Answer:
Activity sequence model = A1B0G1A0B0P3F16A1B0P1A0
Tn= 10(1 + 1 + 3 + 16 + 1 + 1) = 10(23) = 230 TMU (8.3 sec)
A 500 turn coil is wound on an iron core. When a 120Vrms 60Hz voltage is applied to the coil, the current is 1A rms. Neglect the resistance of the coil. Determine the reluctance of the core. Given that the cross-sectional area of the core is 5cm2 and the length is 20cm, determine the relative permeability of the core material.
Answer:
R = 7.854 x 10⁵ anpert turns / Wb
Relative permeability = 405.3
Explanation:
Detailed explanation is given in the attached document.
Answer: 85398.16, 405.28473.
Explanation:
We are given that the number of turns on the core is 500 2 , the cross sectional area is :
A=5cm^{2}({1meter}/{100cm})^{2} =0.0005meters^{2}
and the length of the core is l=20cm ({1meter}/{100cm})= 0.2meters .
In this solution, we are meant to neglect the resistance of the coil , and the current through the coil is I=1amPrms when the voltage applied across it is:
V=120voltsrms at f=60Hz. From this, We can calculate the inductance(L) whiof the coil (a coil have an inductance value of one Henry when an electromotive force of one volt is induced in the coil were the current flowing through the said coil changes at a rate of one ampere/second).
The natural frequency of the applied voltage is:
ω =2π f=2π(60)=120π{radians}/{second} .
The inductive reactance of the coil is equal to X=ω L=120π L . We then know that current is :
I=V/X
=I =20/120πL
L=120/120π
=1/π henries .
For reluctance (R) (which is a unit measuring the opposition to the flow of magnetic flux within magnetic materials and is analogous to resistance in electrical circuits). Looking at the relationship between inductance and reluctance . You will note that it is :
L=n^2/R .
We can use this relationship to find reluctance for our closed iron core coil :
L=1/π = 500^2/R
R=500^2π =785398.16{amps}/{volt-seconds}}
We can therefore use the other equation for reluctance.
R=1/μ A= 1/μA or μRA
To calculate the relative permeability of the core :
R=500^2π
=0.2/(4π ×10^-7)/ μR(0.0005)
μR=0.2/(4π ×10^-7)/ 500^2π(0.0005)
= 405.28473
An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tensile force of 1 kN is applied axially to this heated rod, what is the expected mechanical response
Answer: the metal will experience a strain of approximately 2.037Mpa
This strain is lesser than if the force was applied at room temperature.
This will reduce internal stress and increase some mechanical properties of the aluminum such as mechanical hardening.
Explanation:
Detailed explanation and calculation and comparison with equivalent tensile stretching at ordinary room temperature is compared.
Write a program named CheckZips that is used by a package delivery service to check delivery areas. The program contains an array that holds the 10 zip codes of areas to which the company makes deliveries. Prompt a user to enter a zip code, and display a message indicating whether the zip code is in the company’s delivery area.
Answer:
# list of 10 zip codes assigned to zip
zips = ["12789", "54012", "54481", "54982", "60007", "60103", "60187", "60188", "71244", "90210"]
# user is prompt to enter zip code and assigned to user_zip
user_zip = input("Enter your zip code: ")
# if else statement to check if user input is available for delivery
# if statement check if user zip is in zip, if it is, it display
# delivery is okay to specified zip
if user_zip in zips:
print("Delivery to {} ok.".format(user_zip))
# else it display no delivery to such zip code
else:
print("Sorry - no delivery to {}.".format(user_zip))
Explanation:
The question doesn't specify programming language to use. Since no programming language was stated, the problem was solved using Python3. List structure is the equivalent of array in Python.
Assumption was also made on the array holding 10 zip codes of areas to which the company make deliveries.
The program first initialized a list of 10 zip codes and assigned it to zips. Then it prompt the user to enter a zip code which is assigned to user_zip.
Then if-else statement is used to check if user inputted zip is available with the zips variable.
If it is available, "Delivery ok" is displayed to the user else "no delivery" is displayed to the user.
Water flows inside a smooth circular thin-walled tube of diameter D = 25 mm at a mass flow rate of 50 g/s. Outside of the tube, air moves in cross flow over the tube at a velocity of V = 20 m/s and a temperature of T[infinity] = 10°C. If the mean temperature of the water is Tm = 50°C, determine (a) The Darcy friction factor for the water flow inside the tube
Answer:
See explaination
Explanation:
We can say that that the The Darcy Friction factor or Equation is a theoretical equation that predicts the frictional energy loss in a pipe based on the velocity of the fluid and the resistance due to friction. It is used almost exclusively to calculate head loss due to friction in turbulent flow.
Please kindly check attachment for the step by step solution of the given problem.
Water flows with a velocity of 3 m/s in a rectangular channel 3 m wide at a depth of 3 m. What is the change in depth and in water surface elevation produced when a gradual contraction in the channel to a width of 2.6 m takes place? Determine the greatest contraction allowable without altering the specified upstream conditions.
Answer: new depth will be 3.462m and the water elevation will be 0.462m.
The maximum contraction will be achieved in width 0<w<3
Explanation:detailed calculation and explanation is shown in the image below
Estimate pressure drop for an estimate of pipe diameter Pressure drop is a function of flow rate, length, diameter, and roughness. Either iterative methods OR equation solvers are necessary to solve implicit problems. For a first guess of a 1 ft diameter pipe, what is the fluid velocity? V = 5.67 ft/s What is the Reynolds number? Re = 96014 What is the pipe relative roughness?
Answer:
Explanation:
By using Bernoulli's Equation:
[tex]\frac{P_1}{P_g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{P_g}+\frac{v_2^2}{2g}+z_2+f\frac{L}{D}\frac{v^2}{2g}[/tex]
where;
[tex]z_1 = z_2 \ and \ v_1 = v_2[/tex]
[tex]P_1 - P_2 = f \frac{L}{D}\frac{1}{2}\rho v^2[/tex]
[tex]P_1-P_2 = \frac{5 \ lb}{in^2}( 144 \frac{in^2}{ft^2})[/tex]
[tex]P_1-P_2 = 720 \frac{lb}{ft^2}[/tex]
[tex]V = \frac{Q}{A} \\ \\ V = \frac{6.684 \ ft^2/s}{\frac{\pi}{4}D^2} \\ \\V = \frac{8.51}{D^2}[/tex]
Density of gasoline [tex]\rho = 1.32 \ slug/ft^3[/tex]
Dynamic Viscosity [tex]\mu[/tex] = [tex]6.5*10^{-6} \frac{lb.s}{ft}[/tex]
[tex]P_1-P_2 = f \frac{L \rho V^2}{2D}[/tex]
[tex]720 = f \frac{L(100)(1.32)}{2D}(\frac{8.51}{D^2})^2[/tex]
D = 1.46 f
[tex]Re, = \frac{\rho VD}{\mu} = \frac{1.32 *\frac{8.51}{D^2} D}{6.5*10^{-6}}[/tex]
[tex]= \frac{1.72*10^6}{D}[/tex]
[tex]\frac{E}{D}= \frac{0.00015}{D}[/tex]
However; the trail and error is as follows;
Assume ; f= 0.02 → D = 0.667ft[tex]\left \{ {{Re=2.576*10^6} \atop {\frac{E}{D}=0.000225}} \right.[/tex] [tex]\right \{ {{f=0.014} \atop {\neq 2}}[/tex]
f = 0.0145 → D = 0.0428 ft [tex]\left \{ {{Re=4.018*10^6} \atop {\frac{E}{D}=0.00035}} \right.[/tex] [tex]\right \{ {{f=0.015} \atop {\neq 0.0145}}[/tex]
f = 0.0156 → D = 0.43 ft [tex]\left \{ {{Re=4.0*10^6} \atop {\frac{E}{D}=0.000348}} \right.[/tex] [tex]f = 0.0156[/tex]
∴ pipe diameter d = 0.43 ft
Given that:
D = 1 ft
[tex]V = \frac{Q}{A} \\ \\ V = \frac{6.684}{\frac{\pi}{4}(1)^2} \\ \\ V = 8.51 \ ft/s[/tex]
[tex]Re = \frac{\rho \ V \ D}{\mu } \\ \\ Re = \frac{1.32 *8.51 *1 }{6.5*10^{-6}}[/tex]
[tex]Re = 1.72 *10^6[/tex]
[tex]\frac{E}{D} = \frac{0.00015}{1} \\ \\ = 0.00015[/tex]
[tex]f = 0.0136[/tex]
[tex]P_2-P_1 = \frac{fL \rho V^2 }{2D}[/tex]
[tex]P_2-P_1 = \frac{0.036(100)(1.32)(8.51)^2 }{2*1}[/tex]
[tex]P_2-P_1 = 65 \frac {lb}{ft^2}[/tex] to psi ; we have:
[tex]P_2-P_1 = 0.45 \ psi[/tex]
A supply fan is operating at 30000 cfm and 4 inch of water with an efficiency of 50%. (a) Calculate the fan power at the current operating condition. (b) Calculate the pressure and power if the supply fan reduces its speed to deliver 20000 cfm.
Answer:
The pressure and power of fan is 1.77 and 11.18 Hp respectively.
Explanation:
Given:
Discharge [tex]Q_{1} = 30000[/tex] cfm
Pressure difference [tex]\Delta P = 4[/tex] inch
Efficiency [tex]\eta = 50\%[/tex]
(A)
From the formula of fan power,
[tex]P _{1} = \frac{Q \Delta P}{6356 \eta}[/tex]
[tex]P_{1} = \frac{30000 \times 4}{6356 \times 0.5}[/tex]
[tex]P_{1} = 37.76[/tex] Hp
(B)
Fan power and pressure is given by,
We know that pressure difference is proportional to the square of discharge.
[tex]\frac{\Delta p_{2} }{\Delta P_{1} } = (\frac{Q_{2} }{Q_{1} } ) ^{2}[/tex]
[tex]\Delta P_{2} = (\frac{20000}{30000} ) \times 4[/tex]
[tex]\Delta P_{2} = 1.77[/tex]
Fan power proportional to the cube of discharge.
[tex]\frac{P_{2} }{P^{1} } = (\frac{Q_{2} }{Q_{1} } )^{3}[/tex]
[tex]P_{2} = \ (\frac{20000}{30000} ) ^{3} \times 37.76[/tex]
[tex]P_{2} = 11.18[/tex] Hp
Therefore, the pressure and power of fan is 1.77 and 11.18 Hp respectively.
A 45-kg iron block initially at 280°C is quenched in an insulated tank that contains 100 kg of water at 18°C. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process. The specific heat of water at 25°C is cp = 4.18 kJ/kg·K. The specific heat of iron at room temperature is cp = 0.45 kJ/kg·K
Answer: −3.46kJ/K
Explanation:
From the question above, we have:
The mass of the block (m) = 45kg
The initial temperature of the block (T1) = 280∘C
The weight of the water (mw) = 100kg
The temperature of water (Tw) = 18∘C
Recall the energy balance equation,
ΔUI = −ΔUw
In this case ΔUI is the internal energy of the iron, while ΔUw is the internal energy of water.
[mcp (T2 − T1)]I = −[mcp (T2 − T1)]w
Here cp is the specific heat at constant pressure.
The specific heat of iron is (cp)I = 0.45kJ/kg⋅K, and the specific heat of water is (cp)w = 4.18kJ/kg⋅K.
Now, we substitute the values in above equation,
[45 × 0.45(T2 − 280)]I = −[100 × 4.18(T2 − 18)]w
[20.25(T2 − 280)] = −[418(T2 − 18)]
20.25T2 − 5,670 = −[418T2 − 7,524]
20.25T2 − 5,670 = −418T2 + 7,524
20.25T2 + 418T2 = 7,524 + 5,670
438.35T2 = 13,194
T2 = 30.1K
Recall, the expression to calculate the total entropy change is given as:
ΔStotal = ΔSI + ΔSw
ΔStotal = [mcpln(T2/T1)]I + [mcpln(T2/T1)]w
Now, we substitute the values in above equation,
ΔStotal = [45 × 0.45ln(297.6/553)]I + [100 × 4.18ln(297.6/291)]w
ΔStotal = −12.55 + 9.09
ΔStotal = −3.46kJ/K
Thus the total entropy change is −3.46kJ/K.
An activated sludge plant receive 5.0 MGD of wastewater with a BOD of 220 mg/L. The primary clarifier removes 35% of the BOD. The sludge is aerated for 6 hr. The food-to-microorganism ratio is 0.30. The recirculation ratio is 0.2. The surface loading rate of the secondary clarifier is 800 gal/day-ft2. The final effluent has a BOD of 15 mg/L. What are the (a) BOD removal efficiency of the activated sludge treatment processes (secondary BOD removal), (b) aeration tank volume, (c) MLSS, and (d) secondary clarifier surface area? If 0.5 pound of oxygen is required for each pound of BOD entering the aeration tank and the density of air is approximately 0.075 lb/ft3, and the air is 20.9% oxygen by volume, calculate air requirements per day.
Answer:
(a) BOD removal efficiency = 89.51%
(b) Aeration tank volume = 4732m³
(c) MLSS = 1706.669 mg/L
(d) Secondary clarifier surface area 6250ft²
(e) Air requirement = 12930.284 lb
Explanation:
See the attached file for explanation. This is continuation from page 3 of the attached file.
Air required = 172403.792 ft3
since, density of air = 0.075lb/ft3
air required = 0.075*172403.792 lb
= 12930.284 lb
A 2-m-long and 3-m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surface is 5 m. The atmospheric pressure is 95 kPa. Considering the atmospheric pressure, the hydrostatic force acting on the top surface of this plate is _____. Solve this problem using appropriate software.
Answer:
864 KN
Explanation:
Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.
Please kindly check attachment for the step by step solution of the given problem.
The aerodynamic behavior of a flying insect is to be investigated in a wind tunnel using a ten-times scale model. It is known that the insect’s velocity depends on its size (characteristic length L), wing flapping frequency ω, surrounding fluid’s density rho and viscosity μ. If the insect flaps its wings 50 times a second when flying at 1.25 m/s, determine the wind tunnel air speed and wing oscillation frequency required for dynamic similarit
Answer:
Explanation:
Write the equation for Reynolds number as follows:
Re = VL/v
For dynamic similarity,
(VL/v)m + (VL/v)p…… (1)
Since, the model and prototype are in same medium, the kinematic viscosity remains same.
From equation (1), we can write
(VL)m = (VL)p
Here, L represents length, and V is the velocity.
Re-write the equation as follows:
Vm = Lp/Lm x Vp
Substitute 1/8 for Lp/Lm and 1.5m/s for Vp .
Vm = 1/8 x 1.5
Vm = 0.1875m/s
Therefore, the wind tunnel air speed is 0.1875m/s.
A single lane highway has a horizontal curve. The curve has a super elevation of 4% and a design speed of 45 mph. The PC station is 105+00 and the PI is at 108+75.
What is the station of the PT?
Answer: 112 + 19.27
Explanation:
Super elevation is an inward transverse slope provided through out the length of the horizontal curve which ends up serving as a counteract to the centrifugal force and checks tendency of overturning. It changes from infinite radius to radius of a transition curve.
Super curve elevation (e) = 4%
4/100= 0.04
e= V^2/gR
Make R the subject of the formula.
egR= V^2
R= V^2/eg
V= 45mph
=45 × 0.44704m/s
=20.1168m/s
g (force due to gravity) =9.81
Therefore,
R= (20.1168)^2/9.81 × 0.04
= 1031.31m
Tangent Length( T) = PI - PC
Tangent Length= 10875 - 10500
=375m
T= R Tan(I/2)
375= 1031.31 × Tan(I/2)
I= 39.96
Also,
L= πRI/180
= 719.27m
Station PT= Stat PC+ L
10500 + 719.27
=11219.27
=112 + 19.27
A car hits a tree at an estimated speed of 10 mi/hr on a 2% downgrade. If skid marks of 100 ft. are observed on dry pavement (F=0.33) followed by 200 ft. on an unpaved shoulder (F=0.28), what is the initial speed of the vehicle just before the pavement skid was begun?
Answer:
[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]
Explanation:
The deceleration of the car on the dry pavement is found by the Newton's Law:
[tex]\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}[/tex]
Where:
[tex]a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g[/tex]
[tex]a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a_{1} = -3.040\,\frac{m}{s^{2}}[/tex]
Likewise, the deceleration of the car on the unpaved shoulder is:
[tex]a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g[/tex]
[tex]a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a_{2} = -2.549\,\frac{m}{s^{2}}[/tex]
The speed just before the car entered the unpaved shoulder is:
[tex]v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}[/tex]
[tex]v_{o} = 18.175\,\frac{m}{s}[/tex]
And, the speed just before the pavement skid was begun is:
[tex]v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}[/tex]
[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]
The initial speed of the vehicle just before the pavement skid was begun is 5284.65 ft/hr.
Dry pavement friction coefficient Fdry = 0.33
Length of skid marks on dry pavement ddry = 100 ft
Friction coefficient on unpaved shoulder Fshoulder = 0.28
Length of skid marks on unpaved shoulder = 200 ft
First, let's calculate the work done on dry pavement:
Work on dry pavement = Fdry × ddry = [tex]0.33 *100[/tex]
= 33 ft·lbf
Work on unpaved shoulder = Fshoulder × dshoulder
= [tex]0.28 * 200[/tex]
= 56 ft·lbf
Total work done = Work on dry pavement + Work on unpaved shoulder = 33 + 56
= 89 ft·lbf
Assuming the car's mass remains constant, and the final speed is 0, we have:
89 ft·lbf = (1/2)m × (10 mi/hr)²
Convert the final speed to feet per hour:
[tex]10 mi/hr = 10 × 5280/3600 = 5280 ft/hr[/tex]
Now, solve for the initial speed:
v = √((2 × 89 ft·lbf) / m)
v ≈ √((2 × 89) / (6.38 × 10⁻⁶)) ft/hr
v ≈ [tex]\sqrt{(27889055.9}[/tex] ft/hr
v ≈ 5284.65 ft/hr
2. Similar to problem 1, assume your computer system has a 32-bit byte-addressable architecture where addresses and data are each 32 bits. It has a 16K-byte (16,384 bytes) direct-mapped cache, but now the block size is 32 bytes. Answer the following question to observe how the design change impacts the cache size. [10pts]
Question:
The question is not complete. The question to answer was not added. See below the possible question and the answer.
a. How many blocks are in the cache with this new arrangement?
b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.
C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?
Answer:
(a) Number of blocks = 512 blocks
(b) Tag is 18
(c) Total size of the cache = 8388608 bytes
Explanation:
a .
block size = 32 bytes
cache size = 16384 bytes
No.of blocks = 16384 / 32
No.pf blocks = 512 blocks
b.
Total address size = 32 bits
Address bits = Tag + Line index +block offset
Block Size = 32 bytes.
So block size = 25 bytes.
Hence Offset is 5
No . of Cache blocks = 512 blocks = 29 blocks
Hence line offset is 9
We know that Address bits = Tag + Line index +block offset
So , 32 =tag+9+5
tag = 32-(9+5)
So Tag is 18
c.
Data bits = 32 bits
Tag=18 bits
Valid bit is 1 bit
so Total cache size = 25+218+20
= 223
=8388608 bytes
Technician A says a limited slip differential can redirect power from a drive wheel that is slipping to the wheel that has traction. Technician B says traction control can redirect power by applying the brake on a drive wheel that is slipping. Who is correct?
Answer:
Both Technician A, and Technician B are correct
Explanation:
The Traction control are found in those modern automobile, it's a part of the electronic stability control and it becomes active once the automobile get acceleration. It helps the tired of the car not to slip when the car speed up.
It functions by making the car wheel to stop spinning through the reduction of power that is transferred to the wheel i.e application of traction on the wheels of the car. when car is moving with acceleration on a road with with little friction, the Traction is used.
During raining or snow when the road become slippery , In the old cars that doesn't have traction control, the gas pedal is feathered. Which helps to function as traction control
5) Initially, the pressure and temperature of steam inside a solid capsule is at 100-pound force per square inch absolute (psia), and 600 degrees Fahrenheit (°F), respectively. Because heat is gradually removed from this container, the pressure inside the capsule drops by the amount of 10-pound force per square inch absolute (psia). Answer the following questions, A. (10 points) The change of entropy per unit mass between the initial and final states B. (5 points) The amount of heat transfer per unit mass for the process C. (5 points) Sketch the T-s diagram for the process, showing the associated values of the thermodynamic properties for states and 2 on your sketch.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the full step by step explanation to your question.
Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell wind profile you identified, calculate the 0-1 km and 0-6 km bulk wind shear values. This means we will find the difference between the surface wind (lowest wind barb on the sounding) and the speed of the wind at 1 km and 5 km. The atmospheric pressure at 1 km above sea level is typically very close to 850 mb. The pressure at 6 km above sea level is very close to 500 mb. Please calculate the 0-1 km and 0-6 km wind shear values in knots (kts). For simplicity, assume that the surface winds are due south easterly, the 850 mb winds are due southerly, and the 500 mb winds are due westerly. Show your work.
Answer:
See explaination
Explanation:
2. 0-1 km shear value: taking winds at 1000mb and 850 mb
15 kts south easterly and 50 kts southerly
Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts
3. 0-6 km shear value: taking winds at 1000 mb and 500 mb
15 kts south easterly and 40 kts westerly
Vector difference 135/15 and 270/40 will be 281/51 kts
please see attachment
Assume the following LTI system where the input signal is an impulse train (i.e.,x(t)=∑????(t−nT0)[infinity]n=−[infinity].a)Find the Fourier series coefficient of x(t). Then find its Fourier transform and sketch the magnitude and phase spectra.b)Sketch the magnitude and phase spectra of the output (i.e., |Y(????)|and∡Y(????)) if the system is a low-pass filter with H(????)={1|????|<3????020other????ise, where ????0=2πT0.c)Sketch the magnitude and phase spectra of the output(|Y(????)|and∡Y(????)) if the system is a high-pass filter with H(????)={1|????|>5????020other????ise, where ????0=2πT0.d)Sketch the magnitude and phase spectra of the outputif the system is a filter with H(????)=11+j????.
Answer:
See explaination
Explanation:
The Fourier transform of y(t) = x(t - to) is Y(w) = e- jwto X(w) . Therefore the magnitude spectrum of y(t) is given by
|Y(w)| = |X(w)|
The phase spectrum of y(t) is given by
<Y(w) = -wto + <X(w)
please kindly see attachment for the step by step solution of the given problem.
Create a shell script (utilities1.sh) that will print a menu of commands to execute. (a) The script will prompt the user for a number as input. The input will be read, the corresponding command will be executed, and you will see the output of the executed command on your screen. The script will then exit normally, and return a value of 0 (exit 0). See the sample run below.
Answer:
Explanation:
check the attached files for solution.
What is the frequency response of the stable, causal LTI system defined by the differential equation:fraction numerator d squared y (t )over denominator d t squared end fraction plus 6 fraction numerator d y (t )over denominator d t end fraction plus 2 y (t )equals fraction numerator d x (t )over denominator d t end fraction plus 4 x (t )Use Matlab syntax for your response, assuming w is the frequency vector. Make sure you use parentheses correctly (try plotting your code in Matlab)
Answer:
Explanation:
first convert difference equation to transfer function form,
apply laplase transform to difference equation
s2Y(s) + 6 * s * y(s) + 2 * y(s) = s * X(s) + 4 * X(s)
(s2 + 6s+2) * y(s) = (s+4)*X(s)Y s)
Lets write below code in matlab command prompt
lets use lodspace to create values from 10^-1 to 10^5 and use freqs to plot frequency response of above system with frequency w
>> n=[1 4];
>> d=[1 6 2];
>> w = logspace(-1,5);
>> freqs(n,d,w)
Attached is the written solution and the MATLAB diagram
3) A mixture of nitrogen and oxygen (xN2=0.7) behaves as an ideal gas mixture. 50 moles of this mixture at 1 bar and 25 °C are fed into an initially-empty, rigid, diathermal vessel causing the pressure in the vessel to reach 1 bar. Assuming the surroundings are also at 25 °C, calculate the heat transfer needed for the gas mixture in the vessel to be at 25 °C. At this temperature, Cp for nitrogen = 1.040 J/(g K) and Cp for oxygen = 0.918 J/(g K).
Answer:
435.032 kj
Explanation:
We can describe Heat transfer as a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems.
Please refer to the attached file for the detailed step by step solution of the given problem
A field sample of an unconfined aquifer is packed in a test cylinder. The length and diameter of the cylinder are 50 cm and 6 cm respectively. The field sample is tested for a period of three minutes under a constant head difference of 16.3 cm. As a result, 45.2 cm3 of water is collected at the outlet. Determine the hydraulic conductivity of the aquifer sample.
Answer:
0.09cm/sec
Explanation:
We are going to describe Hydraulic conductivity as a measure of the ease with which water flows through sediments, determining renewal rates of water, dissolved gases, and nutrients.
See attachment for a detailed solution.
The hydraulic conductivity of the given aquifer sample packed in a test cylinder is; 0.027 cm/s
What is the hydraulic Conductivity?
