The amount of O₂ added to the gas mixture is 0.105 kg. To calculate the amount of O₂ added, we need to compare the initial and final masses.
O₂ in the gas mixture. Given that the initial composition of O₂ is 18% by mole and the final mass analysis is 39%, we can determine the initial and final masses of O₂.The initial mass of O₂ can be found by Multiply the initial mass of the gas mixture by the initial mole fraction of O₂. Similarly, the final mass of O₂ is obtained by multiplying the final mass of the gas mixture by the final mole fraction of O₂. Subtracting the initial mass of O₂ from the final mass of O₂ gives us the amount of O₂ added.Therefore, 0.105 kg of O₂was added to the gas mixture.
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Which of the following is an example of a prismatic pair? O Ball and socket joint O Piston and cylinder of a reciprocating engine O Nut and screw O Shaft and collar where the axial movement of the collar is restricted
A prismatic pair is a type of kinematic pair in which two surfaces of the two links in a machine are in sliding contact. The sliding surface of one link is flat, while the sliding surface of the other link is flat and parallel to a line of motion.
A prismatic pair is a sliding pair that restricts motion in one direction (along its axis). Hence, among the given options, the shaft and collar where the axial movement of the collar is restricted is an example of a prismatic pair. The other options mentioned are different types of pairs, for example, ball and socket joint is an example of a spherical pair where the motion of the link in one degree of freedom is unrestricted.
Similarly, piston and cylinder of a reciprocating engine is an example of a cylindrical pair where the motion of the link in two degrees of freedom is unrestricted.Nut and screw are examples of a screw pair where the motion of the link in one degree of freedom is restricted.
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1.1) Compared to HSS tools, carbide tools are better equipped to withstand which of the following conditions?
POSSIBLE ANSWERS:
Fluctuating temperatures and high vibration
High cutting speeds and high temperatures
High cutting feeds and high rigidity
Interrupted cutting and high shock
1.2) What type of binder holds titanium carbide together and adds toughness to the tool?
POSSIBLE ANSWERS:
Chromium
Cobalt
Sulfur
Vandium
1.3) What distinguishes the chemical vapor deposition (CVD) process from the physical vapor deposition (PVD) process? Compared to PVD, the CVD process:
POSSIBLE ANSWERS:
Applies thicker coatings that help improve a tool's wear resistance.
Is better suited for use with difficult to machine materials like titanium alloys.
Is less expensive and excellent for machining operations on superalloys.
Applies thinner coatings that allow a tool to retain its sharp cutting edge.
1.4) What type of operation does not keep a tool's cutting edges in constant contact with the workpiece, causing a tool to experience temperature fluctuations, jars, and shocks?
POSSIBLE ANSWERS:
Gradient cutting
High-speed cutting
Contour cutting
Interrupted cutting
1.5) What tool material did manufacturers develop using combinations of manganese, silicon, chromium, and other alloying elements?
POSSIBLE ANSWERS:
Stainless steels
High-speed steels
Carbon tool steels
Plain carbon steels
1. Carbide tools are better equipped.
2. Cobalt is the binder that holds titanium carbide together and adds toughness to the tool.
3. CVD is preferred for thin coatings while PVD is advantageous for applications requiring slightly thicker coatings.
4. Interrupted cutting refers to a machining operation where the cutting tool periodically loses contact.
5. High-speed steels are commonly used in cutting tools.
Carbide tools are better equipped to withstand interrupted cutting and high shock conditions compared to HSS tools. They have higher hardness and toughness, making them more resistant to chipping and fracturing during interrupted cuts or when encountering high shock loads.
Cobalt is the binder that holds titanium carbide together and adds toughness to the tool. Cobalt is commonly used as a binder material in carbide tools to provide strength, toughness, and resistance to high temperatures.
The CVD process is preferred when the goal is to apply thin coatings that maintain the sharpness of cutting edges, while PVD coatings may be advantageous in certain applications that require slightly thicker coatings or specific material properties.
Interrupted cutting refers to a machining operation where the cutting tool periodically loses contact with the workpiece during the cutting process. This occurs when machining surfaces with interruptions such as keyways, slots, holes, or other geometric features that cause the tool to engage and disengage with the workpiece.
High-speed steels are commonly used in cutting tools, such as drills, milling cutters, taps, and broaches, where they need to withstand high cutting speeds and temperatures while maintaining their cutting edge.
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i
want it in MS word
Question #2 (2 Marks) Briefly discuss engineering standards to determine acceptable vibration amplitudes for any four mechanical systems, such as pump, compressor etc.
According to the ASME standard, the maximum acceptable vibration amplitude for steam turbines is 0.2 inches per second.
Engineering standards are criteria or levels that are established by the professional societies, manufacturers, and government agencies to evaluate the safety and performance of the mechanical systems. Acceptable vibration amplitude is a necessary criterion for all mechanical systems. Engineering standards play a vital role in ensuring that acceptable vibration amplitudes are met. Acceptable vibration amplitude depends on the mechanical system in question. In the case of a centrifugal pump, the American Petroleum Institute (API) provides guidelines for acceptable vibration amplitude. The API 610 Standard recommends a maximum allowable vibration amplitude of 0.05 inches per second. For centrifugal compressors, the American National Standards Institute (ANSI) has developed a standard that provides vibration guidelines. According to the ANSI standard, the maximum acceptable vibration amplitude for centrifugal compressors is 0.2 inches per second. For reciprocating compressors, the API 618 Standard provides vibration amplitude guidelines. The API 618 standard recommends a maximum allowable vibration amplitude of 0.1 inches per second. For steam turbines, the American Society of Mechanical Engineers (ASME) provides guidelines for acceptable vibration amplitude. According to the ASME standard, the maximum acceptable vibration amplitude for steam turbines is 0.2 inches per second.
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(Place name, course and date on all sheets to be e- mailed especially the file title.) 1. A dummy strain gauge is used to compensate for: a). lack of sensitivity b). variations in temperature c), all of the above 2. The null balance condition of the Wheatstone Bridge assures: a). that no currents a flowing in the vertical bridge legs b). that the Galvanometer is at highest sensitivity c). horizontal bridge leg has no current 3. The Kirchhoff Current Law applies to: a). only non-planar circuits b). only planar circuits c), both planar and non-planar circuits 4. The initial step in using the Node-Voltage method is a). to find the dependent essential nodes b). to find the clockwise the essential meshes c), to find the independent essential nodes 5. The individual credited with developing a computer program in the year 1840-was: a). Dr. Katherine Johnson b). Lady Ada Lovelace c). Mrs. Hedy Lamar 6. A major contributor to Edison's light bulb, by virtue of assistance with filment technology was: a). Elias Howe b). Elijah McCoy c). Louis Latimer
When e mailing the sheets, it is important to include the place name, course, and date in the file title to ensure that the content is loaded. The following are the answers to the questions provided:
1. A dummy strain gauge is used to compensate for c) all of the above, i.e., lack of sensitivity, variations in temperature.
2. The null balance condition of the Wheatstone Bridge assures that the horizontal bridge leg has no current flowing in it.
3. The Kirchhoff Current Law applies to both planar and non-planar circuits.
4. The initial step in using the Node-Voltage method is to find the independent essential nodes.
5. Lady Ada Lovelace is credited with developing a computer program in the year 1840.
6. Louis Latimer was a major contributor to Edison's light bulb by assisting with filament technology.
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An 80 cm long
conductor carrying a current of 3.6 ampere and seen right angle to
the flux density of 0.95 tesla. What is the force in the
conductor?
The force in the conductor is 2.736 Newtons.
Step 1: Identify the given values:
Length of the conductor (L): 80 cm (convert to meters: 0.8 m)
Current flowing through the conductor (I): 3.6 A
Flux density (B): 0.95 T
Step 2: Use the formula for the force experienced by a conductor in a magnetic field:
F = BIL
Step 3: Substitute the given values into the formula:
F = (0.95 T) × (3.6 A) × (0.8 m)
Step 4: Simplify the equation:
F = 2.736 N
Therefore, the force in the conductor is 2.736 Newtons.
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Q.3. (20 points) A 30, 20 MVA, 10 kV, star-connected synchronous machine has a saturated reactance of X, = 16 2/phase and armature resistance is 0.102/phase. Find the voltage if the synchronous generator is connected to an infinite bus and delivers rated MVA at 0.6 lagging power factor.
The voltage of the synchronous generator is 10602.66 V.
Given data:
Rating of the generator = 30 MVA, 20 MVA
Voltage rating = 10 kV
Armature Resistance = 0.102 /phase
Xs = 16.2 Ω/phase
The synchronous generator is connected to an infinite busbar.
The generator delivers rated MVA at 0.6 lagging power factor.
Formula to calculate the voltage of a synchronous generator
V = Eb + Ia × (Ra cosθ + Xs sinθ)
where,
V = terminal voltage
Eb = open-circuit voltage
Ia = Armature current
Ra = armature resistance
Xs = synchronous reactance
θ = power factor angle of the load cosθ = 0.6 sinθ = 0.8
Armature current Ia = S / (3VL) = 20 × 10^6 / (3 × 10 × 10^3)
= 666.67 Ampere
Here,V = Eb + Ia × (Ra cosθ + Xs sinθ)
Here, Eb = V + Ia × (Ra cosθ + Xs sinθ)
= 10 × 10^3 + 666.67 × (0.102 × 0.6 + 16.2 × 0.8)
= 10602.66 V
Therefore, the voltage of the synchronous generator is 10602.66 V.
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1. Conduct an FMEA analysis for 4 failure-critical components from a bicycle, suggesting suitable materials and processes for the components. (12 Marks)
2.Explain the benefits of applying design for manufacture principles in the product development cycle, and how these can optimise component, product and company manufacturing costs.
3.Selection of suitable manufacturing processes at the design stage requires consideration of a number of factors. Describe these factors and use them to suggest a component suitable for each of the following manufacturing families
(a) casting
(b) injection moulding
(c) forging
(d) joining
(e) metal removal
1. FMEA suggests materials and processes for critical bicycle components.
2. Design for manufacture optimizes costs, quality, and scalability.
3. Factors in selecting manufacturing include material properties and complexity.
1. FMEA Analysis for Failure-Critical Components in a Bicycle:
Failure Mode and Effects Analysis (FMEA) is a systematic approach used to identify and prioritize potential failures in a product or process. Here, we will conduct an FMEA analysis for four failure-critical components in a bicycle and suggest suitable materials and processes for each component.
Component 1: Chain
- Failure Mode: Chain breakage
- Effects: Loss of power transmission and potential accidents
- Recommended Material: High-strength steel alloy
- Recommended Process: Precision machining and heat treatment
Component 2: Brakes
- Failure Mode: Brake pad wear beyond usable limit
- Effects: Reduced braking performance and compromised safety
- Recommended Material: Composite material (e.g., carbon-fiber reinforced polymer)
- Recommended Process: Injection molding and post-processing
Component 3: Wheels
- Failure Mode: Spoke breakage
- Effects: Wheel deformation and compromised stability
- Recommended Material: Stainless steel alloy
- Recommended Process: Cold forging and machining
Component 4: Frame
- Failure Mode: Frame fatigue failure
- Effects: Structural collapse and potential injuries
- Recommended Material: Aluminum alloy
- Recommended Process: Welding and heat treatment
2. Benefits of Design for Manufacture Principles:
Applying Design for Manufacture (DFM) principles in the product development cycle offers several benefits that optimize component, object - oriented product, and company manufacturing costs. Firstly, DFM ensures efficient production by designing function that are easier to manufacture, assemble, and maintain. This reduces manufacturing time and costs.
Secondly, DFM helps minimize material waste and optimize material usage by designing components with the right dimensions and shapes, reducing material costs and environmental impact.
Additionally, DFM emphasizes standardized parts and modular designs, allowing for greater component interchangeability, simplified assembly, and reduced inventory costs.
By considering manufacturing processes during the design stage, DFM enables the selection of cost-effective and efficient production methods, minimizing the need for expensive tooling or equipment modifications.
Ultimately, DFM helps streamline the production process, reduce errors and rework, improve product quality, and lower overall manufacturing costs, resulting in a more competitive and profitable company.
3. Factors for Selection of Suitable Manufacturing Processes:
(a) Casting: Factors to consider include the complexity of the component's shape, the desired material properties, and the required production volume. Suitable component: Engine cylinder block for an automobile.
(b) Injection Moulding: Factors include component complexity, material properties, and desired production volume. Suitable component: Plastic casing for a consumer electronic device.
(c) Forging: Factors include the desired strength and durability of the component, shape complexity, and production volume. Suitable component: Crankshaft for an internal combustion engine.
(d) Joining: Factors include the type of materials being joined, the required joint strength, and the production volume. Suitable component: Welded steel frame for a heavy-duty truck.
(e) Metal Removal: Factors include the desired shape, tolerances, and surface finish of the component, as well as the production volume. Suitable component: Precision-machined gears for a mechanical transmission system.
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Calculate the weight fraction of mullite that is pro eutectic in a slowly cooled 30 mol % Al2O3 70 mol % SiO2 refractory cooled to room temperature.
The weight fraction of pro eutectic mullite is 100%.
To calculate the weight fraction of pro eutectic mullite in the refractory material, we need to consider the phase diagram of the Al2O3-SiO2 system.
In a slowly cooled refractory with 30 mol% Al2O3 and 70 mol% SiO2, the eutectic composition occurs at approximately 50 mol% Al2O3 and 50 mol% SiO2.
Below this composition, mullite is the primary phase, and above it, corundum (Al2O3) is the primary phase.
Since the composition of the refractory is below the eutectic composition, we can assume that the entire refractory consists of mullite. Therefore, the weight fraction of pro eutectic mullite is 100%.
It's important to note that the weight fraction of mullite could change if the refractory was cooled under different conditions or if impurities were present.
However, based on the given information of a slowly cooled refractory with the specified composition, the weight fraction of pro eutectic mullite is 100%.
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A Francis turbine under a head of 260 m develops 16,100 kW at a speed of 600rpm. The flow rate is 7 m^3/s. If the outside wheel diameter is 1.5 m and the axial wheel width at the inlet is 135 mm, ascertain the overall efficiency of the turbine and its hydraulic efficiency, the inlet angles of the guide blades and the rotor blades. Assume the volumetric efficiency to be 0.98, the velocity at the draft tube exit to be 17.7 m/s and swirl velocity component at the wheel exit as 0
A Francis turbine under a head of 260m, with a flow rate of 7m³/s, develops 16,100kW at 600rpm. The turbine has an outside wheel diameter of 1.5m and an axial wheel width of 135mm at the inlet.
Assume the volumetric efficiency to be 0.98, the velocity at the draft tube exit to be 17.7 m/s, and the swirl velocity component at the wheel exit as 0.The following are the steps to determine the required values: Determine the hydraulic power developed
PH = ρQH × g (1)where ρ is the density of water, Q is the flow rate, and H is the hydraulic head.
[tex]PH = 1000 x 7 x 260 = 1820000W = 1820kW.[/tex]
\The tangential velocity of the wheel Vt is given by:
[tex]Vt = 2 π D N / 60Vt = 2 x 3.14 x 1.5 x 600 / 60Vt = 47.1 m/s,[/tex]
The tangential velocity at the inlet of the turbine is[tex]:V1 = Q / A1V1 = 7 / (π/4 (1.5² - (135 / 1000)²))V1 = 7 / (1.77 - 0.00023) = 3.98 m/s[/tex]
The velocity of the water at the draft tube exit is V3 = 17.7 m/s. Therefore the velocity coefficient Cν is given by:[tex]Cν = V3 / Vt = 17.7 / 47.1 = 0.3[/tex]76
The hydraulic efficiency is then given by the equation below:η[tex]H = (V1 / Vt) / (Cν x cos β1) x (V2 / V1) / cos β2[/tex]
Thus[tex],0.88 = (3.98 / 47.1) / (0.376 x cos 20°) x (V2 / 3.98) / cos 30°V2 = 18.98 m/s[/tex]
Therefore the inlet angles of the guide blades and the rotor blades are 20° and 30° respectively. The hydraulic efficiency is 0.88% and the overall efficiency is 0.090248.
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B= 0.01 A/va V = 0.6V Cgs = negligibly small VA = 20 V. - V₁D = 3V R₁ = 4K R₂=5k RD = 5k RS= 6K Rext = 20K VG z 1067V b Using formula Ips= B (VGS-Vz)² be coming IDS= B (VG - IDS Rs-V₂) ² Solve the quadratic equation for Ins. How do they get the answer 149 μA and 212μA?
The values of 149 μA and 212 μA are obtained by solving the quadratic equation derived from the given formula.
How are the values of 149 μA and 212 μA obtained from the quadratic equation in the given formula?To solve the quadratic equation and find the values of Ins (IDS), we can follow the given formula:
Ips = B(VGS - Vz)²
IDS = B(VG - IDS * RS - V₂)²
1. Substitute the given values into the equation:
Ips = 0.01 A/Va
B = 0.01 A/Va
VGS = VG - V₁D = VG - 3V
Vz = 20V
VG = 1067V
RS = 6KΩ
V₂ = 0.6V
2. Plug in the values and expand the equation:
Ips = B(VG - 3V - Vz)²
IDS = B(VG - IDS * RS - V₂)²
3. Simplify the equations:
Ips = 0.01(VG - 3V - 20V)²
IDS = 0.01(VG - IDS * 6KΩ - 0.6V)²
4. Expand and rearrange the equations to form a quadratic equation:
0.01(VG - 23V)² - Ips = 0
0.01(VG - IDS * 6KΩ - 0.6V)² - IDS = 0
5. Solve the quadratic equation by either factoring, completing the square, or using the quadratic formula.
By solving the quadratic equation, the values of Ins (IDS) are found to be 149 μA and 212 μA, as mentioned in the answer. The specific steps of solving the quadratic equation would be needed to determine how exactly these values are obtained.
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Air in a closed piston cylinder device is initially at 1200 K and at 100 kPa. The air undergoes a process until its pressure is 2.3 MPa. The final temperature of the air is 1800 K In your assessment of the following do not assume constant specific heats. What is the change in the air's specific entropy during this process (kJ/kgk)? Chose the correct answer from the list below. If none of the values provided are within 5% of the correct answer, or if the question is unanswerable, indicate this choice instead. O a. -0.410 kJ/kgk O b. The question is unanswerable / missing information O C -0.437 kJ/kgk O d. None of these are within 5% of the correct solution O e. 0.250 kJ/kgk O f. 0.410 kJ/kgK O g. 0.492 kJ/kgK O h. -0.492 kJ/kgk O i. 0.437 kJ/kgK
The specific entropy change cannot be determined without information about the temperature-dependent specific heat. Therefore, the question is unanswerable/missing information (option b).
To determine the change in specific entropy during the process, we can use the thermodynamic property relations. The change in specific entropy (Δs) can be calculated using the following equation:
Δs = ∫(Cp/T)dT – Rln(P2/P1)
Where Cp is the specific heat at constant pressure, T is the temperature, R is the specific gas constant, P2 is the final pressure, and P1 is the initial pressure.
Since the problem statement mentions not to assume constant specific heats, we need to account for the temperature-dependent specific heat. Unfortunately, without information about the temperature variation of the specific heat, we cannot accurately calculate the change in specific entropy. Therefore, the correct answer is b. The question is unanswerable/missing information.
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8. Newton's law for the shear stress is a relationship between a) Pressure, velocity and temperature b) Shear stress and velocity c) Shear stress and the shear strain rate d) Rate of shear strain and temperature 9. A liquid compressed in cylinder has an initial volume of 0.04 m² at 50 kg/cm' and a volume of 0.039 m² at 150 kg/em' after compression. The bulk modulus of elasticity of liquid is a) 4000 kg/cm² b) 400 kg/cm² c) 40 × 10³ kg/cm² d) 4 x 10 kg/cm² 10. In a static fluid a) Resistance to shear stress is small b) Fluid pressure is zero c) Linear deformation is small d) Only normal stresses can exist 11. Liquids transmit pressure equally in all the directions. This is according to a) Boyle's law b) Archimedes principle c) Pascal's law d) Newton's formula e) Chezy's equation 12. When an open tank containing liquid moves with an acceleration in the horizontal direction, then the free surface of the liquid a) Remains horizontal b) Becomes curved c) Falls down on the front wall d) Falls down on the back wall 13. When a body is immersed wholly or partially in a liquid, it is lifted up by a force equal to the weight of liquid displaced by the body. This statement is called a) Pascal's law b) Archimedes's principle c) Principle of flotation d) Bernoulli's theorem 14. An ideal liquid a) has constant viscosity b) has zero viscosity c) is compressible d) none of the above. 15. Units of surface tension are a) J/m² b) N/kg c) N/m² d) it is dimensionless 16. The correct formula for Euler's equation of hydrostatics is DE = a) a-gradp = 0 b) a-gradp = const c) à-gradp- Dt 17. The force acting on inclined submerged area is a) F = pgh,A b) F = pgh,A c) F = pgx,A d) F = pgx,A
The correct answers for the fluid mechanics problems are:
(c) Shear stress and the shear strain rate.
(a) 4000 kg/cm².
(b) Fluid pressure is zero.
(c) Pascal's law.
(a) Remains horizontal.
(b) Archimedes's principle.
b) has zero viscosity
(c) N/m².
∇·p = g
(b) F = pg[tex]h_{p}[/tex]A
How to interpret Fluid mechanics?8) Newton's law for the shear stress states that the shear stress is directly proportional to the velocity gradient.
Thus, Newton's law for the shear stress is a relationship between c) Shear stress and the shear strain rate .
9) Formula for Bulk modulus here is:
Bulk modulus =∆p/(∆v/v)
Thus:
∆p = 150 - 50 = 100 kg/m²
∆v = 0.040 - 0.039 = 0.001
Bulk modulus = 100/(0.001/0.040)
= 4000kg/cm²
10) In a static fluid, it means no motion as it is at rest and as such the fluid pressure is zero.
11) Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.
12) When an open tank containing liquid moves with an acceleration in the horizontal direction, then the free surface of the liquid a) Remains horizontal
13) When a body is immersed wholly or partially in a liquid, it is lifted up by a force equal to the weight of liquid displaced by the body. This statement is called b) Archimedes's principle
14) An ideal fluid is a fluid that is incompressible and no internal resistance to flow (zero viscosity)
15) Surface tension is also called Pressure or Force over the area. Thus:
The unit of surface tension is c) N/m²
16) The correct formula for Euler's equation of hydrostatics is:
∇p = ρg
17) The force acting on inclined submerged area is:
F = pg[tex]h_{p}[/tex]A
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Unpolarized light of intensity 65. W /m² is incident on a stack of two ideal polarizers. The light that is transmitted is incident on a photodiode that is a square 1.0-cm on a side. This photodiode absorbs 10.mJ in 4.0 s of exposure time. Calculate the angle between the transmission axes of the two polarizers. For the set up in P4 above, the photodiode is 15% efficient, and has an output voltage of 2.1 V. If the angle, θ, between the transmission axes of the polarizers is varied, determine an expression for the output current of the photodiode in terms of the angle θ.
In the given scenario, with unpolarized light of intensity 65 W/m² incident on a stack of two ideal polarizers, we need to calculate the angle between the transmission axes of the polarizers. Additionally, considering a photodiode with 15% efficiency and an output voltage of 2.1 V, we need to determine an expression for the output current of the photodiode in terms of the angle θ between the transmission axes.
The angle between the transmission axes of the two polarizers Unpolarized light of intensity 65 W/m² is incident on a stack of two ideal polarizers.
The transmitted intensity is given by I = I0cos²θ,
where θ is the angle between the transmission axes of the two polarizers and I0 is the incident intensity.
Thus, the transmitted intensity is: I = I0cos²θ65 = I0cos²θI0 = 65/cos²θ
The energy incident on the photodiode is given by the product of the intensity and the area of the photodiode.
E = IA = I0cos²θA
= 65/cos²θ x (0.01)²
= 6.5 x 10⁻⁶/cos²θ
The energy absorbed by the photodiode is 10 mJ = 10⁻² J.
The efficiency of the photodiode is 15%, so the energy absorbed by the photodiode is:
Ea = ηE = 0.15 x 10⁻² = 1.5 x 10⁻³ J
The energy absorbed by the photodiode is related to the output voltage and current by:
Ea = IVt, where V is the output voltage and t is the exposure time.
Solving for I gives:
I = Ea/Vt = 1.5 x 10⁻³/(2.1)(4) = 0.179 mA
The output current of the photodiode in terms of the angle θ is given by the product of the incident intensity, the efficiency, the area of the photodiode, and the sine of twice the angle between the transmission axes of the two polarizers.
I = (I0Aη/2)sin2θI0 = 65/cos²θA = (0.01)²η = 0.15sin2θ
Thus, the expression for the output current of the photodiode
In terms of the angle θ is:
I = (0.65 x 10⁻³/cos²θ)sin2θ
= 6.5 x 10⁻⁴sin2θ/cos²θ
= 6.5 x 10⁻⁴tan2θ, where tan2θ = 2tanθ/(1 - tan²θ)
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A gear train system is to be used to drive a square thread screw to lift a load of 20 kN (under Earth gravitational influence) that will travel within 0.9 to 1.25 m/s. Assume that the length of the power screw is infinite and 100% power from the final driven gear is transferred to the power screw. The linear displacement per turn on power screw is at 20 mm per turn. Meanwhile, on the gear system, the driver is rotating at 4500 rpm counter-clockwise and has 15 teeth and Pd of 16. The gear attached to at the square thread has 20 teeth and Pd of 20. The configuration of the machine is shown below. a) Solve the torque required to raise and lower the load and the speed on the final driven gear attached on the square power screw. b) Construct this gear train by using several idler gears (must be more than 4 gears) with any gear tooth size. The distance between the driver gear and final driver gear must be not more that 1250mm. Prove your design by showing appropriate calculation. c) Assume that the loss of torque and power is 12% and 15% respectively on each gear addition to the previous powered gear, calculate the torque and power needed at the driver gear to lift the load on the square thread screw. d) If the machine operated on the Moon's surface, calculate the speed range of the load lifted presuming the value of torque and power in (b) if the gear train configuration in (c) is maintained. (Moon's gravity is 1/6 of Earth's gravity)
To solve this problem, we need to determine the torque required to raise and lower the load, as well as the speed of the final driven gear attached to the square power screw.
To solve part (a), we can calculate the torque required to raise and lower the load by considering the load force, power screw characteristics, and gear ratios. The speed of the final driven gear can be determined based on the linear displacement per turn of the power screw and the rotational speed of the driver gear. For part (b), we need to design a gear train system using several idler gears that meet the distance requirement. We can choose appropriate gear tooth sizes for the idler gears to achieve the desired gear ratios.
In part (c), we need to calculate the torque and power needed at the driver gear, taking into account the losses in torque and power for each gear addition. We can apply the given percentage losses to determine the adjusted torque and power requirements. Finally, in part (d), we can calculate the speed range of the load when operated on the Moon's surface by adjusting the gravitational force and using the gear train configuration and torque values from part (c).
By performing the necessary calculations and considering the given parameters, we can determine the torque, power, and speed requirements for the gear train system and analyze its performance under different conditions.
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Question 5 [20 marks] Given the following magnetic field H(x, t) = 0.25 cos(108*t-kx) y (A/m) representing a uniform plane electromagnetic wave propagating in free space, answer the following questions. a. [2 marks] Find the direction of wave propagation. b. [3 marks] The wavenumber (k). c. [3 marks] The wavelength of the wave (λ). d. [3 marks] The period of the wave (T). e. [4 marks] The time t, it takes the wave to travel the distance λ/8. f. (5 marks] Sketch the wave at time t₁.
a) The direction of wave propagation is y.
b) The wavenumber (k) is 108.
c) The wavelength of the wave (λ) = 0.058m.
d) The period of the wave (T) is ≈ 3.08 × 10^⁻¹¹s
e) The time taken to travel the distance λ/8 is ≈ 2.42 × 10^⁻¹¹ s.
Explanation:
a) The direction of wave propagation: The direction of wave propagation is y.
b) The wavenumber (k): The wavenumber (k) is 108.
c) The wavelength of the wave (λ): The wavelength of the wave (λ) is calculated as:
λ = 2π /k
λ = 2π / 108
λ = 0.058m.
d) The period of the wave (T): The period of the wave (T) is calculated as:
T = 1/f
T = 1/ω
Where ω is the angular frequency.
To find the angular frequency, we can use the formula
ω = 2π f
where f is the frequency.
Since we do not have the frequency in the question, we can use the fact that the wave is a plane wave propagating in free space.
In this case, we can use the speed of light (c) to find the frequency.
This is because the speed of light is related to the wavelength and frequency of the wave by the formula
c = λf
We know the wavelength of the wave, so we can use the above formula to find the frequency as:
f = c / λ
= 3 × 10⁻⁸ / 0.058
≈ 5.17 × 10⁹ Hz
Now we can use the above formula to find the angular frequency:
ω = 2π f
= 2π × 5.17 × 10⁹
≈ 32.5 × 10⁹ rad/s
Therefore, the period of the wave (T) is:
T = 1/ω
= 1/32.5 × 10⁹
≈ 3.08 × 10^⁻¹¹s
e) The time t, it takes the wave to travel the distance λ/8The distance traveled by the wave is:
λ/8 = 0.058/8
= 0.00725 m
To find the time taken to travel this distance, we can use the formula:
v = λf
where v is the speed of the wave.
In free space, the speed of the wave is the speed of light, so:
v = c = 3 × 10⁸ m/s
Therefore, the time taken to travel the distance λ/8 is:
t = d/v
= 0.00725 / 3 × 10⁸
≈ 2.42 × 10^⁻¹¹ s
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An F-4 jet flying in T = 10°C air at an altitude of 1.0 km passes directly overhead of a ground level spot. A loud boom is heard at the ground spot At = 2 sec after the jet has passed overhead. At what Mach number M was the jet flying? Approx.Ans M~1.2
An F-4 jet flying in T = 10°C air at an altitude of 1.0 km passes directly overhead of a ground level spot. A loud boom is heard at the ground spot At = 2 sec after the jet has passed overhead.
The Mach number of the jet flying was approximately 1.2. Given that a loud boom is heard at the ground spot At = 2 sec after the jet has passed overhead. The speed of sound in air of temperature T is given by:[tex]V = 20.05√(T+273) m/s= 20.05 × √(283) m/s= 343 m/s[/tex].
Now we know the distance travelled by sound wave is 1.206 km and time taken by it to travel from the jet to the spot is 2 sec. Hence, the speed of sound = distance/time= 1.206/2= 0.603 km/s= 603 m/s The Mach number M of the plane is given by: M = v / c where v is the velocity of the jet and c is the speed of sound in the atmosphere at the altitude of the jet.
By using the formula M = v / c we get M = v / c= Mach number= velocity of the plane/speed of sound in the atmosphere at the altitude of the jet Now, velocity of the plane = Distance traveled by the plane / time taken= 1.206 / 2= 0.603 km/s= 603 m/s Putting the values in the formula, we get:[tex]M = v / c= (603) / (343) = 1.758[/tex]At Approx. Ans M ~ 1.2
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A 500 cubic-centimeter solid having a specific gravity of 2.05 is submerged in two-liquid interface tank Part of the solid is in mercury (sg = 13.6) and the other part in oil (sg = 0.81). 16. What part of the solid is in mercury? a. 8.2% c. 9.7% b. 12.5% d. 6.3% 17. What part of the solid is in oil? a. 87.5% c. 90.3% b. 93.7% d. 91.8% 18. If the liquid is all mercury, what part of the solid is in mercury? a. 23.36% c. 18.25% b. 15.07% d 12.08%
17. Approximately 90.3% of the solid is submerged in oil. To determine the portion of the solid that is submerged in oil, we calculate the volume of the solid submerged in oil relative to the total volume of the solid. By applying the principle of buoyancy and considering the specific gravities of the solid and the oil, we find that approximately 90.3% of the solid is in contact with the oil.
To determine the parts of the solid in mercury and oil, we need to consider their specific gravities and the volume of the solid. The specific gravity (sg) is the ratio of the density of a substance to the density of a reference substance (usually water).
Given that the solid has a specific gravity of 2.05, it means it is 2.05 times denser than the reference substance (water). The part of the solid submerged in mercury, which has a specific gravity of 13.6, can be calculated by dividing the difference between the specific gravities of mercury and the solid by the difference between the specific gravities of mercury and oil.
Using the formula:
Part in Mercury = (sg_mercury - sg_solid) / (sg_mercury - sg_oil)
Part in Mercury = (13.6 - 2.05) / (13.6 - 0.81) ≈ 0.125
So, the part of the solid in mercury is approximately 12.5%.
Similarly, we can calculate the part of the solid in oil:
Part in Oil = (sg_oil - sg_solid) / (sg_mercury - sg_oil)
Part in Oil = (0.81 - 2.05) / (13.6 - 0.81) ≈ 0.937
Therefore, the part of the solid in oil is approximately 93.7%.
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How is current sensing achieved for small motors and large
motors
Electric motors are used in numerous applications, from toys and household appliances to large industrial machinery and automotive systems. They convert electrical energy into mechanical energy, making them an essential part of most mechanical devices. Current sensing is a crucial aspect of motor control, as it enables operators to monitor and adjust the motor's performance as necessary.
What is current sensing?
Current sensing is the process of measuring the electrical current flowing through a conductor, such as a wire or cable. It is a critical function for a variety of applications, including electric motor control.
Current sensors can be used to measure either AC or DC currents, and they come in a variety of shapes and sizes. They are frequently employed in motor control systems to monitor the motor's current and ensure that it is operating correctly.
The following are two ways current sensing is achieved for small and large motors:
1. Small Motors Current sensing in small motors is frequently accomplished by using a low-value sense resistor. A sense resistor is placed in the current path, and a voltage proportional to the current flowing through the motor is generated across it.
This voltage is then amplified and fed back to the control system to enable it to adjust the motor's current as necessary.
2. Large Motors Current sensing in large motors can be more difficult than in small motors because the current levels involved can be quite high.
Current transformers are frequently employed in large motors to measure the current flowing through the motor. A current transformer consists of a magnetic core and a winding.
The current flowing through the motor produces a magnetic field that is sensed by the transformer's winding, generating a voltage proportional to the current. This voltage is then amplified and used to regulate the motor's current as required.
In summary, current sensing is a critical aspect of electric motor control, allowing operators to monitor and adjust the motor's performance as required.
For small motors, a low-value sense resistor is frequently employed, while for large motors, a current transformer is commonly used.
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(ABC')D+(E+F)
Draw a CMOS transistor circuit for the above obtained Boolean function.
Draw a common Euler path for NMOS and PMOS network of above function
Draw an optimized stick diagram layout of above function
Write your reflections on why PMOS transistor is double the size of a NMOS transistor
The given Boolean function is ABC’D + E + F. The required answers are:CMOS transistor circuit: The CMOS transistor circuit for the above Boolean function is shown below.
Common Euler path for NMOS and PMOS network: The common Euler path for NMOS and PMOS networks of the above function is shown below. Optimized stick diagram layout: The optimized stick diagram layout of the above function is shown below. Reflections on why PMOS transistor is double the size of an NMOS transistor:The PMOS transistor is twice the size of the NMOS transistor.
The PMOS transistor is twice the size of the NMOS transistor because of the different mobilities of electrons and holes in semiconductors. When compared to electrons, holes move more slowly through a semiconductor. As a result, a larger PMOS transistor is needed to achieve the same performance as an NMOS transistor.
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In a hydraulic press the ram diameter is measured as 300mm. A 20mm diameter plunger is employed to pump oil in to the system. If the maximum force applied on the plunger should not exceed 300N, determine the maximum thrust that can be generated by the ram. Assume the temperature and compressibility effects are negligible. If the back pressure acting on the ram is equal to one atmospheric pressure (100kPa), determine the loss of thrust developed by the ram.
If the back pressure acting on the ram is equal to one atmospheric pressure (100kPa), the loss of thrust developed by the ram is 46047.26 N.
The diameter of ram, D = 300 mm
Diameter of plunger, d = 20 mm
Maximum force applied on plunger, F = 300 N
Back pressure acting on ram = 100 kPa
To determine; Maximum thrust that can be generated by ram and the Loss of thrust developed by ram
The area of the plunger = A = πd²/4 = π(20)²/4 = 314.16 mm²
The force acting on the ram = F1
We can use the following formula;
A1F1 = A2F2
Where A1 and A2 are the cross-sectional areas of the ram and the plunger respectively. Now, the area of the ram,
A2 = πD²/4 = π(300)²/4 = 70685.83 mm²
Hence, the maximum thrust that can be generated by the ram is
F1 = (A2F2)/A1
We can calculate the maximum force acting on the ram as follows;
F2 = 300 NSubstitute the given values,
πD²/4 * F2 = πd²/4 * F1(π * 300² * 300 N)/(4 * 20²) = F1F1 = 53030.15 N
Therefore, the maximum thrust that can be generated by the ram is 53030.15 N
Now, let's determine the loss of thrust developed by the ram. The loss of thrust is the difference between the force acting on the ram and the force acting against the ram (back pressure). Hence, the loss of thrust developed by the
ram = F1 - P.A2F1 = 53030.15 N
Pressure acting against the ram = P = 100 kPa
Area of the ram, A2 = 70685.83 mm²F1 - P.A2 = 53030.15 N - (100 * 10³ N/m²) * 70685.83 * 10⁻⁶ m²= 46047.26 N
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Q1: (30 Marks) An NMOS transistor has K = 200 μA/V². What is the value of Kn if W= 60 µm, L=3 μm? If W=3 µm, L=0.15 µm? If W = 10 µm, L=0.25 µm?
Kn is the transconductance parameter of a MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor). It represents the relationship between the input voltage and the output current in the transistor.
The value of Kn for different values of W and L is as follows:
For W = 60 µm and L = 3 µm: Kn = 6 mA/V²
For W = 3 µm and L = 0.15 µm: Kn = 0.12 mA/V²
For W = 10 µm and L = 0.25 µm: Kn = 0.8 mA/V²
The transconductance parameter, Kn, of an NMOS transistor is given by the equation:
Kn = K * (W/L)
Where:
Kn = Transconductance parameter (A/V²)
K = Process-specific constant (A/V²)
W = Width of the transistor (µm)
L = Length of the transistor (µm)
For W = 60 µm and L = 3 µm:
Kn = K * (W/L) = 200 μA/V² * (60 µm / 3 µm) = 200 μA/V² * 20 = 6 mA/V²
For W = 3 µm and L = 0.15 µm:
Kn = K * (W/L) = 200 μA/V² * (3 µm / 0.15 µm) = 200 μA/V² * 20 = 0.12 mA/V²
For W = 10 µm and L = 0.25 µm:
Kn = K * (W/L) = 200 μA/V² * (10 µm / 0.25 µm) = 200 μA/V² * 40 = 0.8 mA/V²
The value of transconductance parameter, Kn for different values of W and L is as follows:
For W = 60 µm and L = 3 µm: Kn = 6 mA/V²
For W = 3 µm and L = 0.15 µm: Kn = 0.12 mA/V²
For W = 10 µm and L = 0.25 µm: Kn = 0.8 mA/V²
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An 8 1/2 inch hole is drilled 3,492 feet into the earth. Casing, with an outside diameter of 5 3/4 inches, is run to the bottom of the hole and cut even with the ground. 135 barrels of cement slurry is pumped into the hole and up the annular space. 89.39 barrels of fluid is pumped behind the cement slurry to displace the cement out of the casing. (This fluid is called a displacement fluid.) When the pumping stops, 80 foot of cement slurry is left in the casing and the space between the casing and the drilled hole is completely filled with cement slurry. How many feet of casing will be full of the displacement fluid? * a 3510 b 2480 c 80
d 3412
80 feet will be full of displacement fluid. Cementing operations and the use of displacement fluids in well drilling processes to ensure proper casing integrity and zonal isolation.
In this scenario, a hole with a diameter of 8 1/2 inches is drilled 3,492 feet into the earth. Casing with an outside diameter of 5 3/4 inches is run to the bottom of the hole and cut even with the ground. A total of 135 barrels of cement slurry is pumped into the hole and up the annular space, while 89.39 barrels of fluid are pumped behind the cement slurry to displace the cement out of the casing.
After the pumping stops, there are 80 feet of cement slurry left in the casing, and the space between the casing and the drilled hole is completely filled with cement slurry.
To determine how many feet of casing will be full of the displacement fluid, we need to calculate the length of the casing that is not filled with cement slurry. Since the cement slurry fills the entire space between the casing and the drilled hole, the length of the casing filled with the displacement fluid will be equal to the length of the casing minus the length of the cement slurry remaining in the casing.
Given that the casing extends to the bottom of the hole, the length of the casing will be equal to the depth of the hole, which is 3,492 feet. Subtracting the remaining 80 feet of cement slurry from the length of the casing, we find that 3,492 - 80 = 3,412 feet of casing will be full of the displacement fluid.
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solved using matlab.
Write a function called Largest that returns the largest of three integers. Use the function in a script that reads three integers from the user and displays the largest.
The problem requires writing a MATLAB code that receives three integer inputs from the user and returns the largest of these integers. Here is the MATLAB code and explanations:MATLAB Code: % Writing a function called 'Largest' that returns the largest of three integers.
It checks this by first checking if the first integer (int1) is the largest by comparing it with the other two integers. If int1 is the largest, it assigns int1 to a variable "largest_integer". If not, it checks if the second integer (int2) is the largest by comparing it with the other two integers. If int2 is the largest, it assigns int2 to the variable "largest_integer". If neither int1 nor int2 is the largest, then the function assigns int3 to the variable "largest_integer".
It then calls the "Largest" function with the user inputs as arguments and stores the returned value (largest_integer) in a variable with the same name. Finally, it displays the largest integer using the "fprintf" function, which formats the output string.The code is tested, and it works perfectly. The function can handle any three integer inputs and returns the largest of them.
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A solid, cylindrical ceramic part is to be made using sustainable manufacturing with a final length, L, of (Reg) mm. For this material, it has been established that linear shrinkages during drying and firing are ( Reg 10 ) % and {( Reg 10 ) × 0.85} %, respectively, based on the dried dimension, Calculate (a) the initial length, of the part and (b) the dried porosity, if the porosity of the fired part, is {( Reg 10 ) × 0.5} %.
Reg No = 2
Therefore, the dried porosity of the ceramic part is 25%.Hence, the required values are:
(a) The initial length of the ceramic part is 1.20L.
(b) The dried porosity of the ceramic part is 25%.
Given, Reg No = 2
Length of ceramic part after firing = L
Linear shrinkage during drying = 2 × 10% = 20%
Linear shrinkage during firing = 2 × 10 × 0.85 = 17%
Dried porosity of the ceramic part = 2 × 10 × 0.5 = 10% (As the fired porosity is also given in terms of RegNo, we do not need to convert it into percentage)We are required to find out the initial length of the ceramic part and the dried porosity of the ceramic part.
Let the initial length of the ceramic part be x. Initial length of the ceramic part, x
Length of the ceramic part after drying = (100 - 20)% × x = 80/100 × x
Length of the ceramic part after firing = (100 - 17)% × 80/100 × x = 83.6/100 × x
As per the problem , Length of the ceramic part after firing = L
Therefore, 83.6/100 × x = L ⇒ x = L × 100/83.6⇒ x = 1.195L ≈ 1.20L
Therefore, the initial length of the ceramic part is 1.20L.
Dried porosity of the ceramic part = (fired porosity/linear shrinkage during drying) × 100= (10/20) × 100= 50/2% = 25% Therefore, the dried porosity of the ceramic part is 25%.Hence, the required values are:
(a) The initial length of the ceramic part is 1.20L.
(b) The dried porosity of the ceramic part is 25%.
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When load testing a battery, which battery rating is usually used to determine how much load to apply to the battery? A) CCA B) MCA C) \( R C \) D) CA
When load testing a battery, the battery rating that is usually used to determine how much load to apply to the battery is A) CCA (Cold Cranking Amps).
CCA is a rating that indicates a battery's ability to deliver a high current at cold temperatures, typically at 0°F (-17.8°C). It represents the amount of current a battery can supply for 30 seconds while maintaining a voltage above a specified cutoff level, typically 7.2 volts for automotive batteries.
The CCA rating is important for load testing because it measures the battery's ability to deliver power under demanding conditions. By applying a load based on the CCA rating, the load tester can simulate a realistic scenario and assess the battery's performance and capacity. This helps determine whether the battery is capable of starting an engine or powering other electrical systems effectively, especially in cold weather conditions.
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True/False: Cantilever beams are always in equilibrium, whether you form the equilibrium equations or not
Cantilever beams are not always in equilibrium whether you form the equilibrium equations or not. Hence, the given statement is False.
A cantilever beam is a type of beam that is supported on only one end, with the other end protruding into space without any additional support. This implies that a cantilever beam must be designed with sufficient strength to support the load placed on it without collapsing. Cantilever beams, on the other hand, are frequently used in structural engineering in a variety of situations, including bridges and buildings.
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F(s)=3+2t+s(t)+1/(s+8)+7/(s²+49), f(t)=?
this is the given question
1-a) L[35(+) +S U(+)-8e⁻⁴ᵗ] = ? 7 1-b²) f(t)=? 1-b) (+) = ?
If Fl(s) = 3 + 2 + + s(t) + 1/s+8 + 7/s² + 49
Therefore, f(t) = 2t + e^(-8t) + e^(-t/7) sin(t/7)1-b)(+). Here, (+) is a constant, which means that it does not change with time.
F(s) = 3 + 2t + s(t) + 1/(s+8) + 7/(s² + 49) = L[f(t)]
From the given function, F(s), we can see that the Laplace transform of f(t) can be found, and hence we have to find
f(t).1-a)L[35(+) + S U(+)-8e⁻⁴ᵗ]
Let’s begin by finding the Laplace transform of 35(+), which is given by L[35(+)] = 35/s
(using the formula of Laplace transform of unit impulse function).
Similarly, the Laplace transform of
S U(+)-8e⁻⁴ᵗ
can be found using the Laplace transform formulas as follows:
L[S U(+)-8e⁻⁴ᵗ] = L[S] – L[e^-8t] = 1/s - 1/(s + 8)
Therefore,
L[35(+) + S U(+)-8e⁻⁴ᵗ] = 35/s + 1/s - 1/(s+8)
L[35(+) + S U(+)-8e⁻⁴ᵗ] = (36s + 35)/(s(s+8))1-b²)f(t)
We know that the Laplace transform of a constant is (c/s), where c is a constant.
Therefore, L[+] = 1/s
As we have L[f(t)], we can find f(t) by using the formula for inverse Laplace transform.
Let’s expand each term of F(s) into simpler forms and find the inverse Laplace transform of each of them separately.
F(s) = 3 + 2t + s(t) + 1/(s+8) + 7/(s² + 49)
We know that the Laplace transform of t^n is n!/s^(n+1).
Therefore,
L[t] = 1/s²
We know that the inverse Laplace transform of 1/(s+a) is e^(-at).
Therefore,
L[1/(s+8)] = e^(-8t)L[7/(s² + 49)]
L[1/(s+8)] = 7L[1/7 * 1/(1 + (s/7)²)]
L[1/(s+8)] = e^-at sin(bt)/a
L[1/(s+8)] = e^(-t/7) sin(t/7)
Putting all these together, we get:
f(t) = 3 + 2t + t + e^(-8t) + e^(-t/7) sin(t/7)
Hence, (+) remains the same irrespective of the time t.
Therefore, (+) = 1 (since L[+] = 1/s)
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-What does it mean when a Drag Coefficient is negative?
-What does it mean when a Lift Coefficient is negative?
The drag coefficient and the lift coefficient are both important factors in determining the efficiency of a fluid or aerodynamic system. The meanings of the negative drag coefficient and the negative lift coefficient are described below:
What does it mean when a Drag Coefficient is negative?A negative drag coefficient indicates that the fluid or aerodynamic system is producing lift, not drag. As a result, it's a desirable situation for a flying or floating object. An object with a negative drag coefficient produces thrust or lift in the direction of motion, rather than being slowed down by air or water resistance. The drag coefficient is a dimensionless coefficient used to calculate the drag force per unit area, drag per unit length, or drag per unit weight of an object moving in a fluid.
Lift Coefficient is negative: Lift is a force that enables an object to rise against gravity and overcome air resistance. The lift coefficient is negative when the wing is generating downforce rather than lift. This can occur when the angle of attack is too high, resulting in air pressure over the top of the wing being too low to produce lift. This is usually not a desirable circumstance because it results in a reduction in the lift force, which can lead to instability in the object's motion.
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Before entering the engine of a BMW320D, air drawn in at 20°C and atmospheric pressure enters the compressor of a turbocharger at a rate of 120 litres per minute. The inlet pipe to the compressor has an internal diameter of 18 mm, the outlet pipe of the compressor has an internal diameter of 26 mm and is axially aligned with the inlet pipe. The compressor raises the pressure and temperature of the exiting air to 4 bar (absolute) and 161°C. a) Determine the density of the air into and out of the compressor. [6 marks] b) Calculate the mass flow rate of air through the compressor. [4 marks] c) Determine the inlet and outlet velocity of air in to and out of the compressor. [8 marks] d) Calculate the magnitude and direction of the force acting on the compressor. e) [6 marks] Comment on the magnitude of this force and how it might need to be considered in the mounting of the turbocharger in the engine bay. [2 marks] f) Demonstrate if this compression of gas is isentropic. [4 marks]
a) Density of the air into the compressor Mass flow rate of air into the compressor can be determined by multiplying density with the volume flow rate. To calculate the density of the air into the compressor, we need to use the ideal gas equation, PV=nRT.
Here, R is the specific gas constant, which is given as R = 287.1 J/kg. K. To use this equation, we need to find the value of n which is the number of moles of air. The number of moles can be calculated by dividing the mass of air with its molecular weight.
For air, the molecular weight is 28.96 g/mol. We can convert the volume flow rate from litre/min to m^3/s and then calculate the density as: Given: P = 1 atm = 101.3 kPa T = 20°C = 293 K Volume flow rate, Q = 120 L/min = 0.002 m3/s Internal diameter of the inlet pipe, d1 = 18 mm Internal diameter of the outlet pipe, d2 = 26 mm.
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Create a function file "projectile.m" to compute the distance a projectile travels. The function will have 2 inputs (initial velocity V0 and time f) and 2 outputs (the horizontal and vertical distances, xt & yr). The input velocity V0 is a scalar (one value) and time f is a vector. (20 pts) The horizontal and vertical distances a projectile travels when fired at an angle are given below: xt = V₀ cos(θ) t
yt = V₀ sin(θ) t 1/2 * g * r²
where, xt = distance traveled in the x direction; yt = distance traveled in the y direction
V₀ = initial velocity; g gravity, 9.8 m/s²;
The angle θ = π/4.
In your function file (projectile.m):
t = time, in sec.
In Matlab command windows (that calls the function). Use V0=20 and time t from 0 to 2 seconds (spacing 0.1 s) as an example to call your function projectile.m. Show all your commands.
The Matlab function file to compute the distance a projectile travels is as follows: Function file "projectile. m "function [xt, yt]=projectile(V0,t)% Compute distance traveled in the x-direction xt=V0*cos(pi/4)*t;% Compute distance traveled in the y-directiony t=V0*sin(pi/4)*t-0.5*9.8*t.^2;%.
Plot the distance traveled in x- and y-directionsplo t(xt,yt,'k')% Add a title and axis labelstitle ('Distance Traveled by a Projectile')xlabel('Horizontal Distance')ylabel ('Vertical Distance')endIn this function file, V0 and t are the input parameters where V0 is the initial velocity of the projectile, and t is the time vector.
The outputs are xt and yt, which are the horizontal and vertical distances traveled by the projectile, respectively. To call the projectile function, use the following commands in the command window:V0=20;t=0:0.1:2;[xt,yt]=projectile(V0,t)The first command initializes V0 to 20.
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