The student's question involves calculating the water level drop at the center of a rotating cylindrical container, which requires understanding of rotational motion and equilibrium in physics.
Explanation:The student is asking about the change in water level in a rotating cylindrical container due to the centrifugal force. In a rotating frame, the water surface forms a parabolic shape and the level at the center drops. To solve this problem, principles of rotational motion and physics are applied. The task is to calculate the drop in the center of the water surface within a cylindrical container rotating at a constant angular speed. This can be determined by setting up the equilibrium of forces acting on a particle of the water in the radial direction and using the relationship between the angular speed, radius, and acceleration in a rotating system.
. A two-dimensional fluid motion is represented by a point vortex of strength Γ set at distance c from an infinite straight solid boundary. Write expressions for the velocity potential and stream function in Cartesian coordinates. Derive an expression for velocity on the boundary. Draw the streamlines and plot the velocity distribution on the boundary when Γ = π and c = 1.
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation:
Design a digital integrator using the impulse invariance method. Find and give a rough sketch of the amplitude response, and compare it with that of the ideal integrator. If this integrator is used primarily for integrating audio signals (whose bandwidth is 20 kHz), determine a suitable value for T.
Answer:
50 μsec
Explanation:
See the attached pictures for detailed answer.
While it is important to gain information on the prospect, it is relatively unimportant to gain any information on the prospect’s organization prior to initiating sales dialogue.ANS:FPTS:1DIF:Difficulty: EasyREF:p. 124-125OBJ:LO: 5-5
Answer:
FALSE.
Explanation:
This statement can be considered false, because it is important to know essential aspects of the prospect before starting the sales dialogue.
In addition to obtaining basic information about the client, it is relevant for the company to know the prospect's organization.
We can define the prospect as the one who, for the company, has the ideal customer profile, but who has not yet shown interest in consuming its products and services.
Therefore, knowing information about it, about its basic and complex characteristics will help the organization to develop sales and marketing strategies aimed at attracting the prospect.
hree large plates are separated bythin layers of ethylene glycol and water. The top plate moves to the right at 2m/s. At what speed and in what direction must the bottom plate be moved to hold the center plate stationary?
Answer: For the center plate to remain stationed in one position without rotating, the bottom plate has to move to the left at a speed of 2m/s, so as to cancel the force acting on it from the top.
The center plate will not move when the bottom plate is moving left in a speed of 2m/s to counter the speed of the top plate, because a body will continue to be at rest if all the forces acting towards the body are equal. The center plate will be at rest because we have directed equal force from the top and bottom of the plate.
(a) Aluminum foil used for storing food weighs about 0.3 grams per square inch. How many atoms of aluminum are contained in one square inch of the foil? (b) Using the densities and atomic weights given in Appendix A, calculate and compare the number of atoms per cubic centimeter in (i) lead and (ii) lithium.
Answer:
note:
solution is attached due to error in mathematical equation. please find the attachment
Thirty-six grams of air in a piston–cylinder assembly undergo a Stirling cycle with a compression ratio of 7.5. At the beginning of the isothermal compression, the pressure and volume are 1 bar and 0.03 m3, respectively. The temperature during the isothermal expansion is 1200 K. Assuming the ideal gas model and ignoring kinetic and potential energy effects, determine the net work, in kJ.
The net work done during the Stirling cycle is 2.7 kJ.
Explanation:In order to determine the net work done during the Stirling cycle, we first need to calculate the initial and final volumes of the air in the piston-cylinder assembly.
Given:
Mass of air, m = 36 g = 0.036 kgInitial pressure, P1 = 1 bar = 100,000 PaInitial volume, V1 = 0.03 m3Compression ratio, r = 7.5Temperature during isothermal expansion, TH = 1200 KThe compression ratio is given by:
r = V1/V2
where V2 is the final volume. Solving for V2, we get:
V2 = V1/r = 0.03/7.5 = 0.004 m3
Next, we can use the ideal gas law to relate the initial and final pressures and volumes:
P1V1/T1 = P2V2/T2
Since the process is isothermal, T1 = T2 = TH = 1200 K.
Solving for P2, we get:
P2 = (P1V1T2)/V2T1 = (100,000 x 0.03 x 1200)/(0.004 x 1200) = 100,000 Pa
Finally, we can calculate the net work done using the formula:
Wnet = (P1V1) - (P2V2)
Substituting the values, we get:
Wnet = (100,000 x 0.03) - (100,000 x 0.004) = 2700 J = 2.7 kJ
A piston-cylinder device contains 0.15 kg of air initially at 2 MPa and 350 °C. The air is first expanded isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed in an isobaric process to the initial state. Determine the boundary work for each process and the net work for the cycle?
Answer:
Isothermal expansion W₁ =-37198.9 J
Polytropic Compression W₂ =-34872.82 J
Isobaric Compression W₃ = -6974.566 J
The net work for the cycle = -79046.29 J
Explanation:
Mass of air = 0.15 kg = 150 g
Molar mass = 28.9647 g/mol
Number of moles = 150 g /28.9647 g/mol = 5.179 moles of air
PV = nRT therefrore V = nRT/(P) = 5.179*8.314*(350+273.15)/(2×10⁶) = 0.0134167 m³
For isothermal expansion we have
P₁V₁ = P₂V₂ or V₂ = P₁V₁/P₂ = 2×10⁶*0.0134167 / (5×10⁵) = 0.0536668 m³
Therefore work done
W₁ = -nRTln(V₂/V₁) = -26833ln(4) = -37198.9 J
Stage 2
Compression polytropically we have
[tex]\frac{P_2}{P_3} = (\frac{V_3}{V_2} )^n[/tex] where P₃ = 2 MPa
Therefore V₃ = [tex](\frac{1}{4} )^{\frac{1}{1.2} }*V_2[/tex] = 1.6904×10⁻² m³
Work = W₂ = [tex]\frac{P_2V_2-P_3V_3}{n-1}[/tex] = -34872.82 J
[tex]\frac{P_2}{P_3} = (\frac{T_2}{T_3} )^\frac{n}{n-1}[/tex] or T₃ = [tex]T_2*(\frac{P_3}{P_2})^\frac{n-1}{n}[/tex] = 785.12 K
Isobaric compression we have thus
Work done W₃ = P(V₁ -V₃) = -6974.566 J
Total work = W₁ + W₂ + W₃ = -37198.9 J + -34872.82 J + -6974.566 J = -79046.29 J
The two aluminum rods AB and AC with diameters of 10 mm and 8 mm respectively, have a pin joint at an angle of 45.°
Determine the largest vertical force P that can be supported at the joint. The allowable tensile stress for the aluminum is 150 MPa.
Answer:
attached below
Explanation:
A successful quality strategy features which of the following elements? engaging employees in the necessary activities to implement quality an organizational culture that fosters quality an understanding of the principles of quality engaging employees in the necessary activities to implement quality and an understanding of the principles of quality engaging employees in the necessary activities to implement quality, an organizational culture that fosters quality, and an understanding of the principles of quality
Answer:
Quality strategy is an integral part of organization's strategy that ensures quality. It involves market and productivity strategies that have a very high significance. A successful quality strategy must endeavour to have features such as; engaging employees in the necessary activities to implement quality, an organizational culture that fosters quality and an understanding of the principles of quality.
All these will enable the organization to stand out and beat competition.
A successful quality strategy features includes engaging the employees:
in the necessary activities to implement qualityin an organizational culture that fosters qualityin understanding the principles of quality.A quality strategy serve as an integral part of organization's strategy that ensures quality.
The quality strategy involves market and productivity strategies that have a very high significance.
Hence, the successful quality strategy features includes engaging the employees in necessary activities to implement quality, organizational culture that fosters quality and understanding the principles of quality.
Therefore, all the option is correct.
Read more about quality strategy
brainly.com/question/24462624
The pistons of a V-6 automobile engine develop 226.5 hp. If the engine driveshaft rotational speed is 4700 RPM and the torque is 248 ft·lbf, determine the shaft power, in hp, and the percentage of the developed power that is transferred to the driveshaft.
Answer:
221.929 hp
98.19%
Explanation:
See attached picture.