We are given;
Head Loss; H = 16.3 cm
Length of cylinder; L = 50 cm
Diameter of cylinder; d = 6 cm
radius; r = 6/2 = 3 cm
time; t = 3 minutes = 180 s
Volume; V = 45.2 cm³
Formula for the hydraulic gradient is;
i = H/L
i = 16.3/50
i = 0.326
Formula for volumetric rate of flow is;
Q = V/t
Q = 45.2/180
Q = 0.25 cm³/s
Formula for area is;
A = πd²/4
A = π × 6²/4
A = 9π
Formula for hydraulic conductivity is;
K = Q/(A * i)
K = 0.25/(9π * 0.326)
K = 0.027 cm/s
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An element representing maximum in-plane shear stress with the associated ________ normal stresses is oriented at an angle of _______ from an element representing the ________ stresses.
Answer: 1) tensile
2) 45 degree orientation
3) principal shear stress
Explanation:
An element representing maximum in-plane shear stress with the associated _tensile normal stresses is oriented at an angle of ____45_degree__ from an element representing the ____principal shear____ stresses.
Researchers compared protein intake among three groups of postmenopausal women: (1) women eating a standard American diet (STD), (2) women eating a lacto-ovo-vegetarian diet (LAC), and (3) women eating a strict vegetarian diet (VEG).GroupMean protein intake (mg)SDnSTD75910LAC571310VEG47176a.Without using MATLAB ANOVA functions, determine if there is a significant difference in mean protein intake between these groups.
Answer:
see explaination
Explanation:
The statistical procedure for comparing the 3 groups was the F test with 2 degrees of freedom in the numerator (groups - 1) and 23 df in the denominator (N total - groups)
It is calculated as SSB/2/(SSW/23) = 10.22754
As Fcrit =
from Excel, finv(.05,2,23) = 3.4221, we can reject the null hypothesis of no difference between groups.
SSW = the sum of std i ^ 2 * (n i - 1)
SSB = the sum of ni (mean i - mean)^2
2. From Excel, we get the pvalue from fdist(10.22754,2,23) = .000664
3. For LSD, we calculate (mean 1 - mean 2)/(s * sqrt(1/n1+1/n2))
This is the pooled s = sqrt(SSW/23)
Then, we found t crit from tinv(.05,23) = 2.068
Making use of the chart i made, We found significant differences between std and lac as well as std and veg, but no significant difference between lac and veg.
A plate in the shape of an isosceles triangle 3 feet high and 4 feet wide is submerged vertically in water, base doward, with the base 5 ft bellow the surface. Find the force exerted by the water on one side of the plate.
Answer:
The force exerted by the water on one side of the plate is F = 24*pg
Explanation:
From the given question, the first step to take is to find he force exerted by the water on one side of the plate.
Solution
Given that:
Let the pressure the at a depth of y ft be = pgy lb/Pa
the area of the atrip is given as = f(y)*delta(y) = 4/3*(y-2)delta(y)
Then
we combine with the range for y as = y E [2 , 5]
Thus,
F = 4/3*pg * integral from (2 , 5) [y(y-2)] dy
Recall that,
p = water density
g= gravity of acceleration
so,
F = 4/3*pg * integral from (2 , 5) [y^2 - 2y]dy]
F = 4/3*pg * [y^3/3 - y^2] [2 , 5]
F = 4/3*pg * [18]
Finally, F = 24*pg
A doubly drained specimen, 2.54 cm in height, is consolidated in the lab under an applied stress. The time for 50 % overall (or average) consolidation is 12 min. (a) Compute the cv value for the lab specimen. (b) How long will it take for the specimen to consolidate to an average consolidation of 90 %? (c) If the final consolidation settlement of the specimen is expected to be 0.43 cm, how long will it take for 0.18 cm of settlement to occur? (d) After 14 minutes, what percent consolidation has occurred at the middle of the specimen?
Answer:
Cv = 0.026 cm²/min
t = 52.60 min
v% = 41.86 %
tv = 0.1375
t = 8.53 min
v = 53.61 %
Explanation:
given data
height = 2.54 cm
50 % consolidation = 12 min
solution
we get here first Cv value that is express as
Tv = [tex]\frac{Cv\times t}{d^2}[/tex] .................1
here Tv for 50% is 0.196
put here value and we get
0.196 = [tex]\frac{Cv\times 12}{\frac{2.54}{2}^2}[/tex]
solve it we get
Cv = 0.026 cm²/min
and
for tv for 90 % consolidation is 0.848
put value in equation 1
0.848 = [tex]\frac{0.026\times t}{\frac{2.54}{2}^2}[/tex]
solve it we get t
t = 52.60 min
and
v% will be here is
v% = [tex]\frac{0.18}{0.43} \times 100[/tex]
v% = 41.86 %
and
tv = [tex]\frac{\pi }{4}\times \frac{4}{100}^2[/tex]
tv = 0.1375
so now put value in equation 1 we get
0.1375 = [tex]\frac{0.026 \times t}{\frac{2.54}{2}^2}[/tex]
solve it we get
t = 8.53 min
and
now put value of t 14 min in equation 1 will be
tv = [tex]\frac{0.026 \times 14}{\frac{2.54}{2}^2}[/tex]
t = 0.225 min
and v will be after 14 min
0.0225 = [tex]\frac{\pi }{4}\times \frac{v}{100}^2[/tex]
v = 53.61 %
You wish to filter out 60 Hz noise (which arises from electrical interference at the frequency of AC current in our electrical grid) using a simple RC circuit, which is one useful form has a transfer function:
G(s) = 1/RCS + 1
a. What value of the product RC should you choose so that amplitude at 60 Hz is attenuated by 90%?
b. For that value of RC, what is the largest frequency that is attenuated by less than 5%?
c. Turn in a Bode plot for your proposed system generated in Matlab.
Answer:
a. 0.02639
b. 1.98H
Explanation:
Please see attachment
Calculate the length of a metal cylinder while it is subjected to a tensile stress of 10,000 psi. You are given the following data: Original length = 1 in Original cross-sectional area = 0.1 in2 Yield strength, σy = 9 ksi Young’s modulus, E = 1000 ksi
Answer:
length of cylinder can not calculated
Explanation:
given data
tensile stress = 10,000 psi
Original length = 1
Original cross-sectional area = 0.1 in²
Yield strength, σy = 9 ksi
Young’s modulus, E = 1000 ksi
solution
we can see that here that applied stress is greater than yield stress of material that is express
1000 ksi > 9 ksi
so here hooks law and strain relation is not working
so length of cylinder can not calculated
as stress applied 10000 psi
Under 10,000 psi tensile stress, the metal cylinder elongates by 0.01 inches, resulting in a final length of 1.01 inches.
To calculate the elongation (change in length) of the metal cylinder under tensile stress, we can use Hooke's Law, which states that the elongation [tex](\( \Delta L \))[/tex] is directly proportional to the applied tensile stress [tex](\( \sigma \))[/tex] and the original length [tex](\( L_0 \))[/tex], and inversely proportional to the Young's modulus [tex](\( E \))[/tex]:
[tex]\[ \Delta L = \frac{\sigma \cdot L_0}{E} \][/tex]
Given:
- Original length [tex](\( L_0 \))[/tex] = 1 in
- Applied tensile stress [tex](\( \sigma \))[/tex] = 10,000 psi
- Young's modulus [tex](\( E \))[/tex] = 1000 ksi = 1,000,000 psi
Substitute the values into the formula:
[tex]\[ \Delta L = \frac{10,000 \times 1}{1,000,000} \][/tex]
[tex]\[ \Delta L = \frac{10,000}{1,000,000} \][/tex]
[tex]\[ \Delta L = 0.01 \, \text{in} \][/tex]
So, the elongation of the metal cylinder under the given tensile stress is 0.01 inches.
To find the final length, we add the elongation to the original length:
[tex]\[ \text{Final length} = \text{Original length} + \Delta L \][/tex]
[tex]\[ \text{Final length} = 1 + 0.01 \][/tex]
[tex]\[ \text{Final length} = 1.01 \, \text{in} \][/tex]
Therefore, the final length of the metal cylinder under the given tensile stress is 1.01 inches.
A hollow aluminum alloy [G = 3,800 ksi] shaft having a length of 12 ft, an outside diameter of 4.50 in., and a wall thickness of 0.50 in. rotates at 3 Hz. The allowable shear stress is 6 ksi, and the allowable angle of twist is 5°. What horsepower may the shaft transmit?
Answer:
Horse power = 167.84 hp
Explanation:
Horsepower is calculated using the formula;
P = T * w
See the attached file for the calculation
Using a set of values from 0 to 5, perform the following unions using union-by-size. Show the result of each union. When sizes are the same, make the second tree a child of the first tree. Notice the finds return roots, and the union will union the roots. union(find(0),find(1)) union(find(3),find(4)) union(find(5),find(1)) union(find(2),find(5)) union(find(3),find(2)) 10 points
Answer:
Explanation:
Please kindly go through the attached file for a step by step approach to the solution of this problem