Answer:
3.527 seconds
Explanation:
The height of the falling water, assuming no friction or air resistance, is given by ...
h(t) = -(1/2)gt² +61
where g is the standard gravity value, 9.80665 m/s².
The the time required for h(t) = 0 is ...
1/2gt² = 61
t² = 2·61/g
t = √(2·61/9.80665) ≈ 3.527 . . . . seconds
As a project manager of Permagam Construction you want to plan renting a fleet of 25 cu yd tractor-scrapers and have them hauling between the pit and a road construction job. The haul road is a rutted dirt road that deflects slightly under the load of the scraper. There is a slight grade of 5% from the pit at the fill location. The return road is level. The haul distance to the dump location is 0.90 miles and the return distance is 0.75 miles. The scrapers are push loaded in the pit. The cycle time for the pusher is 1.5 minutes and the cycle time for the scrapers is 8 minutes. Assume that the weight on the wheels is 75 tons (full) and 50 tons
(empty). (Use Tables 14.1 and 14.2)
What are the rolling resistances and grade resistance?
What are the effective grades?
How many scrapers you recommend to be rented? Explain.
What is the production of the system in case 4 scrapers would be rented?
Answer:
See explaination
Explanation:
Rolling resistance which in some occassions can be called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface.
In order to calculate our rolling resistance, there should be a force.
Please kindly check attachment for the step by step solution of the given problem.
Select all of the true statements.
A. Diodes are used in voltage regulators and limiters.
B. Different diode models are used to replace the real diode with a simpler version that approximates the i-v characteristics of the real diode.
C. In the ideal diode plus voltage source model, the forward bias region is characterized by VD
Answer:
Diodes consists of two-terminal electronic parts which conducts current mainly in one direction. It has low resistance in one direction and high resistance in the other direction. These are correct statements about a Diode.
A. Diodes are used in voltage regulators and limiters.
C. In the ideal diode plus voltage source model, the forward bias region is characterized by VD
Explanation:
Diodes
The purification of hydrogen gas by diffusion through a palladium sheet. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m2 at 500°C. Assume a diffusion coefficient of 1.0 × 10-8 m2/s and that the concentrations at the high and low pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per m3 of palladium. Assume steady state conditions.
Answer:
The mass of the hydrogen for one hour would be 0.0039168 kg/hr
Explanation:
The concentration and the distance at concentration point is calculated before calculating the mass of hydrogen. The attached images show a clear explanation;
The mass of the hydrogen for one hour would be 0.0039168 kg/hr. The derivation of the diffusion coefficient has been attached in the image below:
According to Fick's law, the mass of a substance dM is proportional to the concentration gradient grad c of this substance as it diffuses in time dt over a surface dF normal to the diffusion direction: dM = D grad c dF dt.
Physically, the diffusion coefficient, therefore, suggests that for a concentration gradient of unity, the mass of the substance diffuses through a unit surface in a unit of time. A square meter per second corresponds to D in the SI system. Physical constants such as temperature, pressure, and the size of the dispersing substance's molecules all affect the diffusion coefficient.
Learn more about diffusion coefficient here:
https://brainly.com/question/31430680
#SPJ6
A square power screw has a mean diameter of 30 mm and a pitch of 4 mm with single thread. The collar diameter can be assumed to be 35 mm. The screw is to be used to lift and lower a load of 7 kN. A coefficient of friction of 0.05 is to be used for friction at the thread and at the collar. Determine the following: (i) Torque required to raise the load, TR, (Equation 8.1) (ii) Torque required to lower the load, TL, (Equation 8.2) (iii) A conservative estimate of self locking condition is to set TL in equation 8.2 to zero and calculate the minimum coefficient of friction to ensure self locking. What is the minimum coefficient of friction to ensure self locking
Answer:
i) The torque required to raise the load is 15.85 N*m
ii) The torque required to lower the load is 6.91 N*m
iii) The minimum coefficient of friction is -0.016
Explanation:
Given:
dm = mean diameter = 0.03 m
p = pitch = 0.004 m
n = number of starts = 1
The lead is:
L = n * p = 1 * 0.004 = 0.004 m
F = load = 7000 N
dc = collar diameter = 0.035 m
u = 0.05
i) The helix angle is:
[tex]tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43[/tex]
The torque is:
[tex]T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm[/tex]
ii) The torque to lowering the load is:
[tex]T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm[/tex]
iii)
[tex]T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\ 0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}[/tex]
Clearing u:
u = -0.016
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
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
Given that the size of Elmer's property is 180 acres, and given that there are 20 volunteers, what would be a reasonable approach to surveying the land for houndstongue?