Ridif bar ABC is supported with a pin at A and an elastic steel rod at C. The elastic rod has a diameter of 25mm and modulus of elasticity E = 200 GPA. The bar is subjected to a uniform load q on span AC and a pointload at B.
Calculate the change in length of the elastic rod.
What is the vertical displacement at point B?
Answer:
the change in length of the elastic rod is 0.147 mm and the vertical displacement of the point b is 0.1911 mm.
Explanation:
As the complete question is not visible , the complete question along with the diagram is found online and is attached herewith.
Part a
Take moments at point A to calculate the tensile force on the steel
rod at point C
[tex]\sum M_A=0\\T_c*2.5-P(2.5+0.75)-q*2.5*2.5/2=0\\T_c*2.5-10(3.25)-5*3.125=0\\T_c=19.25 kN\\[/tex]
Calculate the change in length of the elastic rod
[tex]\Delta l=\dfrac{T_c*L}{A*E}\\\Delta l=\dfrac{19.25\times 10^3*0.75}{\pi/4\times 0.025^2*200\times 10^9}\\\Delta l=0.000147 m \approx 0.147 mm[/tex]
So the change in length of the elastic rod is 0.147 mm.
As by the relation of the similar angles the vertical displacement of the point B is given as
[tex]\Delta B=\Delta l\dfrac{AC+BC}{AC}[/tex]
Here AC=2.5 m
BC=0.75 m
Δl=0.147 mm so the value is as
[tex]\Delta B=\Delta l\dfrac{AC+BC}{AC}\\\Delta B=0.147 mm\dfrac{2.5+0.75}{2.5}\\\Delta B=0.147 mm\dfrac{3.25}{2.5}\\\Delta B=0.1911 mm[/tex]
So the vertical displacement of the point b is 0.1911 mm.
Design a database suitable for a university registrar. This database should include information about students, departments, professors, courses, which students are enrolled in which courses, which professors are teaching which courses, student grades, TA's for a course (TA's are students), which courses a department offers, and any other information you deem appropriate. Note that this question is more free-form than the questions above, and you need to make some decisions about multiplicities of relationships, appropriate types, and even what information needs to be represented.
Answer:
Hello there, see step by step explanation for answers
Explanation:
A database design for a University Registrar. The following requirements are for designing a Database Schema.
It should include:
1. Information about Students .
2. Information about Departments.
3. Information about Professors.
4. Information about courses.
5 . student Grades.
6. TA's for a course.
7. Department offering different courses.
For Designing Database for Registrar System We need to define various
1. Entity: Student, Course, Instructor, Course offering.
2. Attributes of Entities:
(a) Student entity has Sid, name, program as its attributes
(b)Course has Course_n, title, credits, and syllabus as its attributes
(c) instructor has iid ,name ,dept, title as its attributes
(d) course offering has section_no ,time, room, year ,semester as its attributes
3. Relationship among various entities
(a) enrolls
(b) teaches
© is offered
E-R Diagram for University Registrar
This Diagram shows student entity enrolls various courses which can be having teaches relationship with instructor. Course is being offered by relationship is offered by course offering .
A database design for a University Registrar. The following requirements are for designing a Database Schema.
It should include:
1. Information about Students .
2. Information about Departments.
3. Information about Professors.
4. Information about courses.
5 . student Grades.
6. TA's for a course.
7. Department offering different courses.
For Designing Database for Registrar System We need to define various
1. Entity: Student, Course, Instructor, Course offering.
2. Attributes of Entities:
(a) Student entity has Sid, name, program as its attributes
(b)Course has Course_n, title, credits, and syllabus as its attributes
(c) instructor has iid ,name ,dept, title as its attributes
(d) course offering has section_no ,time, room, year ,semester as its attributes
3. Relationship among various entities
(a) enrolls
(b) teaches
© is offered
E-R Diagram for University Registrar
This Diagram shows student entity enrolls various courses which can be having teaches relationship with instructor. Course is being offered by relationship is offered by course offering .
Gas at a temperature of 250°C with a convective coefficient 75 W/m2·K flows through a packed bed of aluminum (2024) spheres that is used as a thermal energy storage system. If the initial temperature of the spheres is 25°C and the diameter of each sphere is 75-mm, find the time, in seconds, required to for a sphere to reach 85% of the maximum possible thermal energy as well as the surface temperature and the centerline temperature at this time, both in °C. Evaluate the properties of the aluminum (2024) at 500 K.
Answer:
See explanation
Explanation:
Airflow t
T¥, h D
Ts
w
Heater
h
(q•, k ) L
(a) Under conditions for which a uniform surface temperature Ts is maintained around the circum- ference of the heater and the temperature Too and convection coefficient h of the airflow are known, obtain an expression for the rate of heat transfer per unit length to the air. Evaluate the
heat rate for Ts = 300°C, D = 20 mm, an alu- minum sleeve (ks = 240 W/m · K), w = 40 mm, N = 16, t = 4 mm, L = 20 mm, Too = 50°C, and h = 500 W/m2 · K.
Using your time efficiency function from HW1, measure the execution times of both insertion and merge sorting algorithms using the attached data files (one with 1,000 integers and the other with 1,000,000 integers ).
Answer:
The data file with 1000 integers
for merge sort the time efficiency is (1000×㏒1000) = 1000 × 3 = 3000
for insertion sort the time efficiency is (1000 × 2) = 2000
The data file with 1,000,000 integers
for merge sort the time efficiency is (1,000,000×㏒1,000,000) = 1,000,000 × 6 = 6,000,000
for insertion sort the time efficiency is (1,000,000 × 2) = 2,000,000
Explanation:
The execution time or temporal complexity refers to how the execution time of an algorithm increases as the size of input increases
For merge sort, the time efficiency is given by the formula
O(n㏒n)For insertion sort the time efficiency is given by the formula
O(n×2)Where n refers to the size of input
declare integer product declare integer number product = 0 do while product < 100 display ""Type your number"" input number product = number * 10 loop display product End While
Full Question
1. Correct the following code and
2. Convert the do while loop the following code to a while loop
declare integer product
declare integer number
product = 0
do while product < 100
display ""Type your number""
input number
product = number * 10
loop
display product
End While
Answer:
1. Code Correction
The errors in the code segment are:
a. The use of do while on line 4
You either use do or while product < 100
b. The use of double "" as open and end quotes for the string literal on line 5
c. The use of "loop" statement on line 7
The correction of the code segment is as follows:
declare integer product
declare integer number
product = 0
while product < 100
display "Type your number"
input number
product = number * 10
display product
End While
2. The same code segment using a do-while statement
declare integer product
declare integer number
product = 0
Do
display "Type your number"
input number
product = number * 10
display product
while product < 100
An enrichment plant has a throughput of 32,000 kg U/day and produces 26,000 kg U as tails. What is the enrichment of the product if the feed is natural uranium and the tails are 0.25%
Answer:
1.10 %
Explanation:
The enrichment of the product can be calculated using the following equation:
[tex] \frac{W}{P} = \frac{x_{p} - x_{f}}{x_{f} - x_{w}} [/tex] (1)
where W: is waste or tails = 26000, P: is the product = 32000, xp: is the wt. fraction of ²³⁵U in product, xf: is the wt. fraction of ²³⁵U in feed = 0.72% and xw: is the wt. fraction of ²³⁵U in waste or tails = 0.25%.
By solving equation (1) for xp, we can find the enrichment of the product:
[tex] x_{p} = \frac {W}{P} \cdot (x_{f} - x_{w}) + x_{f} [/tex]
[tex] x_{p} = \frac {26000}{32000} \cdot (0.72 - 0.25) + 0.72 = 1.10% [/tex]
Therefore, the enrichment of the product is 1.10 %.
I hope it helps you!
100 kg of refrigerant-134a at 200 kPa iscontained in a piston-cylinder device whose volume is 12.322 m3. The piston is now moved until the volume is one-half its original size. This is done such that thepressure of the refrigerant-134a does not change. Determine (a) the final temperatureand (b) the change in the specific internal energy.
Without specific property data for refrigerant-134a at the given conditions, the final temperature and change in specific internal energy cannot be calculated directly. Typically, such tasks require referencing property tables or using equations of state for refrigerant-134a.
Explanation:The question concerns the process of compressing a refrigerant-134a in a piston-cylinder device while keeping the pressure constant. Given that 100 kg of refrigerant-134a at 200 kPa is initially contained within a volume of 12.322 m3 and then is compressed to half its volume without changing the pressure, we are tasked with determining the final temperature and the change in specific internal energy of the refrigerant.
Since the pressure remains constant during the compression, this scenario can be classified as an isobaric process. However, without specific data such as the initial temperature or specific internal energy values for refrigerant-134a at the given states, we cannot directly calculate the final temperature and change in specific internal energy. Typically, these calculations would require consulting refrigerant-134a property tables or equations of state appropriate for refrigerant-134a to find values at the specified initial and final volumes while maintaining constant pressure.
To find the final temperature, one would typically use the relationship between pressure, volume, and temperature given by the ideal gas law or the specific equations pertaining to refrigerant-134a. The change in specific internal energy could then be determined using specific heat capacities if assuming an ideal gas behavior, or more accurately, through the refrigerant-134a property tables looking at the specific internal energy values at the initial and final states.
Consider the Poisson trip generation model in Example 8.4. Suppose that a household has five members with an annual income of $150,000 and lies in a neighborhood with a retail employment of 320. What is the expected number of peak-hour shopping trips, and what is the probability that the household will make more than one peak-hour shopping trip.
Process: (1) Calculate the Poisson parameter (2) Determine the probability of making zero trips and one trip; (3) subtract the values from #2 from 1 to determine probability
Answer:
The solution is given in the attachments.
A parallel plate capacitor has a separation of 2x10 m and free space between the plates. A 10 V battery is connected across the plates and then removed without disturbing the charge on the plates. The plates are now allowed to come together toa separation of 10% m without disturbing the charge on the plates. Fringing fields can be ignored. A) What is the voltage across the plates? B) How has the energy stored in the capacitor changed?
Answer:
a) 5 V.
b) Energy also become half.
Explanation:
See attached picture.
An insulated piston-cylinder device initially contains 0.16 m2 of CO2 at 150 kPa and 41 °C. Electric resistance heater supplied heat for 10 mins. During this procedure the volume has doubled while pressure stayed the same. Considering electric resistance heater running on 110 V, calculate the needed current in A ((Give your answer with three decimals, and do NOT enter units!!!).
Answer:
I=0.3636
Explanation:
See the attached picture for explanation.
A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at −30°C by rejecting its waste heat to cooling water that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The refrigerant enters the condenser at 1.2 MPa and 65°C and leaves at 42°C. The inlet state of the compressor is 60 kPa and −34°C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator
Answer:
A) The quality of the refrigerant at the evaporator inlet = 0.48
B) The refrigeration load = 5.39 kW
C) COP = 2.14
Explanation:
A) From the refrigerant R-144 table I attached,
At P=60kpa and interpolating at - 34°C,we obtain enthalpy;
h1 = 230.03 Kj/kg
Also at P= 1.2MPa which is 1200kpa and interpolating at 65°C,we obtain enthalpy ;
h2 = 295.16 Kj/Kg
Also at P= 1.2MPa which is 1200kpa and interpolating at 42°C,we obtain enthalpy ;
h3 = 111.23 Kj/Kg
h4 is equal to h3 and thus h4 = 111.23 Kj/kg
We want to find the refrigerant quality at the evaporation inlet which is state 4 and P= 60 Kpa.
