Letm(t) = sinc2(t) andmh(t) be its Hilbert transform. Denote the upper side band modulatedwaveform asxUSB(t) =Am(t) cos(2000πt)−Amh(t) sin(2000πt)
(a) Sketch the Fourier transforms ofm(t) andmh(t).
(b) Sketch the Fourier transformX(f) of the complex baseband ofxUSB(t) with respectto carrier frequency 1000.
Answers
Answer:
The Hilbert transform we use in mathematics it usually apply on a function consisting of a real variable and produces another function of a real variable.
Explanation:
Related Questions
A brick of 203 x 102 x 57 mm in dimensions is being burned in a kiln to 1100°C, and then allowed to cool in a room with ambient air temperature of 30°C and convection heat transfer coefficient of 5 W/m2·K. If the brick has properties of rho=1920 kg/m3,Crho= 790 J/Kg·K, and k = 0.90 W/m·K, determine the time required to cool the brick to a temperature difference of 5°C from the ambient air temperature.
Answers
Answer:
407 minutes
Explanation:
Step 1: Calculate the volume of the brick
[tex]V = 0.203 X 0.102 X 0.057[/tex]
V = 0.0012 m³
Step 2: Calculate the surface area of the brick
A= 2[(0.203 X 0.102) +(0.203 X 0.057) +(0.102 X0.057)] = 0.08 m²
Step 3: calculate the characteristic length
[tex]L_{C} =\frac{V}{A}[/tex]
[tex]L_{C} = \frac{0.0012}{0.08}[/tex] = 0.015 m
Step 4: calculate the biot number
[tex]B_{i} = \frac{hL_{c} }{k}[/tex]
[tex]B_{i} = \frac{5X0.015 }{0.9}[/tex] = 0.083
⇒Since [tex]B_{i}[/tex] ∠ 0.1, the lumped system analysis is applicable. Then cooling time is determined from
[tex]b = \frac{hA}{\rho c_{p}V } = \frac{h}{\rho c_{p}L_{c} }[/tex]
[tex]b = \frac{h}{\rho c_{p}L_{c} }[/tex]
[tex]b = \frac{5}{1920 X 790 X 0.015}[/tex]
b = 0.0002197 s⁻¹
[tex]\frac{T(t) -T_{o}}{T_{i} - T_{o}} =e^{-bt}[/tex]
[tex]\frac{5}{1100 - 30} =e^{-0.0002197t}[/tex]
Take natural log of both sides
-5.3659 = -0.0002197t
t = 24,424 seconds = 407 minutes
The required time "7 hours".
Air temperature:
Dimension of brick[tex]= 203\times 102\times 57 \ mm\\\\[/tex]
kiln Temperature [tex]\ T_t = 1100^{\circ}\ C \\\\[/tex]
Air temperature of Ambient [tex]\ T_{\infty} = 30^{\circ}\ C\\\\[/tex]
Heat transmission coefficient by convection:
[tex]\to h=5\ \frac{W}{m^2}\ K\\\\[/tex]
Properties of Bricks:
[tex]\to \rho = 1920\ \frac{kg}{m^3}\\\\ \to C_p = 790\ \frac{J}{kg-K}\\\\ \to k=0.9 \frac{W}{m-K}\\\\[/tex]
Calculating the temperature difference:
[tex]\to T_t -T_{\infty} = 5^{\circ}\ C\\\\[/tex]
Where t represents the amount of time needed to cool the brick for a temperature of
[tex]\to T_t = 35^{\circ}\ C\\\\[/tex]
We are familiar with the lumped system analysis for energy balance.
