a)

Two MCR output instructions are programmed to control entire section of a program. The section of program which needs to be controlled begins with one MCR instruction and the other MCR instruction at the end. When the first MCR instruction becomes true, all the outputs which are present in between the two MCR instructions will act as per the logic and when the first MCR instruction is false, then all the non retentive outputs will be de-energized even though the input instructions present in that rung are true. The retentive outputs present in the MCR zone will retain their previous state. Understand this with an example.

Moment of inertia about x-axis:

I₁= 1/12 (200) (12)³ + (200) (12) (104)²

 = 25.9872×10⁶ mm⁴

I₂= 1/2 (8) (196)³ = 5.0197×10⁶ mm⁴

I₃= I₁ = 25.9872 × 10⁶mm⁴

I = I₁ + I₂ + I₃

 = 56.944×10⁶ mm⁴ = 56.944×10⁻⁶ m⁴

σ = MC/I  with C= 1/2(220) = 110mm = 0.110m

M = Iσ /C with σ= 155× 10⁶ Pa

Mₓ = (((56.944×10⁻⁶) ( 155× 10⁶))/ 0.110) = 80.2×10³N.m

Mₓ = 80.2kN.m

Wide flange beams are excellent for uses requiring this added strength, such as structural supports in buildings, bridges, and other structures.

The letter W, the nominal depth in inches, and the weight in pounds per foot are used to identify wide flange beams.

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The volume flow rate of air through the venturi meter is approximately 69.4 cubic feet per second.

To determine the volume flow rate of air flowing through the venturi meter, we can use the principle of conservation of mass.

Since air is an incompressible fluid, the mass flow rate at location 1 (entrance) must be equal to the mass flow rate at location 2 (throat).

Therefore, we can express the volume flow rate Q as:
\(Q = A_1V_1\)

= \(A_2V_2\)
where \(A_1\) and \(A_2\) are the cross-sectional areas of the entrance and throat respectively, and \(V_{1}\) and \(V_{2}\) are the velocities of air at the two locations.

To calculate the velocities, we can use Bernoulli's equation which states that the sum of the pressure,

kinetic energy and potential energy per unit mass of a fluid is constant along a streamline.

Neglecting potential energy and kinetic energy changes (since the diameter changes only slightly), we have:
\(P_1 +\frac{1}{2}\)ρ\(V_1^2 = P2 + \frac{1}{2}\)ρ\(V_2^2\)
where \(P_1\) and \(P_2\) are the pressures at the entrance and throat respectively, and ρ is the density of air.

Rearranging this equation, we get:
\(V1 = \sqrt(\frac{2\times (P_1 - P_2)}{/\rho} ) \times (\frac{A_2}{A_1)}^2\)

Substituting the given values, we get:
\(V1 = \sqrt\frac{(2\times (12.2 - 11.8)}{(144*0.0765))} \times \frac{(1.8}{2.6)}^2\)
  ≈ 37.4 ft/s
Using the equation for volume flow rate, we get:
\(Q = A_1V_1\)

\(= \pi \times (\frac{2.6}{2})^2\times 37.4\)
  ≈ \(69.4 ft^3/s\)

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To determine the horizontal and vertical components of reaction at point A and the reaction at point B on the beam, we can use the principles of static equilibrium.

Let's break down the problem step by step:

1. Draw a free body diagram of the beam. Label the known forces and reaction components.

2. Since the beam is supported by a pin at point A, the reaction at point A will only have a vertical component (RAy) and no horizontal component (RAx).

3. The strut BC exerts a force on the beam at point B. This force can be broken down into horizontal (FBx) and vertical (FBy) components.

4. Apply the equations of equilibrium. The sum of all the horizontal forces acting on the beam should be equal to zero, and the sum of all the vertical forces acting on the beam should also be equal to zero.

5. Set up the equations:

Sum of horizontal forces: FBx - RAx = 0
Sum of vertical forces: RAy + FBy = 0

6. Substitute the known values and solve the equations to find the unknowns. The value of RAx will be zero since there is no horizontal component at point A.

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Answer:

Can I be it plzzzzz!?

Explanation:

I need points

Answer:

day if you workout without Zero billing that means you're not sweating. Sweating you're not losing anything that means you have zero outcomes

Explanation:

1. LRT1 (L1), LRT2 (L2), and MRT3 (L3) have elevated viaducts, stations, ramps, and tunnels along their alignments.

2. The track system of L1, L2, L3, and PNR consists of elevated tracks supported by viaducts, equipped with rails, sleepers, switches, crossings, and signaling systems.

The alignments of LRT1 (L1), LRT2 (L2), and MRT3 (L3) feature various types of structures. These include elevated viaducts, which are elevated platforms that support the tracks and allow trains to run above ground level.

Additionally, there are stations along the alignment where passengers board and alight from the trains. Ramps are constructed to provide smooth transitions between different levels of the track system.

The track systems of L1, L2, L3, and PNR consist of elevated tracks supported by viaducts, ensuring stability and providing a platform for trains to run on. These tracks are laid with rails and sleepers, which maintain the alignment and support the weight of the trains. The system includes switches and crossings to allow trains to change tracks or merge with other lines.

Signaling systems are implemented to control train movements and ensure safe operations. This comprehensive track system facilitates the smooth and efficient functioning of the LRT and MRT lines, as well as the Philippine National Railways (PNR).

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The major evolutionary innovation that led to the appearance of reptiles and their radiation into continental interiors was the development of amniotic eggs.

An amniotic egg is a type of egg that has a protective membrane called an amnion that surrounds the developing embryo. This allowed reptiles to lay their eggs on land rather than in water. This made it possible for reptiles to colonize new habitats and environments, including the interiors of continents, which were previously inaccessible to amphibians and other aquatic animals. Because they did not have to return to water to lay their eggs, reptiles were able to live and reproduce in a much wider range of habitats, including deserts, grasslands, and forests. This was a major evolutionary innovation that allowed reptiles to become one of the most successful groups of animals on Earth.

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The variation of temperature in the wall is T(x)= (q_0x) /(kL) + T_0. The heat flux q_n is -28.1 W/m^2 and the temperature of the left surface of the wall is T_0 = 300 degree C.

B. To obtain the variation of temperature in the wall, we need to use the heat conduction equation, which states that the rate of heat flow through a solid is proportional to the temperature gradient.

The equation is given by q_x = -kdT/dx. Since we know the heat flux q_0 at the left surface, we can integrate this equation from x = 0 to x = L to obtain the temperature distribution T(x) in terms of q_0 and the thermal conductivity k.

On the right surface, we can use the equation for heat transfer by convection and radiation, which is given by

q_n = h(T_6 - T_surrounding) + esigma(T_6^4 - T_surrounding^4)

where h is the convection heat transfer coefficient, e is the emissivity, sigma is the Stefan-Boltzmann constant, T_6 is the surface temperature and T_surrounding is the surrounding temperature.

Plugging in the given values we can find out the value of heat flux q_n and the temperature of the left surface of the wall (at x = 0).

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The range of frequencies that can pass through a low-pass filter is known as its bandwidth.

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To determine the magnitude of the principal strains and their orientation with respect to the 0 degrees gage, you will need to use the equations for the transformation of rectangular strain gage rosette readings.

The principal strains can be calculated using the formula:

ε1,2 = (E + E90)/2 ± sqrt[(E - E90)^2/4 + ao^2]

Plugging in the given values, we get:

ε1 = (1200 - 400)/2 + sqrt[(1200 + 400)^2/4 + 2000^2] = 1500 μm/m
ε2 = (1200 - 400)/2 - sqrt[(1200 + 400)^2/4 + 2000^2] = -700 μm/m

To determine their orientation, we use the formula:

tan 2θ = 2ao/(E - E90)

Plugging in the given values, we get:

tan 2θ = 2(2000)/(1200 - (-400)) = 4

θ = 63.43°

Therefore, the magnitude of the principal strains are 1500 μm/m and -700 μm/m, with orientations of 63.43° and 153.43° counterclockwise from the 0 degrees gage orientation.

To represent this state of strains on a Mohr circle, we plot the two principal strains on the horizontal axis and the vertical axis, with the center of the circle at the origin. Then, we draw a circle with a radius equal to the average of the two principal strains. Finally, we draw lines from the center of the circle to the points representing the principal strains, which intersect the circle at the points where the normal and shear strains can be read off.

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Answer:

the types of social responsibility are as follows

1)Environmental Responsibility

2)Ethical Responsibility

3)Philanthropic Responsibility

4)Economic Responsibility

Social responsibility is an obligation every person has to do to uphold a balance among the economy as well as the environment. Social responsibility refers to maintain the balance between the environment and the budget.

It affects not only corporate governments but also everybody whose activity impacts the climate.

The Social responsibility towards the environment are as follows

1)Defend the environment, create balanced use of natural ingredients, and take important steps to stop pollution from the several processes adopted by the industries.

2)Encourage the viable use of raw resources and natural materials. Introduce strategies for continuously enlightening our environmental duty.

Introduce rules for continuous development in their ecological performance.

Obey with the law in force on conservational protection, and with other willingly recognized promises.

3)To encourage, by on-the-job practice, the participation of all our workers, independently and based on the team, in environmental safety problems.

4)To unite with the liable Agencies as well as Governmental Authorities.

To proceed with environmental protection courses.

5)To talk and report on our eco-friendly management struggles.

Implement and sustain the necessities of a Chain of Custody.

6)Stop trade and acquisition of illegally obtained wood.

7)Apply and continue a due industry system to avoid the acquisition and sale of criminally harvested timber and its products.

The new resistance of the wire is 2.4.

It should be noted that to solve for the new resistance of the wire after it has been drawn under pressure, we need to find the new cross-sectional area of the wire. We are told that the diameter of the wire has been reduced to 50% of its original value.

The cross-sectional area of a circle is proportional to the square of its diameter, so if the diameter is reduced by a factor of 2, the area is reduced by a factor of 4. Therefore, the new cross-sectional area of the wire is 1/4 of its original value.

It should be noted that R' = 16R

New resistance will be:

= 16 × 0.15

= 2.4

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The gain-bandwidth product (GBW) of an op-amp is typically estimated using the following expression:

GBW [Hz] = A0 * gm / (2 * pi * Cc)

It should be noted that A0 is the open-loop gain of the op-amp, gm is the transconductance of the input stage, and Cc is the value of the compensation capacitor.

This expression represents the frequency at which the product of the open-loop gain and the closed-loop bandwidth of the op-amp is equal to unity. It is a measure of the maximum frequency at which the op-amp can operate as an amplifier with stable feedback.

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Convective acceleration is the acceleration experienced by a fluid element as it is transported from one point to another by a moving flow, and is proportional to the velocity gradients.

The convective acceleration in the x direction can be calculated using the formula a_conv,x = u * ∂u/∂x + v * ∂u/∂y + w * ∂u/∂z. Given the velocity distribution, we have u=ax, v=by, and w=cxy. Taking partial derivatives, we get ∂u/∂x=a, ∂u/∂y=0, and ∂u/∂z=0. Substituting these values into the formula, we get a_conv,x = ax * a + by * 0 + cxy * 0 = a²x.
Convective acceleration (x-direction) = u(∂u/∂x) + v(∂u/∂y) + w(∂u/∂z)

Given the velocity distribution: u = ax, v = by, and w = cxy

First, we need to find the partial derivatives of u:

∂u/∂x = a (as u only depends on x)
∂u/∂y = 0 (as u does not depend on y)
∂u/∂z = 0 (as u does not depend on z)

Now, substitute the values into the convective acceleration formula:

Convective acceleration (x-direction) = (ax)(a) + (by)(0) + (cxy)(0)
Convective acceleration (x-direction) = a²x
So, the convective acceleration in the x-direction is:

B. A²x

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The expression for the pressure gradient is given below: (image)

Pressure gradient denotes the alteration in atmospheric pressure over an indicated distance. Essentially, it measures how quickly the pressure changes regarding distance, figuring by dividing the variation of pressure by the range for which such alteration happens.

Regarding meteorology, the force and direction of winds closely associate to pressure gradient's influence. There exists a flow of air from high amounts of pressure towards low ones; thereby, a greater difference of pressure over a given distance results on far stronger winds causing effects.

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The anticipated shape of the boundary layer development over the long side of the plate is initially laminar, transitioning to turbulent flow at a certain distance from the leading edge. The boundary layer thickness at the end of the laminar section and the percentage of the plate area covered by a laminar boundary layer can be determined. The total drag force on one side of the plate can also be calculated.

The boundary layer is the thin layer of fluid that develops near the surface of an object as it moves through a fluid medium. In this case, the air flowing over the thin flat plate creates a boundary layer. Initially, the boundary layer is laminar, characterized by smooth and ordered flow. As the air flows further along the plate, the boundary layer may undergo transition and become turbulent, which is characterized by chaotic and unpredictable flow patterns.

To sketch the anticipated shape of the boundary layer development, we would start with a thin laminar boundary layer near the leading edge of the plate. This layer would gradually increase in thickness as the air flows along the plate due to the shear stress between the slower-moving air near the surface and the faster-moving free stream air. Eventually, at a certain distance from the leading edge, the laminar boundary layer will transition to turbulent flow.

The distance from the leading edge of the plate where the flow becomes turbulent can be determined using the Reynolds number. The Reynolds number (Re) is a dimensionless parameter that relates the inertial forces to the viscous forces in the flow. For flow over a flat plate, the critical Reynolds number for transition from laminar to turbulent flow is typically around 5 × 10^5. By calculating the Reynolds number using the given flow conditions, the distance at which the flow becomes turbulent can be determined.

The boundary layer thickness at the end of the laminar section can be estimated using the empirical Blasius solution for laminar boundary layers. It is given by the formula: δ = 5.0 × (x/Re_x)^0.5, where δ is the boundary layer thickness, x is the distance along the plate, and Re_x is the Reynolds number at that distance. By calculating the boundary layer thickness using this formula, we can determine the value at the end of the laminar section.

The percentage of the plate area covered by a laminar boundary layer can be estimated by dividing the laminar boundary layer thickness by the plate's height (1.5m) and multiplying by 100.

To calculate the total drag force on one side of the plate, we need to consider both the skin friction drag and the pressure drag. The skin friction drag is caused by the shear stress between the boundary layer and the plate's surface, while the pressure drag is caused by the pressure difference between the front and rear ends of the plate. The total drag force can be calculated by integrating the skin friction drag and the pressure drag along the length of the plate using appropriate formulas.

In conclusion, the anticipated shape of the boundary layer over the plate starts with a laminar boundary layer that transitions to turbulent flow at a certain distance from the leading edge. The distance of transition, boundary layer thickness at the end of the laminar section, percentage of laminar boundary layer coverage, and the total drag force can be calculated using relevant formulas and flow conditions.

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In SI units using the expression appropriate is:

To express this answer in SI units using an appropriate prefix, we need to convert pum to meters and ms to seconds.

1 pum = 10^-12 meters

1 ms = 10^-3 seconds

So, 11.8889 pum/ms = (11.8889)(10^-12 meters/10^-3 seconds) = 11.8889 x 10^-9 meters/second = 11.8889 nm/s

Answer: 11.9 nm/s (to three significant figures)


(b) (34 mm)(0.0763 Ms)/263 mg = 0.00984673 mm*Ms/mg

To express this answer in SI units using an appropriate prefix, we need to convert mm to meters, Ms to seconds, and mg to kilograms.

1 mm = 10^-3 meters

1 Ms = 10^6 seconds

1 mg = 10^-6 kilograms

So, 0.00984673 mm*Ms/mg = (0.00984673)(10^-3 meters)(10^6 seconds)/(10^-6 kilograms) = 9.84673 x 10^6 meters*seconds/kilogram = 9.84673 Ms*m/kg

Answer: 9.85 Ms*m/kg (to three significant figures)


(c) (4.78 mm)(263 Mg) = 1257.14 mm*Mg

To express this answer in SI units using an appropriate prefix, we need to convert mm to meters and Mg to kilograms.

1 mm = 10^-3 meters

1 Mg = 10^6 kilograms

So, 1257.14 mm*Mg = (1257.14)(10^-3 meters)(10^6 kilograms) = 1257.14 x 10^3 meters*kilograms = 1257.14 km*kg

Answer: 1257 km*kg (to three significant figures)

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19.15JK−1mol−1 is the change in entropy due to the expansion of 1 mole of an ideal gas at 25oC is subjected to expand reversibly ten times of its initial volume.

Entropy is a notion in science and a quantifiable physical characteristic that is frequently connected to a condition of disorder, unpredictability, or uncertainty. The phrase and its underlying idea are employed in a wide range of disciplines, from classical thermodynamics—where they were initially identified—through statistical physics, which uses them to describe nature at the molecular level, to the foundations of information theory. It has numerous applications in physics and chemistry, biological systems and how they relate to life, cosmology, economics, sociology, weather science, the study of climate change, and information systems, particularly the transfer of information via telecommunication. Scottish engineer and scientist William Rankine first used the terms thermodynamic function and heat potential to describe the thermodynamic notion in 1850.

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Explanation:

I think is SAE standard J2810

The solution to the differential equation is

y (k)= (-1/3  ) * 1\(^{k}\) +(1/3) * 2\(^{k}\)  

= (  -1/3) +(2/3) * 2\(^{k}\)

To solve the   given differential equation Y(k+2) - 3y (k+1) +2y(k) = 0 with the initial conditions y(0) = 0 and y(1) = 1, we can use the method of characteristic roots.

Let's assume the solution has the form y(k)=   r\(^{k}\). Substituting this into the differential equation, we get   -

\(r^k+2\) -  + \(2r^k\) = 0

Dividing through by \(r^k\), we have

r² - 3r + 2 = 0

This    is a quadratic equation,which can be factored as

(r - 1  ) (r - 2) =0

So, we have two characteristic roots  

r1 = 1 and r2= 2.

