Answer:
Period = 0.2857 nanoseconds
Explanation:
We are told that frequency = 3.5 GHz
This is simply 3.5 × 10^(9) Hz
Now, from wave equations, Period is given by the formula;
Period = 1/frequency
Thus;
Period = 1/(3.5 × 10^(9))
Period = 0.2857 × 10^(-9) seconds
From conversions, we can simplify the answer.
1 second = 10^(-9) nanoseconds
Thus, 0.2857 × 10^(-9) seconds = 0.2857 × 10^(-9) × 10^(-9) nanoseconds = 0.2857 nanoseconds
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
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
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.
Learn more about the power wrench here:
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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%
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.
What is the capacity of the machine in batches?
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
The displacement volume of an internal combustion engine is 3 liters. The processes within each cylinder of the engine are modeled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by P1=95kPa, T1=22oC, and V1 = 3.17 liters.
Determine the net work per cycle, in kJ, the power developed by the engine, in kW, and the thermal efficiency, if the cycle is executed 1000 times per min.
Answer:
1) The power developed by the engine is 14705.7739 kW
2) The thermal efficiency is approximately 61.5%
Explanation:
The given parameters are;
P₁ = 95 kPa
T₁ = 22°C
V₁ = 3.17 liters
The cutoff ratio = 2.5
Displacement volume = 3 liters
The number of times the cycle is executed per minute = 1000 times per minute
We have;
The displacement volume = V₁ - V₂ = 3 l
V₁ = 3.17 l
V₂ = 3 - 3.17 = 0.17 l
Compression ratio = V₁/V₂ = 3.17/0.17 ≈ 18.65
P₂/P₁ = P₂/(95 kPa) = (V₁/V₂)^(k) = 18.65^1.4
P₂ = (95×18.65^(1.4)) ≈ 5710.5 kPa
T₂/T₁ = (V₁/V₂)^(k - 1)
T₂/(295 K)= (18.65)^(1.4 - 1)
T₂ = 295 * (18.65)^(1.4 - 1) = 950.81 K
The cutoff ratio = V₃/V₂ = 2.5
T₃ = T₂ × V₃/V₂ = 2.5 * 950.81 K = 2377.025 K
[tex]Q_{in}[/tex] = [tex]C_p[/tex]×(T₃ - T₂) = 1.006 × (2377.025 - 950.81) = 1,434.77 kJ/kg
T₄ = T₃ × (V₃/V₄)^(k-1) =
Therefore,
[tex]T_4 = T_3 \times \left (\dfrac{r_c}{r} \right )^{k - 1} = 2377.025 \times \left( \dfrac{2.5}{18.65} \right )^{1.4 - 1} \approx 1064 \ K[/tex]
T₄ ≈ 1064 K
[tex]Q_{out}[/tex] = [tex]-C_v \times (T_4 - T_1)[/tex]
[tex]C_v = C_p/k = 1.006/1.4 \approx 0.7186 \ kJ/kg[/tex]
∴ [tex]Q_{out}[/tex] = 0.7186×(1064 - 295) = 552.6034 kJ/kg
1) The net work = [tex]Q_{in}[/tex] - [tex]Q_{out}[/tex] = 1,434.77 kJ/kg - 552.6034 kJ/kg ≈ 882.17 kJ/kg
The number of cycle per minute = 1000 rpm
The number of cycle per minute = 1000 rpm/60 = 16.67 cycles per second
The power developed by the engine = The number of cycles per second × The net work of the engine
Therefore;
The power developed by the engine = 16.67 cycles/second × 882.17 kJ/kg
The power developed by the engine = 14705.7739 kW
2) Efficiency, [tex]\eta _{th}[/tex], is given as follows;
[tex]\eta _{th} = \dfrac{Q_{in}-Q_{out}}{Q_{in}} \times 100 = 1 - \dfrac{Q_{out}}{Q_{in}} \times 100= 1 - \dfrac{552.6034}{1434.77}\times 100 \approx 61.5\%[/tex]
Therefore, the thermal efficiency ≈ 61.5%.
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
12. You need to be at the lift controls whenever the lift is in motion A) True B)False
the answer to that would be a) true
While out on the International Space Station, an engineer was able to gather a sample of a new type of unidentified rock. What knowledge will the engineer use to predict the potential of this new material?
Answer:
The engineer will conduct a variety of tests, including chemical, mechanical, electrical, and physical examinations, to determine the potential of the new material.
Explanation:
They will need to test the material, this will also help to determine its malleability.
Hope this helps!
What are the common approximations made in the analysis of heat exchangers?
Answer: making sure that they are up to date
Explanation:
parallel circuits???
PLEASE GIVE BRAINLIST
A parallel circuit has two or more paths for current to flow through. Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source.
HOPE THIS HELPED
Transmission lines that join two Balancing Authority Areas are known as
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
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
You are designing an airplane to carry liquid cargo that will slosh and move side to side in its container. This could make the plane unstable. What type of airfoil would you use for the wing? Why. Answer in a full sentence or more
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.
Help me for this question
What are materials engineers trying to discover when they study different materials? Whether or not materials corrode and oxidize how materials perform and deteriorate if certain materials have been used in the past and at which historical ages which metals conduct heat
Answer:
Material engineers study various materials to discover the reason and cause of its existence, the chemical properties, how long it has been there, and how it impacts human life.
Explanation:
Material engineer is an engineering discipline that focuses to improve human life by studying the environment and the various elements or materials it holds.
It uses the power of pure science to test and analysis its findings, documents its features and exposes them to the world. Materials like metals and other elemental forms were all tested by these engineers to determine its history and chemical and biological use.
Answer:
how materials perform and deteriorate
Explanation:
Two wooden members of uniform rectangular cross section are joined using a simple glued scarf splice. The maximum allowable shearing stress and maximum allowable normal stress in the glued splice is 50 MPa and 100 MPa, respectively. The cross-section area of the glued member is 400 mm2. (a) What should the value of the angle be to achieve maximum load Fmax? (b) What is the magnitude of the maximum load Fmax?
Answer:
a). α = 26.57
b). Maximum load is 50 .kN
Explanation:
a).
The normal force is given by
N = σ A cosec β
where, σ is the normal stress
A is the cross sectional area
Similarly, shear force is given by
S= τ A cosec β
where, τ is the shearing stress
Now from the figure,
tan β = S/N
= τ/σ
Therefore, [tex]$\beta = \tan^{-1}(2)$[/tex] = 63.43
α = 90 - β = 26.57
b).
The normal force is given by
[tex]$N=(100\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43$[/tex]
[tex]$N=44.78\times 10^3$[/tex] N
We have
[tex]$\Sigma F_y=0$[/tex]
∴ N - F sin β = 0
⇒ F = N / sin β
= [tex]$\frac{44.72\times 10^3}{\sin(63.43)} = 50\times 10^3 N$[/tex]
Similarly,
The shear force is given by
S = τ A cosec β
= [tex]$(50\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43 = 22.36\times 10^3 N$[/tex]
[tex]$\Sigma F_x=0$[/tex]
∴ S - F cos β = 0
⇒ F = S / cos β
[tex]$\frac{22.36\times 10^3}{\cos(63.43)} = 49.99\times 10^3 N$[/tex]
Therefore, force is 50 kN.
how do you make coke for steel?
What is a specialized accreditation? A. evaluation of the quality of instruction B. evaluation of a particular program C. evaluation of students studying in an organization D. evaluation of recreational facilities in an organization
Answer:
B. evaluation of a particular program
Explanation:
Before students enrol into any given discipline, they should first ensure that their program of choice is well accredited. In a specific program, specialized accreditation has the function of telling would be students if the program meets up with academic standards in the field
Accreditation is a way of assessing faculty and curriculum quality of schools to make sure that they are up to academic standards and are also preparing students to future success in the field.
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.
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
Wheel grinders need be equipped with an
Answer:
wheel guard
Explanation:
to protect our hands and reduce spark
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 ✈
What do we need to build a car?
Which statement demonstrates the most scientific observation?
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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