What qualitative effect would you expect each of the following to have on the P50 of hemoglobin (5 pts) a) Increase in pH from 7.0 to 7.4 Answer ___
b) Increase in Pco2 from 25 mmHg to 35 mm Hg Answer___
c) Dissociation into monomer polypeptide chains Answer ____
d) Decrease in 2,3-BPG concentration from 8 mM to 4 mM in red cells Answer___

Answer:

C. Count Noun

Explanation:

Answer:

- the weight of the gravel is 1183.56 kg/m³

- the weight of sand is 588.98 kg/m³

- the weight of water is 165.784 kg/m³

Explanation:

 Given the data in the question;

mixture content of the gravel  = 1173 kg/m³

sand content = 582 kg/m³

free water content = 157 kg/m³

moist of gravel = 0.9%

absorption of gravel = 1.5%

moist of sand = 1.2%

absorption of sand = 1.5%

masses of gravel, sand, and water per cubic meter that should be used at the job site ?

To get the weight of moist gravel, we use the following equation;

Weight of moist gravel  = content of gravel × ( 1 + moist of gravel )

we substitute

Weight of moist gravel  = 1173 × ( 1 + 0.9%)

Weight of moist gravel  = 1173 × ( 1 + 0.009)

Weight of moist gravel  = 1173 × 1.009

Weight of moist gravel  = 1183.56 kg/m³

Therefore, the weight of the gravel is 1183.56 kg/m³

To get weight of the sand, we use the following formula;

Weight of sand = sand content × ( 1 + moist of sand )

we substitute

Weight of sand = 582 × ( 1 + 1.2% )

Weight of sand = 582 × ( 1 + 0.012 )

Weight of sand = 582 × 1.012

Weight of sand = 588.98 kg/m³

Therefore, the weight of sand is 588.98 kg/m³

To get the weight of mix water ; we use the following equation;

[tex]Mix-Water_{weight[/tex] = water content - gravel content( moist of gravel - absorption of gravel) - sand content( moist of sand - absorption of sand )

so we substitute

[tex]Mix-Water_{weight[/tex] = 157 - 1173( 0.9% - 1.5%) - 582( 1.2% - 1.5% )

[tex]Mix-Water_{weight[/tex] = 157 - 1173( 0.009 - 0.015) - 582( 0.012 - 0.015 )

[tex]Mix-Water_{weight[/tex] = 157 - 1173( -0.006 ) - 582( -0.003 )

[tex]Mix-Water_{weight[/tex] = 157 + 7.038 + 1.746

[tex]Mix-Water_{weight[/tex] = 165.784 kg/m³

Therefore, the weight of water is 165.784 kg/m³

Two different kinds of losses can happen in a three-phase induction motor. These losses are variable losses that are constant or fixed losses. Fixed or Constant Losses Consta.

Stator losses, including as stator iron loss and stator copper loss, are financed in part by this power input. Rotor input is provided with the leftover power, or (input electrical power - stator losses). Thus, rotor input P2 equals Pin minus stator losses (stator copper loss and stator iron loss).

Fixed losses, Copper losses, and stray losses are a few types of losses that can occur in three-phase induction motors. Core losses, bearing friction losses, brush friction losses (in wound rotor motors), and windage losses make up the majority of these losses.

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Disc-type automotive braking systems mechanically apply the brake pads against the disc rotor when the actuator causes the control balls to move up the ball ramp.

Automotive disc braking systems are a safe and reliable form of braking. They are designed to stop the vehicle quickly and safely. These braking systems work by applying pressure from the brake discs to the brake discs mounted on the wheel axle.

This in turn causes the discs to press against the brake calipers, which in turn are applied to the discs. This action results in safe and reliable braking.

These braking systems also have a mechanical parking function, which is activated by the actuator lever. When the lever is activated, the control balls are moved up the ball ramp, which causes the brake discs to be applied to the rotor to ensure safe braking.

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The standard overload protection for a 2 hp, 115 V, single phase motor has a nameplate rating of 21.50 A shall not be rated greater than 21.50 A amperes.

What is motor?
Motor is an electro-mechanical device which converts electrical energy into mechanical energy. It is widely used in industrial and domestic applications to produce rotational motion for machines. Motors are available in a variety of sizes and types, ranging from small DC motors to large AC synchronous motors. Motors are used to power a wide range of machines, from small fans and pumps to large industrial machines such as elevators, cranes, and conveyors. Motors can be powered by electricity, batteries, or fuel, and can be used to move objects in different directions. Motor control systems can be used to control the speed and direction of a motor, allowing for precise and efficient control of a machine. Motors are essential components of modern machines and are used in a wide variety of devices such as electric vehicles, robots, and home appliances.

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Answer:

Programming paradigms are different approaches to programming. Each paradigm is concerned with the way code is written, structured, and organized. The main paradigms are procedural, object-oriented, functional, and logical. Each of these has its own set of rules and principles that guide how code is written and executed. Procedural programming focuses on implementing algorithms and breaking tasks down into smaller steps. Object-oriented programming focuses on breaking tasks down into objects and classes. Functional programming focuses on writing code that is easy to read and maintain. Logical programming focuses on using logic to solve problems.

Explanation:

Hope It Help You

Answer:

A programming paradigm is a style or approach to programming, characterized by the way in which the programmer writes the code. There are several different programming paradigms, including:

Explanation:

Imperative programming: This is the most traditional style of programming, in which the programmer writes code that specifies a series of steps to be carried out in order to solve a problem. Imperative programming languages include C, C++, and Java.

Object-oriented programming: In this style of programming, the programmer writes code in the form of objects, which are self-contained units that combine data and behavior. Object-oriented programming languages include Java, Python, and C#.

Functional programming: In functional programming, the programmer writes code in the form of functions, which are self-contained units that take input and produce output, without modifying any external state. Functional programming languages include Haskell, Lisp, and ML.

Logic programming: In this style of programming, the programmer writes code in the form of logical statements, and the language's runtime system is responsible for finding a solution that satisfies all of the statements. Logic programming languages include Prolog and Mercury.

Declarative programming: In declarative programming, the programmer specifies what the desired result should be, rather than how to achieve it. The language's runtime system is responsible for finding a way to produce the desired result. Declarative programming languages include SQL and regular expressions.

The minimum length of the curve meeting the SSD requirements is 643.72 ft.

What is SSD?

From the given question, we have the grades, [tex]g_{1}=+1.25%[/tex], [tex]g_{2}=-2.25%[/tex], and the design speed, [tex]v=65[/tex] mi/h

So, [tex]SSD=vt+\frac{v^{2} }{2g(f \pm \frac{n}{100}) }\\\implies SSD=1.47(65)(2.5)+\frac{65^{2} }{30(0.347826 \pm \frac{1.25}{100}) }\\\implies SSD=630'[/tex]

The length of the vertical curve is given as, [tex]L=2(SSD)-\frac{2158}{A}[/tex]

When SSD>L, we have\

[tex]L=2(630)-\frac{2158}{-2.25-1.25}\\\implies L=1876.57 ft[/tex]

When SSD<L, we have

[tex]L=\frac{A(SSD)^{2} }{2158} \\\implies L=\frac{(-2.25-1.25)(630)^{2} }{2158} \\\implies L=643.72 ft[/tex]

Hence, the minimum length of the vertical curve is 643.721 ft.

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Option A is correct because the details of the procedures should be explained before they are practised at this age level.

Options B, C, and D are incorrect because, while these steps are necessary, students must first understand the procedures.

How to teach procedures to your students?

The key to making your classroom run smoothly and efficiently is to teach procedures. If you take the time to teach your class how to do things, I guarantee you will have more time to focus on academics in the long run.

STEP 1 – PROCEDURES START WITH YOU

Before you can teach your students procedures, you must first envision how you want your classroom to look, sound, and feel on a daily basis. This means that you must consider these issues before the start of the school year.

STEP 2 – PUT YOUR PROCEDURES INTO SIMPLE STEPS

Once you've decided on the procedure that will get you to your ideal classroom, it's time to break it down into simple, easy-to-follow steps. When you do this, keep your students in mind. The language and number of steps will differ if you teach Pre-K or Kindergarten versus 3rd, 4th, or 5th grades. I've always found the K.I.S.S. method to be effective. Keep it as simple as possible.

STEP 3 – TEACH THE PROCEDURES

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Answer: Droughts, pumping effects, and rainfall variations affect the groundwater level height.

Explanation: If a well is pumped at a faster rate than the aquifer around it is recharged by precipitation or other underground flow, then water levels in the well can be lowered.

The  difference in voltage across the two loads is 6 V.

In a series circuit, the total resistance is equal to the sum of the individual resistances. In this case, the total resistance of the circuit is 2 Ω + 4 Ω = 6 Ω.

The current in a series circuit is the same through all of the loads. This means that the same amount of voltage is dropped across each load. Since the power source has a voltage of 12 V, and the total resistance of the circuit is 6 Ω, the current in the circuit can be calculated using Ohm's Law:

I = V/R

= 12 V / 6 Ω

= 2 A.

Since the same current flows through both loads and the same amount of voltage is dropped across each load, there is no difference in voltage across the two loads. Both loads experience a voltage drop of:

12 V / 2

= 6 V.

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Explanation:

A research report is a well-crafted document that outlines the processes, data, and findings of a systematic investigation. It is an important document that serves as a first-hand account of the research process, and it is typically considered an objective and accurate source of information.

(1) to build knowledge and facilitate efficient learning, (2) to understand various issues, (3) to know the truth and prove lies, and (4) to seek opportunities among others

The pressure of seawater will be 2.14*107 N/m2 and the density will be 8.53*103 kg/m3.

Pressure is a type of force applied to an object by another object over a surface area. Pressure is measured in units of force per unit area and is usually expressed in terms of pascals (Pa).

The bulk modulus of elasticity, K, is a measure of the stiffness of a material and is defined as the ratio of bulk stress to the resulting strain. In this problem, we need to calculate the density and pressure at a depth of 2500 m, based on the given bulk modulus and 98 kPa pressure at the surface.
To solve this problem, we will use the equation K = -dp/dρ, where dp is the change in pressure with depth and dρ is the change in density with depth. Since we are given the bulk modulus and pressure on the surface, we can calculate the change in density with depth as follows:
dρ = -K*dp/dz
where dp = rgdz and rg is the gravity of the Earth. Substituting these values, we get:
dρ = -2.34*109*98*103/(9.8*2500) = -7.78*103 kg/m3
This means that for every metre of depth, the density of seawater decreases by 7.78*103 kg/m3. Therefore, at a depth of 2500 m, the density of seawater will be:
ρ = 1030 - 7.78*103*2500 = 8.53*103 kg/m3
Similarly, we can calculate the pressure at a depth of 2500 m using the equation P = P0 + rgdz, where P0 is the pressure at the surface. Substituting the values, we get:
P = 98*103 + 9.8*2500*8.53*103 = 2.14*107 N/m2
Therefore, at a depth of 2500 m, the pressure of seawater will be 2.14*107 N/m2 and the density will be 8.53*103 kg/m3.

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Answer:

5.31 × 10⁻⁴⁰ Fm²

Explanation:

Using the Clausius-Mossotti equation

(ε - 1)/(ε + 2) = Nα/3ε₀ where ε = dielectric constant = 1.0015, ε₀ = 8.854 × 10⁻¹² F/m, N = number of molecules of SF6 per unit volume and α = electronic polarizability

So, α = (ε - 1)3ε₀/[(ε + 2)N]

Also

P = (N/NA)RT where P = pressure = 1 atm = 1.013 × 10⁵ Pa, N = number of molecules of SF6 per unit volume, NA = Avogadro's constant = 6.022 × 10²³ /mol, R = molar gas constant = 8.314 J/mol-K and T = room temperature = 25° = 273 + 25 = 293 K

So, N = PNA/RT

= 1.013 × 10⁵ Pa × 6.022 × 10²³ /mol ÷ (8.314 J/mol-K × 293 K)

= 6.100286 × 10²⁸ Pa/mol ÷ (2436.002 J/mol)

= 0.0025 × 10²⁸ molecules/m³

= 2.5 × 10²⁵ molecules/m³

So,

α = (ε - 1)3ε₀/[(ε + 2)N]

α = (1.0015 - 1)× 3 × 8.854 × 10⁻¹² F/m/[(1.0015 + 2)2.5 × 10²⁵ molecules/m³]

α = (0.0015)26.562 × 10⁻¹² F/m/[(3.0015)2.5 × 10²⁵ molecules/m³]

α = 0.039843 × 10⁻¹² F/m/[7.50375 × 10²⁵ molecules/m³]

α = 0.00531 × 10⁻³⁷ Fm²

α = 5.31 × 10⁻⁴⁰ Fm²

The next step in the process is to crystallize the juice. The pale yellow juice is heated until the temperature reaches about 80°C, at which point the sucrose in the juice begins to form into crystals.