Answer:
Weed Mapping Application is the most reasonable approach for surveying.
Answer:
The most reasonable approach to survey the property would be a GPS survey using a dual-frequency GPS system.
Explanation:
Surveying makes use of coordinates to determine positions of objects, it is the first step towards developing a property. There are different types of survey which are specific to various needs and sizes of a property.
GPS SURVEY
GPS survey makes use of the global positioning system to determine positions of a property or object. It is quite accurate and also an efficient type of survey based on its flexibility.
In the illustration, the property size is 180 acres and 20 volunteers are expected to work on it. To carry out this survey, a station whose coordinate is known normally referred to the master station is used to get the coordinates of the remote stations. The volunteers are stationed at 20 remote stations and the surveyor at the master station makes use of dual-frequency GPS system to pick the coordinates of the remote stations. The dual-frequency GPS system would be used because it covers more distance and for a survey as massive as 180 acres this is reasonable.
For the survey to be carried out seamlessly by the 20 volunteers, a dual-frequency GPS system would be used for the GPS survey.
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
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
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
Steam enters an insulated pipe at 200 kPa and 200°C and leaves at 150 kPa and 150°C. The inlet-to-outlet diameter ratio for the pipe is D1/D2 = 1.80. Determine the inlet and exit velocities of the steam.
Answer:
Inlet and exit velocities are 143.71 m/sec and 465.697 m/sec
Explanation:
At inlet of the pipe
[tex]P_1=150kPa[/tex] and [tex]T_1=150^{\circ}C[/tex]
At this pressure and temperature from steam table.
[tex]h_1=2870kj/kg[/tex] and [tex]s_1=7.508kj/kgK[/tex]
At pressure [tex]P_2=200kPa[/tex] and [tex]T_2=200^{\circ}C[/tex]
By steam table from interpolation method.
[tex]h_2=2776.38+(2768.80-2776.38)(\frac{150-100}{200-100})[/tex]
[tex]h_2=2772.59kj/kg[/tex]
[tex]Q=A_1V_1=A_2V_2[/tex]
[tex]D_1^2V_1=D_2^2V_2[/tex]
We have given [tex]\frac{D_1}{D_2}=1.80[/tex]
[tex]\frac{V_1}{V_2}=\frac{D_2^2}{D_1^2}[/tex]
[tex]\frac{V_1}{V_2}=(\frac{1}{1.80})^2[/tex]
[tex]V_1=0.3086V_2[/tex]
Now energy equation in the pipe
[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}[/tex]
[tex]2870.7\times 10^3+\frac{0.3086V_2^2}{2}=2772.59\times 10^3+\frac{V^2}{2}[/tex]
[tex]V_2=465.697m/sec[/tex]
[tex]V_1=0.3086\times 465.697=143.714m/sec[/tex]
A cylinder fitted with a movable piston contains water at 3 MPa with 50% quality, at which point the volume is 20 L. The water now expands to 1.2 MPa as a result of receiving 600 kJ of heat from a large source at 300◦C. It is claimed that the water does 124 kJ of work during this process. Is this possible?
Answer:
The process is possible:
Explanation:
We are going to find out if the entropy generated is greater than 0, if it is greater than 0, then the process is feasible. If it is not, the process is not feasible.
