expansion occurs at 250K.
a) The coefficient of performanceThis is calculated as
COP =[tex]\frac{T_2}{T_1-T_2}=\frac{250}{325-250} =3.33[/tex]
b) Heat Transfer in isothermal expansion[tex]Q = RT_2In(\frac{p_2}{p_1})[/tex]
Therefore; In isothermal process du = 0
R = 8.314 AkJ/K.mol
Q = 8.314 * 250 In(4/2)
Q = 1440.71kJ/K.mol
c) Work InputW[tex]_i_n[/tex]=[tex]RT_1In(\frac{p_2}{p_1})\\W_i_n=8.314*325In(4/2)\\W_i_n=1872.92kJ/K.mol[/tex]
The work input is 1872.92kJ/K.mol
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Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance heater with 84.5% efficiency supplies heat to the pot. How many minutes will it take to boil 6.03 kg of water?
Answer:
t = 6179.1 s = 102.9 min = 1.7 h
Explanation:
The energy provided by the resistance heater must be equal to the energy required to boil the water:
E = ΔQ
ηPt = mH
where.
η = efficiency = 84.5 % = 0.845
P = Power = 2.61 KW = 2610 W
t = time = ?
m = mass of water = 6.03 kg
H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)
[tex]t = \frac{1.362\ x\ 10^7\ J}{2205.45\ W}[/tex]
t = 6179.1 s = 102.9 min = 1.7 h
Apart from the type of emergency, what factors affect the decision on weather to evacuate or shelter in place?
Answer:
. 1. Type of building
2. Location of emergency
3. Extent of emergency
Explanation:
1. The kind of building in which people find themselves could be a factor that can be used to make this decision. a lot of buildings can be easily affected by disasters such as explosions or tornadoes the extent of the effect is dependent on how the building is constructed. in some situations it is better to shelter in, while in others it is best to evacuate.
2. Another factor to be considered is the location or area where this is happening it is good to consider this so that people can be safely moved given that help can be easily accessed or if best to stay in.
3. The last is the extent of what is happening. The risk involved is one way of making the choice to evacuate or to stay.
The factors that affect the decision on whether to evacuate or shelter in place include:
Type of building.Location of emergency.Extent of emergencyIt should be noted that the building where an individual lives play a vital role during emergencies. Buildings that have poor foundations can easily be affected during emergencies.
Another factor that should be considered is the location where the emergency is taking place. Lastly, the extent of the emergency can determine if the person should stay or not.
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In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2 and a Y value of 1.0. An in-service stress level of 200 MPa has been calculated. What is the length of a surface crack (in mm) that will lead to fracture
Answer:
[tex]l=24mm[/tex]
Explanation:
From the question we are told that:
Plane strain fracture toughness of [tex]T=55 MPa-m1/2[/tex]
Y value [tex]Y=1.0[/tex]
Stress level of[tex]\sigma =200 MPa[/tex]
Generally the equation for length of a surface crack is mathematically given by
[tex]l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2[/tex]
[tex]l=\frac{1}{3.142}(\frac{55}{1*200})^2[/tex]
[tex]l=0.024m[/tex]
Therefore
in mm
[tex]l=24mm[/tex]
Which of the following would make a column more likely to buckle?
Increase its ultimate compressive strength
Increase its length
Increase its Young’s modulus
Increase its cross-sectional area
Increase its area moment of inertia
True or false: Increasing a material’s ultimate compressive strength makes it less likely to crush under its own weight (assuming all other things equal).
True
False
True or false: The area moment of inertia of an object only depends on its shape, not its size.
True
False
Answer:
1) B: Increase its length
2) True
3) True
Explanation:
1) Columns are compressive members and are subjected to primarily compressive stresses.
Now, there is what we call slender Ness ratio in columns which is basically used to check the ability of a column to resist buckling.
The formula is;
Slenderness ratio = Effective length of column/radius of gyration
Thus, the longer the column the more the Slenderness ratio and the more likely it is to buckle.
Thus, increasing the length is what makes columns likely buckle.
2) Compressive strength is the ability of a material to withstand loads that may reduce size or make the material crush under load.
Now, increasing the compressive strength simply means more ability to withstand loads that may lead to crushing under load.
Ultimate compressive strength is the maximum amount of compressive stress that a material can take before it crushes under load. Thus, increasing the ultimate compressive strength means it is less likely to crush under its own weight.
3) There are different factors that affect moment of inertia and they are;
- the mass of the body
- axis of rotation of the body
- shape and size of the body.
However, for area moment of inertia, what is most relevant to us is the shape of the body in question since we are dealing with area and not how big it is.
Determine the resistance values for a voltage divider that must meet the following specifications:_______.
a. The current drawn from the source under unloaded condition is not to exceed 5 mA.
b. The sourcevoltage is to be 10 V, and the required outputs are to be 5 V and 2.5 V.
c. Sketch the circuit.
d. Deter-mine the effect on the output voltages if a load is connected to each tap one at a time.
Answer:
i) when circuit is unloaded : R1 + R2 = 2kΩ.
ii) when 5V output voltage is applied : R1 = 1 kΩ , R2 = 1 kΩ
iii) when 2.5 v output voltage is applied : R1 = 1500 Ω, R2 = 500 Ω
iv) when: R1 = 1 kΩ , R2 = 1 kΩ is connected in parallel output voltage < 5 V
When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V
Explanation:
Current drawn from source under loaded condition ≤ 5 mA
source voltage = 10 v , required output = 5 v , 2.5 v
attached below is the sketch of the circuit
Resistance values
i) when the circuit is unloaded
Req = R1 + R2 = 2 kΩ ( Req = Vs / I = 10 / 5*10^-3 = 2 kΩ )
ii) when output voltage = 5 v
we will apply voltage divider rule
R1 = 1 kΩ ,
R2 = 1 kΩ
iii) When the output voltage = 2.5 v
applying voltage divider rule
R1 = 1500 Ω
R2 = 500 Ω
iv) when the load is connected to each tap one at a time
i.e. when the resistance are in parallel
when: R1 = 1 kΩ , R2 = 1 kΩ is connected in parallel output voltage < 5 V
When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V
attached below is the detailed solution to the given problem
A force measuring instrument comes with a certificate of calibration that identifies two instrument errors and assigns each an uncertainty at 95% confidence over its range. Provide an estimate of the instrument design-stage uncertainty.
Resolution: 0.25 N
Range: 0 to 100 N
Linearity error: within 0.20 N over range
Hysteresis error: within 0.30 N over range
Answer:
[tex]U=\pm 0.382N[/tex]
Explanation:
From the question we are told that:
Resolution: 0.25 N
Range: 0 to 100 N
Linearity error: within 0.20 N over range
Hysteresis error: within 0.30 N over range
Generally the equation for Stage Uncertainty is mathematically given by
[tex]U=\sqrt{u_0^2+u_T^2}[/tex]
Where
[tex]u_0=Zero\ order\ uncertainty[/tex]
[tex]u_0=\pm 0.5*0.25[/tex]
[tex]u_0=\pm=0.125[/tex]
And
u_T=Total instrumental Uncertainty
[tex]u_T=\sqrt{l_e^2+h_e^2}[/tex]
Where
l_e=Error of linearity
h_e=Error due to hysteresis
Hence
[tex]u_T=\sqrt{0.20^2+0.30^2}[/tex]
[tex]u_T=\pm 0.36[/tex]
Therefore
[tex]U=\sqrt{(0.125)^2+0.36^2}[/tex]
[tex]U=\pm 0.382N[/tex]
A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
122 cm
182 cm
82 cm
22 cm
108.2 cm
Answer: (b)
Explanation:
Given
Original length of the rod is [tex]L=100\ cm[/tex]
Strain experienced is [tex]\epsilon=82\%=0.82[/tex]
Strain is the ratio of the change in length to the original length
[tex]\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm[/tex]
Therefore, new length is given by (Considering the load is tensile in nature)
[tex]\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm[/tex]
Thus, option (b) is correct.
