You have been asked to improve the fuel efficiency of an
automobile by 20%. Convert this request into engineering criteria.
What changes might be made to the automobile to achieve this
objective?
In y

Generally speaking, surfaces in the open have a R value of 1. A surface outside should be 32°F or less to the degree.

The maximum BTU/Hr/Ft2 you can have for a 1°F temperature differential at the surface is 150+32, which is 182°F.

182 is the insulation's "R" value.

The equation I = Q t can be used to calculate the amount of charge that goes through a given location in a wire over a certain period of time, t, and hence determine the electric current, I, in that wire. Working through a few instances will allow us to practise applying this equation. The formula V = IR should be known to you. By dividing the voltage by the resistance in the equation—I = V/R—you may quickly determine the current value.

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(a) To find the time when the velocity is again zero, we set the velocity equal to 0 and solve for t:

0 = v0 + at = 9t - 3t^3

Solving for t, we find that t = ±1.

(b) To find the position and velocity when t = 4 s, we first need to find the velocity equation by integrating the acceleration equation:

v = ∫a = ∫(9 - 3t^2)dt = 9t - t^3 + C

Using the initial condition v(0) = 0, we find C = 0 and

v(t) = 9t - t^3

Next, we integrate v(t) to find the position equation:

x = ∫v = ∫(9t - t^3)dt = 3t^2 - t^4 + C

Using the initial condition x(0) = 5, we find C = 5 and

x(t) = 3t^2 - t^4 + 5

Evaluating x(t) and v(t) at t = 4 s, we have:

x(4) = 3(4^2) - 4^4 + 5 = 51

v(4) = 9(4) - 4^3 = -36

(c) To find the total distance travelled by the particle from t = 0 to t = 4 s, we need to find the distance between the initial and final positions:

d = x(4) - x(0) = 51 - 5 = 46 ft

So the total distance travelled by the particle from t = 0 to t = 4 s is 46 ft.

Answer:

For most uses you'll want your water heated to 120 F(49 C) In this example you'd need a demand water heater that produces a temperature rise and it will take about 2 hours

Answer:

Plzzzz answer fast......

Answer:

If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

Explanation:

Reynolds number is an important dimensionless parameter in fluid mechanics.

It is calculated as;

\(R_e__N} = \frac{\rho vd}{\mu}\)

where;

ρ is density

v is velocity

d is diameter

μ is viscosity

All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.

In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

The theoretical volume flow rate through the venturi meter can be calculated by using the Bernoulli's equation, principle of continuity, and given pressure difference and diameters.

To calculate the theoretical volume flow rate through the venturi meter, we can use the Bernoulli's equation and the principle of continuity.

First, we need to determine the velocity at the throat of the venturi meter. Since the flow is incompressible, the equation of continuity tells us that the velocity at the throat is inversely proportional to the area of the throat.

Using the formula for the area of a circle (A = πr²), we can find the ratio of the areas of the throat (A₂) to the pipeline (A₁): A₂/A₁ = (d₂/2)² / (d₁/2)²

Substituting the given diameters, we get: A₂/A₁ = (100/250)² = 0.16

From Bernoulli's equation, we know that the pressure difference (ΔP) is related to the velocity difference (ΔV) as: ΔP = ρ/2 * (ΔV)², where ρ is the density of the fluid.

We can rearrange this equation to solve for ΔV: ΔV = √(2 * ΔP / ρ)

Given that the pressure difference is 0.63 m of mercury and the specific gravity of oil is 0.9 (which implies ρ = 0.9 * ρ_water), we can calculate the velocity difference at the throat.

Next, we can use the principle of continuity to relate the velocity at the throat (V₂) to the theoretical volume flow rate (Q): Q = A₂ * V₂

By substituting the known values, including the calculated velocity difference, we can determine the theoretical volume flow rate through the venturi meter.

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The probability that the winner was from the united states, given that she or he won a gold medalist is 35/ 133 which is 0.2631. We can calculate it in the following manner.

Given inputs are as follows:

U.S. ---> 35 39 29 104

Russia ---> 27 27 38 92

China ---> 32 17 14 63

Australia ---> 17 16 16 49

Others ---> 133 136 153 422

So the probability that the winner was from the united states, given that she or he won a gold medalist is 35/ 133 which is 0.2631

Probability is a branch of mathematics that deals with the measurement and analysis of random events. It is used to quantify the likelihood or chance of an event occurring, based on a set of possible outcomes. Probability is expressed as a number between 0 and 1, with 0 indicating that an event is impossible and 1 indicating that an event is certain. For example, the probability of flipping a coin and getting heads is 0.5, since there are two possible outcomes (heads or tails) and the probability of each outcome is equal. Probability theory is widely used in many fields, including statistics, economics, engineering, physics, and computer science, among others.

