Answer:results for h should on the same order of magnitude of the value I provided. If it isn't, check your units. Update to you calculated values for h and resolve your model.
Explanation:
Match the example to the model type it represents.
1. The client complains about the way the keyboard feels
1.mock-up
2. The engineering team tests how the tire treads on a new SUV perform on 2.various road conditions
preproduction model
3. The engineering team performs tests on the efficiency of the manufacturing process used for a recumbent bicycle
3.presentation model
Answer:
represnt
Explanation:
Because the mechanism of creep deformation is different from the mechanism of slip in most metal deformation processes, one of the fundamental relationships between microstructure and mechanical properties of metals is reversed for creep deformation compared with normal deformation. Is it:________.
A. The Hume-Rothery Rules
B. The Hall-Petch Relation
C. The Schmid Equation
Answer:
B. The Hall-Petch Relation
Explanation:
The Hall-Petch relation indicates that by reducing the grain size the strength of a material is increased up to the theoretical strength of the material however when the material grain size is reduced below 20 nm the material is more susceptible to creep deformation and displays an "inverse" Hall-Petch Relation as the Hall-Petch relation then has a negative slope (k value)
The Hall-Petch relation can be presented as follows;
[tex]\sigma_y[/tex] = [tex]\sigma_0[/tex] + k·(1/√d)
Where;
[tex]\sigma_y[/tex] = The strength
σ₀ = The friction stress
d = The grain size
k = The strengthening coefficient
The model equation for the reverse Hall-Petch effect is presented here as follows;
[tex]\sigma_y[/tex] = 10.253 - 10.111·(1/√d)
A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per million Btus of heat into the plant. Suppose the utility encourages its customers to replace their 75-W incandescents with 18-W compact fluorescent lamps (CFLs) that produce the same amount of light. Over the 10,000-hr lifetime of a single CFL.
Required:
a. How many kilowatt-hours of electricity would be saved?
b. How many 2,000-lb tons of SO2 would not be emitted?
c. If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?
Answer:
a) 570 kWh of electricity will be saved
b) the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF
c) $1.296 can be earned by selling the SO₂ saved by a single CFL
Explanation:
Given the data in the question;
a) How many kilowatt-hours of electricity would be saved?
first, we determine the total power consumption by the incandescent lamp
[tex]P_{incandescent}[/tex] = 75 w × 10,000-hr = 750000 wh = 750 kWh
next, we also find the total power consumption by the fluorescent lamp
[tex]P_{fluorescent}[/tex] = 18 × 10000 = 180000 = 180 kWh
So the value of power saved will be;
[tex]P_{saved}[/tex] = [tex]P_{incandescent}[/tex] - [tex]P_{fluorescent}[/tex]
[tex]P_{saved}[/tex] = 750 - 180
[tex]P_{saved}[/tex] = 570 kWh
Therefore, 570 kWh of electricity will be saved.
now lets find the heat of electricity saved in Bituminous
heat saved = energy saved per CLF / efficiency of plant
given that; the utility has 36% efficiency
we substitute
heat saved = 570 kWh/CLF / 36%
we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)
so
heat saved = 570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (
heat saved = 5.4 × 10⁶ Btu/CLF
i.e eat of electricity saved per CLF is 5.4 × 10⁶
b) How many 2,000-lb tons of SO₂ would not be emitted
2000 lb/tons = 5.4 × 10⁶ Btu/CLF
0.6 lb SO₂ / million Btu = x
so
x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ / million Btu )] / 2000 lb/tons
x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]
x = 3.24 × 10⁶ / 2 × 10⁹
x = 0.00162 ton/CLF
Therefore, the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF
c) If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?
Amount = ( SO₂ saved per CLF ) × ( rate per CFL )
we substitute
Amount = 0.00162 ton/CLF × $800
= $1.296
Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.
Which option identifies the section of a project charter represented in the following scenario?
Updated POS terminals will be available to the following five departments by July 31, 2015.
O project assumptions
O project deliverables
O project constraints
O project requirements
A three-story structure is to be constructed over an 8000-m2 site. The initial subsurface exploration indicates the presence of sinkholes and voids due to dissolution of the limestone formation. The predominant soil type is a silty fine sand grading to a fine sand with seams of sandy clay. The design indicates that shallow foundations can be used for this project provided the soils were made more homogeneous as far as load support and no voids were present within the depth up to 7.6 m below the ground surface. Assume groundwater is not a concern. Dynamic compaction is proposed to improve the ground. The local contractor doing dynamic compaction has a 15-ton tamper with the diameter of 2.0 m and the height of 1.4 m. You are requested to conduct the preliminary design for this dynamic compaction project including drop height, spacing, number of drops, number of passes, estimated crater depth, and settlement
Answer:
a) 24.07 m
b) 4 m
c) 14 number of drops
d) p = number of passes
e) Dcd = 2.27
0.69 m
Explanation:
Given data:
Depth ( D )= 7.6 m below ground surface
dynamic compaction ( w ) = 15-ton , diameter of tamper = 2.0 m , thickness = 1.4 m
Determine :
A) drop height ( H )
D = n √wH
therefore H = 361 / 15 = 24.07 m
where : D = 7.6 m , n = 0.4 , w = 15
B) Drop spacing
drop spacing = average of ( 1.5 to 2.5 ) * diameter of tamper
= 2 * 2.0m = 4 m
C) number of drops
since the applied energy for fine grained soils and day fills range from 250 - 350 kj/m^2 the number of drops can be calculated using the relation below
AE = [tex]\frac{NWHP}{SPACING ^2}[/tex]
w = 15, H = 24.07 , Np = ? , AE = 300 kj/m^2
∴ Np = 4800 / 361.05 = 13.3
the number of drops at one pass = 14
D) number of passes
p = number of passes
E) estimated crater depth and settlement
crater depth ( Dcd ) = 0.028 [tex]N_{d} ^{0.55} \sqrt{wtIt}[/tex]
Nd = 14 , wt = 15, It = 24.07
therefore : Dcd = 2.27
estimate settlement is within 3 to 5% therefore the improved settlement
= 2.27 * 0.04 * 7.6 = 0.69 m
A water jet pump involves a jet cross-sectional area of 0.01 m^2, and a jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross-sectional area associated with the jet and entrained streams is 0.075 m^2. These two fluid streams leave the pump thoroughly mixed with an average velocity of 6 m/s through a cross-sectional area of 0.075 m^2. Determine the pumping rate (i.e., the entrained fluid flowrate) involved in liters/s.
Answer:
the entrained fluid flowrate is 150 liters/s
Explanation:
Given the data in the question;
we determine the flow rate of water though the jet by using the following expression;
Q₂ = A₂ × V₂
where Q₂ is the flow rate of water though the jet, A₂ is the cross sectional area of the jet( 0.01 m² ) and V₂ is the jet velocity( 30 m/s )
so we substitute
Q₂ = 0.01 m² × 30 m/s
Q₂ = 0.3 m³/s
Next we determine the flow rate of water through the pump by using the following expression
Q₃ = A₃ × V₃
where Q₃ is the flow rate of water though the pump, A₃ is the cross sectional area of the pump( 0.075 m² ) and V₃ is the average velocity of mixing( 6 m/s )
so we substitute
Q₃ = 0.075 m² × 6 m/s
Q₃ = 0.45 m³/s
so to calculate the flow pumping rate of water into the water jet pump, we use the expression;
Q₁ + Q₂ = Q₃
we substitute
Q₁ + 0.3 m³/s = 0.45 m³/s
Q₁ = 0.45 m³/s - 0.3 m³/s
Q₁ = 0.15 m³/s
we know that 1 m³/s = 1000 Liter/second
so
Q₁ = 0.15 × 1000 Liter/seconds
Q₁ = 150 liters/s
Therefore, the entrained fluid flowrate is 150 liters/s
You are designing a hydraulic power takeoff for a garden tractor. The hydraulic pump will be directly connected to the motor and supply hydraulic fluid at 250 psi for use by accessories. In order for the tractor to maintain normal operation, the maximum power the hydraulic system can use is limited to 11 hp. For what maximum hydraulic flow rate in gallons per minute (gpm) should you design
Answer:
required flow rate is 75.44 gal/min
Explanation:
Given the data in the question;
Power developed = 250 psi = 1.724 × 10⁶ Pa
hydraulic power W = 11 hp = 11 × 746 = 8206 Watt
now, Applying the formula for pump power
W = pgQμ
where p is density of fluid, Q is flow rate, μ is heat and W is power developed;
W = pgQμ
W = pgμ × Q
W = P × Q -------- let this be equ 1
so we substitute in our values;
8.2027 kW = 1.724 × 10⁶ Pa × Q
Q = 8206 / 1.724 × 10⁶
Q = 4.75986 × 10⁻³ m³/sec
We know that, 1 cubic meter per seconds = 15850.3 US liquid gallon per minute, so
Q = 4.75986 × 10⁻³ × 15850.3 gallon/min
Q = 75.44 gal/min
Therefore, required flow rate is 75.44 gal/min
A closed vessel of volume 80 litres contains 0.5 N of gas at a pressure of 150 kN/m2. If the gas is compressed isothermally to half its volume, determine the resulting pressure.
Answer:
The resulting pressure of the gas when its volume decreases is 300 kN/m².
Explanation:
Given;
initial volume of the gas, V₁ = 80 L
number of moles of the gas, n = 0.5 moles
initial pressure of the gas, P₁ = 150 kN/m² = 150 kPa
Determine the constant temperature of the gas using ideal gas equation;
PV = nRT
where;
R is ideal gas constant = 8.315 L.kPa/K.mol
T is the constant temperature
[tex]T = \frac{P_1V_1}{nR} \\\\T = \frac{150.kPa \ \times \ 80 .L}{0.5 .mol \ \times \ 8.315(L.kPa/mol.K)} \\\\T = 2,886.35 \ K[/tex]
When the gas is compressed to half of its volume;
new volume of the gas, V₂ = ¹/₂ V₁
= ¹/₂ x 80L = 40 L
The new pressure, P₂ is calculated as;
[tex]P_2V_2 = nRT\\\\P_2 = \frac{nRT}{V_2} \\\\P_2 = \frac{0.5 \times 8.315\times 2886.35}{40} \\\\P_2 = 300 \ kPa = 300 \ kN/m^2[/tex]
Therefore, the resulting pressure of the gas when its volume decreases is 300 kN/m².
What's the ampacity of a No. 10 type TW copper Wire in a raceway containing six wires and located in an area where the ambient Temperature is 50 C
A. 15.1 A
B. 8.6 A
C. 13.9 A
D. 10.8 A
A cylindrical 1040 steel rod having a minimum tensile strength of 865 MPa (125,000 psi), a ductility of at least 10%EL, and a final diameter of 6.0 mm (0.25 in.) is desired. Some 7.94 mm (0.313 in.) diameter 1040 steel stock, which has been cold worked 20% is available. Describe the procedure you would follow to obtain this material. Assume that 1040 steel experiences cracking at 40%CW
Answer:
procedure attached below
The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw
Explanation:
Given data:
Minimum tensile strength = 865 MPa
Ductility = 10%EL
Desired Final diameter = 6.0 mm
20% cold worked 7.94 mm diameter 1040 steel stock
Describe the procedure you would follow to obtain this material.
assuming 1040 steel experiences cracking at 40%CW
attached below is a detailed procedure of obtaining the material
The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw
Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim working on now? A. design development B. schematic design C. mechanical D. structural
Answer:
B. schematic design
Explanation:
This correct for Plato/edmentum
Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.
What is a schematic design?A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.
The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.
The schematic design consists of a rough sketch with markings and measurements.
Therefore, the correct option is B. schematic design.
To learn more about schematic design, refer to the link:
https://brainly.com/question/14959467
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Draw the ipo chart for a program that reads a number from the user and display the square of that number ???Anyone please
Answer:
See attachment for chart
Explanation:
The IPO chart implements he following algorithm
The expressions in bracket are typical examples
Input
Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module
Processing
Assign variable to the input number (x)
Calculate the square (x = 5 * 5)
Display the result (25) ----> This will be passed to the output module
Output
Display 25
Air flows through a heating duct with a square cross-section with 9-inch sides at a speed of 6.1 ft/s. Just before reaching an outlet in the floor of a room, the duct widens to assume a square cross-section with sides equal to 13 inches. Compute the speed of the air flowing into the room (in ft/s), assuming that we can treat the air as an incompressible fluid.
Answer:
2.9237 ft/s
Explanation:
Given the data in the question;
A₁ = 9-inch × 9-inch = 81 in² = 81 / 144 = 0.5625 ft²
V₁ = 6.1 ft/s
A₂ = 13 in × 13 in = 169 in² = 1.17361 ft²
v₂ = ?
using the the equation if continuity
( Rate of volumetric flow is constant )
A₁V₁ = A₂V₂
we substitute
0.5625 ft² × 6.1 ft/s = 1.17361 ft² × V₂
3.43125 ft³/s = 1.17361 ft² × V₂
V₂ = 3.43125 ft³/s / 1.17361 ft²
V₂ = 2.9237 ft/s
Therefore, the speed of the air flowing into the room is 2.9237 ft/s
The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable horizontal tray is set 1m below the surface of the water, a) determine the percent removal of particles having a settling velocity of 1m/h. b) Could the removal efficiency of the clarifier to be improved by moving the tray? If so, where should the tray be located and what would be the maximum removal efficiency? c) What effect would moving the tray have if the particle settling velocity were equal to 0.5m/h?