[tex]P_{1} = 3 MPa[/tex]
[tex]x_{1} = 50 % = 0.5[/tex]
[tex]V_{1} = 20 L = 0.02 m^{3}[/tex]
[tex]P_{2} = 1.2 MPa[/tex]
[tex]T_{H} = 300^{0} C = 573 K[/tex]
Received heat energy, [tex]Q_{12} = 600 kJ[/tex]
Work done, [tex]W_{12} = 124 kJ[/tex]
At state 1, using the steam table:
[tex]T_{1} = T_{s} = 233.9^{0} C\\v_{f1} = 0.001216 m^{3} /kg\\v_{fg1} = 0.06546m^{3} /kg\\u_{f1} = 1004.76 kJ/kg\\u_{fg1} = 1599.34 kJ/kg\\s_{f1} = 2.6456 kJ/kg-K\\s_{fg1} = 3.5412kJ/kg-K[/tex]
[tex]v_{1} = v_{f1} + x_{1} * v_{fg1}[/tex]
[tex]v_{1} = 0.001216 + 0.5*(0.06546)\\v_{1} = 0.03395 m^{3} /kg[/tex]
[tex]M = \frac{V_{1} }{v_{1} } \\M = 0.02/0.03395\\M = 0.5892 kg[/tex]
[tex]u_{1} = u_{f1} + x_{1} * u_{fg1}\\u_{1} = 1004.76 + 0.5*1599.34\\u_{1} = 1804.43 kJ/kg[/tex]
[tex]s_{1} = s_{f1} + x_{1} * s_{fg1}\\s_{1} = 2.6456 + 0.5*3.5412\\s_{1} = 4.4162 kJ/kg[/tex]
[tex]Q_{12} = m(u_{2} - u_{1} ) + W_{12} \\600 = 0.5892(u_{2} -1804.43) + 124\\[/tex]
Solving for u₂
[tex]u_{2} = 2612.3 kJ/kg[/tex]
Since P₂ = 1.2 MPa, u₂ = 2612.2 kJ/kg,
then from steam table, T₂ = 200°C, S₂ = 6.5898 kJ/kg-K
The entropy generated will be:
[tex]\triangle S = m(S_{2} -S_{1} ) - \frac{Q_{12} }{T_{H} }\\ \triangle S= 0.5892(6.5898 - 4.4162) - \frac{600 }{573 }\\ \triangle S =0.233 kJ/K[/tex]
Since ΔS > 0, this process is possible
Answer:
Yes it is possible
Explanation:
Attached is the solution
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
A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mechanics is applicable. Each particle has a mass m and is free to perform one-dimensional oscillations about its equilibrium position. Calculate the heat capacity of this system of particles at this temperature in each of the following cases:
Answer:
the restoring force is = 3/4NKT
Explanation:
check the attached files for answer.
// global variables int a = 5, b = 6, c = 7, d = 8; void sub2() { int a = 0, b = 3; //local variables { int b = 5; //local variable System.out.println("a=" + a); System.out.println("b=" + b); System.out.println("c=" + c); System.out.println("d=" + d); } } void sub1() { int a = 2, b = 4, c = 1; //local variables sub2(); } void main() { int a = 1, b = 2, c = 3, d = 4; //local variables sub1(); }\
Answer:
1)
Static scoping
A=0
B=5
C=3
D=4
2) For dynamic scoping
A=2
B=5
C=1
D=8
(A) Develop the activity sequence model and determine the normal time for the following work activity:
An assembly worker on a production line obtains an Allen key within reach, positions it 15 cm (6 in) onto a bolt head, cranks it 7 times to seat the bolt, and then sets the key aside.
(B) Express the MTM-1 motion elements in (a) as one or more MOST activity sequence models with index numbers.
(i) Determine the normal times in TMUs for these sequence activity models.
(ii) What is the total time for this (these) sequence activity model(s) in secs?
Answer:
Activity sequence model = A1B0G1A0B0P3F16A1B0P1A0
Tn= 10(1 + 1 + 3 + 16 + 1 + 1) = 10(23) = 230 TMU (8.3 sec)
This is a multi-part question. Once an answer is submitted, you will be unable to return to this part As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time t Know that the rotor starts from rest at t = 0 and that after 10 s the rotor has completed 20 revolutions.
Determine the angular velocity at t20 s. (You must provide an answer before moving on to the next part)
The angular velocity is [ ] rpm.