What is sucrose?
Sucrose is a type of sugar composed of glucose and fructose molecules bound together. It is a naturally occurring disaccharide found in many plants, where it serves as a form of energy storage. Sucrose is the most common sugar in the human diet and is used in a wide variety of food products. It is found in many processed foods and can be added to foods to enhance flavor. Sucrose is usually extracted from sugar cane or sugar beets, then processed into granulated sugar, which is used in baking or to sweeten drinks.

These crystals are then separated from the liquid in a centrifuge to produce raw sugar crystals. The remaining liquid, known as molasses, is then processed further to remove the remaining sucrose, creating a more viscous dark syrup. The raw sugar crystals are then washed and dried, and are ready to be packaged and sold as raw sugar.

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Answer:

i like icecream

Explanation:

The next step in the process is to purify the juice. This is done by adding calcium hydroxide which removes impurities and non-sucrose components. The juice is then heated and concentrated in vacuum pans.

What is vacuum?
Vacuum is an empty space devoid of matter. It is a space that is completely void of any matter, including molecules, atoms, and subatomic particles. It has zero pressure and is a state of matter considered to be more rare than a gas, liquid, or solid. Vacuums can be generated artificially in a laboratory setting or can occur naturally in the absence of air or other gasses. Vacuum environments are used in many scientific and industrial applications and are especially important in the field of physics, where they are used to study the behavior of matter on a microscopic scale.

This process drives off the water and forms a thick syrup containing 85-95% sucrose. The syrup is then seeded with sugar crystals and further heated in a process called inversion. Finally, the sugar is packaged for sale or further processed into other forms of sugar such as brown sugar, powdered sugar, and so on.

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Complete Question:

Find the capacitive reactance of a 0.1 microfarad capacitor with frequency 50Hz and at 200Hz​.

Answer:

I. Capacitive reactance = 31826.86 Ohms.

II. Capacitive reactance = 7956.72 Ohms.

Explanation:

Given the following data;

Capacitance = 0.1 uF = 0.0000001 Farad

Frequency = 50 Hz and 200 Hz

To find the capacitive reactance;

Mathematically, the capacitive reactance of an electronic circuit is given by the formula;

[tex] X_{c} = \frac {1}{2 \pi fc} [/tex]

Where;

Substituting into the formula, we have;

I. At frequency, f = 50 Hz

[tex] X_{c} = \frac {1}{2 * 3.142 * 50 * 0.0000001} [/tex]

[tex] X_{c} = \frac {1}{314.2 * 0.0000001} [/tex]

[tex] X_{c} = \frac {1}{0.00003142} [/tex]

[tex]X_{c} = 31826.86[/tex]

II. At frequency, f = 200 Hz

[tex] X_{c} = \frac {1}{2 * 3.142 * 200 * 0.0000001} [/tex]

[tex] X_{c} = \frac {1}{1256.8 * 0.0000001} [/tex]

[tex] X_{c} = \frac {1}{0.00012568} [/tex]

[tex]X_{c} = 7956.72[/tex]

Answer:Gearbox design calculations involve a variety of factors, including the type of gearbox, the size and type of gears, the torque and speed requirements, the materials used, and the environmental conditions. The calculations also involve the selection of the appropriate gear ratios, the selection of the appropriate bearings, and the selection of the appropriate lubrication system. Additionally, the calculations must take into account the safety and reliability of the gearbox.

Explanation:

Answer:

A Bipolar Junction Transistor, or BJT, is a solid-state device in which the current flow between two terminals (the collector and the emitter) is controlled by the amount of current that flows through a third terminal (the base).

The main basic function of a BJT is to amplify current it will allow BJTs are used as amplifiers or switches to produce wide applicability in electronic equipment include mobile phones, industrial control, television, and radio transmitters. There are two different types of BJTs are available, they are NPN and PNP.

a) The power transmitted by the belt, is 5.53 hp and (b) the work done in one minute of sanding, in Btu is 37500 Btu.

What is power?
Power is the ability to influence people, events, and outcomes. It is a fundamental element in the human experience, and it has been studied by scholars and philosophers for centuries. Power can manifest itself in many forms, such as political, economic, military, cultural, or social. It is often used to control or dominate others, and it is used to create and maintain systems of inequality and oppression. Power can also be used for good, such as providing people with the resources and ability to make positive changes in their lives and the world around them.

a) Power Transmitted by the Belt = (15 bf) x (1500 ft/min) x (0.2) = 4500 bf-ft/min = 4500 ft-lb/min = 7500 Btu/s = 5.53 hp

b) Work Done in One Minute of Sanding = (15 bf) x (1500 ft/min) = 22500 ft-lb = 37500 Btu

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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:

0.4 Dinas*s/cm^2

Explanation:

Tenemos una viscosidad:

V = 0.04 N*s/m^2

Y queremos reescribir esto en Dinas*s/cm^2

Primero transformemos la unidad del denominador, es decir, tenemos que pasar de 1/m^2 a 1/cm^2

Para ello, usamos que:

1m = 100cm

entonces:

(1m/100cm) = 1

Si elevamos ambos lados al cuadrado, obtenemos:

(1m/100cm)^2 = 1

Ahora podemos multiplicar el valor de la viscosidad por esto (que es igual a 1)

V = 0.04 N*s/m^2*((1m/100cm)^2 = 0.00004 N*s/cm^2

Ahora debemos convertir de Newtons a Dinas

Sabemos que:

1 N = 100,000 dinas

1 = (100,000 dinas/1N)

Entonces, de vuelta podemos multiplicar nuestra viscosidad por  (100,000 dinas/1N), que es igual a 1 (asi que no cambia el valor, solo sirve para cambiar las unidades)

0.00004 N*s/cm^2 = (100,000 dinas/1N)*(0.00004 N*s/cm^2)

                                 = (100,000 dinas)*(0.00004 s/cm^2)

                                 = 0.4 Dinas*s/cm^2

Answer:

5.6 × 10⁻²[1 + 4.643 × 10⁻⁷exp (x/18)] V/cm

Explanation:

The total current density, J = drift current density, J' + diffusion current density, J"

J = J' + J"

J' = nμeE where n = electron concentration =  10¹⁶ exp (-x/18) cm³, μ = electron mobility = 960 cm²/V-s, e = electron charge = 1.602 × 10⁻⁹ C and E = electric field

J" = eDdn/dx where e = electron charge = 1.602 × 10⁻¹⁹ C, D = diffusion coefficient = 25 cm²/s and dn/dx = concentration gradient = d10¹⁶ exp (-x/18) cm³/dx = (-10¹⁶/18)exp (-x/18) cm³

So,

J = J' + J"

J = nμeE + eDdn/dx

E = (J - eDdn/dx)/nμe

Since J = 40 A/cm², substituting the values of the variables into the equation, we have

E = (J - eDdn/dx)/nμe

E = [40 A/cm² - 1.602 × 10⁻⁹ C × 960 cm/V-s × (-10¹⁶/18)exp (-x/18) cm²] ÷ (10¹⁶exp (-x/18) cm³ × 960 cm²/V-s × 1.602 × 10⁻⁹ C)

E = [40 A/cm² - 1537.92 × 10⁻⁹ C-cm/V-s × (-10¹⁶/18)exp (-x/18) cm²] ÷ (10¹⁶exp (-x/18) cm² × 1537.92 × 10⁻⁹  C-cm³/V-s)

E = [40 A/cm² + 85.44 × 10⁷ C-cm/V-s exp (-x/18) cm²] ÷ (exp (-x/18) cm² × 1537.92 × 10⁷  C-cm³/V-s)

E = [40 A/cm² ÷ (exp (-x/18) cm² × 1537.92 × 10⁷  C-cm³/V-s)] + [85.44 × 10⁷ C-cm/V-s exp (-x/18) cm² ÷ (exp (-x/18)  × 1537.92 × 10⁷  C-cm²/V-s]

E = [0.026 × 10⁻⁷ A/cm²exp (x/18)    V-s/C-cm³] + 0.0056]

E = [0.026 × 10⁻⁷exp (x/18) V/cm] + 0.0056]

E = 0.056[0.026 × 10⁻⁷exp (x/18) V/cm]/0.056 + 1]

E = 0.056[4.643 × 10⁻⁷exp (x/18) V/cm] + 1]

E = 5.6 × 10⁻²[1 + 4.643 × 10⁻⁷exp (x/18)] V/cm

About 296.5 N of anchoring force is required to keep the elbow in place. About 134.8 is the direction of the anchoring force.

The speed of an object is its magnitude (or value), which is the velocity. The item is traveling in the direction indicated by the velocity vector. Imagine a circle (or, better yet, draw one) and an object traveling along the path it defines.

Units of mass times length over time squared are used to express the strength of a force. The most used unit in metric measurements is the newton (N), which is equal to one-kilogram times one meter over one second squared.

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__________________________________________________________

Hello! In this question, I will answer the first part of the question, in which we will determine the x, y, and z components for the reaction at joint A in the ball-and-socket.
__________________________________________________________

Explanation:

To better understand what the question entails, we will start off by drawing the free-body diagram to understand the direction of our components. The image is attached below. This will help us solve for our components.

__________________________________________________________

Solve:

To start, we will first solve for the weight of the sign, we will use the formula:

[tex]W=mg[/tex]

Whereas:

Plug in our values into the formula and solve:

[tex]W = 90\cdot9.81=882.9 N[/tex]

Since we now know our weight value, we can solve for our force in the cable BC, expressed as a vector. We will use the formula:

[tex]\bold{T}_{BC}=T_{BC}(\frac{\bold{r}_{BC}}{r_{BC}})[/tex]

Where:

Plug in our values into the formula and solve:

[tex]\bold{T}_{BC}=T_{BC}\bigg(\cfrac{\text{[i - 2j+ 2k] ft}}{\sqrt{(1)^2+(-2)^2+(2)^2}}}\bigg)\\\\\bold{T}_{BC}=T_{BC}\bigg(\cfrac{\text{[i - 2j+ 2k] ft}}{3}}}\bigg)\\\\\text{Simplify}\\\\\bold{T}_{BC}=\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k[/tex]

Now, let us solve for our force in the cable BD, also expressed as a vector. Use the formula:

[tex]\bold{T}_{BD}=T_{BD}(\frac{\bold{r}_{BD}}{r_{BD}})[/tex]

Use the same steps from solving for our vector force of cable BC. Plug in the values and solve:

[tex]\bold{T}_{BD}=T_{BD}\bigg(\cfrac{\text{[-2i - 2j+ k] ft}}{\sqrt{(-2)^2+(-2)^2+(1)^2}}}\bigg)\\\\\bold{T}_{BD}=T_{BD}\bigg(\cfrac{\text{[-2i - 2j+ k] ft}}{3}}}\bigg)\\\\\text{Simplify}\\\\\bold{T}_{BD}=-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j+\frac{1}{3} T_{BD}k[/tex]

We now need to find the moment at A at equilibrium (0). This is known as:

[tex]\sum M_{a}=0 \Rightarrow\bold{r}_{B}\times(\bold{T}_{BC}+\bold{T}_{BD}+\bold{W})=0[/tex]

Whereas:

Plug in values into the equation:

[tex]2\text{j}\times\big[\big(\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k\big)+\big(-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j+\frac{1}{3} T_{BD}k\big)\big]\\+\text{j}\times-882.9\text{k}=0[/tex]

Find the moment about the z-axis to zero. This is known as:

[tex]\sum M_{z}=0\\\\-\frac{2}{3}T_{BC}+\frac{4}{3}T_{BD}=0\\\\\text{Simplify}\\\\T_{BC}=2T_{BD}[/tex]

Now, find the moment about the x-axis to zero. This is known as:

[tex]\sum M_{x}=0\\\\\frac{4}{3}T_{BC}+\frac{2}{3}T_{BD}-882.9=0\\\\\text{We know }T_{BC}=2T_{BD}\text{, plug in }2T_{BD}\text{ in }T_{BC}\\\\\frac{4}{3}\times2T_{BD}+\frac{2}{3}T_{BD}-882.9=0\text{ Solve for }T_{BD}\\\\\frac{10}{3}T_{BD}=882.9\\\\T_{BD}=264.87\text{ N}[/tex]

Now, let us calculate the tension in wire BC.