[tex]\to \ln(\frac{T_t-T_{\infty}}{T_i-T_{\infty}}) = \frac{hA_s}{\rho V C_p} t \\\\\\[/tex]
Calculating the brick surface area:
[tex]\to A_s = 2(ab + bc +ca)[/tex]
[tex]= 2(0.203\times 0.102 +0.102 \times 0.057 +0.057\times 0.203)\\ \\= 2(0.020706 +0.005814 +0.011571)\\\\ = 2 \times 0.038091 \\\\ =0.076182 \ m^2\\\\[/tex]
Calculating the volume:
[tex]\to V = abc[/tex]
[tex]= 0.057 \times 0.203 \times 0.102 \\\\ = 0.00118\ m^3\\\\[/tex]
We know that:
[tex]\to \ln(\frac{T_t-T_{\infty}}{T_i-T_{\infty}}) = - \frac{hA_s}{\rho V C_p} t \\\\\to \ln(\frac{5}{1100-30}) = \frac{5\times 0.076}{1920 \times 790 \times 0.00118 } t \\\\\to -5.3659=-2.128 \times 10^{-4}\ t\\\\\to t =\frac{-5.3659}{-2.128 \times 10^{-4}}\\\\\to t= 2.521423 \times 10^{4}\ s\\\\\to t=25214.23 \ s\\\\\to t=7.0039527778 \ h\\\\[/tex]
Therefore, the final answer is "7 hours".
Find out more information about air temperature here:
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A tool chest has 950 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B. The surface has a coefficient of friction of 0.3. Determine the value of the horizontal force P necessary to cause motion of the chest to the right, and determine if the motion is sliding or tipping. The value of P is N. The motion is .
Answers
The value of P needed to cause motion is 142.5 N, and it's a sliding motion.
Multiply the coefficient of static friction (0.3) by the normal force (475 N) to find the maximum static friction force.
Static friction equals
= [tex]0.3 * 475 N[/tex]
= 142.5 N.
To initiate motion, the applied force (P) must overcome static friction. Thus,
P=142.5 N.
If P was less than s, the chest wouldn't move. Since P equals fs, the motion is sliding. Thus, the value of P needed to cause motion is 142.5 N, and it's a sliding motion.
A steam turbine in a power plant receives 5 kg/s steam at 3000 kPa, 500°C. Twenty percent of the flow is extracted at 1000 kPa to a feed water heater, and the remainder flows out at 200 kPa. Find the two exit temperatures and the turbine power output.
Answers
Answer:
The temperature of the first exit (feed to water heater) is at 330.15ºC. The second exit (exit of the turbine) is at 141ºC. The turbine Power output (if efficiency is %100) is 3165.46 KW
Explanation:
If we are talking of a steam turbine, the work done by the steam is done in an adiabatic process. To determine the temperature of the 2 exits, we have to find at which temperature of the steam with 1000KPa and 200KPa we have the same entropy of the steam entrance.
In this case for steam at 3000 kPa, 500°C, s= 7.2345Kj/kg K. i=3456.18 KJ/Kg
For steam at 1000 kPa and s= 7.2345Kj/kg K → T= 330.15ºC i=3116.48KJ/Kg
For steam at 200 kPa and s= 7.2345Kj/kg K → T= 141ºC i=2749.74KJ/Kg
For the power output, we have to multiply the steam flow with the enthalpic jump.
The addition of the 2 jumps is the total power output.
The concrete canoe team does some analysis on their design and calculates that they need a compressive strength of 860 psi. They test several mix designs and chose one that has a sample mean of 900 psi with a standard error of the mean of 10 psi. The team wishes to be very conservative in their design and calculates a one-sided 99% confidence interval on the lower limit. What is this value?
Answers
Answer:
874 psi
Explanation:
Given a sample mean (x') = 900,
and a standard error (SE) = 10
At 99% confidence, Z(critical) = 2.58
That gives 99% confidence interval as,
x' ± Z(critical) x SE = 900 ± 2.58 x 10
The value of the lower limit is,
900 - 25.8 = 874.2
≈ 874 psi
A mitochondrial membrane complex consisting of ATP synthase, adenine nucleotide translocase (ATP–ADP translocase), and phosphate translocase functions in oxidative phosphorylation. Adenine nucleotide translocase, an antiporter located in the inner mitochondrial membrane, moves ADP into and ATP out of the matrix. Phosphate translocase is also located in the inner mitochondrial membrane. It transports H + ions and phosphate H 2 PO − 4 ions into the matrix. The energy derived from the movement of H + ions down an electrochemical gradient from the intermembrane space into the matrix is used to drive the synthesis of ATP. How many H + ions must be moved into the matrix for the synthesis of 1 ATP? number of H + ions:'
Answers
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
True/False
If a contractor chooses to use their own quality system, they must demonstrate compliance with the established military standard through formal third party certification. [Identify DoD policy regarding Basic Quality Systems and the role of ISO 9001.