The general solution is given by   -

y(k) = A   * \(r1^k\) +B * \(r2^k\)

Applying the initial conditions, we have -

y(0) = A * 1⁰ + B* 2⁰ =   A + B

= 0 → A = -B

y(1) =A * 1¹ + B * 2¹

=    A + 2B = 1

Solving these equations    simultaneously,we find A = -1/3 and B = 1/3.

Therefore, the solution to the differential equation Y(k+2) - 3y(k+1) + 2y(k) = 0 with the initial conditions y(0) = 0 and y(1) = 1 is  -

y(k )   = (-1/3) * 1\(^{k}\) +(1/3) * 2\(^{k}\)

= (-1/3) +   (2/3) * 2\(^{k}\)

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Bus: linear arrangement Round in the form of a ring Systems are arranged in the form of a star using wire in this topology. Mesh: A non-synchronized arrangement of systems.

This topology is referred to as a star bus topology since it incorporates different star topologies into a single bus. A bus or star topology is comparable to the widely used tree topology in networks.

The most popular topology in home and office networks is the star topology because it is so simple to deploy, operate, and troubleshoot. A redundant network of connections between nodes forms a mesh topology.

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Answer:

u need to plan it out

Explanation:

u need to plan it out

Answer:

use the turn 1 degrees option and put a repeat loop on it

Explanation:

u can add sound in ur loop

The first activity in test preparation is to determine what the test will be about. (True)

Tests are a necessary part of school life because you'll eventually need to demonstrate how much you've learned. You're not alone if you often panic before exams and frantically study, but there are more effective ways to get ready.

Here, we've compiled some advice on how to prepare for your upcoming test, including scheduling regular study sessions, coming up with creative ways to learn the material, and taking care of yourself the night before and the day of the test.

Check the exam date and plan your study sessions backwards in time. If your schedule permits it, aim to complete two to three study sessions per week for at least a few weeks before the test. Reduce the amount of time before the exam if your exams are too close together.

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Answer:

d) "Napoleon Declares Himself Holy Roman Emperor"

Explanation:

The French Revolution is defined as a period of the major social upheaval which began in the year 1787 and lasted till year 1799. This revolution completely redefined the the very nature of the political power in France. and also the relationship between the rulers of France and the people they governed.

The 1789 Estates-General was the 1st meeting since year 1614 of the French Estates-General. It is a general assembly which represents the French estates of  realm.

During the French revolution was at peak, the National Assembly issued the Declaration of the rights of the man to the public.

Maximilien Robespierre was considered to be one of the most influential figure and most important statesman during the French Revolution.

Thus all the options (a),(b) and (c) are headlines about the French Revolution, except option (d).

2% glutaraldehyde is a disinfectant that is often used. preferred agent for semi-critical objects that can't be exposed to such adaptable fiberoptic sigmoidoscopy.

The small incision medical procedure known as a sigmoidoscopy examines the large intestine from the rectum to the sigmoid colon, which is the closest portion of the colon. There are two different forms of sigmoidoscopy: rigid sigmoidoscopy employs a rigid equipment while flexible sigmoidoscopy uses a flexible endoscope. The chosen procedure is usually flexible sigmoidoscopy. Although they are not the same, a colonoscopy and a sigmoidoscopy are related procedures. While a colonoscopy looks at the entire big bowel, a sigmoidoscopy just looks at the sigmoid, which is the farthest portion of the colon. The doctor can view the interior of the large intestine from the rectum to the sigmoid, or left side of the colon, using flexible sigmoidoscopy. The process can be used by doctors to identify the source of constipation, abdominal pain, or diarrhea. In the descending colon and rectum, they also utilize it to check for benign and malignant polyps as well as early indications of cancer. The doctor can see intestinal bleeding, inflammation, abnormal growths, and ulcers in the descending colon and rectum during flexible sigmoidoscopy.

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Answer:

4

Explanation:

A cylinder pressure has an outer diameter of 200 mm  then minimum value of wall thickness t=4 mm.

Given:

D=200 mm

P=4 MPa

t= Wall thickness

maximum shear stress=25 MPa

We know that

hoop stress  σh\(=\frac{pd}{2t}\)

Longitudinal stress σl=\(=\frac{pd}{4t}\)

So maximum shear tress in plane Tmax=(σh-σl)/2

\(Tmax=\frac{pd}{8t}\)

Now by putting the value

\(25=\frac{4*200}{8t}\)

So   t= 4mm

The minimum value of wall thickness t=4 mm.

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