The exhaust gas recirculation valve position sensor is a linear potentiometer mounted on top of the EGR valve.
True
False
Answers
False. The position sensor of the Exhaust Gas Recirculation (EGR) valve is not mounted on top of the EGR valve.
The position sensor of the EGR valve is a linear potentiometer that measures the position of the valve, but it is typically mounted elsewhere in the EGR system. It may be mounted on the EGR control solenoid or on the intake manifold. The EGR system is designed to reduce the amount of nitrogen oxide emissions from the engine by recirculating some of the exhaust gases back into the intake manifold.
The EGR valve and its associated position sensor are key components of this system, and they work together to regulate the flow of exhaust gases and ensure that the engine is operating efficiently and emitting as few pollutants as possible.
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Related Questions
Centrifugal pump delivers water against a net head of 14. 5 m and at designed speed of 1000 rpm. The vanes are curved back at an angle of 300 with the periphery. The impeller diameter is 300 mm and outlet width 50 mm. Determine the discharge of the pup, if the manometric efficiency is 95 %
Answers
The discharge of the pump is approximately 0.0744 cubic meters per second.
To determine the discharge of the centrifugal pump, we need to consider the head, impeller diameter, outlet width, and the manometric efficiency.
Given:
Net head (H) = 14.5 m
Impeller diameter (D) = 300 mm = 0.3 m
Outlet width (W) = 50 mm = 0.05 m
Manometric efficiency (η) = 95% = 0.95
The discharge (Q) can be calculated using the following formula:
Q = (π/4) * D^2 * W * N / (g * H * η)
where:
π = 3.14159 (pi)
D = Impeller diameter
W = Outlet width
N = Speed of the pump in revolutions per minute (rpm)
g = Acceleration due to gravity (9.81 m/s^2)
H = Net head
η = Manometric efficiency
Substituting the given values into the formula:
Q = (3.14159/4) * (0.3)^2 * 0.05 * 1000 / (9.81 * 14.5 * 0.95)
Simplifying the equation:
Q ≈ 0.0744 m^3/s
Therefore, the discharge of the pump is approximately 0.0744 cubic meters per second.
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A 250KVA 11000volts /50kHz single phase transformer has 80 turns on the secondary. Calculate:
A) approximate values of the primary and secondary currents
B)approximate number of primary turns
C) the maximum value of the flux
Answers
Student question: A 250KVA 11000 volts /50kHz single phase transformer has 80 turns on the secondary. Calculate: B) approximate number of primary turns C) the maximum value of the flux in 200 words
B) To find the approximate number of primary turns, we can use the transformer turns ratio formula:
Turns ratio = Primary turns / Secondary turns = Primary voltage / Secondary voltage
First, we need to calculate the secondary voltage. We can use the power formula:
Power = Voltage × Current
250 KVA = 11000 V × I
I = 250,000 VA / 11,000 V = 22.73 A
Now, we can find the secondary voltage:
Secondary voltage = Power / Current = 250,000 VA / 22.73 A = 11,000 V
Now, we can calculate the turns ratio:
Turns ratio = 11,000 V / 11,000 V = 1
Therefore, the approximate number of primary turns is:
Primary turns = Turns ratio × Secondary turns = 1 × 80 = 80 turns
C) To calculate the maximum value of the flux, we use the formula:
Flux = Voltage / (Frequency × Turns)
Since the frequency is given in kHz, we need to convert it to Hz:
50 kHz = 50,000 Hz
Now, we can find the maximum value of the flux:
Flux = 11,000 V / (50,000 Hz × 80 turns) = 0.00275 Wb
So, the maximum value of the flux in the transformer is approximately 0.00275 Weber (Wb).
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What are the differences between bulk deformation processes and sheet metal processes?.
Answers
Bulk deformation processes are applied to achieve large deformation of metal materials, resulting in a change in their overall shape. On the other hand, sheet metal processes primarily affect the surface layers of metal sheets to achieve a specific shape, size, and finish.
This fundamental difference results in several other notable differences between these two processes, which are discussed below.Bulk deformation processes:This process is used to transform the overall shape of metal materials. The material is placed in a die and subjected to high stress and pressure to achieve the desired deformation. This process often involves heating the metal to high temperatures to enhance its ductility. This process is also referred to as "forging" or "forming" processes.
Some of the common bulk deformation processes include:
Forging: This process is applied to metal materials at high temperatures to change their overall shape.
Rolling: This process involves reducing the thickness of a metal material by compressing it between two rollers.
Extrusion: This process is used to produce uniform cross-section shapes and is applied to create pipes, tubes, and other structures.
Sheet metal processes:This process primarily affects the surface of metal sheets, resulting in a specific shape, size, and finish. Sheet metal is thinner than bulk metals, making it easy to shape and deform without heating it. Sheet metal is used in many applications, such as building facades, ductwork, automotive bodies, and machine casings.
Some of the common sheet metal processes include:
Cutting: This process is used to cut a piece of sheet metal to the desired shape and size.
Bending: This process is applied to a sheet metal workpiece to achieve a specific shape.
Rolling: This process is used to reduce the thickness of sheet metal while maintaining its shape and size.
Drawing: This process is applied to sheet metal to create specific shapes, such as cups or cans.
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EXAMPLE 3.6 The possible existence of an optimum insulation thickness for radial systems is suggested by the presence of competing effects associated with an increase in this thickness. In partic- ular, although the conduction resistance increases with the addition of insulation, the con- vection resistance decreases due to increasing outer surface area. Hence there may exist an insulation thickness that minimizes heat loss by maximizing the total resistance to heat transfer. Resolve this issue by considering the following system. 3.3. Radial Systems 139 1. A thin-walled copper tube of radius r; is used to transport a low-temperature refrigerant and is at a temperature T; that is less than that of the ambient air at T. around the tube. Is there an optimum thickness associated with application of insulation to the tube? 2. Confirm the above result by computing the total thermal resistance per unit length of tube for a 10-mm-diameter tube having the following insulation thicknesses: 0, 2, 5, 10, 20, and 40 mm. The insulation is composed of cellular glass, and the outer surface convection coefficient is 5 W/m²K SOLUTION Knoun: Radius r; and temperature T; of a thin-walled copper tube to be insulated from the ambient air. Find: 1. Whether there exists an optimum insulation thickness that minimizes the heat transfer rate. 2. Thermal resistance associated with using cellular glass insulation of varying thickness. Schematic: = 5 W/m2K Alr -Insulation, Assumptions: 1. Steady-state conditions. 2. One-dimensional heat transfer in the radial (cylindrical) direction. 3. Negligible tube wall thermal resistance. 4. Constant properties for insulation. 5. Negligible radiation exchange between insulation outer surface and surroundings. Properties: Table A.3, cellular glass (285 K, assumed): k = 0.055 W/m-K. Analysis: 1. The resistance to heat transfer between the refrigerant and the air is dominated by con- duction in the insulation and convection in the air. The thermal circuit is therefore 7. 21 Inlaira 2KR where the conduction and convection resistances per unit length follow from Equations 3.33 and 3.9, respectively. The total thermal resistance per unit length of tube is then In(rin) 2π& 2π/h RO where the rate of heat transfer per unit length of tube T.-T RE An optimum insulation thickness would be associated with the value of r that minimized q' or maximized Rice Such a value could be obtained from the requirement that DR = 0 dr Hence 1=0 2 mkr 2wrh or To determine whether the foregoing result maximizes or minimizes the total resis- tance, the second derivative must be evaluated. Hence dR 1 2 krah or, at r = k/h. + dr² RO! 1 1 >0 dr? (k/h2k 2k) 2nk/h2 Since this result is always positive, it follows that r= klh is the insulation radius for which the total resistance is a minimum, not a maximum. Hence an optimum insulation thickness does not exist. From the above result it makes more sense to think in terms of a critical insulation radius Por which maximizes heat transfer, that is, below which q' increases with increasing r and above which a' decreases with increasing r. 2. With h = 5 W/m²K and k = 0.055 W/m-K, the critical radius is 0.055 W/m = 0.011 m 5 W/m²K Hence r>r; and heat transfer will increase with the addition of insulation up to a thickness of For - - r;= 0.011 -0.005) m = 0.006 m The thermal resistances corresponding to the prescribed insulation thicknesses may be calculated and are plotted as follows: cond Ri(m/w) Com 2 1 0 0 6 10 20 30 40 50 - (mm) Comments: 1. The effect of the critical radius is revealed by the fact that, even for 20 mm of insula- tion, the total resistance is not as large as the value for no insulation. 2. Ifr: re, any addition of insulation would increase the total resistance and therefore decrease the heat loss. This behavior would be desirable for steam flow through a pipe, where insu- lation is added to reduce heat loss to the surroundings. 3. For radial systems, the problem of reducing the total resistance through the application of insulation exists only for small diameter wires or tubes and for small convection coeffi- cients, such that >1,For a typical insulation (k = 0.03 W/m-K) and free convection in air (h 10 W/mK), ra = (k/h) 0.003 m. Such a small value tells us that, normally, :>rand we need not be concerned with the effects of a critical radius. 4. The existence of a critical radius requires that the heat transfer area change in the direction of transfer, as for radial conduction in a cylinder (or a sphere). In a plane wall the area per- pendicular to the direction of heat flow is constant and there is no critical insulation thick- ness (the total resistance always increases with increasing insulation thickness).
Answers
Yes, it is possible that an optimum insulation thickness exists for radial systems due to the competing effects associated with an increase in insulation thickness.
What happened after increasing insulation thickness?
On one hand, increasing the thickness of insulation can improve the overall insulation performance of the system by reducing heat loss and increasing energy efficiency. This is because thicker insulation offers more resistance to heat flow and can therefore reduce the amount of heat lost from the system.
On the other hand, increasing insulation thickness also increases the cost of the system, and can add weight and size to the overall design, which can be a disadvantage in certain applications.
Therefore, there is likely to be an optimal insulation thickness that balances the benefits of improved insulation performance with the added costs and drawbacks of thicker insulation. Finding this optimal thickness requires careful consideration of the specific application, as well as factors such as the insulation material used, the desired level of energy efficiency, and the overall cost and size constraints of the system.
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Are you?
Yes
No
omg secret message
Answers
Answer:
are you wht
didn't understand the question
Answer:
Yes/No
Explanation:
NANI?!
the _____ is to visit the tree nodes level by level. first visit the root, then all children of the root from left to right, then grandchildren of the root from left to right, and so on.
a. inorder traversal
b. preorder traversal
c. postorder traversal
d. breadth-first traversal
Answers
The breadth-first traversal is used to visit the tree nodes level by level. First, visit the root, then all children of the root from left to right, then grandchildren of the root from left to right, and so on.
Breadth-first traversal, also known as level-order traversal, is a method of traversing or visiting the nodes of a tree or graph in a breadth-first manner. It explores all the nodes at the current level before moving on to the nodes at the next level.
In breadth-first traversal, the algorithm starts at the root node and visits all the immediate children of the root node from left to right. Then, it moves on to visit the grandchildren of the root node, again from left to right. This process continues, visiting nodes level by level until all nodes have been visited.
The main idea behind breadth-first traversal is to visit the nodes in the order of their distance from the root. This traversal strategy ensures that nodes at the same level are visited before moving on to the next level.
Breadth-first traversal is typically implemented using a queue data structure. The algorithm enqueues the root node and iteratively dequeues nodes from the front of the queue, visiting them and enqueuing their children. This process continues until the queue becomes empty, indicating that all nodes have been visited.
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wing structures are subject to: group of answer choices aerodynamic loads fuel loads engine loads landing gear loads inertial loads of structural mass inertial loads of non-structural mass
Answers
wing structures are subject to: Inertial loads of structural mass and non-structural mass, aerodynamic loads, fuel loads, engine loads, landing gear loads, and inertial loads.
Wings are airfoils that produce lift when they are moved quickly through the air. They come in a variety of sizes and forms. Different wing designs might offer specific desirable flight characteristics. Control at various operating speeds, the amount of lift produced, balance, and stability all change as the shape of the wing changes. Either the wing's leading and following edges are straight or curved, or one edge is straight while the other is curved. To make the wing smaller at the tip than at the base, where it joins the fuselage, one or both edges may be tapered. Wing tips can be pointy, rounded, or even square.
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A venture tube is used to measure the flow rate of a liquid in a pipe (liquid density is 800 kg/m3). The pipe has a diameter of 10 cm and the smallest diameter of the venture has a diameter of 4 cm. A manometer with a manometer fluid of mercury (specific weight of 133 kN/m3) is used to calculate the flow rate which is connected to the venture section such that one leg is far upstream and the second leg is at the minimum diameter of the venture tube. If the flow rate is 0.05 m3/s determine the elevation change in the manometer fluid.
a. 14.6 m
b. 9.28 m
c. 4.64 m
d. 2.32 m
Answers
Answer:
\(\triangle h=4.935m\)
Explanation:
From the question we are told that:
Liquid density \(\rho=800\)
Diameter of pipe \(d=4cm \approx 0.004m\)
Diameter of venture \(d=10cm \approx 0.010m\)
Specific weight of mercury P_mg \(133 kN/m^3\)
Flow rate \(r=0.05 m^3/s\)
Area A:
\(A_1=\frac{\pi}{4}0.1^2\\A_1=0.00785m^2\\A_2=\frac{\pi}{4}0.04^2\\A_2=0.001256m^2\\\)
Generally the Bernoulli's equation is mathematically given by
\(\frac{P_1}{\rho_1g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}\\\)
Where
\(V_1=\frac{r}{A_1} \\\\ &V_1=\frac{r}{A_2}\)
Therefore
\(P_1-P_2=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})\)
Generally the equation for pressure difference b/w manometer fluid is given as
\(P_1-P_2=(p_mg-pg)\triangle h\)
Therefore
\((p_mg-pg)\triangle h=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})\)
\(\triangle h=\frac{\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})}{(p_mg-pg)}\)
\(\triangle h=\frac{\frac{(800)(0.05)^2}{2}(\frac{(0.1)^2-(0.4)^2}{(0.1)^2(0.04)^2})}{(1.33*10^3-800*9.81)}\)
\(\triangle h=4.935m\)
Therefore elevation change is mathematically given by
\(\triangle h=4.935m\)
A kite is an airfoil that uses the wind to produce a lift. Held in place by a string, a kite can remain aloft indefinitely. The diamond kite is a particularly simple kite. As the wind passes under the lower surface of the diamond kite, it bends away from the surface and slows down. The air pressure under the lower surface is
Answers
Answer:
a) results in an increased pressure.
b) results in a pressure lower than the atmospheric pressure.
c) The direction of the overall force is upwards.
d) other forces are the weight, drag, and tension on the string.
e) the extra weight due to the tail adds drag and keeps the nose of the kite pointing upwards and the kite generally stable.
Explanation:
a) As the wind passes under the lower surface of the diamond kite, it slows down. The slowing down of the wind under the lower surface results in an increased air pressure under the lower surface which now exceeds the pressure at the upper surface of the diamond kite, generating more lift upwards.
b) As the wind passes over the upper surface of the diamond kite, turbulence occurs resulting in pressure lower than the atmospheric pressure. This results in a reduce air drag on the kite.
c) The direction of the overall force exerted by the air on the kite is upwards.
d) Other force that acts on the kite in the air are; gravity force (due to the weight of the kite), drag (due to air flow around the kite), and tension force on the string from you. Once at the maximum height, the approximate net force on the kite is zero (provided the kite is stable at this height).
e) The extra weight of the kite due to the tail adds drag at the bottom of the kite, keeping the nose pointing upwards, and keeps the kite stable.
What is the purpose of the lines painted on the shop floor?
Answers
A network technician is conducting maintenance on the bonded cabled links to a single logical channel. What can be used to auto-negotiate the bonded link between the switch ports and the end system, detects configuration errors, and recovers from the failure of one of the physical links?
A. ARP
B. LACP
C. Neighbor Discovery Protocol
D. MAC Address Tables
Answers
A network technician can use the Link Aggregation Control Protocol (LACP) to auto-negotiate the bonded link between the switch ports and the end system. LACP is a part of the IEEE 802.3ad standard, which allows multiple physical connections to be combined into a single logical channel, improving bandwidth and providing redundancy.
1)LACP works by exchanging control packets between the network devices to detect configuration errors and establish a bond between the links. This protocol also monitors the status of each physical link, ensuring that if one fails, traffic is automatically redistributed across the remaining active links. This allows for seamless recovery without affecting network performance.
2)Additionally, MAC Address Tables play a vital role in the operation of the network switches. These tables store the MAC addresses of connected devices and their corresponding switch ports. By using MAC Address Tables, switches can make informed decisions on forwarding or filtering data packets to maintain efficient network traffic flow.
3)In summary, LACP is an essential tool for auto-negotiating bonded links, detecting configuration errors, and recovering from physical link failures in a network. It works alongside MAC Address Tables to ensure efficient and reliable network communication.
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In a __________, a high-pressure hose is hand-threaded in the spark plug hole of the cylinder to be tested and then connected to the compression gauge.
Answers
All of the following are derived units except ________.
Answers
The option that is not a derived units is known to be option E) kg.
What are derived units and examples?The units that are known to be used for any form of derived quantities are said to be called the derived units.
They are:
The meter per second (distance)Mole per cubic meterVolumeNote that this unit are derived because they are known to be derived because they have to be solved for using different ways. The others such as kg m-3 is one that need to be derived to arrive at it.
Therefore, based on the above, one can say that The option that is not a derived units is known to be option E) kg because it is one that cannot be derived,
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The following are derived units EXCEPT
Options
A) kg m-3
B)N
C)Ns
D)m3
E)kg
applications in different isolated vir- tual areas Question 14 (E) None of the above What best describes authorization? Question 17 (A) An acceptance of the provided identification Examples of the principle Use choke points are information (Select all that apply) (A) Having one administrator account for vari- (B) Activation of security models ous admin tasks (C) Checking the account to make sure it is valid (B) Having a firewall to inspect every incoming and outgoing packet (D) A process of defining what resources a user needs and the type of access to those re- (C) Having perimeter fences and security sources guards to monitor people in and out (E) None of the above (D) Having a spam detector to examine every incoming and outgoing emails of an organi sation Question 15 (E) None of the above
Answers
A process of defining what resources a user needs and the type of access to those re-sources is authorization.
Examples of the principal having a firewall to inspect every incoming and outgoing packet. Having perimeter fences and security sources guards to monitor people in and out. Thus, option 1. (d), 2.(b) and (d) is correct.
Authorization is the process of determining whether a verified user or process is permitted access to a certain resources or system in accordance with the security policy. A person or process must go through authentication before being allowed access to protected networks and systems.
Unauthorized access or data breaches can be avoided by using a firewall to inspect each incoming and outgoing packet. In order to monitor persons entering and exiting a certain area, security guards and perimeter fences can be deployed, which can aid in preventing unauthorized access or security breaches.
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Your question is incomplete, but most probably the full question was.
What best describes authorization?
(A) An acceptance of the provided identification information.
(B) Activation of security model's our admin tasks
(C) Checking the account to make sure it is valid
(D) A process of defining what resources a user needs and the type of access to those re-sources.
(E) None of the above
Examples of the principal Use choke points are (Select all that apply):
(A) Having one administrator account for various tasks.
(B) Having a firewall to inspect every incoming and outgoing packet.
(C) Having perimeter fences and security sources guards to monitor people in and out.
(D) Having a spam detector to examine every incoming and outgoing emails of an organi sation.
(E) None of the above
A cyclist is turning the pedals of his bicycle at a speed of 60RPM while exerting a torque of 1 lb-ft. The input sprocket has 40 teeth. What is the number of teeth on the output sprocket, if the driven wheel is rotating at 240 RPM?
Answers
Answer:
10
Explanation:
The speed of the chain is the same for both sprockets.
vᵢ = vₒ
ωᵢ rᵢ = ωₒ rₒ
The sprockets have the same pitch, so the radius is proportional to the number of teeth. So we can say:
ωᵢ nᵢ = ωₒ nₒ
Plugging in values:
(60 rpm) (40) = (240 rpm) nₒ
nₒ = 10
The output sprocket has 10 teeth. This makes sense, if the output sprocket is 4 times smaller, it will turn 4 times faster than the input sprocket.
Answer: it is 10 teeth and the guy under me already explained soo
Explanation:
According to Ref. 213/91, fire extinguishing equipment can be frozen True or False
Answers
False. Fire extinguishing equipment cannot be frozen according to Ref. 213/91.
According to Ref. 213/91, fire extinguishing equipment cannot be frozen. Fire extinguishers are essential safety devices designed to combat fires effectively. They contain pressurized agents that are specifically formulated to extinguish different types of fires. Freezing temperatures can significantly impair the functionality of fire extinguishers and render them ineffective in emergency situations.
When fire extinguishing equipment freezes, several issues can arise. First, the contents of the extinguisher may expand as they freeze, potentially leading to ruptures or leaks in the container. This can cause the extinguisher to malfunction or become hazardous when used. Second, freezing temperatures can affect the performance of the extinguishing agent itself. Certain agents, such as water-based solutions, can solidify or lose their effectiveness when exposed to extreme cold.
It is crucial to store fire extinguishers in suitable environments that are above freezing temperatures. This ensures that the equipment remains in optimal condition and is ready for immediate use during emergencies. Regular inspections and maintenance are also essential to identify any signs of damage or deterioration that may compromise the functionality of fire extinguishers.
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Project cost is ETB 30,000 and the cash inflows are ETB 10,000, the life of the project is 5
years. Calculate the pay-back period
Answers
The payback period is the length of time it takes for a project to recoup its initial cost. To calculate the payback period, we need to determine how many years it will take for the cash inflows to equal the initial project cost.
How to calculate payback period?
In this case, the project cost is ETB 30,000 and the annual cash inflows are ETB 10,000.
Payback period = Project cost / Annual cash inflows
Payback period = 30000/10000
Payback period = 3 years
It will take 3 years for the project to recoup its initial cost of ETB 30,000 with annual cash inflows of ETB 10,000.
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An electric train is moving north at 100 mph and a wind is blowing to the west at 10 mph. Which way does the smoke blow?
Answers
there is no smoke its a electric train
Which of the following best describes empathy?
the understanding of the feelings and beliefs of others
the lack of pride or boastfulness
the courage to speak up with one’s ideas
the possession of honesty and high morals
Answers
Answer:
the first one is the correct answer
Answer:
the first one would be correct
Explanation:
An LTI system has the frequency response function H()- 1/(jw +3). Compute the output if the input is: (a) x(t) 3 (b) x(t) 3V2cos(3t) (c) x(t)-5cos(4t) (d) x(t) ö(t) (e) x(t) u(t) (f) x(t)1 (a) At to 1.2, for input (a), find the output of the system: Submit Answer Tries o/3 (b) At to-1.2, for input (b), find the output of the system: Submit Answer Tries 0/3 (c) At to-1.2, for input (c), find the output of the system
Answers
The output of the LTI system for the given inputs can be found by convolving the input signal with the impulse response function. The exact numerical values depend on the specific frequency response function and input signals provided.
How to compute LTI system output?To find the output of the LTI system for the given inputs, we need to perform the convolution of the input signal with the impulse response of the system. In this case, the frequency response function H(s) is given as:
H(s) = 1 / (js + 3)
To compute the output for each input, we'll use the inverse Laplace transform to obtain the impulse response function h(t), and then perform the convolution with the respective input signals.
(a) For input x(t) = 3:
Since x(t) = 3 is a constant input, the output y(t) will be the product of the input and the impulse response function:
h(t) = InverseLaplaceTransform(H(s)) = InverseLaplaceTransform(1 / (js + 3)) = e^(-3t)
Therefore, the output y(t) = x(t) * h(t) = 3 * e^(-3t)
To find the output at t = 1.2, substitute the value of t into the expression:
y(1.2) = 3 * e^(-3 * 1.2)
(b) For input x(t) = 3V2cos(3t):
Using Euler's formula, we can rewrite the input as x(t) = 3√2 * (1/2) * (e^(j3t) + e^(-j3t)):
x(t) = (3√2/2) * e^(j3t) + (3√2/2) * e^(-j3t)
To find the output, we'll first compute the response to each complex exponential term separately using the frequency response function H(s):
For the term e^(j3t):
H(j3) = 1 / (j(j3) + 3) = 1 / (-3 + 3j)
For the term e^(-j3t):
H(-j3) = 1 / (-j(-j3) + 3) = 1 / (-3 - 3j)
Now, we can calculate the response for each term and take the real part of the result to obtain the output y(t):
y(t) = Re{H(j3) * (3√2/2) * e^(j3t)} + Re{H(-j3) * (3√2/2) * e^(-j3t)}
To find the output at t = -1.2, substitute the value of t into the expression:
y(-1.2) = Re{H(j3) * (3√2/2) * e^(j3(-1.2))} + Re{H(-j3) * (3√2/2) * e^(-j3(-1.2))}
(c) For input x(t) = -5cos(4t):
We can express the cosine function in terms of complex exponentials using Euler's formula:
cos(4t) = (1/2) * (e^(j4t) + e^(-j4t))
Substituting this into the input expression:
x(t) = -5 * (1/2) * (e^(j4t) + e^(-j4t))
Now, similar to part (b), we'll calculate the response to each complex exponential term separately using the frequency response function H(s):
For the term e^(j4t):
H(j4) = 1 / (j(j4) + 3) = 1 / (-4 + 3j)
For the term e^(-j4t):
H(-j4) = 1 / (-j(-j4) + 3) = 1 / (-4 - 3j)
The output y(t) can be calculated as:
y(t) = Re{H(j4) * (-5/2) * (e^(j4t)} + Re{H(-j4) * (-5/2) * e^(-j4t)}
To find the output at t = -1.2, substitute the value of t into the expression:
y(-1.2) = Re{H(j4) * (-5/2) * e^(j4(-1.2))} + Re{H(-j4) * (-5/2) * e^(-j4(-1.2))}
Note: To obtain the exact numerical values, you need to substitute the values of H(jw) and H(-jw) into the expressions and perform the calculations.
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IV. An annealed copper strip 9 inches wide and 2.2 inches thick, is rolled to its maximum possible draft in one pass. The following properties of annealed copper are given: strength coefficient is 90,000 psi; true strain at the onset of non-uniform deformation is 0.45; and, engineering strain at yield is 0.11. The coefficient of friction between strip and roll is 0.2. The roll radius is 14inches and the rolls rotate at 150 rpm. Calculate the roll-strip contact length. Calculate the absolute value of thetrue strain that the strip undergoes in this operation. Determine the average true stress of the strip in theroll gap. Calculate the roll force. Calculate the horsepower required.
Answers
Answer:
13.9357 horse power
Explanation:
Annealed copper
Given :
Width, b = 9 inches
Thickness, \($h_0=2.2$\) inches
K= 90,000 Psi
μ = 0.2, R = 14 inches, N = 150 rpm
For the maximum possible draft in one pass,
\($\Delta h = H_0-h_f=\mu^2R$\)
\($=0.2^2 \times 14 = 0.56$\) inches
\($h_f = 2.2 - 0.56$\)
= 1.64 inches
Roll strip contact length (L) = \($\sqrt{R(h_0-h_f)}$\)
\($=\sqrt{14 \times 0.56}$\)
= 2.8 inches
Absolute value of true strain, \($\epsilon_T$\)
\($\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$\)
Average true stress, \($\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$\) Psi
Roll force, \($L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$\)
= 788,900 lb
For SI units,
Power = \($\frac{2 \pi FLN}{60}$\)
\($=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$\)
= 10399.81168 W
Horse power = 13.9357
Hi im ***ar and im doing sculptural but what should it be about star wars or Marvel
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Answer:
Cool I think u should do Marvel first
A cylindrical bar of metal having a diameter of 20.5 mm and a length of 201 mm is deformed elastically in tension with a force of 46300 N. Given that the elastic modulus and Poisson's ratio of the metal are 60.5 GPa and 0.33, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.
Answers
Answer:
a) The amount by which this specimen will elongate in the direction of the applied stress is 0.466 mm
b) The change in diameter of the specimen is - 0.015 mm
Explanation:
Given the data in the question;
(a) The amount by which this specimen will elongate in the direction of the applied stress.
First we find the area of the cross section of the specimen
A = \(\frac{\pi }{4}\) d²
our given diameter is 20.5 mm so we substitute
A = \(\frac{\pi }{4}\) ( 20.5 mm )²
A = 330.06 mm²
Next, we find the change in length of the specimen using young's modulus formula
E = σ/∈
E = P/A × L/ΔL
ΔL = PL/AE
P is force ( 46300 N), L is length ( 201 mm ), A is area ( 330.06 mm² ) and E is elastic modulus (60.5 GPa) = 60.5 × 10⁹ N/m² = 60500 N/mm²
so we substitute
ΔL = (46300 N × 201 mm) / ( 330.06 mm² × 60500 N/mm² )
ΔL = 0.466 mm
Therefore, The amount by which this specimen will elongate in the direction of the applied stress is 0.466 mm
(b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.
Using the following relation for Poisson ratio
μ = - Δd/d / ΔL/L
given that Poisson's ratio of the metal is 0.33
so we substitute
0.33 = - Δd/20.5 / 0.466/201
0.33 = - Δd201 / 20.5 × 0.466
0.33 = - Δd201 / 9.143
0.33 × 9.143 = - Δd201
3.01719 = -Δd201
Δd = 3.01719 / - 201
Δd = - 0.015 mm
Therefore, The change in diameter of the specimen is - 0.015 mm
Which term is defined as the study of sound waves and their behaviors and interactions? propulsion acoustics hydraulics thermodynamics
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Answer: i believe it’s acoustics
Explanation:
Question 2 Resilience engineering is concerned with adverse external events that can lead to system failure. Resilient systems are flexible and adaptable so that they can cope with the unexpected. As a software engineer you need to educate system developers of four characteristics as outlined by Hollnagel (2010) that reflect the resilience of an organisation. Make sure to also include an example for each characteristic
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These four characteristics, as outlined by Hollnagel (2010), are critical in developing resilient systems that can cope with unexpected situations and adverse external events. Software engineers must educate system developers about these characteristics to develop more resilient systems.
Resilience engineering is about developing adaptable systems that can handle unexpected situations and are able to cope with the effects of adverse external events that might cause system failure. As a software engineer, you need to inform system developers about the following four characteristics that reflect the resilience of an organization, as described by Hollnagel (2010):Maintainability: This characteristic reflects the degree to which a system can be maintained or repaired after it has been damaged. In other words, it assesses the system's ability to remain in good working order or quickly recover from damage. An example of maintainability would be the ability to quickly repair an engine that has been damaged during an accident.Flexibility: This characteristic reflects the degree to which a system can be modified or adapted to cope with changing circumstances. Flexibility is essential for resilience because it enables a system to respond to new challenges and adapt to different circumstances. An example of flexibility would be the ability to change the specifications of a car to adapt to different driving conditions.Redundancy: This characteristic reflects the degree to which a system can continue to function even if some of its components fail. Redundancy is important because it ensures that the system can continue to operate even if one or more components are not working properly. An example of redundancy would be having a backup generator in case the primary generator fails.Responsiveness: This characteristic reflects the degree to which a system can respond to changing circumstances or threats. Responsiveness is important because it enables a system to quickly and effectively respond to unexpected events. An example of responsiveness would be the ability of an air traffic control system to quickly respond to changing weather conditions to ensure the safety of airplanes in the area.
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An engineer wants to design a knee replacement.Identify one criterion the design should meet.
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A "wet" combustion test kit measures which two substances?A. Soot and carbon monoxideB. Carbon dioxide and oxygenC. Carbon monoxide and oxygenD. Carbon monoxide and carbon dioxide
Answers
The correct response is B. Carbon dioxide and oxygen. A "wet" combustion test kit measures Carbon dioxide and oxygen.
One carbon atom is covalently doubly linked to two oxygen atoms in every carbon dioxide molecule, giving it the chemical formula CO2. At room temperature, it exists in the gaseous state. Carbon dioxide functions as a greenhouse gas in the atmosphere by being transparent to visible light while absorbing infrared radiation. As of May 2022, it was a trace gas in the Earth's atmosphere, up from pre-industrial levels of 280 parts per million (ppm), or roughly 0.04% by volume. These elevated CO2 concentrations and climate change are mostly brought about by the burning of fossil fuels. Groundwater, lakes, ice caps, and ocean all contain carbon dioxide because it is soluble in water. Ocean acidification is brought on by the formation of carbonate, primarily bicarbonate (HCO 3), when carbon dioxide dissolves in water.
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(6) effects of urbanization on the urban hydrology
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Answer:
123456
Explanation:
to produce cooling the refrigerant must
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Answer:
The compressor
Explanation:
to compress the low-pressure dry gas refrigerant from the evaporator and raise its pressure and temperature to that of the condenser, to produce flow around the system.
A simply supported wood beam is subjected to uniformly distributed load q. The width of the beam is 6 in. And the height is 8 in. Determine the normal stress and the shear stress at point C. Show these stresses on a sketch of a stress element at point C
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The question is missing some important details but I will guide you on how to solve for normal stress
How to solve for normal stressThe point in the beam has got a normal stress (σ) which can be calculated by taking the ratio of the bending moment (M) and section modulus (S):
σ = M/S
To ascertain the section modulus of a rectangular cross-section, the following equation is employed:
S = (b * h^2) / 6
In answer to the supplicated inquiry, within this example, b stands for the width of the beam (6 in), and h represents the height of the beam (8 in). Hence, by accounting for these dimensions, S can be calculated.
The effect of a uniformly distributed load (q) on an elicited bending moment (M) can be determined through utilization of the pertinent beam equations, that can be derived once the location of point C, as well as the length of the demonstrated beam is known.
Having notably acknowledged M and S, it is possible to estimate the normal stress (σ) at point C.
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