Answer:
1.4KN 4KN - m C 3KN 0.6m A 0 B 0.6m R 0.6m 0.6m 30° R Resolving Forces Horizontally, we get X=0 Rx
Dry air does NOT contain
Explanation:
Dry air doesn't contain water vapor .
A detailed image of a brain scan with height, width, and depth is an example of a(n) 3D _________ model.
Answer:
is a mathematical representation of something three-dimensional.
Explanation:The typical base of a the model is a 3D mesh; the structural build consists of polygons.
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
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
The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphere. An orifice of diameter D with a smooth entrance (i.e., negligible losses) is open at the bottom. Develop a relation for the time required for the tank (a) to empty halfway (5-point) and (b) to empty completely (5-point).
Answer:
a. The time required for the tank to empty halfway is presented as follows;
[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]
b. The time it takes for the tank to empty the remaining half is presented as follows;
[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]
The total time 't', is presented as follows;
[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]
Explanation:
a. The diameter of the tank = D₀
The height of the tank = H
The diameter of the orifice at the bottom = D
The equation for the flow through an orifice is given as follows;
v = √(2·g·h)
Therefore, we have;
[tex]\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}[/tex]
[tex]\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}[/tex]
Where;
P₁ = P₂ = The atmospheric pressure
z₁ - z₂ = dh (The height of eater in the tank)
A₁·v₁ = A₂·v₂
v₂ = (A₁/A₂)·v₁
A₁ = π·D₀²/4
A₂ = π·D²/4
A₁/A₂ = D₀²/(D²) = v₂/v₁
v₂ = (D₀²/(D²))·v₁ = √(2·g·h)
The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;
dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh
We have;
[tex]dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh[/tex]
The time for the tank to drop halfway is given as follows;
[tex]\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]
[tex]t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]
[tex]t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]
[tex]t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]The time required for the tank to empty halfway, t₁, is given as follows;
[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]
(b) The time it takes for the tank to empty completely, t₂, is given as follows;
[tex]\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]
[tex]t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)[/tex]
[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]
The time it takes for the tank to empty the remaining half, t₂, is presented as follows;
[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]
The total time, t, to empty the tank is given as follows;
[tex]t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}[/tex]
[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]
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:
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)
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 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
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
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 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
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.
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:
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
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
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:
write to change past tense
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
If an improvement creates no significant change in a product’s performance, then it is a(n) design improvement.
Answer:
Following are the responses to these question:
Explanation:
They might believe that it was an enhanced layout because the quality is not updated. For instance, its new XS Max iPhone does have a better display than the iPhone X, however, the performance wasn't enhanced. It also has the same processor or graphic cards however a bigger pixel every centimeter ratio. When its output AND is not altered, the layout doesn't change, basically the very same item.
Answer:
If an Improvement creates no significant change in a product’s performance, then it is a(n) Superficial design improvement.
Explanation:
DIFFERENT BREED!
Write a python program to get the following output. 1-----99 2-----98 3-----97 . . . . . . 98-----2 99-----1
Answer:
i dont know th answer can u help ?
Explanation:
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
#SPJ5
Choose two other elements from the periodic table that you predict should react to form something like table salt
Please pleassssss helppp
I give branlistttttt