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The purpose of the low ambient control on a commercial air conditioner is to ensure proper operation and performance of the unit in low ambient temperature conditions.

Low ambient control is a feature present in many commercial air conditioning systems, especially those used in colder climates. It is designed to address the challenges that arise when the ambient temperature drops below the normal operating range of the air conditioner. When the temperature is low, the refrigerant in the system may not evaporate properly, causing issues such as compressor damage, reduced efficiency, or system shutdown.

The low ambient control function helps prevent these problems by regulating the refrigerant flow and adjusting the operation of the air conditioner. It may involve methods such as fan speed control, refrigerant temperature sensing, or utilizing a special valve to maintain appropriate pressures and temperatures in the system. By implementing low ambient control, the air conditioner can continue to operate effectively even in colder temperatures, ensuring comfort and reliability in commercial spaces.

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Tech B is correct. tech b says shift valves are spool valves that direct the flow of hydraulic oil to a clutch or band.

Shift valves are special valves that control the movement of hydraulic oil to turn on or off a clutch or band in an automatic transmission system. They make sure that the hydraulic pressure goes to the right parts to start changing gears or doing other things with the transmission.

There are other valves that control transmission pressure, but they are used for different reasons than the main line pressure regulation valves.

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Answer:

6 amps

Explanation:

current =v÷r

12v÷2 ohms= 6amps

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

\($ C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2) $\)

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

\($ T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$\)

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

\(C_p = 520 \:\: J/kg.K\)

So, the inlet temperature of argon is

\($ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$\)

\($ T_{in} = \frac{1}{2} \times 119 + 300$\)

\($ T_{in} = 360K $\)

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

\(C_p = 5193 \:\: J/kg.K\)

So, the inlet temperature of helium is

\($ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$\)

\($ T_{in} = \frac{1}{2} \times 12 + 300$\)

\($ T_{in} = 306K $\)

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

\(C_p = 1039 \:\: J/kg.K\)

So, the inlet temperature of nitrogen is

\($ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$\)

\($ T_{in} = \frac{1}{2} \times 60 + 300$\)

\($ T_{in} = 330K $\)

Note: Answers are rounded to the nearest whole numbers.

The inverse Laplace transforms for the given functions F(s) are as follows:

a. f(t) = 2e^(-t) - 3e^(-6t)

b. f(t) = e^(-2t) * (cos(t) + sin(t))

c. f(t) = (1/2)e^(-t) * (3cos(2t) + 2sin(2t))

d. f(t) = 1 + e^(-2t) * (cos(t) + sin(t))

e. f(t) = c^(-1) * (1 - e^(-t))

The inverse Laplace transforms of the given functions are calculated to find the corresponding functions in the time domain. Each case represents a different form of the Laplace transform and requires specific techniques to find the inverse. The resulting inverse Laplace transforms are expressed as functions of time (t). These inverse transforms allow us to understand the behavior of the original functions F(s) in the time domain and provide insights into their properties and characteristics.

a. Inverse Laplace Transform for f(t) = 2e^(-t) - 3e^(-6t) (Case 1):

To find the inverse Laplace transform, we need to identify the corresponding function in the time domain. The inverse Laplace transform of F(s) = 2e^(-t) - 3e^(-6t) can be found using the properties and formulas of Laplace transforms.

Applying the linearity property, the inverse Laplace transform can be split into two parts:

f(t) = L^(-1){2e^(-t)} - L^(-1){3e^(-6t)}

Using the formula for the inverse Laplace transform of e^(-at), where a is a constant:

L^(-1){e^(-at)} = δ(t - a), where δ(t) is the Dirac delta function.

Therefore, we have:

f(t) = 2δ(t - 1) - 3δ(t - 6)

b. Inverse Laplace Transform for f(t) = e^(-2t)sin(t) (Case 2):

Using the properties and formulas of Laplace transforms, we can find the inverse Laplace transform for F(s) = e^(-2t)sin(t).

The inverse Laplace transform of e^(-2t)sin(t) can be obtained by using the complex frequency shift property and the Laplace transform of sin(t):

L^(-1){e^(-as)F(s)} = sin(t - a)

Therefore, we have:

f(t) = sin(t - (-2)) = sin(t + 2)

c. Inverse Laplace Transform for f(t) = e^(-t) - 4e^(-4t) + e^(-4t)t (Case 3):

To find the inverse Laplace transform for F(s) = e^(-t) - 4e^(-4t) + e^(-4t)t, we will use the linearity property and the formulas for the inverse Laplace transforms of e^(-at) and te^(-at).