Answer:
a) 35%
b) yes it can be improved by moving the tray near the top
Tray should be located ( 1 to 2 meters below surface )
max removal efficiency ≈ 70%
c) The maximum removal will drop as the particle settling velocity = 0.5 m/h
Explanation:
Given data:
flow rate = 8000 m^3/d
Detention time = 1h
depth = 3m
Full length movable horizontal tray : 1m below surface
a) Determine percent removal of particles having a settling velocity of 1m/h
velocity of critical sized particle to be removed = Depth / Detention time
= 3 / 1 = 3m/h
The percent removal of particles having a settling velocity of 1m/h ≈ 35%
b) Determine if the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray and the maximum removal efficiency
The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the
Total Maximum removal efficiency
= percent removal[tex]_{above}[/tex] + percent removal[tex]_{below}[/tex]
= ( d[tex]_{a}[/tex],v[tex]_{p}[/tex] ) . [tex]\frac{d_{a} }{depth}[/tex] + ( d[tex]_{a}[/tex],v[tex]_{p}[/tex] ) . [tex]\frac{depth - d_{a} }{depth}[/tex] = 100
hence max removal efficiency ≈ 70%
c) what is the effect of moving the tray would be if the particle settling velocity were equal to 0.5m/h?
The maximum removal will drop as the particle settling velocity = 0.5 m/h
An infinite cylindrical rod falls down in the middle of an infinite tube filled with fluidat a constant speed V (terminal velocity). The density of the rod and the fluid are different.Assume that the pressure field is hydrostatic.(a)[5pts] Solve for the velocity profileas a function of rin terms of V and the other variables.(b)[2pts] Calculate upward force per unit length of the rod from the fluid wall shear stress on the rod.(c)[2 pts] Calculate upward force per unit length of the rod from bouyancy.(d)[1pts] Calculate V.VR1
Answer:
the speed of your poop
Explanation:
write to change past tense
Resistors are used to reduce current flow, adjust signal levels to divide voltages, bias active elements and terminate transmission line.true or false
Answer:
True
Explanation:
Those are the exact uses of a resistor
hi i am a teacher and i want to say to stop using brainly we strictly banned brainly but our students keep using it
Answer:
Then miss try to get your school to banned it for your district like forbid then from the web site then it will not allow them to use it. then sometimes use it to help other people who needs help like teachers do or we need help with something we dont know so we ask for help for a better understanding of it.
Explanation:
^-^ hope you have a good day miss or sir
Answer:
So I would love to say that you and your team at your school had banned brainly and why are you using brainly here, then you shouldn't use brainly, and brainly all students around the world gets help even you is one.
A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The normal ground water level was at the ground surface. Observation wells were located at distances of 10 and 25 ft. from the pumping well. At a discharge of 761 ft3 per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft. Compute the hydraulic conductivity in ft. /sec.
Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.
Explanation:
Determine the convection heat transfer coefficient, thermal resistance for convection, and the convection heat transfer rate that are associated with air at atmospheric pressure in cross flow over a cylinder of diameter D = 100 mm and length L = 2 m. The cylinder temperature is Ts = ° 70 C while the air velocity and temperature are V = 3 m/s and T[infinity] = 20°C, respectively. Plot the convection heat transfer coefficient and the heat transfer rate from the cylinder over the range 0.05 m ≤ D ≤ 0.5 m.
Answer:
attached below
Explanation:
Attached below is a detailed solution to the question above
Step 1 : determine the Reynolds number using the characteristics of Air at 45°c
Step 2 : calculate the Nusselt's number
Step 3 : determine heat transfer coefficient
Step 4 : calculate heat transfer ratio and thermal resistance
Repeat steps 1 - 4 for each value of diameter from 0.05 to 0.5 m
attached below is a detailed solution
A cylindrical specimen of some metal alloy 10 mm in diameter and 150 mm long has a modulus of elasticity of 100 GPa. Does it seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm
Answer:
N0
Explanation:
It does not seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm
Given data :
Diameter ( d ) = 10 mm
length ( l ) = 150 mm
elasticity ( ∈ ) = 100 GPa
longitudinal strain ( б ) 200 MPa
Poisson ratio ( μ ) ( assumed ) =0.3
Assumption : deformation totally elastic
attached below is the detailed solution to why it is not reasonable .
The Sd value = 0.08 > the calculated Sd value ( 6*10^-3 ) hence it is not reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen