A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a stress state greater than 200 N/mm2. If the linkage is to be constructed from solid round stock, what is the minimum required diameter?

Answer:

The landscape architect uses knowledge of legal requirements to inform design choices.

Explanation:

The landscape architect doesn’t only draw on knowledge of law and government to design grounds for public buildings, but will draw on that knowledge quite often on many projects. The landscape architect’s knowledge of design is not purely intuitive but is a product of study and might involve research. The landscape architect’s knowledge of law and government, however, does come into play in making design choices that are consistent with the law and government regulations.

Answer:

Tack coat is a sprayed application of an asphalt binder upon an existing asphalt or Portland cement concrete pavement prior to an overlay, or between layers of new asphalt concrete.

Explanation:

It is proved from the below that the temperature of the egg varies from the shell to the center of the egg.

Using the 2-dimensional finite element method, we will need to find the Biot number of the system by using the formula:

\(\mathbf{Bi = \dfrac{h_c r_o}{k} }\)

According to the Heisler chart, the curves in the graph represent a range of values for the inverse of the Biot number,

where;

\(\mathbf{Bi = \dfrac{h_c r_o}{k} }\)

here;

By developing nodes based on assumptions;

\(\mathbf{Bi = \dfrac{1700 \times 0.025}{0.682} }\)

Bi = 62.32

The inverse of Biot number is:

\(\mathbf{=\dfrac{1}{Bi}}\)

\(\mathbf{=\dfrac{1}{62.32}}\)

= 0.016

Similarly, the Fourier number \(\mathbf{F_o}}\) is calculated by using the expression;

\(\mathbf{F_o = \dfrac{\alpha t }{r_o^2} }\)

\(\mathbf{F_o = \dfrac{k t }{\rho c_pr_o^2} }\)

where;

\(\mathbf{F_o = \dfrac{0.682 \times 15 \times 60 }{958.4 \times 4211 \times 0.025^2} }\)

\(\mathbf{F_o = 0.243}\)

Using the 2-dimensional finite element method, the temperature ratio is:

\(\mathbf{\dfrac{T(0,t) -T_{\infty}}{T_o -T_{\infty}} = 0.1 }\)

T(0,t) = 100 + 0.1(4-100)

T(0,t) = 100 + 0.1(-96)

T(0,t) = 100 - 9.6

T(0, t) = 90.4° C

Therefore, we can conclude that it is proved that the temperature of the egg varies from the shell to the center of the egg.

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The inequality which can be used to determine s, the number of sets of spoons Christopher could buy for the restaurant to have enough spoons is

35 + 10s ≥ 75

An inequality is simply used to show the relationship between the expressions that aren't equal.

The restaurant needs at least 75 spoons and there are currently 35 spoons and each set on sale contains 10 spoon.

Let each set be represented as s. This will be illustrated as:

35 + (10 × s) ≥ 75

35 + 10s ≥ 75

Collect like terms

10s ≥ 75 - 30

10s ≥ 45

Divide

s ≥ 45/10

s ≥ 4.5

The restaurant must have at least 5 sets.

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The inequality that can be used to determine s, the number of sets of spoons Christopher could buy for the restaurant, is 35 + 10s ≥ 75.

Inequality is a term used to show two relations between two values that are not equal to each other.

Given, the restaurant needs at least 775 spoons. They have 355 spoons and each set contains 10 spoons.

Then the expression is ;

35 + (10 × s) ≥ 75

35 + 10s ≥ 75

Like terms are extracted

10s ≥ 75 - 30

10s ≥ 45

s ≥ 45/10

s ≥ 4.5

Thus, the inequality in the number of sets of spoons is 35 + 10s ≥ 75.

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The pressure inside the oil droplet is 0.1 N/m².

A compound drop consists of an oil droplet inside a water droplet.

To calculate the pressure inside the oil droplet if the interfacial tension between water and oil is 25 mN/m, we will use the Laplace pressure equation.Laplace's law relates the pressure difference across the interface of two immiscible fluids to the surface tension and the curvature of that interface. For a droplet, it is given by:P = 2γ/R,

where P is the Laplace pressure, γ is the surface tension, and R is the radius of curvature of the interface.For a compound drop, there are two radii of curvature to consider: the curvature of the oil-water interface, and the curvature of the water-air interface. We can calculate the pressure inside the oil droplet using the radii of curvature of the oil-water interface as follows:We know that the diameter of the oil droplet is 1 mm, so the radius of curvature (R) of the oil-water interface is 0.5 mm.

The diameter of the water droplet is 3 mm, so the radius of curvature (r) of the water-air interface is 1.5 mm.

Using the Laplace pressure equation, the pressure inside the oil droplet can be calculated as:

P = 2γ/R=2x25x10^-3/0.5x10^-3=100 mN/m²=0.1 N/m²

Therefore, the pressure inside the oil droplet is 0.1 N/m².

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Answer: i believe it’s acoustics

Explanation:

The goal of the Water Power Corporation is to start operations that involve dumping waste into navigable waters. The business is required by the Clean Water Act to install specific machinery. before beginning operations.

What is operation?
Operation is the process of carrying out or performing a task or set of tasks. It is the execution of a plan or an action. Operations involve working with people, equipment, or technology to manage resources and create goods and services. Operations are essential to any organization, regardless of its size or type. They can range from simple tasks such as phone calls and mailings to complex tasks such as manufacturing parts and shipping them to customers. Operations are at the heart of any company’s success; they are the lifeblood of any business. They encompass the entire process from initial planning and design through production and quality control to the delivery of goods and services. Operations are responsible for ensuring that all aspects of the organization, from financial management to customer service, are running smoothly.

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a. The maximum thickness of the copper nozzle is 3.3 mm

b. The maximum thickness of the steel nozzle is 0.054 mm

The question has to do with heat transfer

Heat transfer is the movement of heat energy from one body to anotrher.

Since the rate of heat loss by the gas equal rate of heat gain by the metal.

The rate of heat loss by gas is P = -hA(T - T') where

The rate of heat gain by metal is given by P' = kA(T - T")/t where

Since P = P', we have that

-hA(T - T') =  kA(T - T")/t

Making t subject of the formula, we have

t = -k(T - T")/h(T - T')

Given that for copper

Substituting the values of the variables into t, we have

t = -k(T - T")/h(T - T')

t = -378 W/m-K(540°C - 150°C)/[2 × 10⁴ W/m²-K(540°C - 2750°C)]

t = -378 W/m-K(390°C)/[2 × 10⁴ W/m²-K(-2210°C)]

t = 147420 W/m/4420 × 10⁴ W/m²

t = 147420 W/m/44200000 W/m²

t = 0.0033 m

t = 3.3 mm

So, the maximum thickness of the copper nozzle is 10.71 cm

Given that for steel

Substituting the values of the variables into t, we have

t = -k(T - T")/h(T - T')

t = -23.2 W/m-K(980°C - 150°C)/[2 × 10⁴ W/m²-K(980°C - 2750°C)]

t = -23.2 W/m-K(830°C)/[2 × 10⁴ W/m²-K(-1770°C)]

t = 19256 W/m/3540 × 10⁴ W/m²

t = 19256 W/m/35400000 W/m²

t = 0.0000544 m

t = 0.0544 mm

t ≅ 0.054 mm

So, the maximum thickness of the steel nozzle is 0.054 mm

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There are a few different ways to implement the legv8 instructions. One way is to use a simple logic circuit, another way is to use a microcontroller.

What exactly is LEGv8?

LEGv8 is a straightforward subset of the ARMv8 AArch64 architecture; it is a 64-bit architecture with 32-bit instructions. It contains 32 registers, each of which is 64 bits wide.

If you want to use a logic circuit, you will need to build up a single cycle datapath. This will require a few different components, including an ALU, a register file, and a control unit. The control unit will decode the instructions and control the flow of data through the datapath. If you want to use a microcontroller, you can find many different legv8 instruction sets online. You will need to find one that is compatible with your microcontroller, and then you can load the instructions onto the microcontroller.

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Answer:

liquifid gas

Explanation:

when gas is introduced to a certain chemical it ends up slowly liquifying itself.

Answer:

some parts of your question is missing attached below is the missing part

Answer :  Fa = 4.46 Ib

Explanation:

use the equation

Z = 0.1 sin2∅

next we will differentiate the equation to get the locus of the velocity

z = 0.2 cos2∅∅

differentiate the equation furthermore to get the locus of acceleration in the horizontal axis

z = -0.4sin2∅(∅)^2 + 0.2cos2∅∅

note : express ∅ as 6 rad.\(s^{-1}\) for angular velocity and ∅ = 0 for angular acceleration

equation above becomes :

Z = - 0.4 sin 2∅ ( 6)^2 + 0.2 cos 2∅(0)

  = - 14.4 sin 2∅ ( acceleration of the follower in horizontal direction )

next calculate The force at the end of A of the follower

Fa - Kx = mz  

note: m = w / g hence : Fa - Kx = w/g z  ------- (2)

w = weight of the spring-held follower = 0.75 Ib

x = compression of the spring = 0.4

k = spring stiffness = 12 Ib/ft

∅ = 45⁰

g = 32.2 ft/s^2

input these values into equation 2

hence : Fa = 4.46 Ib ( force at the end A of the follower )

Answer:

Visual connectivity refers to the tangible aspects of a space; extent to which a place can be viewed from other places. It is believed that the design properties of a spatial layout of an atrium leaves unobstructed views horizontally and vertically.

Explanation:

The apparent power is the product of the voltage and the current, so the apparent power is 480 VAC x 15 A = 7,200 VA. The power factor is 0.77 lagging, so the reactive power is 7,200 x 0.77 = 5,544 VAR. The actual power supplied to the motor is the product of the apparent power and the efficiency, which is 7,200 x 0.82 = 5,824 W.

The power produced by the motor is 5,824 W divided by the efficiency, so the power produced by the motor is 5,824/0.82 = 7,121 W or 9.5 HP.
The impedance, resistance, and reactance of each phase of the motor depending on the type of motor and the design of the windings. Generally, the impedance is determined by the resistance and the reactance, so the impedance of each phase can be calculated using Ohm's law. The resistance and reactance are usually determined experimentally using an ohmmeter or an AC clamp meter.

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Three common controls used in central electric heat applications are thermostats, contactors, and sequencers.The correct answer is option B.

Thermostats play a crucial role in controlling the temperature in a central electric heating system. They detect the ambient temperature and send signals to the heating system to turn on or off accordingly.

Thermostats allow users to set their desired temperature and maintain it within a specific range, ensuring comfort and energy efficiency.

Contactors are electrical switches that control the flow of electricity to the heating elements. They are responsible for connecting or disconnecting the power supply to the heating system based on the signals received from the thermostat.

Contactors are essential for ensuring the safe and efficient operation of the electric heat system.

Sequencers are used to control the sequencing or timing of the electric heating elements. They ensure that the heating elements turn on and off in a specific order to prevent excessive power consumption and ensure even heating.

Sequencers also help protect the electrical system from overload conditions by controlling the activation of heating elements in a coordinated manner.

In conclusion, the correct answer is B. Thermostats, contactors, and sequencers are the three common controls used in central electric heat applications.

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The total resistance of a circuit that contains two 100 resistors connected in parallel is equal to 50 Ohms.

In Science, resistance can be defined as an opposition to the flow of current in an electric circuit. Additionally, the standard unit of measurement of the resistance of an electric component is Ohms, which can either be converted to kilo-ohms or mega-ohms.

Mathematically, the total equivalent resistance of two or more resistors in an electrical circuit that are connected in parallel is giving by this mathematical expression;

Rt = R₁R₂/R₁ or Rt = R/n

Where:

Substituting the given parameters into the formula, we have;

Rt = 100/2

Rt = 50 Ohms.

In conclusion, the total equivalent resistance is equal to 50 Ohms.

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According to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.

They are experts in construction projects, supporting the design of plans, estimating costs, planning work methods, etc.

Research can facilitate the work of building technicians in the following ways:

Therefore, we can conclude that according to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.

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(a) Dynamic Programming Algorithm to find the MCS length for X and Y:

The common substring refers to a string that occurs in two or more strings. The Longest Common Substring (LCS) is a problem that involves finding the longest substring that appears in two or more strings. In this case, the substrings are required to be contiguous in the original string.

To find the Maximum Common Substring (MCS) length for the given string X and Y, we need to use the following dynamic programming algorithm:

Initialize a 2-D array (Table) with all values 0.

For i = 1 to n, where n is the length of string X:

For j = 1 to m, where m is the length of string Y:

if X[i] == Y[j]:

Table[i][j] = Table[i-1][j-1] + 1

if Table[i][j] > length:

length = Table[i][j]

Here, the Table[i][j] represents the length of the longest common suffix of X[1..i] and Y[1..j]. If X[i] and Y[j] are not the same, Table[i][j] is set to 0. The length variable stores the maximum length of all the entries in the Table. The length of the longest common substring is stored in the variable length.

(b) Time and Space Complexity:

The dynamic programming algorithm for the longest common substring takes O(nm) time and O(nm) space in the worst-case scenario. Here, n and m are the lengths of the two input strings X and Y, respectively. Hence, the time and space complexity of the algorithm is O(nm).

(c) Dynamic Programming Tables:

The dynamic programming tables for the strings X = ABCDEFG and Y = ABCEDEFG

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Answer:

20368.917N

Explanation:

Frictional force (F) is the product of the Coefficient of friction and the normal reaction.

F = μN

Coefficient of friction, μ = 0.2

Normal reaction = MgCosθ

Mass, m = 12 tonnes = 12 * 1000 = 12000 kg

N = 12000 * 9.8 * cos30

N = 101844.58

F = 0.2 * 101844.58

F = 20368.917N

Answer:

Maximum concentration for Pb is only 41%. we get 64.3% which exceeds the maximum concentration. Therefore it is not possible.

Explanation:

let's first determine the mass fraction of the phases as follows

\(W_{a}=\frac{m_{a} }{m_{a}+m_{p_{g2}p_{b} } }\)

Given that masses are \(m_{a}=7.39kg\) and \(m_{p_{g2}p_{b} }=3.81kg\)

Thus, \(W_{a} =\frac{7.39}{7.39+3.81}=0.659\)

\(W_{M_{g2Pb} } =1-W_{a}=1-0.659=0.341\)

Now determine the concentration

\(W_{a}=\frac{C_{0}-C_{a} }{C_{b}-C_{a} }\)

At 81% PB and 70% Pb only point at which \(M_{g2Pb}\) exists, we can solve

\(0.659=\frac{70-81}{C_{b}-81 }\)

\(C_{b}=\)64.3%

Refer text book page no. 326, figure 9.20, from figure we can say maximum concentration for Pb is only 41%. we get 64.3% which exceeds the maximum concentration.

Answer:

Haha I'm a great guy but my friend has been in a day of the day and a lot to be able and I'm happy holi and a lot to the world of the day and day to

Answer:

What about it?

Explanation:

Answer:

a)  358.8 KJ/kg

b)  0.0977  KJ/K- kg

c)  83.28%

Explanation:

N2 at 300 k. ( use the properties of N2 at 300 k (T1) )

Cp = 1.04 KJ/kg-k , Cv = 0.743 KJ/Kgk ,  R = 0.1297 KJ/kgk , y = 1.4 ,

Given data:

T2 = 645 k

P1 = 1 bar , P2 = 10.5 bar

a)Determine the work input in KJ/Kg of N2 flowing

Winput = h2 - h1  = Cp( T2 - T1 ) = 1.04 ( 645 - 300 ) = 358.8 KJ/kg

b) Determine the rate of entropy in KJ/K- kg   of N2 flowing

Rate of entropy ( Δs ) = Cp*InT2/T1 - R*In P2/P1

                                    = 1.04 * In (645/300) - 0.1297 * In ( 10.5 / 1 )

                                    = 0.0977  KJ/K- kg

c) Determine isentropic compressor efficiency

Isentropic compressor efficiency = 83.28%

calculated using the relation below

( h'2 - h1 ) / ( h2 - h1 ) = ( T'2 - T1 ) / ( T2 - T1 )

where T'2 = 587.314

Answer:

Deburring Tool

Explanation:

A deburring tool is used in order to debur the PVC pipes. They are mostly used for the plastic pipes.

After the PVC pipes are cut, there are burrs on the pipe surface. To remove these burrs, a deburring tool is used. It removes the burrs form the edges of the PVC pipes that results from grinding, cutting, milling, drilling, etc.

The deburring tools are made from high speed steels.

In a Beam-and-Girder system, the steel members are finished through a process called painting or coating, which provides protection against corrosion and enhances the steel's appearance.

The finishing process of steel members in a Beam-and-Girder system involves several steps:
1. Surface preparation: Before painting or coating, the steel members' surfaces are cleaned and prepared to remove any contaminants, such as rust, dirt, or grease. This can be done through methods like sandblasting or power tool cleaning.
2. Primer application: After surface preparation, a primer is applied to the steel members. The primer acts as a base layer that improves the adhesion of the paint or coating and enhances its protective properties.
3. Intermediate coat (optional): In some cases, an intermediate coat may be applied to provide additional protection, such as resistance to corrosion or abrasion.
4. Finish coat: The final step in the process is the application of the finish coat, which gives the steel members their desired appearance and provides an additional layer of protection. The finish coat can be of various types, such as epoxy, polyurethane, or acrylic.

The finishing of steel members in a Beam-and-Girder system is essential for protection against environmental factors and maintaining their structural integrity. The process involves surface preparation, primer application, and the application of a finish coat, which may also include an intermediate coat for added protection.

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In a Beam-and-Girder system, steel members are finished using a process that involves cleaning, priming, and painting or applying a protective coating.

The first step in finishing steel members in a Beam-and-Girder system is cleaning them to remove any dirt, rust, or contaminants. This can be done using methods such as sandblasting, power brushing, or solvent cleaning.

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Frederick Taylor created a series of guidelines known as the "motion study principles" to boost productivity at work. Time study, motion study, fatigue study, etc technique conduct research are the four divisions into which they fall.

Motion studies are the study of actions like lifting, putting things down, standing, changing direction, and other actions that are made when performing routine work.

In order to execute the task effectively in less time, unnecessary moves are attempted to be removed. The following three headings comprise the fundamentals: Making use of the human body. The layout of the workplace. Machinery and tools design.

Motion study principles will be the which will help in designing or putting films into a particular order. Also how the routine works are being made is suggested.

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A: Copper has a higher elastic modulus than aluminum.

B: The density of copper is closer to that of aluminum than it is to iron.

C: Bronze is an alloy of copper and zinc.

D: Copper and its alloys form a green tarnish over time, consisting of sulfides and carbonates.

E: Copper is relatively resistant to corrosion by neutral and even mildly basic water, making it useful for freshwater plumbing applications.

Answer:

30 mm is the minimum thickness that must be applied.

Explanation:

Given the data in the question;

Using Fourier's equation. the heat rate is  

q = kA(ΔT/Δx)

where

A is the surface area, we must consider all surfaces through which the heat can dissipate through

i.e 2×2 for one wall gives you 4m²,

there are 5 walls, so we will  have 20m² for surface area.

k is thermal conductivity of the styrofoam ( 0.030 W/m K)    

q is the heat loss (500 W  )

ΔT is the Temperature difference ( 35 - 10) = 25°C

Δx  = ?

So we substitute

500 = (0.030)(20)(25/Δx)

500 = 0.6 (25/Δx)

500 = 15 / Δx

Δx = 15 / 500

Δx = 0.03 m = 30 mm

Therefore, 30 mm is the minimum thickness that must be applied.