Answer:
60 rpm
Explanation:
At t = 0,
Angular speed = 0
At t = 10 sec
Angular speed = 20/10 = 2 rev/s
Average speed = (2 - 0)/2 = 1 rev/s
= 1 x 60 = 60 rpm
Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.28 mm and that has a tip radius of curvature of 0.002 mm when a stress of 1430 MPa is applied.
Answer:
theoretical fracture strength = 16919.98 MPa
Explanation:
given data
Length (L) = 0.28 mm = 0.28 × 10⁻³ m
radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m
Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa
solution
we get here theoretical fracture strength s that is express as
theoretical fracture strength = [tex]s_{0} \times \sqrt{\frac{L}{r} }[/tex] .............................1
put here value and we get
theoretical fracture strength = [tex]1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }[/tex]
theoretical fracture strength = [tex]16919.98 \times 10^6[/tex]
theoretical fracture strength = 16919.98 MPa
A four‐lane freeway (two lanes in each direction) operates at flow rate of 1700 during the peak hour. It has 11‐ft lanes, 4‐ft shoulders, and there are three ramps within three miles upstream of the segment midpoint and four ramps within three miles downstream of the segment midpoint. The freeway has only regular users. There are 8% heavy trucks, and it is on rolling terrain with a peak‐hour factor of 0.85. It is known that 12% of the AADT occurs in the peak hour and that the directional factor is 0.6.
What is the freeway’s AADT?
Answer:
Please see the attached file for the complete answer.
Explanation:
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
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.
4. A 25 km2 watershed has a time of concentration of 1.6 hr. Calculate the NRCS triangular UH for a 10-minute rainfall event and plot it. Determine the runoff hydrograph for a 30-minute storm where there is 4 cm of runoff in the first 10 min, 2.5 cm of runoff in the second 10 min, and 2 cm of runoff in the third 10 min. Plot the runoff hydrograph for each 10-minute rainfall excess along with the aggregated (total) runoff hydrograph on the same axes. Also report the peak flow of the aggregated runoff hydrograph. Also report the total volume of runoff. You do not need to report any tables of data.
Answer:
The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s
Peak flow of the aggregated runoff hydrograph is 420.58 m³/s
The total volume of runoff is 2125000 m³/s
Explanation:
We have
A = 25 km²
tr = 10 min = 1/6 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr
Qp = 2.08×25/1.043 = 49.84 m³/s
Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr
Since the area is
Time (min) Runoff (cm) Volume of runoff m³
0 0 0
10 4 1000000 m³
20 2.5 625000 m³
30 2 500000 m³
Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s
For the 1st 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×4/1.043 = 197.92 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
For the 2nd 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×2.5/1.043 = 123.7 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
For the 3rd 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×2.5/1.043 = 98.96 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
Peak flow of aggregate runoff is given by
Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s
Total volume of runoff is given by
Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s
A certain process requires 3.0 cfs of water to be delivered at a pressure of 30 psi. This water comes from a large-diameter supply main in which the pressure remains at 60 psi. If the galvanized iron pipe connecting the two locations is 200 ft long and contains six threaded 90o elbows, determine the pipe diameter. Elevation differences are negligible.
Answer:
diameter of the pipe = 0.4932ft.
Explanation:
assuming d = 0.4932
Re = 3.16 x 10∧5/0.4932
= 6.4 x 10 ∧5
E/d = 0.0005/0.4932
= 0.0010 from moody chat t = 0.02
if 0.02 is beign substituted in equation 2 we will get the same required diameter of the pipe which is 0.4932ft.
check the attachment for better explanation.thanks
Determine the average power, complex power and power factor (including whether it is leading or lagging) for a load circuit whose voltage and current at its input terminals are given by: v(t) = 100cos(377t-30)v, i(t) = 2.5cos(377t-60)A
Answer:
Average power = 108.25W
Complex power = 125W
PF = 0.866
Explanation:
Check attachment for step by step instructions.
Answer:
Average power: 108.25 W
Complex power: 125 VA
Power factor: 0.866
Refer below for the explanation.
Explanation:
Refer to the picture for brief explanation.
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.
Calculate the headloss through a filter bed consisting of 30.0 in. of stratified sand with the gradation given below. Assume a filtration rate of 5.0 gpm/ft2, a clean bed porosity of 0.42 and a temperature of 40i F.
U.S. Sieve Number 50
40 30 20 18 16
Sieve Opening (mm) 0.297
0.420 0.590 0.840 1.000 1.190
Percent Passing (by wt.) 0.10
6.50 22.00 76.00 90.00 98.00
Answer:
1.5258m
Explanation:
Please see attachment
A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500 N (10,000 lbf) force, producing only elastic deformation. Calculate the resulting strain. Assume elastic modulus of Cu to be 110GPa. (Points: 5).
Answer:
The resulting strain is [tex]1.39\times 10^{-3}[/tex].
Explanation:
A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm
Force, F = 44,500 N
Th elastic modulus of Cu to be 110 GPa
The resulting strain is given by the formula as follows :
[tex]\epsilon=\dfrac{F}{AE}[/tex]
E is elastic modulus of Cu is are of cross section
[tex]\epsilon=\dfrac{44500}{15.2\times 19.1\times 10^{-6}\times 110\times 10^9}\\\\\epsilon=1.39\times 10^{-3}[/tex]
So, the resulting strain is [tex]1.39\times 10^{-3}[/tex].
When will stemuless checks come I was told not to file my taxes because I get a pension check every month direct deposit money into my bank account is this true I don't have to do my taxes because of this
The IRS and the U.S Department of the treasury declared that social security recipients are not required to file a simple tax return to receive stimulus payments under the CARES Act.
Explanation:
Due to the impact of corona virus problem, the CARES Act calls for stimulus payment to be sent to Americans based on their gross income.
The Social security recipients are not required to file a tax return and do not take action and they will receive the payments directly to their bank accounts.
The Automatic payments will begin by next week. The eligible taxpayers who filed tax returns for 2019 or 2018 and chose direct deposit for their refund will automatically receive a stimulus payment of $1,200 for individuals or $2,400 for married couples and $500 for each qualifying child.
Your program this week will have the same output as lab 10, except that instead of redirecting the filename for the input file, you will get the filename from the command-line, and use a file pointer to open the file. You will also dynamically allocate space in memory after reading in the first value from the file indicating the number of lean proteins that are in the file.
Answer:
Check the explanation
Explanation:
defs.h
#ifndef DEFS_H
#define DEFS_H
#include <stdio.h>
#include <stdlib.h>
typedef struct
{
char item[20];
char quantity[10];
int calories;
float protein;
float carbs;
float fats;
} food;
int size;
void printArray(int size, food arr1[]);
#endif
arrayProcessing.c
#include "defs.h"
void printArray(int size, food arr1[])
{
int i = 0;
printf("\nFOOD ITEM\t\tQUANTITY\tCALS\tPRO\tCARBS\tFAT");
for (i = 0; i < size; i++)
{
printf("\n%i.%s", i + 1, arr1[i].item);
printf("\t\t%s", arr1[i].quantity);
printf("\t\t%i", arr1[i].calories);
printf("\t%.2f", arr1[i].protein);
printf("\t%.2f", arr1[i].carbs);
printf("\t%.2f\n", arr1[i].fats);
}
}
lab12.c
#include "defs.h"
int main(int argc, char *argv[])
{
int i = 0;
FILE *inFile;
inFile = fopen(argv[1], "r");
if(inFile==NULL){
fprintf(stderr, "File open error. Exiting program\n");
exit(1);
}
fscanf(inFile, "%i", &size);
food *arr = (food *)malloc(sizeof(food)*size);
for (i = 0; i < size; i++)
{
fscanf(inFile, "\n%[^\n]s", arr[i].item);
fscanf(inFile, "%s", arr[i].quantity);
fscanf(inFile, "%i", &arr[i].calories);
fscanf(inFile, "%f", &arr[i].protein);
fscanf(inFile, "%f", &arr[i].carbs);
fscanf(inFile, "%f", &arr[i].fats);
}
printArray(size, arr);
return 0;
}
Kindly check the Output below,