Answer:
On July 13, 1787, Congress enacts the Northwest Ordinance, structuring settlement of the Northwest Territory and creating a policy for the addition of new states to the nation. ... In 1781, Virginia began by ceding its extensive land claims to Congress, a move that made other states more comfortable in doing the same
Answer:
They divided up the Northwest Territory into acre squares and sold them.
Explanation:
Which of the following is a disadvantage of using a resistor in place of an inductor in a power-supply filter.
A The output DC voltage will be lower
B. The life of the capacitors will be shorter
C. A resistor will cost much more than an inductor
D. A resistor weighs much more than an inductor
Answer: A. The output DC voltage will be lower
Explanation:
Using a resistor in a power-supply filter instead of an inductor will lead to a lower DC voltage output as resistors reduce voltage.
It would therefore be ideal to use an inductor as it does not lower DC voltage but inductors are expensive and can be quite large which is why it is more common to see resistors used in power-supply filter circuits.
entor" by
What type of signal word is used in this sentence?
need and
en who was
generalization
description
thought
feeling
Answer:
generalization
Explanation:
Please mark me brainliest I need to level up
The reversible and adiabatic process of a substance in a compressor begins with enthalpy equal to 1,350 kJ/kg, and ends with enthalpy equal to 3,412 kJ/kg. If the compressor efficiency is 0.85, find the actual specific work required by the compressor to operate, in kJ/kg.
Answer:
the actual specific work required by the compressor to operate is 2425.88 kJ/kg
Explanation:
Given that;
h₁ = 1350 kJ/kg
h₂₅ = 3412 kJ/kg
compressor efficiency П_ise = 0.85
we know that;
compressor efficiency П_ise = isentropic work / actual work
П_ise = (h₂₅ - h₁) / (h₂ - h₁ )
so
0.85 = (h₂₅ - h₁) / (actual work )
Actual work = (h₂₅ - h₁) / 0.85
Actual work = (3412 - 1350) / 0.85
Actual work = 2062 / 0.85
Actual work = 2425.88 kJ/kg
Therefore the actual specific work required by the compressor to operate is 2425.88 kJ/kg
If a poems has a regular rhythm throughout the poem, it has: PLEASE HELP MEH I WILL GIVE YOU BRAINLEIST!!
A. tone
B. imagery
C. irony
D. meter
Answer:
D, meter.
Explanation:
Rhythm is associated with meter, which identifies units of stressed and unstressed syllables.
If a poem has a regular rhythm throughout the poem, it has a meter. Option D is correct.
What is meter in poem?Meter, which distinguishes between stressed and unstressed syllables, is related to rhythm. The fundamental rhythmic framework of a stanza or a line of poetry is known as meter.
The number of feet in the poem serves as a measure of the poem's meter, which is the rhythm of the language.
Many traditional poem forms call for a certain verse meter or a group of meters that alternate in a specified pattern. Prosody refers to both the study of meters and other types of versification, as well as their practical application.
Therefore, option D is correct.
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A road is constructed at the capital cost of $6 million. At the end of Year 10, major improvements are to be made costing $17 million. At the end of Year 25, a replacement and upgrade is to be done at a cost of $29 million. At the end of year 40, the federal government issues a one-time tax credit in the amount of $12 million.
Over a 50-year analysis period (assuming a 10% interest rate) what is the annualized cost to the nearest dollar?
Answer:
The annualized cost is:
$299,272.
Explanation:
a) Data and Calculations:
Year 0 Capital cost of road construction = $6 million
Year 10 Major improvements cost = $17 million
Year 25 Replacement and upgrade cost = $29 million
Year 40 Federal government one-time tax credit = $12 million
Period of project analysis = 50 years
Cost of capital (discount rate) = 10%
Annualized cost at present value costs:
Amount spent Discount Factor Present value
$6 million 1 $6,000,000
$17 million 0.386 6,562,000
$29 million 0.092 2,668,000
($12 million) 0.0222 (266,400)
Total cost $14,963,600
Annualized cost = $14,963,600/50 = $299,272
b) The annualized cost for the road construction project, which is the annualized value of the net present costs of $14,963,600, is divided by 50. Before obtaining the net present costs, the cash outflows, including the tax credit, are discounted to their present values, using the discount rate of 10%. And then, the average of the cost is obtained by dividing the total net present cost into 50 years.
These are sites that allow you to upload and download media content such as images, audio, and video
calculate the quantities of materials required for the first class brickwork in 1:6 cement mortar for 10 cu.m. assume the suitable data.
Which of the following is a basic type of weld? O Groove O Lap O Edge O Corner
A structural component is fabricated from an alloy that has a plane strain fracture toughness of 45 MPa.m1/2. It has been determined that this component fails at a stress of 300 MPa when the maximum length of a surface crack is 0.95 mm. What is the maximum allowable surface crack length (in mm) without fracture for this same component exposed to a stress of 300 MPa and made from another alloy with a plane-strain fracture toughness of 57.5 MPa.m1/2
Answer:
1.5510 mm
Explanation:
Plane strain fracture toughness = 45 MPa√m
failing stress ( б ) = 300 MPa
maximum length of surface crack ( a )= 0.95 mm
Determine maximum allowable surface crack length ( in mm )
we will make use of this relationship for Design stress equation to determine the value of Y
б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex] --------- ( 1 )
k = 45 MPa√m
б = 300 MPa
a = 0.95 mm
y = ?
From equation 1 make Y subject of the equation ( also substitute values into equation 1 above )
hence ; y = 2.7457
Now determine maximum allowable surface crack when component is exposed to a stress of 300 MPa and made from another alloy with plane-strain fracture toughness of 57.5 MPa√m
we will apply the equation
б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex] --------- ( 2 )
K = 57.5 MPa√m
б = 300 MPa
y = 2.7457
a ( maximum allowable surface crack ) = ?
from equation make a subject of the equation
a = [tex]\frac{1}{\pi } (\frac{k}{\alpha y} )^{2}[/tex]
a = [tex]\frac{1}{\pi } (\frac{57.5}{300*2.7457} )^2[/tex] = 1.5510 mm
Which tool is used for cutting bricks and other masonry materials with precision?
Answer:
A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.
Explanation:
Answer:
Masonry Saw
Explanation:
Masonry Saws can be used to cut brick, ceramic, tile and or stone.
I need help with part (C). Pleasee help me. It’s due in a few hours.
Answer:
u do the same thing as part B but only add 100 k, I think, cuz I'm still in middle school but I mean if u see it asks u to do the same thing as B but C says that instead, u do it at half pressure and 100 k is higher temp so what its asking is to repeat b but the twist is u do it at half pressure and 100 k is the higher temp
hope this helps :)
Laminated steel parts are designed:
A. to reduce noise and vibration.
B.for higher energy absorption.
C.to allow good part formability,
D. to increase tensile strength.
Answer:
to increase tensile strength.
Explanation:
Steel as part of the materials used in the construction industries, or other manufacturing industries could be defective if it was not appropriately manufactured. This defects could be due to foreign materials in it, scratches, blisters etc which would lead to the tensile strength being reduced.
In-order to overcome this, there is need to carry out a technique called lamination. This technique is the process of manufacturing the steel in multiple layers so that the composite materials improves the strength of the steel and its stability. This helps to prevent its failure when used in construction.
2. What is the main job of a cylinder head?
OA. Contain the rapid increase in combustion chamber temperature
OB. Contain the rapid increase in combustion chamber pressure
OC. Prevent engine oil from getting past the pistons
OD. Hold the Head Gasket in place
Grade/Exit
Answer:
Explanation:
The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder
The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to carry a minimum load of 13 N applied evenly to the top of the span. The maximum load for each member is 4.5 Newtons in compression and tension. Each member is 4.2 cm long (ball center to ball center). All joints must be fully constrained. Determine the max value of P before the truss fails. The section that can hold the most weight will receive bonus points.
Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
= [tex]\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12[/tex] = 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2
How do you describe sound? (SELECT ALL THAT APPLY.) PLEASE HELP AND SELECT ALL THAT APPLY PLEASE!! A. Sound waves have to have a medium to travel through. B. The volume of a sound is known as amplitude. C. Loud sounds have high amplitude and vibrate with more energy than soft sounds. D. Sound waves are compression waves that cause energy transfer in air molecules.
Answer:
Sound waves are compression waves that cause energy transfer in air molecules
Sound waves have to have a medium to travel through
Loud sounds have high amplitude and vibrate with more energy than soft sounds
Explanation:
Sound waves is a form of energy composed of compression and rare factions. Sound waves are compression waves that cause energy transfer in air molecules.
Sound is an example of a mechanical wave hence it requires a material medium for propagation.
The amplitude of a sound wave determines its loudness or volume. A larger amplitude implies that we will have a louder sound, and a smaller amplitude means that we will have a softer sound.
How many kg moles of Sodium Sulphate will contain 10 kg of
Sodium?
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
(as per my thinking)
Answer:
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as ______.
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as cold forging.
What is forging?Forging is a metalworking process that involves shaping metal with localized compressive forces. A hammer or a die is used to deliver the blows.
Forging is frequently classified according to temperature: cold forging, warm forging, or hot forging.
The goal of forging is to make metal parts. Metal forging produces some of the most durable manufactured parts available when compared to other manufacturing methods.
Minor cracks and empty spaces in the metal are filled as the metal is heated and pressed.
Cold forging is the process of deforming a metal material at room temperature using extremely high pressure. The slug is placed in a die and compressed by a press until it fits into the desired shape.
Thus, the answer is cold forging.
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What is the name given to the vehicles that warn motorists about oversized loads/vehicles?
a) Pilot Car
b) Advanced Car
c) Trail Car
d) Leader Car
The car that is used to warn drivers about oversized loads is the Pilot Car.
What is a Pilot Car?The Pilot Car is also called Escort Car. It is a vehicle used to warn other vehicles of the presence of an over-sized vehicle.
The role of Pilot vehicle operators is to warn road users (motorists) to be cautious of over-sized loads or vehicles.
The cars are used to guide motorists that are making use of roads in construction sites.
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using credit reduces future income
Answer:
lol
Explanation:r
Discuss why TVET Institutions need advice of the business community in order
to provide good programmes.
Answer:
Without the indispensable advice of the business community, TVET Institutions will be unable to cover the gap in career knowledge required by the business community. To develop workers who possess the knowledge and skills required by today's business entities, there is always the continual need for the educational institutions (gown) to regularly meet the business community (town). This meeting provides the necessary ground for the institutions to develop programs that groom the workforce with skills that are needed in the current workplace. Educational institutions that do not seek this important advice from the business community risk developing workers with outdated skills.
Explanation:
TVET Institutions mean Technical and Vocational Education and Training Institutions. They play an important role in equipping young people to enter the world of work. They also continue to develop programs that will improve the employability of workers throughout their careers. They regularly respond to the changing labor market needs, adopt new training strategies and technologies, and expand the outreach of their training to current workers while grooming the young people for work.
what is heat unit?in X ray machine
Answer:
joule
Explanation:
heat is express in joules in x Ray equipment
Technician A says that the B-pillars aid in resisting roof crush.
Technician B says that the IIHS rates vehicles to resist roof crush to five times the weight of the vehicle.
Who is right?
A only
O B only
Both A and B
Neither Anor B
Answer:
The answer is "Both A and B" are right
Explanation:
During the previous twenty years car producers have made significant advances in planning vehicle structures that give more noteworthy tenant insurance in planar accidents (Lund and Nolan 2003). Be that as it may, there has been little agreement with respect to the significance of rooftop strength in rollover crashes, just as the best strategy for surveying that strength. In 2006 one-fourth of lethally harmed traveler vehicle tenants were associated with crashes where vehicle rollover was considered the most hurtful occasion (Protection Establishment for Expressway Wellbeing, 2007). Numerous lethally harmed tenants in rollovers are unbelted, and some are totally or mostly launched out from the vehicle (Deutermann 2002).
There is difference concerning how underlying changes could influence launch hazard or the danger of injury for inhabitants who stay in the vehicle, paying little mind to belt use.
Answer:A Only
Explanation:
I Took the test
Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs. If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot fro the seesaw to be balanced.
Given :
Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs.
To Find :
If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot from the seesaw to be balanced.
Solution :
We know, for seesaw to balance :
[tex]m_1gd_1=m_2gd_2[/tex]
Here, [tex]d_1\ and \ d_2[/tex] is distance from origin from pivot point.
Putting all the values in the equation, we get :
[tex]50\times g\times 3=100\times g\times d_2\\\\d_2=1.5\ feet[/tex]
Therefore, distance of right child from pivot to balance the seesaw is 1.5 feet.
Hence, this is the required solution.
Find R subscript C and R subscript B in the following circuit such that BJT would be in the active region with V subscript C E end subscript equals 5 V and I subscript C equals 25 m A V subscript C C end subscript equals 15 space V comma space V subscript D 0 end subscript equals 0.7 space V comma space beta equals 100 comma space V subscript A equals infinity.. Ignore the early effect in biasing calculations.
Answer: Rc = 400 Ω and Rb = 57.2 kΩ
Explanation:
Given that;
VCE = 5V
VCC = 15 V
iC = 25 mA
β = 100
VD₀ = 0.7 V
taking a look at the image; at loop 1
-VCC + (i × Rc) + VCE = 0
we substitute
-15 + ( 25 × Rc) + 5 = 0
25Rc = 10
Rc = 10 / 25
Rc = 0.4 k
Rc = 0.4 × 1000
Rc = 400 Ω
iC = βib
25mA = 100(ib)
ib = 25 mA / 100
ib = 0.25 mA
ib = 0.25 × 1000
ib = 250 μAmp
Now at Loop 2
-Vcc + (ib×Rb) + VD₀ = 0
-15 (250 × Rb) + 0.7 = 0
250Rb = 15 - 0.7
250Rb = 14.3
Rb = 14.3 / 250
Rb = 0.0572 μ
Rb = 0.0572 × 1000
Rb = 57.2 kΩ
Therefore Rc = 400 Ω and Rb = 57.2 kΩ
A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures of (a) 500 kPa and (b) 784 kPa.
Answer:
a) for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b) for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Explanation:
Given that;
A₂ = 0.001 m²
P₁ = 1 MPa
T₁ = 360 K
k = 1.4
P₂ = 500 Kpa
(1000/500)^(1.4-1 / 1.4) = 360 /T₂
2^(0.4/1.4) = 360/T₂
1.219 = 360 / T₂
T₂ = 360 / 1.219
T₂ = 295.32 K
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
we substitute
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
1.005 × 360 = 1.005 × 295.32 + v₂²/2000
v₂ = 360.56 m/s²
p₂v₂ = mRT₂
500 × (0.001 × 360.56) = m × 0.287 × 295.32
m = 2.127 kg/s
so Mach Number = V₂ / Vc
Vc = √( kRT) = √( 1.4 × 287 × 295.32) = 344.47 m/s
So Mach Number = V₂ / Vc = 360.56 / 344.47 = 1.046
Therefore for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b)
AT P₂ = 784 kPa
(1000/784)^(1.4-1 / 1.4) = 360/T₂
T₂ = 335.82 K
now
V₂²/2000 = 1.005( 360 - 335.82)
V₂ = 220.45 m/s
P₂V₂ = mRT₂
784 × (0.001 × 220.45) = m( 0.287) ( 335.82)
172.83 = 96.38 m
m = 172.83 / 96.38
m = 1.793 kg/s
just like in a)
Vc = √( kRT) = √( 1.4 × 287 × 335.82) = 367.32 m/s
Mach Number = V₂ / Vc = 220.45 / 367.32 = 0.6
Therefore for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Following are the
Given:
[tex]A_2 = 0.001\ m^2\\\\P_1 = 1\ MPa\\\\T_1 = 360\ K\\\\k = 1.4\\\\P_2 = 500\ Kpa\\\\[/tex]
To find:
Flow rate of mass, and Mach number
Solution:
For point a)
Using formula:
[tex]\to P^{\frac{r-1}{r}} \alpha T\\\\\to (\frac{1000}{500})^(\frac{1.4-1}{1.4}) = \frac{360}{T_2}\\\\\to 2^(\frac{0.4}{1.4}) = \frac{360}{T_2}\\\\\to 1.219 = \frac{360}{T_2}\\\\\to T_2 = \frac{360}{1.219}\\\\\to T_2 = 295.32\ K\\\\[/tex]
[tex]\to C_pT_1 + \frac{V_1^2}{2000} = C_pT_2 + \frac{V_2^2}{2000}\\\\\to 1.005 \times 360 = 1.005 \times 295.32 + \frac{v_2^2}{2000}\\\\\to v_2 = 360.56 \ \frac{m}{s^2} \\\\\to p_2v_2 = mRT_2\\\\\to 500 \times (0.001 \times 360.56) = m \times 0.287 \times 295.32\\\\\to m = 2.127\ \frac{kg}{s}\\\\[/tex]
Mach Number [tex]= \frac{V_2}{V_c}\\\\[/tex]
[tex]\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 295.32)} = 344.47 \ \frac{m}{s}\\\\[/tex]
[tex]\to \frac{V_2}{V_c} = \frac{360.56}{344.47} = 1.046[/tex]
For point b)
[tex]\to P_2 = 784\ kPa\\\\\to (\frac{1000}{784})^{(\frac{0.4}{1.4})} = \frac{360}{T_2}\\\\\to T_2 = 335.82\ K\\\\[/tex]
now
[tex]\to \frac{V_2^2}{2000} = 1.005( 360 - 335.82)\\\\\to V_2 = 220.45 \frac{m}{s}\\\\\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 335.82)} = 367.32\ \frac{ m}{s}\\\\\to c= \frac{220.45}{ 367.32} = 0.6\\[/tex]
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A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maximum shear stress is 450 N/cm2 when the maximum bending stress is 1500 N/cm2. Calculate the value of the centrally applied point load W and the span L. The overall height of the I section is 29 cm.
Answer:
W = 11,416.6879 N
L ≈ 64.417 cm
Explanation:
The maximum shear stress, [tex]\tau_{max}[/tex], is given by the following formula;
[tex]\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )[/tex]
[tex]t_w[/tex] = 1 cm = 0.01
h = 29 cm = 0.29 m
[tex]h_w[/tex] = 25 cm = 0.25 m
b = 15 cm = 0.15 m
[tex]I_c[/tex] = The centroidal moment of inertia
[tex]I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )[/tex]
[tex]I_c[/tex] = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴
Substituting the known values gives;
[tex]I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}[/tex]
[tex]I_c[/tex] = 1.2257083[tex]\bar 3[/tex] × 10⁻⁴ m⁴
From which we have;
[tex]4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )[/tex]
Which gives;
W = 11,416.6879 N
[tex]\sigma _{b.max} = \dfrac{M_c}{I_c}[/tex]
[tex]\sigma _{b.max}[/tex] = 1500 N/cm² = 15,000,000 N/m²
[tex]M_c[/tex] = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²
From Which we have;
[tex]M_{max} = \dfrac{W \cdot L}{4}[/tex]
[tex]L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417[/tex]
L ≈ 0.64417 m ≈ 64.417 cm.
determine the values of the viscous damping coefficient c for which the system has a damping ratio of 0.5 and 1.5
Answer:
7.47 lb. s/ft
22.42 lb. s/ft
Explanation:
Without mincing words let us dive straight into the solution to the above problem.
STEP ONE: calculate or determine the frequency.
The frequency can be calculated by making use of the formula given below;
frequency, w = √k/m. Thus, frequency = √ 12 × 15/ [40/32.2] = 12 rad/s.
STEP TWO: Determine or calculate for the viscous damping coefficient, c for damping ratio of 0.5 and 1.5 respectively.
The viscous damping coefficient, c for 0.5 = [ 12 × 0.5 × 2 ×[ 40/32.2] / 2= 7.47 lb. s/ft.
The viscous damping coefficient, c for 1.5= [ 12 × 1.5 × 2 ×[ 40/32.2] / 2 = 22.42 lb. s/ft.
trevor moves a magnetic toy train away from a magnet that cannot move. what happens to the potential energy in the system of magnets during the movement?
Answer:a
Ieieksdjd snsnsnsnsksks
An open-circuit wind tunnel draws in air from the atmosphere through a well-contoured nozzle. In the test section, where the flow is straight and nearly uniform, a static pressure tap is drilled into the tunnel wall. A manometer connected to the tap shows that static pressure within the tunnel is 45 mm of water below atmospheric. Assume that the air is incompressible, and at 25 C, 100 kPa absolute. Calculate the air speed in the wind-tunnel test section
Answer:
Air speed in the wind-tunnel [tex]v_{2}[/tex] = 27.5 m/s
Explanation:
Given data :
Manometer reading ; p1 - p2 = 45 mm of water
Pressure at section ( I ) p1 = 100 kPa ( abs )
temperature ( T1 ) = 25°C
Pw ( density of water ) = 999 kg/m3
g = 9.81 m/s^2
next we apply Bernoulli equation at section 1 and section 2
p1 - p2 = [tex]\frac{PairV^{2} _{2} }{2}[/tex] ---------- ( 1 )
considering ideal gas equation
Pair ( density of air ) = [tex]\frac{P}{RT}[/tex] ------------------- ( 2 )
R ( constant ) = 287 NM/kg.k
T = 25 + 273.15 = 298.15 k
P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2
substitute values into equation ( 2 )
= 100 * 10^3 / (287 * 298.15)
= 1.17 kg/m^3
Also note ; p1 - p2 = PwgΔh ------- ( 3 )
finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )
[tex]\frac{PairV^{2} _{2} }{2}[/tex] = PwgΔh
[tex]V^{2} _{2}[/tex] = [tex]\frac{2*999* 9.81* 0.045}{1.17}[/tex] = 753.86
[tex]v_{2}[/tex] = 27.5 m/s
According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
You've asked an Incomplete question, lacking options. I answered based on the existing O*NET report.
Answer:
high school diploma
Explanation:
According to the Occupational Information Network (O*NET), most people who are Licensing Examiners and Inspectors typically have a high school diploma.
In other words, they do not seek to acquire a post-secondary school education.
Answer:
B
Explanation:
According to edge its answer B
associate's degree or on-the-job experience
got it right as a lucky guess as the O*net site is updated but edge doesn't bother to update their questions or links.
Click this link to view O*NET’s Education section for Licensing Examiners and Inspectors. According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
bachelor’s degree
associate's degree or on-the-job experience .......This is the correct answer.
some college, no degree
associate degree