Compute the minimum length of vertical curve that will provide 220 m stopping sight distance for a design speed of 110 km/h at the intersection of a -3.50% grade and a +2.70% grade.
i have made notes and saved it as a pdf u can take it to answer question and make ur concept good
The minimum length of vertical curve that will provide 220 m stopping sight distance is; 458.8 m
We are given;
Stopping sight distance; S = 220 m
Design Speed; V = 110 km/h
Intersection grade 1; G1 = +2.7
Intersection Grade 2; G2 = -3.5
From the AASHTO Table attached, we can trace the value of the radius of vertical curvature for the given stopping sight distance and design speed.From the table, at S = 220 m and V = 110 km/h, we can see that;
Radius of vertical curvature; K = 74
Now, the difference in grade given is;A = G1 - G2
A = 2.7 - (-3.5)
A = 2.7 + 3.5
A = 6.2
Formula for the minimum length of vertical curve is;L = KA
Thus;
L = 74 × 6.2
L = 458.8 m
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Why water parameters of Buriganga river vary between wet and dry seasons?
Explain.
Steam enters a nozzle operating at steady state at 20 bar, 2808C, with a velocity of 80 m/s. The exit pressure and temperature are 7 bar and 1808C, respectively. The mass flow rate is 1.5 kg/s. Neglecting heat transfer and potential energy, determine (a) the exit velocity, in m/s. (b) the inlet and exit flow areas, in cm2
Answer:
A) 600.8 m/s
B) (i) 22.5 cm^2 (ii) 7.11 cm^2
Explanation:
Given data :
P1 = 20 bar , T1 = 2808°C
P2 = 7 bar , T2 = 1808°C
mass flow rate = 1.5 kg/s
Using the superheated vapor region in Table A-4
h1 = 2976.4 KJ/kg , v1 = 0.1200 m^3/kg
h2 = 2799.1 KJ/kg , v2 = 0.2847 m^3/kg
A) calculate exit velocity ( m/s )
given that we are to neglect heat transfer and potential energy
V2 ( exit velocity ) = ( V1^2 + 2 (√h1 - h2 )
= [ (80)^2 + 2 ( √ 2976.4 - 2799.1 )
= 600.8 m/s
B) calculate the inlet and exit flow areas ( cm^2 )
i) Inlet flow area
A1 = ( m * v1 ) / V1
= ( 1.5 * 0.1200 )/ 80 = 22.5 cm^2
ii) exit flow area
A2 = ( m * v2 ) / V2
= ( 1.5 * 0.2847 ) / 600.8 = 7.11 cm^2
Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and a 1k ohm load are connected the DC output voltage is 14V. What is... The value of the capacitor? The value of the peak to peak ripple voltage?
Answer:
[tex]V_{pp}=2V[/tex]
Explanation:
Source Voltage [tex]V= 120V[/tex]
Frequency [tex]f=60Hz[/tex]
Peak output voltage [tex]Vp=15V[/tex]
Peak Output Voltage with filter [tex]V_p'=14V[/tex]
Generally the equation for Peak to peak voltage is mathematically given by
[tex]V_p'=V_p-\frac{V_{pp}}{2}[/tex]
Therefore
[tex]V_{pp}=2(V_p-v_p')[/tex]
[tex]V_{pp}=2(15-14)[/tex]
[tex]V_{pp}=2V[/tex]
In heavy traffic areas you should wave pedestrians across the street if there is no crosswalk
In heavy traffic areas, you should wave pedestrians across the street if there is no crosswalk: False.
What is a crosswalk?A crosswalk can be defined as the marked or specially paved part of a road that is characterized by heavy traffic, so as to enable pedestrians have right of way to cross the street because drivers are required by traffic law to stop for them.
However, a driver or other road users in heavy traffic areas shouldn't wave pedestrians across the street if there is no crosswalk
Read more on traffic laws here: https://brainly.com/question/22768531
A pinion and gear pair is used to transmit a power of 5000 W. The teeth numbers of pinion
and gear are 20 and 50. The module is 5 mm, the pressure angle is 20o
and the face width is 45 mm. The
rotational speed of pinion is 300 rev/min. Both the pinion and the gear material are Nitralloy 135 Grade2 with a hardness of 277 Brinell. The quality standard number Qv is 5 and installation is open gearing
quality. Find the AGMA bending and contact stresses and the corresponding factors of safety for a
pinion life of 109
cycles and a reliability of 0.98
Answer:
mark me as a brainleast
Explanation:
209781
zener shunt regulator employs a 9.1-V zener diode for which VZ = 9.1 V at IZ = 9 mA, with rz = 40 and IZK= 0.5 mA. The available supply voltage of 15 V can varyas much as ±10%. For this diode, what is the value of VZ0?For a nominal load resistance RL of 1 k and a nominal zenercurrent of 10 mA,what current must flow in the supply resistorR? For the nominal value of supply voltage, select a valuefor resistor R, specified to one significant digit, to provideat least that current. What nominal output voltage results?For a ±10% change in the supply voltage, what variationin output voltage results? If the load current is reduced by50%, what increase in VO results? What is the smallest valueof load resistance that can be tolerated while maintainingregulation when the supply voltage is low? What is the lowestpossible output voltage that results? Calculate values for theline regulation and for the load regulation for this circuit usingthe numerical results obtained in this problem.
Answer:
[tex]V_z=9.1v[/tex]
[tex]V_{zo}=8.74V[/tex]
[tex]I=10mA[/tex]
[tex]R=589 ohms[/tex]
Explanation:
From the question we are told that:
Zener diode Voltage [tex]V_z=9.1-V[/tex]
Zener diode Current [tex]I_z=9 .A[/tex]
Note
[tex]rz = 40\\\\IZK= 0.5 mA[/tex]
Supply Voltage [tex]V_s=15[/tex]
Reduction Percentage [tex]P_r= 50 \%[/tex]
Generally the equation for Kirchhoff's Voltage Law is mathematically given by
[tex]V_z=V_{zo}+I_zr_z[/tex]
[tex]9.1=V_{z0}+9*10^{-3}(40)[/tex]
[tex]V_{zo}=8.74V[/tex]
Therefore
[tex]At I_z-10mA[/tex]
[tex]V_z=V_{z0}+I_zr_z[/tex]
[tex]V_z=8.74+(10*10^{-3}) (40)[/tex]
[tex]V_z=9.1v[/tex]
Generally the equation for Kirchhoff's Current Law is mathematically given by
[tex]-I+I_z+I_l=0[/tex]
[tex]I=10mA+\frac{V_z}{R_l}[/tex]
[tex]I=10mA+\frac{9.1}{0}[/tex]
[tex]I=10mA[/tex]
Therefore
[tex]R=\frac{15V-V_z}{I}[/tex]
[tex]R=\frac{15-9.1}{10*10^{-3}}[/tex]
[tex]R=589 ohms[/tex]
A investor will invest in mutual fundwith a probability of 0.6, will invest in government fundwith a probability of 0.3, and will invest in both fundswith a probability of 0.15. Find the probability that the investor will invest in either mutual fundor government fund.
Answer:
0.75
Explanation:
From this question above we have the following information
A = probability of investment in mutual fund= 0.6
B = probability of investment in government fund = 0.3
C = probability of investing in both the mutual fund and the government= 0.15
Where to find the probability of this investor investing in either of these two
= Prob(a) + prob(b) - prob(c)
= 0.6 + 0.3 - 0.15
= 0.9 - 0.15
= 0.75
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Answer:
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Explanation: