A machine with five states requires three state variables. In a machine with three state variables, there are up to eight available states, leaving five unused states.
The number of available states in a machine with n state variables can be calculated using the formula 2^n. In this case, the machine has three state variables, so the number of available states is 2^3 = 8. However, the machine with five states requires only three state variables, which means that it utilizes only three out of the eight available states.
Therefore, there are five unused states remaining in the machine. These unused states do not have any assigned values or represent any specific conditions or behaviors in the system. They are simply the additional states that are not required for the machine's operation with the given number of state variables.
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Airtight plugs are installed in the ends of large sizes of PVC conduit during bending to_____________.
Airtight plugs are installed in the ends of large sizes of PVC conduit before bending to A. prevent the conduit from collapsing when heated.
What is Airtight plugs?Airtight plugs is necessary in the PVC conduit so as to be able to avoid the collapse of conduit.
In this case, Airtight plugs are installed in the ends of large sizes of PVC conduit before bending to A. prevent the conduit from collapsing when heated.
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A 50-mm cube of the graphite fiber reinforced polymer matrix composite material is subjected to 125-kN uniformly distributed compressive force in the direction 2, which is perpendicular to the fiber direction (direction 1). The cube is constrained against expansion in direction 3. Determine:
a. changes in the 50-mm dimensions.
b. stresses required to provide constraints.
Answer:
hello some parts of your question is missing attached below is the missing part
answer :
A) Determine changes in the 50-mm dimensions
The changes are : 0.006mm compression in y-direction
0.002 mm expansion in x and z directions
B) the stress required are evenly distributed
Explanation:
Given data :
50-mm cube of graphite fiber reinforced polymer matrix
subjected to 125-KN force in direction 2,
direction 2 is perpendicular to fiber direction ( direction 1 ) and cube is constrained against expansion in direction 3
A) Determine changes in the 50-mm dimensions
The changes are : 0.006mm compression in y-direction
0.002 mm expansion in x and z directions
B) the stress required are evenly distributed
attached below is the detailed solution
The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a particle) is 2 m above the road. Calculate the speed v of the truck.
Answer:
19.81 m/s
Explanation:
The total acceleration of the truck (a) is due to the centripetal acceleration and as a result of the linear acceleration. Therefore the total acceleration (a) is given by:
[tex]a^2=a_n^2+a_t^2\\\\where\ a_n=centripetal\ acceleration=\frac{v^2}{r},a_t=linear \ acceleration\\\\But\ since\ the \ speed\ is \ constant, the \ linear \ acceleration(a_t)\ would\ be\ 0.\\\\a^2=a_n^2+a_t^2\\\\a^2=a_n^2\\\\a=a_n=\frac{v^2}{r} \\\\v^2=ar\\\\v=\sqrt{ar} \\\\a=0.4g=0.4*9.81,r=98\ m+2\ m=100\ m\\\\v=\sqrt{0.4*9.81*100} \\\\v=19.81\ m/s[/tex]
For some alloy, the yield stress is 345-MPa (50,000-psi) and the elastic modulus (E) is 103-GPa (15x106 psi). What is the maximum load that may be applied to a specimen with a cross-sectional area of 130-mm2 (0.2-in2) without plastic deformation
The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is; 35535 N
How to find Elastic Modulus?We are told that for an alloy, the yield stress is 345-MPa and the elastic modulus (E) is 103-GPa.
Now, we want to find the maximum load that may be applied to a specimen with a cross-sectional area of 130-mm² without plastic deformation. Thus;
We are given the parameters;
Yield Stress; σ = 345 Mpa = 345 * 10⁶ Pa
Elastic Modulus; E = 103 GPa = 103 * 10⁹ Pa
Cross sectional Area; A = 130 mm² = 103 * 10⁻⁶ m²
Formula for stress without Plastic deformation is;
σ = F_max/Area
where;
σ is stress
F_max is maximum force
Area is Area
Thus making maximum force the subject of the formula gives;
F_max = σ * A
Plugging in the relevant values for stress and area gives us;
F_max = 345 * 10⁶ * 103 * 10⁻⁶
F_max = 35535 N
The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is gotten to be 35535 N
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how skateboards works?
Answer
The skateboarder applies pressure to the trucks and gives/releases pressure on the levers. Second, the wheels and the axles are also examples of simple machines. They help the skater ride, spin, grind, and do a bunch of other radical movements on a skateboard.:
Explanation:
The demand schedules for Jones, Smith, and other buyers are shown in the table below:
Draw the 4 demand curves for i) Jones ii) Smith iii) Other buyers and iv) All buyers.
The demand curve is the graphical representation of the relationship between the price of a good and the quantity demanded for a given period of time.
What is a demand schedule?A demand schedule is a table which shows the quantity demanded of a good or service at different price levels.
A demand schedule can be graphed as a continuous demand curve on a chart where the Y-axis represents the price and the X-axis represents quantity.
Here, a typical representation, the price will appear on the left vertical axis, the quantity demanded on the horizontal axis.
Note that the complete information wasn't found and an overview was given.
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True or false? if i were to hook up an ac voltage source to a resistor, the voltage drop across the resistor would be in phase with the current in the circuit.
Answer: True
Explanation:
How are new slip rings attached to the rotor core during alternator rebuilding?
The new slip rings is been attached to the rotor core during alternator rebuilding by b. Soldered on.
What Soldering ?Soldering serves as the way of joining different types of metals together by melting solder.
This is been used on new slip rings to get it attached to the rotor core during alternator rebuilding .
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CHECK THE COMPLETE QUESTION BELOW:
How are new slip rings attached to the rotor core during alternator rebuilding?
a. Machine clamped
b. Soldered on
c. Crimped on
d. epoxy glue
I have an AC waveform with the voltage peak value of 100 V. I'm trying to find the RMS value of the voltage
Answer:
98 x 100=9,00
Explanation:
Consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the pressure drop will _____.
A +7.5% grade meets a horizontal grade on a section of a rural mountainous highway. If the length of the crest vertical curve formed in that section is 300 ft long, determine the safe operating speed on the highway.
Using the formula of comfort condition it is possible to determine the safe operating speed on the highway.
What is a deflection angle?In physics, we define angular deviation as the angle resulting from the deflection of light when reflected from a prism. A prism has the ability to separate white light into several other colors, which are reflected by the object.
The comfort condition formula is given by: [tex]L=2(\frac{NV^{3} }{C} )^{\frac{1}{2} }[/tex]
Where, L is the length of the vertical curve, N is the grade, V is the safe speed operation and assuming the standart value for C (0,6 m/s3) it is possible to calculate:
[tex]C = 0,6 m/s^{3} = 1,968 ft/s^{3}\\ N= N1 - N2 = 7,5-0 = 7,5% = 0,075[/tex]
Changing the values in the formula:
[tex]L=2(\frac{NV^{3} }{C}){\frac{1}{2} } \\300 = 2(\frac{0,075.V^{3} }{1,968})^{\frac{1}{2} } \\V = 83,89 ft/s[/tex]
So, the safe operating speed on the highway is 83,89 ft/s.
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Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 399 MPa causes the specimen to plastically elongate to a length of 54.4 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 47.7 mm to a length of 57.8 mm.
Answer:
The answer is "583.042533 MPa".
Explanation:
Solve the following for the real state strain 1:
[tex]\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}[/tex]
Solve the following for the real stress and pressure for the stable.[tex]\sigma_{r1}=K(\varepsilon_{r1})^{n}[/tex]
[tex]K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}[/tex]
Solve the following for the true state stress and stress2.
[tex]\sigma_{r2}=K(\varepsilon_{r2})^n[/tex]
[tex]=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa[/tex]
Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligible medium resistance. The cake solids (dry basis) per volume of filtrate was 20 g/liter. It is desired to operate a larger rotary vacuum filter (diameter 8 m and length 12 m) at a vacuum pressure of 80 kPA with a cake formation time of 20 s and a cycle time of 60 s. Determine the filtration rate in volumes/hr expected for the rotary vacuum filter.
Answer:
5.118 m^3/hr
Explanation:
Given data:
viscosity of cell broth = 5cP
cake resistance = 1*1011 cm/g
dry basis per volume of filtrate = 20 g/liter
Diameter = 8m , Length = 12m
vacuum pressure = 80 kpa
cake formation time = 20 s
cycle time = 60 s
Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter
attached below is a detailed solution of the question
Hence The filtration rate in volumes/hr expected for the rotary vacuum filter
V' = ( [tex]\frac{60}{20}[/tex] ) * 1706.0670
= 5118.201 liters ≈ 5.118 m^3/hr
It is possible to design a reactor where the scy conductor and the nitrogen/ammonia electrode operate at different temperatures. Which combination of temperatures is expected to give the best results?.
The combination of temperatures that is expected to give the best result is option A: SCY temperature higher than electrode temp.
Why the case above?The answer is option A because the proton conductivity of SCY is known to be one that often goes up along with increasing temperature, and the favorability of reaction goes down or reduces with total temperature.
Hence, It is said to be good for one to to keep the SCY conductor at a higher temperature and as such, The combination of temperatures that is expected to give the best result is option A: SCY temperature higher than electrode temp.
See options below
. Which combination of temperatures is expected to give the best results?
A.
SCY temperature higher than electrode temperature
B.
SCY temperature lower than electrode temperature
C.
SCY temperature the same as electrode temperature
D.
The temperature of the components does not make a difference.
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Explain any three ways research can facilitate the work of building technicians
According to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.
What are building technicians?They are experts in construction projects, supporting the design of plans, estimating costs, planning work methods, etc.
Research can facilitate the work of building technicians in the following ways:
Conception, design and planning of the project.Selection of materials for floor, wall and ceiling systems.Carry out the programming of works, budgets and analysis of unit prices.Therefore, we can conclude that according to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.
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The high-pressure fuel pumps used in gasoline direct-injection (GDI) systems are powered by ________.
Answer:
Camshaft
Explanation:
A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16.5 s later. The descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude. Assume that g = 32.2 ft/s2.
Determine
(a) the speed v1 of the rocket at the end of powered flight,
(b) the maximum altitude reached by the rocket.
Answer:
[tex]u = 260.22m/s[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Explanation:
Given
[tex]S_0 = 89.6ft[/tex] --- Initial altitude
[tex]S_{16.5} = 0ft[/tex] -- Altitude after 16.5 seconds
[tex]a = -g = -32.2ft/s^2[/tex] --- Acceleration (It is negative because it is an upward movement i.e. against gravity)
Solving (a): Final Speed of the rocket
To do this, we make use of:
[tex]S = ut + \frac{1}{2}at^2[/tex]
The final altitude after 16.5 seconds is represented as:
[tex]S_{16.5} = S_0 + ut + \frac{1}{2}at^2[/tex]
Substitute the following values:
[tex]S_0 = 89.6ft[/tex] [tex]S_{16.5} = 0ft[/tex] [tex]a = -g = -32.2ft/s^2[/tex] and [tex]t = 16.5[/tex]
So, we have:
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2[/tex]
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45[/tex]
[tex]0 = 89.6 + 16.5u- 4383.225[/tex]
Collect Like Terms
[tex]16.5u = -89.6 +4383.225[/tex]
[tex]16.5u = 4293.625[/tex]
Make u the subject
[tex]u = \frac{4293.625}{16.5}[/tex]
[tex]u = 260.21969697[/tex]
[tex]u = 260.22m/s[/tex]
Solving (b): The maximum height attained
First, we calculate the time taken to attain the maximum height.
Using:
[tex]v=u + at[/tex]
At the maximum height:
[tex]v =0[/tex] --- The final velocity
[tex]u = 260.22m/s[/tex]
[tex]a = -g = -32.2ft/s^2[/tex]
So, we have:
[tex]0 = 260.22 - 32.2t[/tex]
Collect Like Terms
[tex]32.2t = 260.22[/tex]
Make t the subject
[tex]t = \frac{260.22}{ 32.2}[/tex]
[tex]t = 8.08s[/tex]
The maximum height is then calculated as:
[tex]S_{max} = S_0 + ut + \frac{1}{2}at^2[/tex]
This gives:
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - 1051.11[/tex]
[tex]S_{max} = 1141.0676[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Hence, the maximum height is 1141.07ft
A centrifugal pump levers water
to a boiter at the rate of 0.05mls
pressure rises across the
is 7 bar, the suction pipe 100m
I'm diameter, and delivering pipe
the
ритр
is 15mm diameter, the outlet
being 0.5m above the intet.
Calculate the efficiency of
the pump
the
kipe, it
power delivered to
the driving shaft is 45 kWatt
Answer:
Industrial users can select from numerous pump technologies for water supply and the handling of other liquid media. A particular favorite is the centrifugal pump. How do centrifugal pumps work? What different delivery principles are there? For which applications are centrifugal pumps especially well-suited?
Here we provide you an overview of the different types of centrifugal pumps. In addition, the operating principles are explained, and typical applications are listed also.
Explanation:
The need for extraction of raw metals for making steel has been reduced due to the?
Answer: Increase in minimills
Explanation:
what is the distance in term of wavelengh between successive minima in the standing wave ratio
Answer:
hi there
Explanation:
An example of a power consuming device would be a(n) ____ while an example of a non-power consuming device would be a(n) ____.
An example of a power consuming device would be a(n) light bulb, a computer device while an example of a non-power consuming device would be a(n) switch or button.
What is power consumption?The quantity of energy utilized per unit of time is known as power consumption. Power use is a significant factor in digital systems.
Power consumption is a limiting factor for the battery life of portable devices like laptop computers and cell phones.
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An isolated pretimed signalized intersection has an approach with a traffic flow rate of 750 veh/h and a saturation flow rate of 3200 veh/h. This approach is allocated 32 seconds of effective green time. The cycle length is 100 seconds. Determine the average approach delay a) 4,6 s b) 30.2 s c) 34.8 s d) 35.0 s
Answer: This approach is allocated 32 seconds of effective green time. The cycle length is 100 seconds. Determine the average approach delay (using Eq. 7.27). A) 34.8 s.
3) Explain how dc machines Can work as motor and generator
The working principle of a DC machine is when electric current flows through a coil within a magnetic field, and then the magnetic force generates a torque that rotates the dc motor. The DC machines are classified into two types such as DC generator as well as DC motor.
In order to live and grow, bacteria need moisture, food, the right temperature, and ______? Fill in the blank
Answer:
acidity, time,oxygen
Explanation:
These are some conditions for the growth of bacteria
With _____, only one criterion must evaluate true in order for a record to be selected and with _____, all criteria must be evaluate true in order for a record to be selected.
a. parameter criteria, double criteria
b. function criteria, IF criteria
c. simple criteria, complex criteria
d. OR criteria, AND criteria
Answer:
d
Explanation:
OR criteria, AND criteria
In an OR criteria, it doesn't need all the records to be true. Just one record is enough and all other criterion becomes true.
In an AND criteria, it's unlike the OR criteria and works in opposite. It needs every member of the record to be true to be able to adjudge the whole record as true.
And as such, we have
With OR criteria, only one criterion must evaluate true in order for a record to be selected and with AND criteria, all criteria must be evaluate true in order for a record to be selected.
Where is the primary area for romanesque architectural sculpture?
Answer:
on the façade, entrance, and column capitals
What are the top 4 solar inventions, how they are used, and how they are better than the original way of powering them
who was part of dempwolf his firm when he first started
Explanation:
Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.
What frequency bands are used primarily for short range distances select all the nearby?
The frequency bands that are used primarily for short-range distances include the following:
The upper portion of the VHF band (225 MHz to 300 MHz).The lower portion of the UHF band (300 MHz to 400 MHz).What is GSM?GSM is an abbreviation for global system for mobile communication and it can be defined as an open and digital cellular network technology that is designed and developed to be used for the transmission of both mobile voice and data services at the same time.
In Computer technology and networking, the global system for mobile communication (GSM) is typically designed and developed to operates at the following bands of frequencies:
850 MHz900 MHz1800 MHz1900 MHzIn this context, we can infer and logically deduce that the frequency bands that are used primarily for short-range distances such as PAN, include the following:
The upper portion of the VHF band (225 MHz to 300 MHz).The lower portion of the UHF band (300 MHz to 400 MHz).Read more on frequency bands here: https://brainly.com/question/19097094
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different lever designs can be engineered to alter the brake pedal effort required of the driver by using different levels of ?
Different lever designs can be engineered and developed to alter the brake pedal effort required of the driver by using different levels of mechanical advantage.
What is mechanical advantage?Mechanical advantage can be defined as a ratio of the output force of a lever to the force acting on it (input force or effort), assuming no losses due to wear, flexibility, tear or friction.
This ultimately implies that, different lever designs can be suitably engineered and developed to alter the brake pedal effort (input force) that is required of the driver, especially by using different levels of mechanical advantage.
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