Answers
Answer:
false jdbebheuwowjwjsisidhhdd
A gas mixture containing 85.0 mole% N2 and the balance n-hexane flows through a pipe at a rate of 100.0 m3/h. The pressure is 2.00 atm absolute and the temperature is 140.0°C.
a. What is the molar flow rate of the gas? kmol/h
b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? °C
c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75.0% of the hexane? °C
Answers
Answer:
a.6531.53 mole/hr
b. 32.76 degC
c. 3.78 deg.c
Explanation:A gas mixture containing 85.0 mole% N2 and the balance n-hexane flows through a pipe at a rate of 100.0 m3/h. The pressure is 2.00 atm absolute and the temperature is 140.0°C.
a. What is the molar flow rate of the gas? kmol/h
b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? °C
c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75.0% of the hexane? °C
Given V= Volume of gas mixture= 100m3/hr=10^5 Lt/hr P= 2.0 atm and T= 100 deg.c =100+273.15= 373.15K
n= moles of mixture= PV/RT , where R=0.08206 L.atm/mole.K
n= 2.0*10^5/0.08206*373.15)=6531.53 mole/hr
b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? A gas mixture containing 85 mole% N2 and the°C
Condensation begins at a point at which the partial pressure of vapor =vapor pressure of liquid at the given temperature
Partial pressure of hexane in the mixture= 0.15*2.0= 0.3 atm
so for saturation to begin, the vapor pressure shoudl correspond to 0.3 atm=30.39 Kpa
Antoine constant for Hexane
lnP (Kpa)= 13.82- 2696/(T-48.833) ( T is in K
ln(30.39 )= 13.8193- 2696/ (T-48.833)
, 3.414= 13,82-2696/(T-48.333)
2696/(T-48.833)= 13.82-3.414=10.405
T-46.833= 2696/10.414=259.08
259.08+46.833 K=305.91k
305.91k-273.15K
32.76C
c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75% of the hexane? °C?
Vapor present in the gas mixture= 25%, Its partial pressure=0.15* 0.25*2.0 =0.075 atm= 7.59Kpa
from ln(7.59)= 13.82- 2696/ (T-48.333)
2.02 = 13.82- 2696/(T-48.333)
2696/(T-48.333)= 11.179
T-48.333= 2696/11.79=228.60
T= 228.6+48.333= 276.93 K= 3.78 deg.c
Which of the following is the definition of keyspace? The range of values that construct a cryptosystem key. The process of applying an algorithm to cleartext (or plaintext) data, resulting in a ciphertext. Provides a stronger cryptographic result with a shorter key. Ensuring that only the intended recipient can read the data.
Answers
Answer:
A
Explanation:
A key space is a set of all arrangements of a key, when given a fixed key size.
It is the total number of possible values of keys in a cryptographic algorithm or other security measures such as a password.
A key which is a byte long (or 8 bits) would consist of 256 possible keys.
Explain why it was important to clean each piece of metal you used with sandpaper prior to doing the experiment. (2) What specific problem might you have encountered had you cleaned the metal surfaces witha soft tissue instead? On the basis of your results from this experiment, what observation(s) did you make that woul allow you to determine the place H2 would take in your relative activity series?
Answers
Answer: Metal has a oxide layer when in air. Before an experiment cleaning it with sand paper will remove the leyer of oxide and ensure truer results. Cleaning with tissue paper will not result in removing the layer of oxide.
Shows how active H2 can be.
Explanation:
Following are the responses to the given question:
For question 1:
Sandpaper must be used to polish the metallic surface utilized for the experimentation. Metals oxidize in the environment, forming an oxide coating on the surface thus rendering them unreactive. As a consequence, sand is required to remove this before continuing with the reaction.For question 2:
Then using soft tissue to cleanse a metal's surfaces, the oxides layer on top remains intact. The metal does not fully react underneath the response circumstances. As a consequence, the experimental data were untrustworthy and incorrect.For question 3:
Unless the metal inside the experiment can generate H2, it's much less reactive than Hydrogen. In the relative series, the metal gets put below the hydrogen.When creating the hydrogen, suggesting that metals are acting as a strong reduction agent. Most metals in the range above hydrogen act as oxidizing agents, while most metals below hydrogen act as reductants.Learn more:
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A 12-ft high retaining wall has backfill of granular soil with an internal angle of friction of 30 and unit weight of 125 pef. What is the Rankine passive earth pressure on the wall?
Answers
The Rankine passive earth pressure on a 12-ft high retaining wall with backfill of granular soil having an internal angle of friction of 30° and a unit weight of 125 pcf is calculated to be 500 psf.
Explanation:The question relates to calculating the Rankine passive earth pressure on a retaining wall that is 12-ft high with backfill of granular soil. The internal angle of friction (φ) provided is 30°, and the unit weight (γ) of soil is 125 pcf (pounds per cubic foot). First, to calculate the passive earth pressure (Ψp), we use the Rankine theory formula: Ψp = γh [1 - sin(φ)]/[1 + sin(φ)], where h is the height of the wall. Substituting the given values, Ψp = 125 * 12 * [1 - sin(30°)]/[1 + sin(30°)]. Since sin(30°) = 0.5, the calculation simplifies to: Ψp = 125 * 12 * [1 - 0.5]/[1 + 0.5], which further simplifies to Ψp = 125 * 12 * 0.5/1.5. Therefore, the Rankine passive earth pressure on the wall amounts to Ψp = 500 psf (pounds per square foot).
Marco is a franchisee with Daggies, a chain of sandwich shops. His business was doing well until several Daggies franchisees got in trouble and were forced to close their shops. Soon afterward, Marco's business deteriorated and he too was forced to close. This is an example of:
Answers
Answer:
The coattail effect
Explanation:
Clearly, the Daggies franchise were a huge part of Marco's success and attracted many customers for the business. This is why Marco deteriorated as soon as many Daggies franchisees closed.
The coattail effect is the phenomenon where an influencing member in a party (franchisee in this case) contributes largely to the success of another, which is the case with Marco and Daggies
"It is better to be a human being dissatisfied than a pig satisfied; better to be Socrates dissatisfied than a fool satisfied. And if the fool, or the pig, are of a different opinion, it is because they only know their own side of the question. The other party to the comparison knows both sides." This is a very famous quotation from Mill. How would you explain what it says to someone else?
Answers
Explanation:
It may be instructive to look at the opposite of the sentence here. Perhaps the smarter creature would be more unhappy when frustrated, recognizing how it gains from happiness relative to the creature with less experience and less knowledge of a situation that does not define it at the moment.
Perhaps the argument is really about the fact that wisdom helps one to hypothetically live in multiple states and a lack of wisdom prevents or fails this possibility.
The form of mechanical weathering that occurs when a magma chamber starts expanding after the overlying volcano has been removed is known as what?
Answers
Answer:
Exfoliation.
Explanation:
Exfoliation is a form of chemical weathering that occurs when sheets of rocks peel from a massive rock's surface. It is also the unloading and sheeting of stress in a rock that expands the magma chamber.
Refrigerant-134a enters a compressor as a saturated vapor at 160 kPa at a rate of 0.03 m3/s and leaves at 800 kPa. The power input to the compressor is 10 kW. If the surroundings at 20°C experience an entropy increase of 0.008 kW/K, determine (a) the rate of heat loss from the compressor, (b) the exit temperature of the refrigerant, and (c) the rate of entropy generation.
Answers
Answer:
a) [tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]
b) T_2 = 36.4 degree C
c) [tex]\Delta S_{gen} = 0.006512 KW/K[/tex]
Explanation:
Given data:
[tex]P_1 = 160 kPa[/tex]
volumetric flow [tex]V_1 = 0.03 m^3/s[/tex]
[tex]P_2 = 800 kPa[/tex]
power input [tex]W_{in} = 10 kW[/tex]
[tex]T_{surr} = 20 degree C[/tex]
entropy = 0.008 kW/K
from refrigerant table for P_1 = 160 kPa and x_1 = 1.0
[tex] v_1 = 0.12355 m^3/kg[/tex]
[tex]h_1 = 241.14 kJ/kg[/tex]
[tex]s_1 = 0.94202 kJ/kg K[/tex]
a) mass flow rate [tex] \dot m = \frac{V_1}{v_1}[/tex]
[tex]\dot m = \frac{0.03}{0.12355} = 0.2428 kg/s[/tex]
heat loss[tex] = T_{surr} \times entropy[/tex]
heat loss[tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]
b) from energy balance equation
[tex]W_{in} 0 \dot Q_{out} = \dot m (h_2 -h_1}[/tex]
[tex]10 - 2.344 = 0.2428 (h_2 - 241.14}[/tex]
[tex]h_2 = 272.67 kJ/kg[/tex]
from refrigerant table, for P_2 = 800 kPa and h_2 = 272.67 kJ/kg
T_2 = 36.4 degree C
c) from refrigerant table P_2 = 800 kPa and h_2 = 272.67 kJ /kg
[tex]s_2 = 0.93589 kJ/kg K[/tex]
rate of entropy
[tex]\Delta S_R = \dot m =(s_2 -s_1)[/tex]
[tex]\Delta S_R = 0.2428 \times (0.93589 -0.94202) = - 0.0014884 kW/K[/tex]
rate of entropy for entire process
[tex]\Delta S_{gen} = \Delta _S_R + \Delta_{surr}[/tex]
[tex]\Delta S_{gen} = 0.0014884 + 0.008 = 0.006512 KW/K[/tex]
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6660 N (1500 lbf). If the length of the rod is 380 mm (15.0 in.), what must be the diameter to allow an elongation of 0.50 mm (0.020 in.)?
Answers
Answer:
7.65 mm
Explanation:
Stress, [tex]\sigma=\frac {F}{A}[/tex] where F is the force and A is the area
Also, [tex]\sigma=E\times \frac {\triangle L}{L}[/tex]
Where E is Young’s modulus, L is the length and [tex]\triangle L[/tex] is the elongation
Therefore,
[tex]\frac {F}{A}= E\times \frac {\triangle L}{L}[/tex]
Making A the subject of the formula then
[tex]A=\frac {FL}{E\triangle L}=\frac {6660\times 380}{110\times 10^{9}\times 0.5}=4.60145\times 10^{-5} m^{2}[/tex]
Since [tex]A=\frac {\pi d^{2}}{4}[/tex] then
[tex]d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4\times 4.60145\times 10^{-5}}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm[/tex]
Describe an E2 mechanism
A. Not stereospecific and not concerted
B. Stereospecific and not concerted
C. Stereospecific and concerted
D. Not stereospecific and concerted
Answers
Answer:
The answer is C): Stereospecific and concerted.
Explanation:
An E2 mechanism is a single step elimination reaction that is both stereospecific and concerted: it is stereospecific because it can convert the components of the reaction into stereoisomeric products; it is concerted because all bonds are broken and formed during the single step.
Your department's laser printer recently began printing a vertical black line near the edge of every printed page. What should you do to resolve the problem?
Answers
Answer:
Replace the toner cartridge
Explanation:
solution
when laser printer print black color liner vertical line print it is very frustrating that condition nearly empty toner cartridge
but first we clean the corona wire for that color line
and Reinstall the toner cartridges after Shake the toner cartridge side to side
if problem not solve than Clean the drum unit
and finally if not solve change the toner cartridge
so as that our problem will be resolve
A) A cross-section of a solid circular rod is subject to a torque of T = 3.5 kNâ‹…m. If the diameter of the rod is D = 5 cm, what is the maximum shear stress? include units.B) The maximum stress in a section of a circular tube subject to a torque is Tmax = 37 MPa . If the inner diameter is Di = 4.5 cm and the outer diameter is Do = 6.5 cm , what is the torque on the section? include units.
Answers
Answer:
[tex]\tau_{max} = 142.6[/tex] MPa
T = 1536.8 N m
Explanation:
Given data:
Torque = 3.5 k N m = 3.5*10^3 N.m
Diameter D = 5 cm = 0.05 m
a) from torsional equation we have
[tex]\frac[T}{J_{solid}} = \frac{\tau_{max}}{D/2}[/tex]
[tex]\frac{T}{\pi/32 D^4} = \frac{\tau_{max}}{D/2}[/tex]
solving for [tex]\tau_{max}[/tex]
[tex]\tau_{max} = \frac{16 T}{\pi D^3} =\frac{16 \times 3.5*10^3}{\pi 0.05^3}[/tex]
[tex]\tau_{max} = 142.6[/tex] MPa
B)
[tex]\tau = 37 MPa = 37 \times 10^6[/tex] Pa
[tex]D_i = 4.5 cm = 0.045[/tex] m
[tex]D_o = 6.5 cm = 0.065[/tex] m
[tex]\frac{T}{J_{hollow}} = \frac{\tau_{max}}{D_o /2}[/tex]
[tex]\frac{T}{(\pi/32) (0.065^4 - 0.045^4)} =\frac{37*10^6}{0.065/2}[/tex]
T = 1536.8 N m
Hydrogen peroxide, H2O2, enters a gas generator at 25 Celsius, 500 kPa, at the rate of 0.1 kg/s and is decomposed to steam and oxygen exiting at 800 K, 500 kPa. The resulting mixture is expanded through a turbine to atmospheric pressure, 100 kPa. Determine the power output of the turbine and the heat transfer rate in the gas generator. The enthalpy of formation of liquid H2O2 is −187 583 kJ/kmol.
Answers
Take a look at the pictures that should help you out.
Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity is 528 m/s. Determine the static pressure and temperature of the air at this state. The properties of air at an anticipated average temperature of 600 K are cp = 1.051 kJ/kg·K and k = 1.376.
Answers
To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as
[tex]T_0 = T+\frac{V^2}{2c_p}[/tex]
Where
T = Static temperature
V = Velocity of Fluid
[tex]c_p =[/tex] Specific Heat
Re-arrange to find the static temperature we have that
[tex]T = T_0 - \frac{V^2}{2c_p}[/tex]
[tex]T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})[/tex]
[tex]T = 672.88K[/tex]
Now the pressure of helium by using the Adiabatic pressure temperature is
[tex]P = P_0 (\frac{T}{T_0})^{k/(k-1)}[/tex]
Where,
[tex]P_0[/tex]= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that
[tex]P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}[/tex]
[tex]P = 0.399Mpa[/tex]
Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.
The simple majority decision rule may generate results that are
Answers
Answer:
Efficient when the marginal benefits of project = marginal costs of project.
Explanation:
Majority Decision Rule:
Majority decision rule is based on the notion of equality. An alternative is selected which has majority of votes. The simple majority decision rule may generate efficient results if the marginal benefits of a project are equal or greater than the marginal costs of the project.
Ruth owns Skyview, a complex under construction that will include commercial and residential suites, and a parking garage. She allows Town Contracting, the contractor, to complete a stage of the project late. Thiswaives Ruth's right to sue for Select one: a. none of the choices. b. any subsequent breaches. c. any past breaches. d. this delay.
Answers
Answer:
Option D
This delay
Explanation:
The contract between Ruth and the contractor is assumed to be oral and since the contractor is supposed to complete a stage of the project when date for completion is over. Ruth can sue the contractor for the delay of the project. The contractor may consequently pay an interest for the delay of the project to Ruth.
A 10 m3rigid insulated tank contains air at 1000 kPa, 1000 K. It is connected to an adiabatic reversible turbine that has an exit of 100 kPa.
The air is now allowed to run out through the turbine until a tank pressure of 100 kPais reached and the flow stops.
a) Find the initial and final mass in the tank.
b) How much work did the turbine produce.
Answers
Answer:
a) initial mass = 34.85 gm, final mass = 3.48gm
b)
Explanation:
At the initial state, the:
Pressure, p1 = 1000kPa,
Temperature, T1 = 1000K,
Volume, V1 = 10 m³
The mass of air in the tank can be evaluated using;
m1 = (p1 x V1) /(R x T1)
m1 = (1000 x 10)/(287 x 1000)
Therefore, m1 = 34.85g
At the final state,
Pressure, p2 = 100kPa,
Temperature, T2 = 1000K,
Volume, V2 = 10 m³
The mass of air in the tank is given by;
m2 = (p2 x V2) / R x T2
m2 = (100 x 10)/ 287 x 1000
Therefore m2 = 3.48 gm
The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without using trigonometric identities. However, you cannot use the phasor technique in all cases. Select the expressions below for which the phasor technique cannot be used to combine the sinusoids into a single expression.
Check all that apply.
O 45 sin(2500t – 50°) + 20 cos(1500t +20°)
O 25 cos(50t + 160°) + 15 cos(50t +70°)
O 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)
O -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
O 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
Answers
The phasor technique for combining sinusoidal functions into a single expression cannot be used when the frequencies within the same expression differ. The expressions 45 sin(2500t – 50°) + 20 cos(1500t +20°) and -100 sin(10,000t +90°) + 40 sin(10,100t – 80°) + 80 cos(10,000t) both contain differing frequencies, and therefore cannot be combined using the phasor technique.
Explanation:In the phasor technique, for the method to be applicable, all sinusoidal functions being combined should have the same frequency. Looking at the expressions given, the phasor technique cannot be applied to these:
45 sin(2500t – 50°) + 20 cos(1500t +20°) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)The reasoning for these exceptions is due to discrepancies in the frequencies of the sinusoidal functions within the same expression; in these two cases, the frequencies 2500t and 1500t, as well as 10,000t and 10,100t, do not match, making the phasor technique inapplicable.
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Consider a model of a wing-body shape mounted in a wind tunnel. The flow conditions in the test section are standard sea-level properties with a velocity of 100 m/s. The wing area and chord are 1.5 m2 and 0.45 m, respectively. Using the wind tunnel force and moment-measuring balance, the moment about the center of gravity when the lift is zero is found to be -12.4 N-m. When the model is pitched to another angle of attack, the lift and moment about the center of gravity are measured to be 3675 N and 20.67 N-m, respectively. Calculate the value of the moment coefficient about the aerodynamic center and the location of the aerodynamic center.
Answers
Answer:
A. -0.003
B. 0.02
Explanation:
Step 1: identify the given parameters
Giving the following parameters
Wing area (S)= 1.5 m²
Wing chord (C) = 0.45 m
Velocity (V) = 100 m/s
moment about center of gravity(Mcg) = -12.4 N-m
at another angle of attack, L = 3675 N and Mcg = 20.67 N-m
Step 2: calculate the value of the moment coefficient about the aerodynamic center (Cmcg)
[tex]q_{∞} =\frac{1}{2}\rho*v^{2}[/tex]
[tex]q_{∞} =\frac{1}{2}1.225*100^{2}[/tex]= 6125 N/m²
[tex]C_{mcg,w} =\frac{M_{cg,w} }{q_{∞}*S*C }[/tex]
[tex]C_{mcg,w} =\frac{-12.4}{6125*1.5*0.45 }[/tex] = -0.003
[tex]C_{mcg,w}= C_{ac,w}= -0.003[/tex] at zero lift
Step 3: calculate coefficient of lift
Cl = L/q*s
Cl = 3675/6125*1.5 = 0.4
Step 4: calculate the location of the aerodynamic center
New moment coefficient about the aerodynamic center (Cmcg):
[tex]C_{mcg} =\frac{20.67}{6125*1.5*0.45}[/tex] = 0.005
[tex]C_{mcg,w} = C_{ac} ,w + C_{l}(h-h_{ac})[/tex]
[tex]h-h_{ac}= \frac{C_{mcg,w} -C_{ac,w}}{C_{l} }[/tex]
[tex]h-h_{ac}= \frac{0.005-(-0.003)}{0.4}[/tex]=0.02
the location of the aerodynamic center = 0.02
1 moles of a gas are initially initially under pressure 4 × 105 Pa at T=300 K. In this problem, you will want to use the following relationships:Definition of Gibbs free energy: G=U−TS+PV Equipartition: U=NDOF2NkT+constEntropy of an ideal gas: S=NklnV+const Ideal gas law: pV=NkTEquation of state for system with excluded volume b: (V−Nb)p=NkT
1)If the gas is ideal, how much does G change when the gas is allowed to expand isothermally until the pressure is reduced to 2 × 105 Pa?
Answer = -1729
2)How much did the chemical potential, ??, change in part (1)? ?final-?initial= ?
3) Now assume the gas has no interaction energy but does have an excluded volume per particle, b=1.9 × 10-28 m3, which means that S = Nk ln(V-Nb) + f(T,N) , where f is some function you won't need to know since T and N don't change. How much does G change when the gas expands isothermally until the pressure is reduced from 4 × 105 Pa to 2 × 105Pa? Gfinal-Ginitial= ?
4) How much did the chemical potential change in the process of part (3)? ?final-?initial= ?
Answers
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
4. Write the command to import the coolfunc() function from the neatomod module in the funpack package and rename it to coolf. Now write the command to import all functions from the neatomod module. What command can we use to obtain a list of all functions inside the module?
Answers
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The four-wheel-drive all-terrain vehicle has a mass of 320 kg with center of mass G2. The driver has a mass of 82 kg with center of mass G1.
If all four wheels are observed to spin momentarily as the driver attempts to go forward, what is the forward acceleration of the driver and ATV?
The coefficient of friction between the tires and the ground is 0.59.
Answers
The mass of the car plus its occupants is calculated using Newton's second law of motion. After determining the net force by subtracting the force of friction from the force exerted by the wheels, it is divided by the given acceleration to find the mass, which is approximately 1027.78 kg.
Explanation:The question pertains to the application of Newton's laws of motion and forces to determine the mass of a vehicle, given the acceleration, force exerted, and the opposing forces of friction. The calculation is based on Newton's second law, which states that the net force acting on an object is equal to the mass of the object times its acceleration (F = ma). To find the mass of the car plus its occupants, we start with calculating the net force (Fnet) which is the difference between the force the wheels exert backward on the road and the force of friction. By applying Newton's second law (Fnet = m × a), we rearrange it to solve for mass (m = Fnet / a).
Net Force Equation: Fnet = Force exerted - Friction
= 2100 N - 250 N
= 1850 N
Now, we can determine the mass using the net force and the given acceleration.
Mass (m) = Fnet / a
= 1850 N / 1.80 m/s2
= 1027.78 kg
Therefore, the mass of the car plus its occupants is approximately 1027.78 kg.
A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype will be operated in seawater. Determine the speed ofthe prototype to ensure Reynolds number similarity‘
Answers
Answer:
Explanation:
Given
scale i.e. [tex]L_r=1:15[/tex]
Using Reynolds number similarity
[tex](Re)_m=(Re)_p[/tex]
[tex](\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p[/tex]
Properties of air
[tex]\nu _{air}=1.57\times 10^{-4} ft/s[/tex]
Properties of sea water
[tex]\nu _{sea}=1.26\times 10^{-5} ft/s[/tex]
[tex]\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )[/tex]
[tex]V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )[/tex]
[tex]V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}[/tex]
[tex]V_p=0.963\ ft/s[/tex]
A rectangular concrete (n=0.013) channel 20 feet wide, on a 2.5% slope, is discharging 400 ft3 /sec into a stilling basin. The stilling basin is also 20 feet wide and has a water depth of 8 ft determined from the downstream channel condition. What is the length of the stilling basin? What is the height of the endsill?
Answers
Answer:
Length of stilling basin = 32.9 feet
Height of end sill = 6.58 feet
Explanation:
Discharge = Q = 400 ft^3 /sec
Slope = 2.5 ft
Width = 20 feet
n = 0.013
we will assume the depth of flow as "d"
Q = 1/n (R)^2/3 (slope)^1/2 A ( here R is the hydraulic Radius)
by substituting the given data in above formula we get:
400 = 1/0.013 * (R)^2/3 * sqrt (2.5/100) * 20d
R = A/P
here, A is the flow area and P is the wetted perimeter
400*0.013 = (20d/(20+20d))^2/3 * sqrt(2.5/100) * 20d
d = 1.42 feet
Depth of stilling channel before the jump will be = d1 = 8 feet
Depth of stilling channel after the jump will be = d2 = 1.42 feet
Length of stilling basin = 5(d2 - d1)
= 5( 8 - 1.42)
Length of stilling basin = 32.9 feet
Now calculating the height of end sill:
Jump height = (8 - 1.42)
Height of end sill = 6.58 feet