Biomaterials - like the materials in an artificial hip - are typically composed of polymers. True or false?

Answer:

Milk Production, Health, and Age.

Explanation:

You want a younger cow when buying cattle so you can have gather more milk from it over it's lifetime. You also want to make sure that it can actually produce milk. Then you want a cow in good health.

The state of plane stress at a point on a body can be depicted by the Mohr's circle below. Point X has coordinates (13,-29) MPa and point Y has coordinates(-18, 29) MPa. The normal and sheer stresses are

б\(_{x'}\)= -14.752080900309004 MPa

б\(_{y'}\) =9.752080900309004 MPa

T\(_{x' y'}\)= -30.5145295492538 MPa

What is a Mohr's Circle?

A two-dimensional graph of the transformation law for the Cauchy stress tensor is called Mohr's circle. Calculations in mechanical engineering for the strength of materials, geotechnical engineering for the strength of soils, and structural engineering for the strength of built structures frequently use Mohr's circle.

Additionally, by breaking down stresses into their vertical and horizontal components, it is utilized to calculate stresses in many different planes. In order to discover the principle planes and the principal stresses in a graphical representation, one of the simplest methods is to use Mohr's circle. These are known as principal planes in which principal stresses are calculated.

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We created a program that alternates between the localization, steering, and navigation modules. The navigation module implements a routing algorithm and selects the global route.

A program is described as a set of instructions that a computer executes to do a certain activity. Study programs are made to fit your specific learning needs and get you ready for more education, training, or work.

Navigation modules are defined as the interactions that let users move between, inside of, and outside of the various pieces of content in your app. Extra parts are frequently included in the modules to make the sensor easier to use.

Thus, we created a program that alternates between the localization, steering, and navigation modules. The navigation module implements a routing algorithm and selects the global route.

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I recommend a solid-state welding approach, specifically friction stir welding, for 4140 sheet in a tempered TMP1 condition.

In liquid-based welding (e.g., Tungsten Inert Gas Welding), the material is melted and then solidified, potentially causing changes to the microstructure and affecting the properties of the heat-treated 4140 sheet. On the other hand, solid-state welding (e.g., Friction Stir Welding) involves no melting, and therefore preserves the original microstructure, maintaining the properties of the base material.

The choice between the two welding methods may depend on the size and geometry of the part. Larger or more complex parts may require alternative approaches to ensure proper welding and minimal distortion. Overall, considering the microstructure and properties of the material, friction stir welding is the more suitable option for a heat-treated 4140 sheet.

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The answer is analyze

Answer:

I dont think you can blur text in google docs

Stairs, ladders, or ramps are required at an elevation break of 19 inches or more.

This is to ensure the safety of those who use the facility, particularly people with disabilities. If there is a level change of more than 19 inches, it is suggested that a ramp or elevator be installed to make it more accessible. Ramps, ladders, and stairs are vital components of any building or construction that has an elevation break of 19 inches or more. Their function is to offer a secure and convenient way of accessing various floors within a facility, whether it's a residential or commercial structure.

The Americans with Disabilities Act (ADA) has laws that ensure that facilities are built to be accessible to individuals with disabilities. For instance, they provide guidelines on the minimum width of a ramp or the steepness of the stairs. This is to ensure that individuals who use a wheelchair or walker can easily navigate the building. If a building has an elevation break of more than 19 inches, then it is required to have a ramp that meets the guidelines of the ADA. This will make it easier for individuals with disabilities to move from one floor to the next.

Moreover, if the building is relatively large and has multiple floors, an elevator is recommended. The elevator will ensure that people with disabilities can quickly move around the building without any issues. In conclusion, stairs, ladders, or ramps are required at an elevation break of 19 inches or more. The ADA guidelines ensure that individuals with disabilities have access to all parts of the building. They stipulate the minimum width, height, and slope of a ramp, as well as the steepness of the stairs. When designing a building, it is crucial to consider these guidelines to make the facility accessible to all.

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Stairs, ladders, or ramps may be required at certain elevation breaks to ensure safe access between different levels.

In physics, when there is a change in elevation, stairs, ladders, or ramps may be required to provide a safe means of accessing different levels. The specific elevation break at which one of these options is required depedependsnds on various factors, such as the height and angle of the change in elevation, as well as any applicable building codes or regulations. For example, building codes may require the use of stairs for elevation breaks of a certain height, but ramps may be required for breaks above a certain threshold that accommodate individuals with mobility impairments.

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Baselines in geodetic control networks are a critical component of modern surveying and mapping. Baselines are defined as the straight-line distance between two points in a geodetic survey, which is used to create a reference system for all other measurements.

The baseline is then used to calculate distances and angles between other points, which can be used to create maps and survey data. Baselines are typically measured using a variety of methods, including satellite-based Global Positioning Systems (GPS), which provide highly accurate measurements. Geodetic control networks are used for a wide range of applications, including construction, mining, land management, and environmental studies.

By providing accurate, reliable data about the earth's surface, these networks are essential for effective management of natural resources and development projects. In summary, baselines in geodetic control networks are the fundamental building blocks that allow surveyors and mapping professionals to create accurate and reliable data about the earth's surface.

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Answer:

2.35 + j8.34 Ω

Explanation:

Voltage = V\(_{L}\) = 240 V rms

supplying power = S\(_{s}\) = 8 kVA

power factor = pf\(_{s}\) = 0.6

Let P₁ represents one load draws 3kW at unity powder factor

The power angle is:

θ\(_{s}\) = cos⁻¹  pf\(_{s}\) = cos⁻¹  0.6 = 53.13°

Complex power supplied source is:

S\(_{s}\) =  S\(_{s}\) < θ\(_{s}\) = 8<53.13° kVA

Complex power for first load:

S₁ = P₁ = 3kVA

Since the power angle of first load is  θ₁ = 0°

According to principle of conservation of AC power, the power of second load is:

S₂ =  S\(_{s}\) - S₁

    = 8<53.13° - 3

    = 6.65<74.29° kVA

Since the second load is a Y connected load the phase voltage:

V\(_{p}\) =  V\(_{L}\) / \(\sqrt{3}\)

    = 240/1.732051

    = 138.564

    = 138.56 V

Complex power of second load:

S₂ = 3 V\(_{p}\)² / Z\(_{p}\)

impedance per phase of the second load:

Z\(_{p}\) =  3 V\(_{p}\)² / S₂

   = 3 (138.56)² /  6.65<74.29°

   = 3(19198.8736) / 6.65<74.29°

   = 57596.6208 / 6.65<74.29°

Z\(_{p}\) = 2.35 + j8.34Ω

The highest possible average flow velocity in the duct is the same at both the inlet and the outlet and can be determined by calculating the velocity using the cross-sectional area of the duct, which is π(0.15 m)².

To determine the highest possible average flow velocity in the duct, we can use Bernoulli's equation, which relates the pressure, velocity, and elevation of a fluid in a streamline.

Bernoulli's equation states:

P₁ + ½ρV₁² + ρgh₁ = P₂ + ½ρV₂² + ρgh₂

Where:

P₁ and P₂ are the pressures at points 1 and 2, respectively,

ρ is the density of the fluid,

V₁ and V₂ are the velocities at points 1 and 2, respectively,

g is the acceleration due to gravity,

h₁ and h₂ are the elevations at points 1 and 2, respectively.

Since the fan is raising the pressure of air by 50 Pa, we can consider P₂ as P₁ + 50 Pa.

To find the highest possible average flow velocity, we need to determine the pressure drop across the fan.

Since the fan is in a horizontal flow section, the elevations at points 1 and 2 are the same, and we can neglect the term ρgh₁ and ρgh₂.

The equation then simplifies to:

P₁ + ½ρV₁² = P₁ + 50 Pa + ½ρV₂²

Simplifying further:

½ρV₁² = 50 Pa + ½ρV₂²

We know that the fan needs to raise the pressure of air by 50 Pa to maintain the flow. Therefore, we can rewrite the equation as:

½ρV₁² = ½ρV₂²

Since the density ρ is the same on both sides, it cancels out, and we are left with:

V₁² = V₂²

Taking the square root of both sides, we get:

V₁ = V₂

This means that the highest possible average flow velocity in the duct occurs when the velocity at the inlet (V₁) is equal to the velocity at the outlet (V₂).

Therefore, the highest possible average flow velocity in the duct is the same at both the inlet and the outlet, and it can be determined by calculating the velocity of the airflow at either location.

To calculate the velocity, we can use the equation for the flow rate (Q) through a duct:

Q = A₁V₁ = A₂V₂

Where:

A₁ and A₂ are the cross-sectional areas of the duct at points 1 and 2, respectively.

The diameter of the duct is given as 30 cm, so the radius (r) is 15 cm or 0.15 m. The cross-sectional areas can be calculated as follows:

A₁ = πr₁² = π(0.15 m)²

A₂ = πr₂² = π(0.15 m)²

Since A₁ = A₂, the velocities V₁ and V₂ will also be equal.

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Answer: True

Explanation:

Answer:

The meole fraction of acetone in the outlet liquid is \(x_1 = 0.0072\)

Explanation:

1.

The schematic diagram to represent this process is shown in the diagram attached below:

2.

the mole fraction of acetone in the outlet liquid is determined as follows:

solute from Basis Gas flow rate \(G_s = 10(1-0.012) =9.88 kmol/hr\)

Let the entering mole be :\(y_1 = 1.2\) % = 0.012

\(y_1 =(\dfrac{y_1}{1-y_1})\)

\(y_1 =(\dfrac{0.012}{1-0.012})\)

\(y_1 =0.012\)

Let the outlet gas concentration be \(y_2\) = 0.1% = 0.001

\(y_2 = 0.001\)

Thus; the mole fraction of acetone in the outlet liquid is:

\(G_s y_1 + L_s x_2 = y_2 L_y + L_s x_1\)

\(9.88(0.012-0.001)=15*x_1\)

\(9.88(0.011) = 15x_1\)

\(x_1 = \dfrac{0.10868}{15}\)

\(x_1 = 0.0072\)

The mole fraction of acetone in the outlet liquid is \(x_1 = 0.0072\)

Answer:

Explanation: 2 is thy answer

Answer:

Yeah the metal part of the seat belt is a better thermal conductor than the plastic part of the seat belt.

Explanation:

The entropy changes in the system when there are constant specific heats and variable specific heats are found out to be

a) ASsys= 2.20 kJ/K

b) ASsys= 1.56 kJ/K respectively.

(a) Constant Specific Heat

To solve the problem, use the following formula:

ASsys = Cv ln(T₂/T₁)+R ln(V₂/V₁)

ASsys = [Cv ln(T₂/T₁)] + [R ln(V₂/V₁)]

where: ASsys= system entropy change

R = 0.287 kJ/kg.K

Cv = 0.717 kJ/kg.K

T₁ = 17 + 273 = 290 K (initial temperature)

P₁ = 120 kPa (constant pressure)

P₂ = 120 kPa (constant pressure)

V₁ = 300 L

= 0.3 m³ (initial volume)

V₂ = V₁ [since V is constant]

= 0.3 m³

T₂ = T₁ + q/Cp

where q= amount of heat added

Cp= specific heat of air at constant pressure

From the given data,

Q= P x V

= 120 kPa x 0.3 m³

= 36 kJ

Q/Δt = 200 W

= 200 J/s

Cp= 1.005 kJ/kg.K

[Table A-2, at 17°C, for air]

T₂ = (q/Cp) Δt + T₁

= (200/1.005) x 900 + 290

= 692 K

Now apply the formula for system entropy change.

ASsys = Cv ln(T₂/T₁)+R ln(V₂/V₁)

= 0.717 x ln(692/290) + 0.287 x ln(0.3/0.3)

= 2.20 kJ/K

(b) Variable Specific Heat

To solve the problem, use the following formula:

ASsys = ∫(Cp/T)dT- R ln(V₂/V₁)

whereASsys= system entropy change

ΔT= T₂ - T₁

= (692-290)

= 402 K

R= 0.287 kJ/kg.K

V₁ = V₂

= 0.3 m³

Using the data from Table A-2, Cps and T can be tabulated as:

Cp (kJ/kg.K)T (K)1.0052731.005284.16 (Note: T₂)1.005692

Apply the trapezoidal rule for the integral to get:

ASsys = ∫(Cp/T)dT- R ln(V₂/V₁)

= 0.287[(1.005273/290) + (1.005284.16/273) + (1.005692/692)]- 0.287 ln(0.3/0.3)

= 1.56 kJ/K

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Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite (\(\% V_{Gr}\)) is determined by the following expression:

\(\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%\)

\(\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%\)

Where:

\(V_{Gr}\) - Volume occupied by the graphite phase, measured in cubic centimeters.

\(V_{Fe}\) - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

\(V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}\)

\(V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}\)

Where:

\(m_{Gr}\), \(m_{Fe}\) - Masses of the graphite and ferrite phases, measured in grams.

\(\rho_{Gr}\), \(\rho_{Fe}\) - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

\(\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%\)

\(\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%\)

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

\(m_{Gr} = \frac{2.5}{100}\times (100\,g)\)

\(m_{Gr} = 2.5\,g\)

\(m_{Fe} = 100\,g - 2.5\,g\)

\(m_{Fe} = 97.5\,g\)

If \(m_{Gr} = 2.5\,g\), \(m_{Fe} = 97.5\,g\), \(\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}\) and \(\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}\), the volume percent of graphite is:

\(\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%\)

\(\% V_{Gr} = 91.906\,\%\)

The volume percent of graphite is 91.906 per cent.

Apply Gauss-Jordan elimination to find the inverse of a matrix.

The closed-loop transfer function can be obtained by multiplying the system's transfer function with the PID controller. The closed-loop poles are the roots of the denominator polynomial of the closed-loop transfer function.

a) To obtain the closed-loop transfer function analytically, you need to consider the transfer function of the system (given in the figure) and the transfer function of the PID controller. Multiply these two transfer functions and simplify the expression to obtain the closed-loop transfer function.

To obtain the closed-loop transfer function with MATLAB, you can use the 'feedback' function. Provide the transfer function of the system and the transfer function of the PID controller as input to the 'feedback' function to obtain the closed-loop transfer function.

b) The closed-loop poles can be determined by finding the roots of the denominator polynomial of the closed-loop transfer function.

c) The bandwidth of the closed-loop system can be determined by identifying the frequency at which the magnitude of the closed-loop transfer function is equal to \(1/sqrt(2)\). This frequency represents the -3 dB point.

d) To develop an equivalent digital control law for the continuous-time PID, you need to discretize the continuous-time PID controller. You can use techniques such as the Z-transform, impulse invariance, or Tustin's method to convert the continuous-time PID controller to its discrete-time equivalent.

e) To obtain the open-loop response for the continuous-time system with MATLAB, you can use the 'step' or 'lsim' function. Provide the transfer function of the system as input to the function and plot the response.

f) To obtain the closed-loop response for the continuous-time system, you can use the 'feedback' function in MATLAB. Provide the transfer function of the system and the transfer function of the PID controller as input to the 'feedback' function, and then use the 'step' or 'lsim' function to plot the response.

g) To obtain the closed-loop response for the digital implementation with MATLAB, you can use the 'c2d' function to discretize the continuous-time transfer functions. Then, use the 'feedback' function with the discrete-time transfer functions to obtain the closed-loop response.

h) The features of each component of the PID control action are as follows:

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Answer:

Discover GEOVIA Surpac, the world´s most popular geology and mine planning software.

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Micromine

Intuitive Mining Solutions Micromine 2020: Optimising underground mine planning The launch of MICROMINE’s latest version of leading exploration and 3D mine design solution, Micromine 2020 will see the introduction of new stope optimisation functiona…

Answer:

a c 1 museum is a patient who by think and pay the the price

The Fourier equation for heat transfer through a flat surface is given by:Q = (k * A * ΔT) / d.Using the given values and calculations for the different layers,

you can substitute the appropriate values into the equations to find the required cooling load of the refrigeration unit. where:Q is the rate of heat transfer,k is the thermal conductivity of the material,A is the surface area, ΔT is the temperature difference across the surface, and d is the thickness of the material. To modify the Fourier equation for the multi-layer surface of the cold store, we need to consider the heat transfer through each layer. We can calculate the overall heat transfer coefficient (U) for the multi-layer wall using the following formula:1/U = (1/h1) + (Δx1/k1) + (Δx2/k2) + ... + (1/hn)

where: h1, h2, ... hn are the surface heat transfer coefficients of each layer,Δx1, Δx2, ... are the thicknesses of each layer, and k1, k2, ... are the thermal conductivities of each layer. To determine the required cooling load of the refrigeration unit, we can use the formula:Q = U * A * ΔT where: Q is the rate of heat transfer,U is the overall heat transfer coefficient of the wall, A is the surface area, and ΔT is the temperature difference between the inside and outside of the cold store.Using the given values and calculations for the different layers, you can substitute the appropriate values into the equations to find the required cooling load of the refrigeration unit.

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The statement is true because the "java_home" environment variable is a required configuration variable for Java applications to run correctly. It is used to point to the location where Java is installed on a computer.

When a Java application is launched, it needs to locate the Java Runtime Environment (JRE) in order to run. The "java_home" environment variable provides the path to the directory where the JRE is located. If the variable is not defined or is defined incorrectly, the application will not be able to find the JRE and will not be able to run.

Therefore, if the "java_home" environment variable is not defined correctly, it is necessary to update it to the correct path to enable Java applications to run on the computer.

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The statement is true. The JAVA_HOME environment variable is crucial for running certain Java-related programs. If it's not correctly defined, issues may arise when running applications developed in Java.

The statement is true. The JAVA_HOME environment variable is indeed crucial for running certain programs, especially those related to Java development. When you install Java Development Kit (JDK) on your system, JAVA_HOME is an environment variable that should point towards the directory where JDK is installed. If it's not defined correctly, you would encounter issues while running Java implemented software. It serves as a reference point for other Java-based applications to locate JDK on your system. For instance, in a Java-based server like Apache Tomcat, the server start-up scripts often need to access tools provided within the JDK, and they use the JAVA_HOME environment variable to locate the right directory.

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The command to obtain the Time Zone, Page File Location(s), and Hotfix(s) information on a Windows XP VM is "systeminfo".

The question is asking for a command that can be run in a Windows XP virtual machine to obtain information about the Time Zone, Page File location(s), and any installed Hotfixes. The command should produce the desired output, not just the name of the binary that can be used.

One possible command that can be used is "systeminfo". This command provides detailed configuration information about the computer and its operating system, including the Time Zone, Page File location(s), and a list of installed Hotfixes. To run the command, open the command prompt by clicking Start > Run and typing "cmd" in the Run box. Then, type "systeminfo" in the command prompt and press Enter. The command will produce a long list of information about the system, including the desired information about the Time Zone, Page File location(s), and installed Hotfixes.

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Answer:

thrust washer

can be used to eliminate rubbing friction of wheel touching frame

Answer:

"Physical" would be the right approach.

Explanation:

So the answer here is just the perfect one.

Yes, according to the question, all major PC operating systems feature a graphical user interface.

What is operating system?
An operating system (OS) is a type of software that manages the hardware and software resources of a computer. It is responsible for the management and coordination of activities and the sharing of the resources of the computer. It acts as an intermediary between the user of a computer and the computer hardware. The operating system allows the user to interact with the hardware and execute programs. It provides a platform on which the applications can run. It provides services such as memory management, device management, task scheduling, security, networking, and file system management. It also provides the user interface, which allows the user to interact with the system.

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To determine the March funnel reading of a new drilling fluid with increased viscosity, given the initial fluid's density and funnel reading, we need to consider the effect of the viscosifying additive on the viscosity. The new fluid's funnel reading can be calculated based on the additive's impact on viscosity.

The March funnel is a device used to measure the viscosity of drilling fluids. The funnel reading indicates the time taken for a fixed volume of fluid to flow through the funnel.

In this case, the density of the drilling fluid remains unchanged after the addition of the viscosifying additive. However, the viscosity of the new fluid increases by a factor of 1.12 compared to the original fluid.

To determine the new funnel reading, we need to consider the relationship between viscosity and the funnel reading. A higher viscosity will result in a longer funnel reading.

Since the new fluid's viscosity is increased by 1.12 times the old viscosity, we can expect the new fluid to have a longer flow time through the March funnel. Therefore, the March funnel reading for the new fluid will be 1.12 times the original funnel reading of 66 seconds.

Calculating 1.12 * 66, we find that the March funnel reading for the new fluid should be approximately 73.92 seconds.

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Answer:

True

Explanation:

An LED test light is a piece of electronic test equipment used to determine the presence of electricity in a piece of equipment under test, making this statement true.