Thus, from the table attached, we see that hf = 3.84 at that pressure and hg = 227.8
Now, to find the quality of the refrigerant, we'll use the formula,
x4 = (h4 - hg) /(hf - hg)
Where x4 is the quality of the refrigerant. Thus;
x4 = (111.23 - 3.84)/(227.8 - 3.84) = 0.48
B) The mass flow rate of the refrigerant can be determined by applying a 1st law energy balance across the
condenser. Thus, the water properties can be obtained by using a saturated liquid at the given temperatures;
So using the first table in the image i attached; interpolating at 18°C; hw1 = hf = 75.54 kJ/kg
Also interpolating at 26°C; hw2 = hf = 109.01 kJ/kg
Now;
(mr) (h2 − h3)= (mw) (hw2 − hw1)
mr is mass flow rate
Making mr the subject, we get;
mr = [(mw) (hw2 − hw1)] /(h2 − h3)
mr = [(0.25 kg/s)(109.01 − 75.54) kJ/kg
] /(295.13 − 111.37) kJ/kg
mr = 8.3675/183.76
mr = 0.0455 kg/s
Formula for refrigeration load is;
QL = mr(h1 − h4)
Thus,
QL = (0.0455 kg/s)(230.03 − 111.37) kJ/kg = 5.39 kW
C) The formula for specific work into the compressor is;
W(in) = [(h2 − h1)] − (Q(in
)/mr)
= (295.13 − 230.03) kJ/kg − (0.450kJ/s
/0.0455 kg/s)
= 65.10 Kj/kg - 9.89 Kj/kg
55.21 kJ/kg
Formula for COP is;
COP = qL
/W(in)
Thus; COP = (h1 − h4)/W(in
)
= [(230.03 − 111.37) kJ/kg
] /55.24 kJ/kg
= 2.14
Water is to be boiled at sea level in a 30-cm-diameter stainless steel pan placed on top of a 3-kW electric burner. If 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water.
Answer:
mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Explanation:
The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
The rate of evaporation will be "2.871 Kg/hour".
Evaporation of water:According to the question,
Rate of heat supplied:
= 60% of 3 kW
= 1.8 kW
= 1.8 KJ/s
Vaporization of water, [tex]\Delta H = 2257 \ KJ/Kg[/tex]
Time taken will be:
= [tex]\frac{2257}{1.8}[/tex]
= [tex]1254 \ s[/tex]
= [tex]\frac{1254}{3600} \ hour[/tex]
= [tex]0.3482 \ hour[/tex]
hence,
The rate of evaporation,
= [tex]\frac{Mass \ of \ water}{Time}[/tex]
= [tex]\frac{1}{0.3482}[/tex]
= [tex]2.871 \ Kg/hour[/tex]
Thus the above answer is right.
Find out more information about evaporation here:
https://brainly.com/question/4406110
An inventor has developed a machine that extracts energy from water and other effluents drained from apartments in a building. The machine essentially consists of a hydraulic turbine placed in the basement (level zero), through which the effluents pass before being dumped into the sewage system.
(a) If the average height of an apartment relative to the location of the turbine is 30 meters and the average flow from each apartment (assume it to be water) is 100 liters per day, calculate the average power produced by the contraption if there are 100 apartments in the building. Assume ambient state to be 300 K, 1 bar.
Answer:
34.06 W.
Explanation:
Assumptions : ideal turbine " no loss of work", no pipe or friction losses, all the available energy of water in converted to useful power.
(A) total volume of water per day in cubic meters:
1 cubic meters=1000 L
average flow from each apartment*total apartments/1000= (100/1000)*100
=10 m^3
total mass of water m= density*volume=1000*10=10000 kg
total energy of water at 30 m height= m*g*h= 10000*9.81*30=2943000 J
if all the available energy is converted to power.
power produced per day=total energy of water / time
time in seconds=24*3600=86400 s
power produced in a day=2943000/86400= 34.06 W
Answer:
34.06 W
Explanation:
Assuming ideal conditions which are assuming no fiction or pipe loss is made along the line of extraction of water
Energy of water at ideal condition ( Eₐ ) = 1000 kg/m³
height given = 30 meters
quantity of water = 100 Liters
to calculate the quantity of water in M³/s ( cubic per second )
= [tex]\frac{100*10^{-3} }{24*3600}[/tex] = 1.157 * [tex]10^{-6}[/tex]
power produced by water ( Pw) = energy of water * quantity of water in m^3/s
= 1.157 * [tex]10^{-6}[/tex] * [tex]10^{3}[/tex] = 1.157 *[tex]10^{-3}[/tex]
since it is an ideal condition all the power produced by water is converted to power produced by the contraception
Pc = ( Pw * g * h ) * n
H = height = 30
g = 9.81
n = number of apartments = 100
Pc =( 1.157 * [tex]10^{-3}[/tex] * 9.81 * 30) *100 = 34.06 W
Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per second. (b) Repeat the calculations if the tube is a nanoscale tube with a dimeter of D= 100 nm.
Answer:
a) [tex]Re_{D} = 111896.745[/tex], b) [tex]Re_{D} = 1.119\times 10^{-7}[/tex]
Explanation:
a) The Reynolds number for the water flowing in a circular tube is:
[tex]Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}[/tex]
Let assume that density and dynamic viscosity at 25 °C are [tex]997\,\frac{kg}{m^{3}}[/tex] [tex]0.891\times 10^{-3}\,\frac{kg}{m\cdot s}[/tex], respectively. Then:
[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]
[tex]Re_{D} = 111896.745[/tex]
b) The result is:
[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]
[tex]Re_{D} = 1.119\times 10^{-7}[/tex]
Design a circuit that will output HIGH when the input is a prime number larger than 2 and smaller than 16. The input should be a 4-bit binary number. Design the circuit using a 4:16 decoder and other gates.
Answer:
The detailed answer to the question is explained in the attached file.
Explanation:
Suppose an op amp has a midband voltage gain of 500,000. If the upper cutoff frequency is 15 Hz, what does the frequency response look like? (Malvino, 20150123) Malvino, A. (20150123). Electronic Principles, 8th Edition [VitalSource Bookshelf version]. Retrieved from vbk://9781259200144 Always check citation for accuracy before use.
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation:
An asphalt mixture is placed and compacted using normal rolling procedures. Two tests are taken from the compacted mixture and checked for density. Core 1 had a density of 150 pcf and an air void content of 6.5 %. Core 2 had a density of 151 pcf. What is the air void content of core 2.
Answer:
Air void content of core 2 = 5.87%
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
In a CNC milling machine, the axis corresponding to the feed rate uses a dc servomotor as the drive unit and a rotary encoder as the feedback sensing device. The motor is geared to a leadscrew with a 10:1 reduction (10 turns of the motor for each turn of the leadscrew). If the leadscrew pitch is 6 mm, and the encoder emits 60 pulses per revolution, determine (a) the rotational speed of the motor and (b) pulse rate of the encoder to achieve a feed rate of 300 mm/min.
Answer:
a) 500 rev/min.
b) 50 Hz.
Explanation:
See the attached pictures.
1- If the elongation of wire BC is 0.3 mm after the force P is applied, determine the magnitude of P. The wire is made of steel with an elastic modulus of 200 GPa and has a diameter of 2 mm.
Answer:
P = 188.496 N.
Explanation:
The axial elongation of the wire is modelled by this equation:
[tex]\delta = \frac{P \cdot L}{(\frac{\pi}{4}\cdot D^{2})\cdot E_{steel}}[/tex]
The force can be calculated by isolating it within the expression:
[tex]P=\frac{\delta \cdot (\frac{\pi}{4}\cdot D^{2})\cdot E_{steel}}{L}[/tex]
Let assume that [tex]L = 1 m[/tex]. Then:
[tex]P =\frac{(0.3 mm)\cdot (\frac{1 m}{1000 mm} )\cdot (\frac{\pi}{4}\cdot [(2 mm)\cdot(\frac{1 m}{1000 mm} )]^{2} )\cdot (200 \times 10^{9} Pa)}{1 m} \\P \approx 188.496 N[/tex]
You are an electrician on the job. The electrical blueprint shows that eight 500-W lamps are to be installed on the same circuit. The circuit voltage is 277V and is protected by a 20-A circuit breaker. A continuous-use circuit can be loaded to only 80% of its rating. Is a 20-A circuit large enough to carry this load
Answer:
I = 14.44A
Explanation:
calculating 80% of the circuit breaker current
I = (80/100)20A = 16A
eight 500W lamps are to be installed on the circuit. assume they are connected in parallel
total power = 500 x 8 = 4000W
power = voltage x current
current = power/voltage = 4000W/277V = 14.4A
The current obtained is less than 80% of the circuit breaker.
20A circuit breaker is large enough to carry the load.