Answer:
15
Explanation:
A compass may be used to create circles, arcs, bisect lines, angle bisectors, and locate midpoints. By using a compass, a protector is also drawn.
What is the compass?A compass, or more correctly, a pair of compasses, is a tool for technical drawing that may be used to draw arcs or circles. It can also be used as dividers to indicate distances, particularly on maps.
A compass, or more correctly, a pair of compasses, is a tool for technical drawing that may be used to draw arcs or circles. It may also be used as dividers to indicate distances, particularly on maps. Compasses can be used for navigation, mathematics, and other things.
Therefore, a protector is sketched using a compass.
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Milton has been tracking the migrating patterns of whales in the northwest Atlantic Ocean for five years. He knows where and when to find them as well as how quickly they move. Which qualification makes Milton successful in his research?
Answer:
knowledge of animal behavior and anatomy
Explanation:
the qualification that will make Milton successful in his research is a knowledge of animal behaviour and also their anatomy. the knowledge of whales behaviour has opened his eyes into their world so he knows to a great deal about them. it is through his knowledge of the behaviour of whales that he's able to get used to their migrating patterns to know where and when to find them. Also, through the body anatomy of whales he knows what their movement is like.
Answer:
knowledge of animal behavior and anatomy
What allows negative feedback to control a system
Answer:
Negative feedback control of the amplifier is achieved by applying a small part of the output voltage signal at Vout back to the inverting ( – ) input terminal via the feedback resistor
Answer:
The system has parts that sense the amount of output
Explanation:
Apex cirtified
Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Steam is then reheated at constant pressure to a temperature of 5008C before it is routed to the second stage, where it exits at 30 kPa and a quality of 97 percent. The work output of the turbine is 5 MW. Assuming the surroundings to be at 258C, determine the reversible power output and the rate of exergy destruction within this turbine.
Answer:
1) The exergy of destruction is approximately 456.93 kW
2) The reversible power output is approximately 5456.93 kW
Explanation:
1) The given parameters are;
P₁ = 8 MPa
T₁ = 500°C
From which we have;
s₁ = 6.727 kJ/(kg·K)
h₁ = 3399 kJ/kg
P₂ = 2 MPa
T₂ = 350°C
From which we have;
s₂ = 6.958 kJ/(kg·K)
h₂ = 3138 kJ/kg
P₃ = 2 MPa
T₃ = 500°C
From which we have;
s₃ = 7.434 kJ/(kg·K)
h₃ = 3468 kJ/kg
P₄ = 30 KPa
T₄ = 69.09 C (saturation temperature)
From which we have;
h₄ = [tex]h_{f4}[/tex] + x₄×[tex]h_{fg}[/tex] = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg
s₄ = [tex]s_{f4}[/tex] + x₄×[tex]s_{fg}[/tex] = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)
The exergy of destruction, [tex]\dot X_{dest}[/tex], is given as follows;
[tex]\dot X_{dest}[/tex] = T₀ × [tex]\dot S_{gen}[/tex] = T₀ × [tex]\dot m[/tex] × (s₄ + s₂ - s₁ - s₃)
[tex]\dot X_{dest}[/tex] = T₀ × [tex]\dot W[/tex]×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)
∴ [tex]\dot X_{dest}[/tex] = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW
The exergy of destruction ≈ 456.93 kW
2) The reversible power output, [tex]\dot W_{rev}[/tex] = [tex]\dot W_{}[/tex] + [tex]\dot X_{dest}[/tex] ≈ 5000 + 456.93 kW = 5456.93 kW
The reversible power output ≈ 5456.93 kW.
. Fume is the product of consumables, base metals, and coatings on metals during arc welding processes (during all cutting, brazing, and welding). A) TrueB) False
Answer:
True
Explanation:
This is true. Welding fumes are a combination of metallic oxides, silicates and also fluorides. the formation of these films are a result of a metal being exposed to heat above its boiling point and a condensation of its vapours into good fine particles. such films are made up of particles from the material that one is welding and also particles from electrode.
A mechanical system comprises three subsystems in series with reliabilities of 98, 96, and 94 percent. What is the overall reliability of the system?
Answer:
The overall reliability of the system is 88%Explanation:
When solving for the reliability of a complex machine, that is a machine that has more than one component, the reliability of the machine is the products of all individual components.
Given the
reliabilities of 98%,
96%, and
94%
Converting to decimals we have
98/100= 0.98
96/100= 0.96
94/100= 0.94
The product of all reliability is
0.98* 0.96 0.94= 0.88
now converting back to percent we have
0.88*100= 88%
Key length is designed to provide desired factor of safety
a. True
b. False
Answer: true
Explanation:
A key is a machine element that us used to connect the element of a rotating machine to a shaft. It should be noted that the key hinders the relative rotation that may take place between the two parts.
Key length is designed to provide desired factor of safety. It should also be noted that the factor of safety shouldn't be much and the key length is typically limited to the hub length.
Is microwave man made
Answer:
yes..?
Explanation:
I mean humans made it? the wave length that microwaves use is not man made, but using that wave length, microwaves were made by man
Which of the following was an effect of world war 2 on agricultural industry
Answer:
Option C..Farmers saught new technology to help with the workload
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4. Which of these is NOT a power wrench?
OA. Impact
OB. Air ratchet
OC. Breaker bar
OD. Air hammer
Answer:
it's C
Explanation:
hopefully that helps you
The Breaker bar is not a power wrench, the power wrench are impact, air ratchet, and air hammer option (C) is correct.
What is a power wrench?To swiftly tighten or loosen nuts, use a power or impact wrench. They are essentially tiny, portable electric or pneumatic motors with high-speed socket wrench rotation.
As we know,
An impact wrench also referred to as an impactor, impact gun, air wrench, air gun, rattle gun, torque gun, or windy gun, is a type of socket wrench power tool that uses a rotating mass to store energy before abruptly releasing it into the output shaft to produce high torque output with little user effort.
Thus, the Breaker bar is not a power wrench, the power wrench are impact, air ratchet, and air hammer option (C) is correct.
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Five kg of water is contained in a piston–cylinder assembly, initially at 5 bar and 240°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 2660 kJ and kinetic and potential energy effects are negligible. Determine the final volume, in m3, and the work for the process, in kJ.
Answer:
Explanation: that’s the answer! 46kg
This ball is technology too! It can be
rolled, kicked, or thrown. Is that a need
or a want?
That is an example of a want because you don't need the ball.
What is the transfer of energy from one object to another?
Answer:
Kinetic Energy
Explanation:
Energy is transferred from one object to another when a reaction occurs. Energy can be transferred from one object to another as heat, light, and motion.
An analog baseband audio signal with a bandwidth of 4kHz is transmitted through a transmission channel with additive white noise. The channel is assumed to be distortionless, and the power spectral density of white noise, No/2 is 10 WHz. An RC low-pass filter with a 3-dB bandwidth of 8 kHz is used at the receiver to limit the output noise power. Calculate the output noise power.
Answer:
2k20
Explanation:
4k ✈
A standard carbon resistor has a gold band to indicate + 5% tolerance. If its resistance is 3,500 , what are the upper and lower limits for its resistance? OA . 3495 - 3505 2 OB. 3300 Q - 3600 0 OC. 3325 N - 3675 OD 3450 - 35500
Answer:
C. 3325 Ω - 3675 Ω
Explanation:
5% of 3500 Ω is ...
0.05 × 3500 = 175
The lower limit is this amount less than the nominal value:
3500 -175 = 3325
The upper limit is the nominal value plus the tolerance:
3500 +175 = 3675
The lower and upper limits are 3325 Ω and 3675 Ω, respectively.
Most hazardous waste can be disposed of in a landfill. A) TrueB) False
Answer:
This one is false
Explanation:
I just took a test on it..
Most hazardous waste can be disposed of in a landfill. Thus, the above statement is true. Thus, option (A) is correct.
What is the meaning of hazardous waste?A hazardous waste is a waste that has characteristics that make it unsafe or allow it to negatively impact the environment or human health.
Wastes classified as hazardous have characteristics that make them unsafe or potentially damaging to the environment or human health. Liquid, solid, confined gas, and sludge are all examples of hazardous wastes.
Any source that could potentially cause injury, damage, or unfavorable health impacts to someone or something is a hazard.
The majority of hazardous garbage can be dumped in landfills. So, the aforementioned claim is accurate. Therefore, it can be concluded that option (A) is correct.
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Which one is NOT a benefit of Shine? 1. Less production downtime 2.Happier employees 3. Improved quality 4. Inventory reduction 5. Customer satisfaction
Answer:
Inventory reduction
Explanation:
Shine is a term that deals with the general cleaning and overall maintenance of the workplace. It is used as a part of an organizational theme involving five steps which are often referred to as 5S and it is basically defined as the mainstay of the visual workplace. They are Sort, Set in order, Shine, Standardize, and Sustain.
Hence, in this situation, considering the option that is not a benefit of Shine the correct answer is "Inventory reduction." This is because Shine is not about reducing the inventory levels. It is Sort and Set-in-Order will help reduce inventory.
The answer choice which is NOT a benefit of Shine is:
D. Inventory reduction
According to the given question, we are asked to show the answer choice which is not a direct benefit of Shine and how Shine is used to increase productivity in an office space.
Shine is a term which is used to refer to the general cleanliness of an office space and is a part of organization which is used to improve quality, make the employees happier, and have a less production downtime, etc
As a result of this, we can see that inventory reduction is not a benefit of Shine.
Therefore, the correct answer is option D
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What is the capacity of the machine in batches?
A boiler is designed to work at 14bar and evaporate 8 kg/s of water. The inlet water to the boiler has a temperature of 400C and at exit the steam is 0.95 dry. The flow velocity at inlet is 10 m/s and at exit 5 m/s and the exit is % m above the elevation at entrance. Determine the quantity of heat required. What is the significance of changes in kinetic and potential energy on the result.
Answer:
Explanation: 2 is thy answer
You are an engineer working in a auto crash test lab. Some members of your team have raised objections against the use of cadavers in the crashes. No reliable data exists that shows that cadavers produces more accurate data, but many coworkers argue that common suggest they do. Others claim that the use of cadavers as mere instruments for our well-being disrespects those individuals, who have not consented to use their bodies for this purpose. The question of whether to continue to use cadavers in the lab is what kind of issue?
Answer: Application.
Explanation:
The question on wether to contine the use of cadavers in the lab for test is being centered around its application. Cadaver which is same as a corpse or dead body is used in crash site during automobil test in lab, some of this cadavers are been disrespected with their applications in the automobile industries because many didn’t consent to be used in those experiments or test.
An eyewash station needs to be located no more than 10 seconds or less than 55 feet away from a work area.
Answer: True is correct
A third-degree burn may look charred and black or dry and white.
Explanation:
An input voltage of 9.2 V is to be converted into its digital counterpart using an analog-to digital converter. The voltage range is 0 to 16 V. The ADC has 4-bit capacity. Determine: (a) What are the number of quantization levels, resolution, and the maximum quantization error of this ADC
Answer:
ask someone u fool hhhhhhhh
i need solution for this question please
Select the right answer
Explanation:
the 7th one answer is beacause mercury is bad at sharing electrons
the 8th one's answer is Rhodium
Answer:
1.E aluminum
2.E all
3. Either d or e leaning to e
Transmission lines that join two Balancing Authority Areas are known as
In an intrinsic semiconductor, the number of free electrons is equal to what?
Answer:
The number of holes.
Explanation:
A semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity. Semiconductors are classified into two main categories;
1. Extrinsic semiconductor.
2. Intrinsic semiconductor.
An intrinsic semiconductor is a crystalline solid substance that is in its purest form and having no impurities added to it. Examples of intrinsic semiconductor are Germanium and Silicon.
In an intrinsic semiconductor, the number of free electrons is equal to the number of holes. Also, in an intrinsic semiconductor the number of holes and free electrons is directly proportional to the temperature; as the temperature increases, the number of holes and free electrons increases and vice-versa.
In an intrinsic semiconductor, each free electrons (valence electrons) produces a covalent bond.
The easiest way to reduce your exposure to dangerous fumes is to keep your head out of a fume plume, making sure that you're breathing the cleanest air possible.
Answer:
keep your head out of a fume plume
Explanation:
According to the guidelines and precaution procedures of OSHA, an acronym for Occupational Safety and Health Administration in the United States of America stated that The easiest way for an individual to reduce his or her exposure to dangerous fumes is to ensure that such person keeps his or her head out of a fume plume.
Hence, in this situation, the correct answer is "keep your head out of a fume plume"
A company that produces footballs uses a proprietary mixture of ideal gases to inflate their footballs. If the temperature of 230 grams [g] of gas mixture in a 15-liter [L] tank is maintained at 465 degrees Rankine [°R] and the tank is pressurized to 135 pound-force per square inch [psi], what is the molecular weight of the gas mixture in units of grams per mole
Answer:
The molecular weight of the gas mixture is 35.38 g/mol.
Explanation:
The molecular weight of the gas can be found using the following equation:
[tex] M = \frac{m}{n} [/tex]
Where:
m: is the mass = 230 g
n: is the number of moles
First, we need to find the number of moles using Ideal Gas Law:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 135 psi
V: is the volume = 15 L
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 465 °R (K = R*5/9)
[tex]n = \frac{PV}{RT} = \frac{135 psi*\frac{1 atm}{14.6959 psi}*15 L}{0.082 L*atm/(K*mol)*465*(5/9) K} = 6.50 moles[/tex]
Finally, the molecular weight of the gas is:
[tex] M = \frac{m}{n} = \frac{230 g}{6.50 moles} = 35.38 g/mol [/tex]
Therefore, the molecular weight of the gas mixture is 35.38 g/mol.
I hope it helps you!
Six forces act on a beam that forms part of a building's
frame. The vector sum of the forces is zero. The magnitudes
|FB| = Fel 20 kN, Fc = 16 kN, and (Fpl = 9 kN.
Determine the magnitudes of FA and FG.
Answer:
FA = 13 kN FG = 15.3 kNExplanation:
write each force in terms of magnitude and directions
Fx = F sin Ф
Fy = F cos Ф
where Ф is to be measured from x axis.
∑F at y = o
FAy + FBy + FCy + FDy + FEy + FGy = 0
∑F at x = o
FAx + FBx + FCx + FDx + FEx + FGx = 0
Let
FA = FA sin (110) + FA cos (110)
FB = 20 sin (270) + 20 cos (270)
FC = 16 sin (140) + 16 cos (140)
FD = 9 sin (40) + 9 cos (40)
FE = 20 sin (270) + 20 cos (270)
FG = FG sin (50) + FG cos (50)
add x and y forces:
FAx + FBx + FCx + FDx + FEx + FGx = 0
FAy + FBy + FCy + FDy + FEy + FGy = 0
FA sin (110) + 0 + 16 sin (140) + 9 sin (40) + 0 + FG sin (50) = 0
FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0
FA sin (110) + 0 + 10.285 + 5.785 + 0 + FG sin (50) = 0
FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0
FA sin (110) + 16.070 + FG sin (50) = 0
FA cos (110) - 45.363 + FG cos (50) = 0
solving for FA, and FG
FA = 13 kN
FG = 15.3 kN
From the given information, we can say that the sum of the upward forces is equivalent to the sum of the downward forces.
From the diagram, equating the component of the upward forces and the downward forces, we have:
[tex]\mathbf{-F_a sin 70 +F_c sin 40+F_d sin 40 + F_g sin50= F_b+F_e}[/tex] --- (1)
Also, the sum of the horizontal positive x-axis as well as the horizontal negative x-axis can be computed as:
[tex]\mathbf{F_g cos 50 +F_d cos 40 = F_c cos 40 +F_a cos 70 --- (2)}[/tex]
If:
[tex]F_B = F_E = 20 \ kN \\ \\ F_c = 16 \ kN \\ \\ F_D= 9\ kN[/tex]
Then, from equation (1), we can have the following:
[tex]\mathbf{F_a sin 70 + 16 sin 40 + 9 sin40 + F_gsin 50 = (20 + 20 )kN}[/tex]
collecting like terms;
[tex]\mathbf{F_a sin 70 + F_gsin 50 = 40 - 16 sin 40 - 9 sin40 }[/tex]
[tex]\mathbf{F_a sin 70 + F_gsin 50 =23.93}[/tex] --- (3)
From equation (2);
[tex]\mathbf{F_g cos 50 + 9cos 40 = 16 cos 40 + F_a cos 70}[/tex]
collecting like terms:
[tex]\mathbf{F_g cos 50 -F_a cos 70 =16cos 40 -9cos 40}[/tex]
[tex]\mathbf{-F_a cos 70 +F_g cos 50 =5.36 ----- (let \ this \ be \ equation (4))}[/tex]
Suppose we equate (3) and (4) together using the elimination method;
[tex]\mathbf{F_a sin 70 + F_gsin 50 =23.93}[/tex] --- (3)
[tex]\mathbf{-F_a cos 70 +F_g cos 50 =5.36 --- (4)}[/tex]
Let's multiply (3) with ( cos 70 ) and (4) with (sin 70);
Then, we have:
[tex]\mathbf{F_a sin 70 cos 70 + F_gsin 50 cos 70 =23.93 cos 70}[/tex]
[tex]\mathbf{-F_a cos 70 sin 70 +F_g cos 50 sin 70 =5.36 sin 70}[/tex]
Adding both previous equations together, we have:
[tex]\mathbf{F_a sin 70 cos 70 + F_gsin 50 cos 70 =23.93 cos 70}[/tex]
[tex]\mathbf{-F_a cos 70 sin 70 +F_g cos 50 sin 70 =5.36 sin 70}[/tex]
[tex]\mathbf{(0 + F_g(sin 50 cos 70 + sin70 cos50) = 23.93 cos 70 + 5.36 sin 70)}[/tex]
[tex]\mathbf{( F_g(0.262 + 0.604)) =(8.19 + 5.04)}[/tex]
[tex]\mathbf{( F_g(0.866)) =(13.23)}[/tex]
[tex]\mathbf{ F_g =\dfrac{(13.23)}{(0.866)}}[/tex]
[tex]\mathbf{ F_g =15.28 \ N}[/tex]
Replacing the value of [tex]\mathbf{F_g}[/tex] into equation (3), to solve for [tex]\mathbf{F_a}[/tex], we have:
[tex]\mathbf{F_a sin 70 + F_gsin 50 =23.93}[/tex]
[tex]\mathbf{F_a sin 70 + 15.28sin 50 =23.93} \\ \\ \mathbf{F_a sin 70 +11.71 =23.93} \\ \\ \mathbf{F_a sin 70 =23.93-11.71 } \\ \\ \mathbf{F_a sin 70 =12.22 } \\ \\ \mathbf{F_a =\dfrac{12.22 }{sin 70}} \\ \\[/tex]
[tex]\mathbf{F_a =13.01 \ N}[/tex]
Therefore, we can conclude that the magnitudes of [tex]\mathbf{F_a}[/tex] and [tex]\mathbf{F_g}[/tex] are 13.0 N and 15.28 N respectively.
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Block B has a mass of 27g and a density of 0.9 g/mL. What is the volume of block B? Will it float in water? (density of water is 1.0 g/mL) Don't forget the units.
A 5000-ft long X-65 pipeline is laid down on seabed with two PLETS (One at each end). The pipe OD=7-in with 0.5-in wall thickness. The pipeline was laid at environmental temperature of 40 °F (As- laid temperature). When pipeline is put into operation, the oil flow was produced at 140 °F. If the thermal expansion coefficient of the pipe material is 6.5*10-/°F and its modulus of elasticity is 30,000 ksi, determine the compressive load applied by the pipeline on a PLET due to its thermal expansion. Assume no temperature change and no seabed friction along the pipeline span.
Answer: 199.1 kip
Explanation:
Given that
Outer diameter is Do = 7 in
Inner diameter Di = ( Do - ( 2×0.5)) = 6 in
Length = 5000 ft = 60000 in
Now change in length of the pipe due to temperature difference
SL = L∝ΔT
= 60000 × 6.5×10^-6(140-40)
SL = 39 in
Also
sL = PL/AE
A = cross sectional area of pipe = π/4(Do^2 - Di^2)
so
P = SL×A×E / L
= (39 × π/4(7^2 - 6^2)×30000) / 60000
= 199.1 kip
compressive load applied by the pipeline on a PLET due to its thermal expansion is 199.1 kip
You have a piece of film paper that is 3 in x 5 in. You fix it inside the back of a pinhole camera with a focal length of 5.5 in. You want to use it to take a picture of your team’s mascot – a giant guinea pig that just barely fits in a 4 ft. tall cube. The picture will be taken directly in front, from a stool that places the aperture 2 ft. above the ground. You have to determine how far away the camera must be from your mascot to get a good portrait that fills up the whole film paper, without cutting any part of him off. How far apart should the camera and the mascot be to take the portrait? Show your work.
Answer:
In order to take a portrait, the distance of the mascot from the camera should be approximately 7.33 feet
Explanation:
The size of the film paper = 3 in. × 5 in.
The focal length of the camera = 5.5 in.
The height and width of the guinea pig = 4 ft.
The height of the aperture above the ground = 2 ft.
Therefore, we have;
Magnification = Height of image/(Height of object)
Withe the 3 in. wide film, we have;
Magnification = 3 in./(4 ft.) = 3 in./(48 in.) = 0.0625
Magnification = Length of camera/(Distance of object from pin hole)
∴ Length of camera/(Distance of object from pin hole) = 0.0625
Length of camera = Focal length of the camera = 5.5 in.
Therefore;
5.5 in./(Distance of object from pin hole) = 0.0625
Distance of object from pin hole = 5.5/0.0625 = 88 inches = 7.33 ft
Therefore, the camera should be approximately 7.33 ft. from the mascot to take a portrait.
In this exercise we have to use the magnification knowledge to calculate the distance that the photograph should be taken, thus we have to:
Distance of the mascot from the camera should be approximately 7.33 feet
To calculate the best distance to take the photo, we have that some information must be taken into account such as:
Size of the film paper: [tex](3)*(5) in[/tex] Focal length of the camera: [tex]5.5 in[/tex] Height and width of the guinea pig: [tex]4 ft[/tex] Height of the aperture above the ground: [tex]2 ft[/tex]
Therefore, we have that the formula of magnification is:
[tex]Magnification = Height \ of \ image/(Height \ of \ object)[/tex]
With the 3 in wide film, we have;
[tex]Magnification = 3 in/(4 ft) \\= 3 in/(48 in) = 0.0625 in[/tex]
Rewriting the magnification formula as:
[tex]Magnification = Length \ of \ camera/(Distance \ of \ object \ from \ pin \ hole)[/tex]
Substituting the values already known we have the equation will be matched as:
[tex]Length\ of \ camera/(Distance\ of\ object \ from \ pin \ hole) = 0.0625\\Length \ of \ camera = Focal \ length \ of \ the \ camera = 5.5 in.[/tex]
Therefore;
[tex]5.5 /(Distance \ of \ object\ from \ pin\ hole) = 0.0625 in\\Distance \ of \ object\ from \ pin \ hole = 5.5/0.0625\\ = 88 inches = 7.33 ft[/tex]
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