[tex]T_{BC}=2T_{BD} = 2(264.87) = 529.74\text{ N}[/tex]

Let us calculate the reaction at A by solving for the equilibrium force equation. The formula is:

[tex]\sum F=0\\\\\bold{F}_A+\bold{T}_{BC}+\bold{T}_{BD}+\bold{W}=0[/tex]

Plug in our known information into the equation and simplify.

[tex]\big[(A_xi+A_yj+A_zk)+(\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k)+(-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j\\+\frac{1}{3} T_{BD}k)-981\text{k}\big]=0\\\\\text{Separate by component}\\\\\big[(A_x+\frac{1}{3}T_{BC}-\frac{2}{3}T_{BD})\text{\bold{i}}+(A_y-\frac{2}{3}T_{BC}-\frac{2}{3}T_{BD})\text{\bold{j}}+(A_z+\frac{2}{3}T_{BC}+\frac{1}{3}T_{BD}\\-981)\text{\bold{k}}\big]=0\\\\\text{Plug in known values}\\[/tex]

[tex]\big[(A_x+\frac{1}{3}\times529.74-\frac{2}{3}\times264.87)\text{\bold{i}}+(A_y-\frac{2}{3}\times529.74-\frac{2}{3}\times264.87)\text{\bold{j}}+(A_z+\frac{2}{3}\times529.74+\frac{1}{3}\times264.87-882.9)\text{\bold{k}}\big]=0[/tex]

Now, we can calculate our reactions.

Calculate the reaction at A in the x-direction.

[tex]\sum F_x=0\\\\A_x+\frac{1}{3}\times529.74-\frac{2}{3}\times264.87=0\\\\\boxed{A_x=0\text{ N}}[/tex]

Calculate the reaction at A in the y-direction.

[tex]\sum F_y=0\\\\A_y-\frac{2}{3}\times529.74-\frac{2}{3}\times264.87=0\\\\A_y-529.74=0\\\\\boxed{A_y=529.74\text{ N}}[/tex]

Calculate the reaction at A in the z-direction.

[tex]\sum F_z=0\\\\A_z+\frac{2}{3}\times529.74+\frac{1}{3}\times264.87-981=0\\\\A_z-441.45=0\\\\\boxed{A_z=441.45\text{ N}}[/tex]

__________________________________________________________

Answer:

[tex]\boxed{\text{A=} < \text{0, 529.74, 441.45} > }[/tex]

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Answer:

105 kips

Explanation:

Determine the factored bearing deformation strength per bolt

we can determine this using the relationship below

Bearing deformation strength = C * Area * Fl  --------- ( 1 )

Fl = factored load = 100 kips

Area ( bearing area ) = diameter of bolt * thickness of end plate

                                  = 0.875 * 0.5 = 0.4375 in^2

C = 2.4 ( constant for bolt holes not drilled at edges )

back to equation 1

2.4 * 0.4375 * 100

= 105 kips

The process of creating sugar from raw sugar cane involves several steps. First, the raw cane is passed through a series of crushers to extract the juice from the cane.

A crusher is a machine designed to reduce large rocks into smaller rocks, gravel, or rock dust.

This step also adds about 5% extra mass of water to the sugar cane to help with the extraction process. The crushed cane and liquid juice are then sent to a filter press, which creates a cake containing 4% of the weight of the cane juice. This cake has a composition similar to the overall non-fiber content of the raw cane. The filtrate is then sent to an evaporator, where enough water is evaporated to obtain a pale yellow juice that is 41% water. Finally, the liquid is crystallized to produce sugar crystals, which are typically about 99.8% sucrose.

To learn more about crusher
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