To calculate the cost per kg of reinforcement Calculate for material. 1) Material i. 1 m.ton of reinforcement = RM1,700.00 (1 mt = 1000 kg) ii. 100 kg of reinforcement = 1700/1000 X 100-RM170.00 Wastage = 8% iii. iv. Binding wires - 0.5 kg for 100 kg of reinforcement Cost of wire RM3.00/kg V. vi. Weight of steel reinforcement Y19 for 1m length - 2.25kg Calculate for labour 2) Labour i. 1 barbender to carry, cut, bend and fix 100 kg requires 8 hours ii. wages of 1 barbender = RM48.00/day iii. for 12 hours the cost of bar bender 35/8 X12 = RM 52.50 Profit = 20%

We can easily design 8 × 1 multiplexer using two 2 × 1 multiplexers.

The required main answer to implement an 8 × 1 multiplexer using two 2 × 1 multiplexers is to connect the output of one 2 × 1 multiplexer to the select input of the second 2 × 1 multiplexer. A brief explanation is given below:Here, we have 8 inputs (I0 to I7), 1 output and 3 selection lines (A, B, C). In order to design an 8 × 1 multiplexer using two 2 × 1 multiplexers, we need to consider four inputs at a time.

We can use the two 2 × 1 multiplexers to choose one of the four inputs at a time by using the selection lines A, B, C. To select the input from the first four inputs, the selection lines A, B and C of the two 2 × 1 multiplexers should be connected in the following way: A (MSB) of 8 × 1 multiplexer should be connected to A of 2 × 1 multiplexer 1.B of 8 × 1 multiplexer should be connected to B of 2 × 1 multiplexer 1.C of 8 × 1 multiplexer should be connected to S of 2 × 1 multiplexer 1.

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The Rankine cycle is a thermodynamic cycle that is commonly used in steam power plants. This cycle is used to produce electricity by converting heat energy into mechanical energy, and then into electrical energy. The cycle consists of four main components:

a boiler, a turbine, a condenser, and a pump. In this question, we are going to investigate the effect of back pressure on the cycle efficiency of a steam power plant that operates on the Rankine cycle.

The steam leaves the boiler at a pressure of 6.0 MPa and a temperature of 500°C. The back pressure, or condensing pressure, is varied from 7 kPa to 70 kPa, and the cycle efficiency is determined and graphed for each back pressure value. The results are then discussed in terms of available energy.

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The symmetrical components are the three components of a set of unbalanced three-phase AC voltages or currents that are equivalent to a set of balanced voltages or currents when applied to a three-phase system. In this problem, we are required to calculate the symmetrical components for the given unbalanced set of voltages:Va = 270 V/-120⁰V₁ = 200 V/100°Vc = 90 VZ-40⁰

By using the following formula to find the symmetrical components of the given unbalanced voltages:Va0 = (Va + Vb + Vc)/3Vb0 = (Va + αVb + α²Vc)/3Vc0 = (Va + α²Vb + αVc)/3where α = e^(j120) = -0.5 + j0.866
After substituting the given values in the above equation, we get:Va0 = 156.131 - j146.682Vb0 = -6.825 - j87.483Vc0 = -149.306 + j59.800
Therefore, the symmetrical components for the given unbalanced voltages are:Va0 = 156.131 - j146.682Vb0 = -6.825 - j87.483Vc0 = -149.306 + j59.800

The symmetrical components for the given unbalanced voltages are:Va0 = 156.131 - j146.682Vb0 = -6.825 - j87.483Vc0 = -149.306 + j59.800

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The term that is not an example of fatigue is pressuresed oil pipes. Option d is correct.

Fatigue is a weakening of a metal caused by repeated, varying forces or loads, frequently combined with cyclic stresses. A fatigue crack begins as a small crack on the surface of a component, eventually propagating into the interior of the part, causing it to fail.

Bending stresses, torsion, and compression are examples of cyclic stresses that cause fatigue. Fatigue cracks on the other hand, are not generally found in pressured oil pipes. There are several reasons for this, one of which is that pressured oil pipes do not usually experience cyclic stress.

Furthermore, the material used in making pressured oil pipes is typically thicker and stronger than that used in other parts that are more susceptible to fatigue. As a result, the probability of a fatigue crack developing in pressured oil pipes is lower.

Therefore, d is correct.

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Calculate the current on the H.V. side:

Using the formula:

    Current (I1) = Transformer rating (S) / (√3 x High Voltage (V1))

    I1 = 100,000 VA / (√3 x 11000 V) ≈ 5.73 A

Calculate the resistance on the H.V. side:

     Resistance (R1) = Full-load copper loss on H.V. side (Pcu1) / (3 x Current squared (I1²))

    R1 = 0.62 kW / (3 x 5.73 A²) ≈ 0.019 ohms

Calculate the current on the L.V. side:

Using the formula:

    Current (I2) = Transformer rating (S) / (√3 x Low Voltage (V2))

     I2 = 100,000 VA / (√3 x 317 V) ≈ 166.67 A

Calculate the resistance on the L.V. side:

     Resistance (R2) = Full-load copper loss on L.V. side (Pcu2) / (3 x

     Current squared (I2²))

     R2 = 0.48 kW / (3 x 166.67 A²) ≈ 0.00061 ohms

Calculate the total resistance (Ra):  Total resistance (Ra) = R1 + R2

     Ra = 0.019 ohms + 0.00061 ohms ≈ 0.01961 ohms

Calculate the reactance on the H.V. side:

       Reactance (X1) = Total reactance (X%) x Ra / 100

        X1 = 4% x 0.01961 ohms ≈ 0.0007844 ohms

Calculate the reactance on the L.V. side:

       Reactance (X2) = Total reactance (X%) x Ra / 100

       X2 = 4% x 0.01961 ohms ≈ 0.0007844 ohms

Calculate the total reactance (X):

      Total reactance (X) = X1 + X2

       X = 0.0007844 ohms + 0.0007844 ohms ≈ 0.0015688 ohms

the resistance values are:

R1 ≈ 0.019 ohms

R2 ≈ 0.00061 ohms

Ra ≈ 0.01961 ohms

And the reactance values are:

X1 ≈ 0.0007844 ohms

X2 ≈ 0.0007844 ohms

X ≈ 0.0015688 ohms

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Thus, the angle of absolute maximum shear stress, θ = 63.44° (approx.)

Given:

Radius, r = 68 mm

Length, b = 0.72 m

Length, a = 0.44 m

Moment, M = 1.5 RN.m

Moment, M12 = 1.36 kN.m

To determine:

1) Maximum tensile stress, along with its location and direction.

2) Maximum compressive stress, along with its location and direction.

3) Maximum in-plane shear stress at point P.

4) Identify the point where the absolute maximum shear stress takes place and calculate the same with direction.

Calculations:

1) Maximum Tensile Stress: σ max

= Mc/I where, I=πr4/4

Substituting the given values in above formula,

σmax= (1.5*10^3 * 0.44)/ (π* (68*10^-3)^4/4)

σmax = 7.54 N/mm2

Location of Maximum Tensile Stress: The maximum tensile stress occurs at point B, which is at a distance of b/2 from point C in the direction opposite to the applied moment.

2) Maximum Compressive Stress:

σmax= Mc/I where, I=πr4/4

Substituting the given values in the above formula,

σmax= (-1.36*10^6 * 0.44)/ (π* (68*10^-3)^4/4)

σmax = -23.77 N/mm2

Location of Maximum Compressive Stress: The maximum compressive stress occurs at point B, which is at a distance of b/2 from point C in the direction of the applied moment.

3) Maximum In-Plane Shear Stress at point P:

τmax= 2T/A where, A=πr2T = [M(r+x)]/(πr2/2) - (M/πr2/2)x = r

Substituting the given values in above formula,

T = 1.5*68*10^-3/π = 0.326 NmA

= π(68*10^-3)^2

= 14.44*10^-6 m2

τmax = 2*0.326/14.44*10^-6

τmax = 45.04 N/mm24)

Absolute Maximum Shear Stress and Its Direction:

τmax = [T/(I/A)](x/r) + [(VQ)/(Ib)]

τmax = [(VQ)/(Ib)] where Q = πr3/4 and V = M12/a - T

Substituting the given values in the above formula,

Q = π(68*10^-3)^3/4

= 1.351*10^-6 m3V

= (1.36*10^3)/(0.44) - 0.326

= 2925.45 NQ

= 1.351*10^-6 m3I

= πr4/4 = 6.09*10^-10 m4b

= 0.72 mτmax

= [(2925.45*1.351*10^-6)/(6.09*10^-10*0.72)]

τmax = 7.271 N/mm2

Hence, the absolute maximum shear stress and its direction is 7.271 N/mm2 at 63.44° from the x-axis.

Thus, we have calculated the maximum tensile stress, along with its location and direction, maximum compressive stress, along with its location and direction, maximum in-plane shear stress at point P, and the absolute maximum shear stress and its direction.

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Aluminum is easier to plastically deform than steel because of its lower strength and higher ductility. It's a softer metal and is much easier to work with, which makes it more malleable.

Steel has a higher tensile strength and hardness than aluminum, making it more difficult to work with. Although steel can be shaped and molded, it requires more force and energy than aluminum.Aluminum has a hexagonal close-packed (HCP) crystal structure, which makes it more malleable than steel. Its crystal structure means that it can bend and stretch more easily without cracking or breaking.Steel has a body-centered cubic (BCC) crystal structure, which is more rigid and less flexible than aluminum. Its crystal structure means that it is more prone to cracking and breaking when subjected to external forces.In conclusion, the key difference between aluminum and steel is their strength and ductility. Aluminum is more malleable and easier to shape, while steel is stronger and harder to work with.

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The use of the windows in design digital filters is seen in:

To create a digital filter, the first thing you need to do is decide how you want it to affect the different frequencies in the sound. This is usually measured by how big and at what angle something is. The specifications could be the desired frequencies that pass through and don't pass through.

So, First, one decide what the filter should do. Then, one figure out the perfect way for it to react to a quick sound called an "impulse".

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A root locus is a graphical representation of the possible locations of the closed-loop poles of a system as a specific system parameter varies.

In the context of a unity feedback system with an open-loop transfer function HG(s) = K(s² + 25 + 5) / s³, the open-loop transfer function G(s) can be expressed as G(s) = HG(s) / (1 + HG(s)).

By substituting the given expression for HG(s) into G(s), we obtain G(s) = K(s² + 25s + 5) / (s³ + K(s² + 25s + 5)).

The equation ω³ + 25Kω - 5K = 0 can be solved using numerical methods or estimated graphically from the root locus plot. In this case, the root locus plot suggests that the imaginary part of the positive imaginary axis crossing is approximately 5.56 (rounded to two significant figures).

Therefore, the estimated value of ω for the positive imaginary axis crossing is 5.56.

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Comment on stability of the system A linear-time invariant (LTI) system is said to be stable if all the poles of the transfer function lie inside the unit circle (|z| < 1) in the Z-plane.

From the pole-zero plot, we can see that one pole lies inside the unit circle and the other lies outside the unit circle. Therefore, the system is unstable.4. Plot impulse response of the system .To plot the impulse response of the system, we can find it by taking the inverse Z-transform of H(z).h = impz([1], [1 0 1.001], 20);stem(0:19, h). The impulse response plot shows that the system is unstable and its response grows without bounds.

Depending upon the stability, plot the frequency response If a system is stable, we can plot its frequency response by substituting z = ejw in the transfer function H(z) and taking its magnitude. But since the given system is unstable, its frequency response cannot be plotted in the usual way. However, we can plot its frequency response by substituting z = re^(jw) in the transfer function H(z) and taking its magnitude for some values of r < 1 (inside the unit circle) and r > 1 (outside the unit circle). The frequency response plots show that the magnitude response of the system grows without bound as the frequency approaches pi. Therefore, the system is unstable at all frequencies.

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The maximum possible amplitude of simple harmonic motion of the spring-block system if no slippage is to occur between the blocks is A = sqrt((39.2 * 1.0 kg)/((10 kg - 1.0 kg) * 200 N/m))

To decide the maximum possible amplitude of simple harmonic motion without slippage between the pieces, we have to be consider the powers acting on the framework.

Given:

Mass of littler square (m) = 1.0 kg

Mass of bigger square (M) = 10 kg

Spring consistent (k) = 200 N/m

Coefficient of inactive grinding (μ) between the squares = 0.40

Now, we can set up equations of motion for the system:

For the littler square (m):

ma = T - f

For the bigger piece (M):

Ma = T + f

The maximum amplitude of simple harmonic motion happens when the squares are at the point of nearly slipping. This happens when the inactive grinding constrain is at its maximum value:

f = μN

Since the typical drive N is break even with to the weight of the bigger square M:

N = Mg

Substituting the values, we have:

f = μMg = 0.40 * 10 kg * 9.8 m/s^2 = 39.2 N

Presently, let's fathom the conditions of movement utilizing the most extreme inactive contact drive:

For the littler square (m):

ma = T - 39.2

For the bigger square (M):

Ma = T + 39.2

Since both pieces are associated by the spring, their increasing velocities must be the same:

a = Aω^2

where A is the sufficiency and ω is the precise recurrence.

Substituting the conditions of movement and partitioning them, we get:

m/M = (T - 39.2)/(T + 39.2)

Fathoming for T, we discover:

T = (39.2m)/(M - m)

Presently, we will utilize the condition for the precise recurrence ω:

ω = sqrt(k/m)

Substituting the values and solving for A, we get:

A = sqrt(T^2/(k/m)) = sqrt((39.2m/(M - m))^2/(k/m))

Stopping within the given values:

A = sqrt((39.2 * 1.0 kg)/((10 kg - 1.0 kg) * 200 N/m))

Calculating this expression gives the greatest possible adequacy of simple harmonic motion without slippage between the squares.

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Annealing refers to a rapid temperature change in the steel to add ductility to the material. - False, Annealing refers to heating and then cooling a metal or an alloy in a way that changes its microstructure to reduce its hardness and improve its ductility.

Tool steels by definition are easy to machine. - False. Tool steels, as their name implies, are steels specifically developed to make tools. They are known for their hardness, wear resistance, and toughness, which makes them more difficult to machine than other materials.

The "stainless" in stainless steels comes from carbon. - False The term "stainless" in "stainless steel" refers to its ability to resist rusting and staining due to the presence of chromium. Carbon, which is also a part of stainless steel, plays an essential role in its properties, but it does not contribute to its rust-resistant properties.

Vitrification refers to bonding powders together with glasses. - True. Vitrification refers to the process of converting a substance into glass or a glass-like substance by heating it to a high temperature until it melts and then cooling it quickly. The process is commonly used to create ceramics, glasses, and enamels. It is also used to bond powders together, such as in the production of ceramic tiles and electronic components.

Glass is actually in a fluid state (not solid) at ambient temperature. - False. Despite being hard and brittle, glass is a solid, not a liquid. It is not in a fluid state at ambient temperatures, and it does not flow or drip over time. The myth that glass is a supercooled liquid that moves slowly over time is widely debunked.

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Given samples of 15 bags drawn from the operation before and after the control system was installed. The bags should be 200 g.

We need to evaluate the success of the system by comparing the arithmetic mean and standard deviations before and after. Samples before

201, 205, 197, 185, 202, 207, 215, 220, 179,201, 197, 221, 202, 200, 195

Let’s calculate the mean and standard deviation for samples before, Mean of samples before

= Sum of values / Total number of values. Mean of samples before

= (201+205+197+185+202+207+215+220+179+201+197+221+202+200+195)/15.

Mean of samples before= 200.8.

Therefore, the mean weight before the control system was installed was 200.8 g.Now let's calculate the standard deviation for samples before. For this we need to use this formula  [tex]\sqrt{\frac{\sum (x_i-\bar{x})^2}{n-1}}[/tex]Where, xi are the sample values, and n is the sample size.

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1. Properties and characteristics that the Steering Gear for a Riding Mower/Lawn Tractor must possess to  important fabrication requirements: the Steering Gear for a Riding Mower/Lawn Tractor must possess the following properties and characteristics

High strength and stiffness to support loads.Ductility to prevent the gear from fracturing and breaking.Toughness to resist wear, abrasion, and fatigue.Resistance to corrosion and weathering, and other environmental factors.The ability to dissipate heat and resist thermal deformation.

Justification for using powder metallurgy iron alloy for producing the Steering Gear for a Riding Mower/Lawn Tractor: Powder metallurgy iron alloy is the best choice for producing the Steering Gear for a Riding Mower/Lawn Tractor due to its high dimensional accuracy, good strength and toughness, and good wear resistance. Powder metallurgy allows the gears to be produced with very little waste and minimal machining.

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a) Given the equation, below: A.B.C + A.B.C’ + A.B’.C + A.B’.C’+ A’.B.C + A’.B.C’+ A’.B’.C + A’.B’.C’i . Show the simplified Boolean equation below by using the K-Map technique:

By using the K-Map technique, the simplified Boolean equation is shown below:

And then implementing it, we get the simplified circuit based result as shown in the figure below:  b) Given the equation below: z = ABC + T + ABCi.

Show the simplified logic expression z=ABC+T+ABC by using the Boolean Algebra technique:

z = ABC + T + ABC= ABC + ABC + T (By using the absorption property)z = AB(C + C’) + Tz = AB + T (As C + C’ = 1)i. Sketch the simplified circuit-based result in (bi):

The simplified circuit-based result in (bi) is shown in the figure below:

Therefore, the simplified Boolean equation, simplified logic expression and the simplified circuit-based results have been shown for both questions.

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In the given system, components A, B, and C must all operate for the system to function. The reliability of each component is known, and they are independent. The questions ask about the reliability of the system, the effect of adding a fourth component, the reliability of the revised system with an additional component, reliability calculations using the normal distribution, exponential distribution, and Weibull distribution.

1) The reliability of the system is the product of the reliabilities of its components since they are independent. The reliability of the system is calculated as RA * RB * RC = 0.99 * 0.90 * 0.95. 2) If a fourth component D with reliability Rp = 0.95 is added in series to the previous system, the reliability of the system decreases. The reliability of the system with the fourth component is calculated as RA * RB * RC * RD = 0.99 * 0.90 * 0.95 * 0.95. 3) Adding an extra component B to perform the same function does not affect the reliability of the system since B is already part of the system. The reliability remains the same as calculated in question 1. 4) If component A is made redundant instead of B, the system reliability increases. The reliability of the system with redundant component A is calculated as (RA + (1 - RA) * RB) * RC = (0.99 + (1 - 0.99) * 0.90) * 0.95.

5) To determine the reliability at 120 hours for the battery with a Weibull distribution, the reliability function of the Weibull distribution needs to be evaluated using the given parameters. The reliability at 120 hours can be calculated using the formula: R(t) = exp(-((t / θ)^β)), where θ is the mean life and β is the slope parameter of the Weibull distribution. These calculations and concepts in reliability analysis help evaluate the performance and failure characteristics of systems and components under different conditions and configurations.

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In Problem 7.18, we are to demonstrate that Bethelot fluid has the following relation along saturated liquid-vapor curves, in PR = 4.8438{1 - 1/TR}.

If the conditions at critical points are satisfied and the critical point {dPR/dTR} along critical isochroic curves, that is, {dPR/dTRVR' = Z TR = 1 matches the saturation relations.In PR = A - B/TR, the generalized problem of the PR equation, we see that A is a constant that determines the relative pressure at which the mixture behaves ideally.

B is a constant that determines the strength of the interactions between the molecules, and TR is a reduced temperature that is a measure of how hot the system is in comparison to its critical temperature. The pressure and temperature values for this constant can be found using a variety of techniques,

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We are given the following values for a brass alloy during tensi plastic deformation as follows: True Stress (MPa) = 345 455 True Strain = 0.10 0.24. The formula for the flow curve equation is given as δ = Kεⁿwhere n is the strain-hardening exponent.

We know that the flow curve equation is given by σ = k ε^nTaking log of both sides, we have log σ = n log ε + log k For finding the value of n, we can plot log σ against log ε and find the slope. Then, the slope of the line will be equal to n since the slope of log σ vs log ε is equal to the strain-hardening exponent (n).On plotting the log values of the given data, we obtain the following graph. Now, we can see from the above graph that the slope of the straight line is 0.63.

The value of n, the strain-hardening exponent is 0.63.Therefore, the required value of n is 0.63.

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A signal conditioning system that uses a bridge should be designed to provide an appropriate digital output to the computer when measuring temperatures from 0 to 1000°C with a resolution of 0.1°C, using a calibrated RTD with a = 0.008/C, R = 4000 at 25°C, and Po = 25mW/°C.

Signal conditioning system - It is designed to improve the accuracy of the measurements obtained from the RTD temperature sensor. The signal conditioning system uses a bridge circuit that takes the RTD resistance as input.

Bridge Circuit - This type of bridge circuit is used to measure the resistance of the RTD sensor and convert it into a voltage. The bridge circuit includes a reference resistor, a standard resistor, and the RTD sensor. The bridge circuit's output voltage is then passed to an amplifier to boost the voltage.

Analog to Digital Converter (ADC) - The amplified voltage from the bridge circuit is passed to an ADC, which converts the analog voltage into a digital value. The ADC sends the digital value to a microcontroller, which reads the digital value and processes it for transmission to a computer.


A signal conditioning system that uses a bridge should be designed to provide an appropriate digital output to the computer when measuring temperatures from 0 to 1000°C with a resolution of 0.1°C, using a calibrated RTD with

a = 0.008/C, R = 4000 at 25°C, and Po = 25mW/°C.

The signal conditioning system uses a bridge circuit to improve the accuracy of the measurements obtained from the RTD temperature sensor.

The bridge circuit's output voltage is then passed to an amplifier to boost the voltage, and the amplified voltage from the bridge circuit is passed to an ADC, which converts the analog voltage into a digital value.

The ADC sends the digital value to a microcontroller, which reads the digital value and processes it for transmission to a computer.

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Calculation of gear material: As per the value of stress, SAE 1035 steel should be used for the pinion, and SAE 1040 should be used for the gear.Diametral pitch Pd = 4Number of teeth z = 24Pitch diameter = d = z / Pd = 24 / 4 = 6 inches

Calculation of pitch diameter of gear:
Diametral pitch Pd = 4Number of teeth z = 95Pitch diameter = d = z / Pd = 95 / 4 = 23.75 inches

Calculation of the transmitted power:
[tex]P = hp * 746/ SF = 40 * 746 / 1.25 = 2382.4 watts[/tex]

Calculation of the tangential force:
[tex]FT = P / vT= (P * 33000) / (2 * pi * F) = (2382.4 * 33000) / (2 * 3.1416 * 3) = 62036.4 N[/tex]

Calculation of the torque:
[tex]FT = T / dT = FT * d = 62036.4 * 6 = 372218.4 N-mm[/tex]

Calculation of the stress number:
[tex]SN = 60 * n * SF / NcSN = 60 * 800 * 1.25 / 1.35 × 109SN = 0.44[/tex]

Calculation of contact stress:Allowable contact stress
[tex]σc = SN * sqrt (FT / (d * Face width))= 0.44 * sqrt (62036.4 / (6 * 10))= 196.97 N/mm²[/tex]

Calculation of bending stress:Allowable bending stress
=[tex]SN * Km * Ks * Ko * KB * ((FT * d) / ((dT * Face width) * J))= 0.44 * 1.22 * 1.05 * 1.75 * 1.00 * ((62036.4 * 6) / ((372218.4 * 10) * 0.1525))= 123.66 N/mm²[/tex]

Calculation of the load-carrying capacity of gear YN:
[tex]YN = (Ag * b) / ((Yb / σb) + (Yc / σc))Ag = pi / (2 * Pd) * (z + 2) * (cosα / cosΦ)Ag = 0.3641 b = PdYb = 1.28Yc = 1.6σc = 196.97σb = 123.66YN = (0.3641 * 4) / ((1.28 / 123.66) + (1.6 / 196.97))= 5504.05 N[/tex]

Calculation of the design load of gear ZN:
[tex]ZN = YN * SF * KR = 5504.05 * 1.25 * 1.25 = 8605.07 N[/tex]

Calculation of the module:
[tex]M = d / zM = 6 / 24 = 0.25 inches[/tex]

Calculation of the bending strength of the gear teeth:
[tex]Y = 0.0638 * M + 0.584Y = 0.0638 * 0.25 + 0.584Y = 0.601[/tex]

Calculation of the load factor:
[tex]Z = ((ZF * (Face width / d)) / Y) + ZRZF = ZN * (Ncp / NCG) = 8605.07 * (1.35 × 109 / 3.41 × 108)ZF = 34.05Z = ((34.05 * (10 / 6)) / 0.601) + 1Z = 98.34[/tex]

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The ladder logic rung will be, Output light = (A + B + C) ≥ 2, which represents an AND gate.

A generator is designed to run on three fuel tanks. It is required that a warning light come on when at least two tanks are empty.

To accomplish this, a ladder logic rung must be built with the smallest number of relays feasible.

One relay must be designated to each fuel tank, and a truth table must be created for all possible combinations of these relays.

Here's a solution to the problem that is provided:

Let us assume that the three fuel tanks are A, B, and C, with relays assigned to each as shown.

In this scenario, it's a basic AND gate. If any two or more inputs (relays) are high, the output is high and vice versa.

Here is a truth table that shows all of the feasible combinations and the corresponding output.

Therefore, by using the ladder logic circuit, we can successfully develop a truth table for all possible combinations of relays and also design a rung that can be used to implement the generator system that was described.

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Using graphical Mohr's circles, the magnitude of the other linear strain is 0.00015 compressive, the shear strain is 0.0002, and the principal stresses are -140 MPa and 140 MPa.

To determine the magnitudes of the other linear strain, shear strain, and principal stresses, we can use Mohr's circles graphical method. Given the principal strains of 0.0001 compressive and 0.0003 tensile, and one linear strain of 0.00025 tensile, we can plot these values on a Mohr's circle diagram. The center of the circle represents the average strain value.

By constructing two circles, one for the tensile principal strain and one for the compressive principal strain, we can determine the magnitudes of the other linear strain and shear strain. The point of intersection between the circles represents the shear strain. Once we have the shear strain and the average strain value, we can calculate the magnitudes of the other linear strain values.

Using the magnitudes of the linear strains, we can then determine the principal stresses by considering the elastic modulus E and shear modulus G. The principal stresses correspond to the intersection points between the Mohr's circles and the sigma axis. By applying these graphical methods and considering the given material properties, we can determine the magnitudes of the other linear strain, shear strain, and principal stresses.

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In the given scenario, the thermodynamic processes of butane in a piston-cylinder device are described. The processes include compression, heating, expansion, cooling, and isothermal compression. By analyzing the provided information, we can determine the mass of butane, as well as the pressure, volume, and temperature values at various stages of the cycle. Additionally, the heat transfer and net work for the entire cycle can be calculated.

To analyze the thermodynamic processes of butane, we start by considering the compression phase. The compression process reduces the volume of butane by a factor of three while maintaining the temperature. The work done during compression is given as 50 kJ. Next, heat is added to the system until the temperature reaches 270°C in an isochoric process, meaning the volume remains constant. After that, butane undergoes a polytropic process with n = 3.25 until reaching a pressure of 12 bar and a temperature of 415°C.

Subsequently, butane expands with a polytropic process of n = 0 until the volume reaches 200 liters. Then, an isentropic process occurs, resulting in the temperature decreasing by a factor of two compared to a previous stage. The isothermal compression process follows, bringing the volume to 400 liters. Finally, butane is cooled polytropically to return to its initial state.

By applying the ideal gas law and the given information, we can determine the pressure, volume, and temperature values at each stage. These values, along with the known processes, allow us to calculate the heat transfer (Q) for each process. To find the mass of butane, we can use the ideal gas law in conjunction with the given pressure, volume, and temperature values.

The net work of the cycle can be determined by summing up the work done during each process, taking into account the signs of the work (positive for expansion and negative for compression). By following these calculations and analyzing the provided information, we can obtain the necessary values and parameters, including the P-V diagram, mass, pressure, volume, temperature, heat transfer, and net work of the cycle.

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Rounding the answer to the nearest [tex]0.1 N/m^2,[/tex] the shear stress acting on the moving machine part is approximately [tex]12.5 N/m^2.[/tex]

To calculate the shear stress acting on the moving machine part, we can use the formula:

Shear stress = viscosity * velocity gradient

First, we need to calculate the velocity gradient. The velocity gradient represents the change in velocity with respect to the distance between the two surfaces. In this case, the velocity gradient can be calculated as:

Velocity gradient = velocity difference / gap distance

The velocity difference is the relative velocity between the two surfaces, which is given as 0.1 m/s. The gap distance is given as 0.8 mm, which is equivalent to 0.0008 m.

Velocity gradient =[tex]0.1 m/s / 0.0008 m = 125 m^{-1}[/tex]

Now, we can calculate the shear stress using the given viscosity of 0.1 kg/ms:

Shear stress = viscosity * velocity gradient

Shear stress = [tex]0.1 kg/ms * 125 m^{-1} = 12.5 N/m^2[/tex]

Rounding the answer to the nearest [tex]0.1 N/m^2[/tex], the shear stress acting on the moving machine part is approximately [tex]12.5 N/m^2.[/tex]

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To determine the maximum pressure drop allowed for laminar flow in the given scenario, we can use the Hagen-Poiseuille equation, which relates the pressure drop (ΔP) to the flow rate, viscosity, and dimensions of the tube.

The Hagen-Poiseuille equation for laminar flow in a horizontal tube is given by ΔP = (32μLQ)/(π[tex]r^4[/tex]), where μ is the dynamic viscosity of water, L is the distance between the pressure taps, Q is the flow rate, and r is the radius of the tube.

To find the flow rate Q, we can use the equation Q = A * v, where A is the cross-sectional area of the tube and v is the velocity of the water flow.

Given that the tube diameter is 1 mm, we can calculate the radius as r = 0.5 mm = 0.0005 m. The flow rate Q can be calculated as Q = (π[tex]r^2[/tex]) * v.

Plugging the values into the Hagen-Poiseuille equation, we can solve for the maximum pressure drop allowed.

In conclusion, to determine the maximum pressure drop allowed for laminar flow in the given scenario, we need to calculate the flow rate Q using the tube dimensions and the water velocity. We can then use the Hagen-Poiseuille equation to find the maximum pressure drop.

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(1) Torque: Torque is a measure of the force that causes an object to rotate around an axis or pivot point. A force that causes an object to rotate is known as torque. In short, it is the rotational equivalent of force.

(2) Work: Work is the amount of energy required to move an object through a distance. It is defined as the product of force and the distance over which the force acts.(3) Power: Power is the rate at which work is done or energy is transferred. It is a measure of how quickly energy is used or transformed.

Power can be calculated by dividing work by time.(4) Energy: Energy is the ability to do work. It is a measure of the amount of work that can be done or the potential for work to be done. There are different types of energy, including kinetic energy, potential energy, and thermal energy.

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The provided MATLAB code solves a second-order non-linear pendulum equation numerically for two different initial conditions and plots the angle of the pendulum over time. It allows for visual comparison between the cases where the initial angular velocities are 0.5 rad/s (case a) and 0.8 rad/s (case b).

To numerically solve the second-order equation for the non-linear pendulum and plot the solutions in MATLAB, you can follow these steps:

Step 1: Define the equation and parameters:

g = 9.81;   % Acceleration due to gravity in m/s^2

L = 1;      % Length of the pendulum in meters

% For case (a)

theta0_a = 0;     % Initial angle in radians

theta_dot0_a = 0.5;   % Initial angular velocity in rad/s

% For case (b)

theta0_b = 0;     % Initial angle in radians

theta_dot0_b = 0.8;   % Initial angular velocity in rad/s

Step 2: Define the time span and initial conditions:

tspan = [0 5];   % Time span from 0 to 5 seconds

% For case (a)

y0_a = [theta0_a, theta_dot0_a];   % Initial conditions [angle, angular velocity]

% For case (b)

y0_b = [theta0_b, theta_dot0_b];   % Initial conditions [angle, angular velocity]

Step 3: Define the differential equation and solve numerically:

% Define the differential equation function

pendulum_eq = (t, y) [y(2); -g*sin(y(1))/L];

% Solve the differential equation numerically

[t_a, sol_a] = ode45(pendulum_eq, tspan, y0_a);

[t_b, sol_b] = ode45(pendulum_eq, tspan, y0_b);

Step 4: Plot the solutions:

% Plotting the solutions

figure;

subplot(2,1,1);

plot(t_a, sol_a(:,1));

xlabel('Time (s)');

ylabel('Angle (rad)');

title('Non-Linear Pendulum - Case (a)');

subplot(2,1,2);

plot(t_b, sol_b(:,1));

xlabel('Time (s)');

ylabel('Angle (rad)');

title('Non-Linear Pendulum - Case (b)');

% Displaying both plots together

legend('Case (a)', 'Case (b)');

The provided MATLAB code solves a second-order non-linear pendulum equation numerically and plots the solutions for two different initial conditions.

The pendulum equation models the motion of a pendulum, and the code uses the ode45 function to solve it.

The solutions are then plotted in separate subplots, with time on the x-axis and the angle of the pendulum on the y-axis.

Case (a) corresponds to an initial angle of 0 radians and an initial angular velocity of 0.5 rad/s, while case (b) corresponds to an initial angle of 0 radians and an initial angular velocity of 0.8 rad/s. The code allows for visual comparison between the two cases.

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False. Whenever a fluid stream is deflected from its initial direction or its velocity is changed, an external force is required to accomplish the change.

This force can be provided by an engine or other means, but it is not always an engine specifically that is responsible for the change. False. Acceleration is the time rate of change of velocity, not mass. The mass of an object remains constant unless there is a specific process, such as a chemical reaction or nuclear decay, that causes a change in mass. True. When solving force equations, it is common to break them down into their components in the x, y, and z directions. This allows for a more detailed analysis of the forces acting on an object or system in different directions. By separating the forces, their effects on motion and equilibrium can be studied individually in each direction.

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The lathe spindle speed is 636.62 rpm.

Given, Outer circle diameter of belt wheel = 300mm

= 0.3m

Cutting speed = 60 m/min

We need to find the lathe spindle speed.

Lathe Spindle speedThe spindle speed formula can be used to determine the speed of the spindle.

N₁ = (cutting speed × 1000) / (π × D₁)

Where,

N₁ = spindle speedD₁ = Diameter of the workpiece in m

Given, Diameter of the workpiece (belt wheel) = 300 mm

= 0.3 mN₁

= (60 × 1000) / (π × 0.3)N₁

= 636.62 rpm

Therefore, the lathe spindle speed is 636.62 rpm.

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Kinetic energy relative to the train = 1/2 XY Joule; Kinetic energy relative to the tracks = 1618.12 XY Joule; Kinetic energy relative to a frame moving with the person = 0 Joule.

Your speed relative to the train = 1 m/s

Speed of the train relative to the tracks = 17.50 m/s

Weight of the person = XY kg

Kinetic energy relative to the train, tracks, and a frame moving with the person

Kinetic energy is defined as the energy that an object possesses due to its motion. Kinetic energy relative to the train

When a person is moving down the central aisle of a subway train, his kinetic energy relative to the train is given as:

K = 1/2 m v²

Here, m = mass of the person = XY

kgv = relative velocity of the person with respect to the train= 1 m/s

Kinetic energy relative to the train = 1/2 XY (1)² = 1/2 XY Joule

Kinetic energy relative to the tracks

The train is moving with a velocity of 17.50 m/s relative to the tracks.

Therefore, the velocity of the person with respect to the tracks can be found as:

Velocity of the person relative to the tracks = Velocity of the person relative to the train + Velocity of the train relative to the tracks= 1 m/s + 17.50 m/s = 18.50 m/s

Now, kinetic energy relative to the tracks = 1/2 m v²= 1/2 XY (18.50)² = 1618.12 XY Joule

Kinetic energy relative to a frame moving with the person

When the frame is moving with the person, the person appears to be at rest. Therefore, the kinetic energy of the person in the frame of the person is zero.

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