Using these formulas, we can express F(s) as follows:

F(s) = 1/(s + 1) - 4/(s + 4) + (1/(s + 4))^2

Applying the inverse Laplace transforms, we have:

f(t) = L^(-1){1/(s + 1)} - L^(-1){4/(s + 4)} + L^(-1){(1/(s + 4))^2}

The inverse Laplace transforms of the individual terms can be found using the respective formulas:

L^(-1){1/(s + a)} = e^(-at)

L^(-1){1/(s + a)^2} = te^(-at)

Therefore, we obtain:

f(t) = e^(-t) - 4e^(-4t) + te^(-4t)

d. Inverse Laplace Transform for f(t) = 1 - e^(-4t) (Case 4):

For F(s) = 1 - e^(-4t), we can directly apply the linearity property and the inverse Laplace transform formula for e^(-at) to find the inverse Laplace transform.

Using the formulas, we have:

f(t) = L^(-1){1} - L^(-1){e^(-4t)}

The inverse Laplace transform of 1 is simply 1, and the inverse Laplace transform of e^(-4t) is e^(-4t).

Thus, we obtain:

f(t) = 1 - e^(-4t)

e. Inverse Laplace Transform for f(t) = Ce^(-t) (Case 5):

For F(s) = Ce^(-t), where C is a constant, we can directly apply the linearity property and the inverse Laplace transform formula for e^(-at) to find the inverse Laplace transform.

Using the formulas, we have:

f(t) = CL^(-1){e^(-t)}

The inverse Laplace transform of e^(-t) is simply e^(-t).

Hence, we obtain:

f(t) = Ce^(-t)

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Answer:

Cave-ins

Explanation:

The term excavation means any form of cuts, depression or trench by removing the surface of the earth. This process is intended primarily for the purpose of construction and maintenance or exploration. In this process there are many hurdles that pose danger to both human life and earth. The excavation workers face the great threat because of cave-ins. The collapsing of the earth's surface and random accidents prove to be very dangerous for the workers.

Answer:

If I am not mistaken I believe it is a higher voltage.

Explanation:

Hope this helps

Answer:

The answers are in the explanation. The pH is 5.91

Explanation:

The CH3NH2 reacts with HCl as follows:

CH3NH2 + HCl → CH3NH3⁺ + Cl⁻

When 200mL of HCl are added, the moles of CH3NH2 and HCl are reacting completely producing CH3NH3+ and Cl-. That means the species present are:

no H+. All reacted

yes H2O. Because the water is present in the solutions of HCl and CH3NH2

yes Cl-. Is a product of the reaction

Yes CH3NH2. Is consumed in the reaction but comes from the equilibrium of CH3NH3+

yes CH3NH3+. Is the other product of the reaction. MAJOR SPECIES

When 300.00mL of HCl are added, 100mL are in excess:

yes H+. Is in excess: H+ + Cl- = HCl in water. MAJOR SPECIES. Determine the pH of the solution.

yes H2O. Is present because the reactants are diluted.  

yes Cl-. Is a product of reaction and comes from HCl.

Yes CH3NH2. The reactant is over but comes from the equilibrium of CH3NH3+

yes no CH3NH3+. Yes. Is a product and remains despite HCl is in excess.

To find the pH:

At equivalence point the ion that determines pH is CH3NH3+. Its concentration is:

0.100L * (0.200mol/L) = 0.0200 moles / 0.300L = 0.0667M CH3NH3+

The equilibrium of CH3NH3+ is:

Ka = Kw/kb = 1x10-14/4.4x10-4 = 2.273x10-11 = [H+] [CH3NH2] / [CH3NH3+]

As both [H+] [CH3NH2] comes from the same equilibrium:

[H+] =  [CH3NH2] = X

2.273x10-11 = [X] [X] / [0.0667M]

1.5159x10-12 = X²

X = 1.23x10-6M = [H+]

As pH = -log [H+]

pH = 5.91

The pH at the equivalent point for this titration is "5.91".

\(CH_3NH_2 = 0.200\ M\\\\ \text{volume} = 100.0\ mL = 0.100\ L\\\\HCl = 0.100\ M\\\\\)

We must now quantify the pH well at the equivalence point.

We know that even at the point of equivalence, moles of acid and moles of the base are equivalent. As such, first, we must calculate the number of moles of the given base.

Calculating the Moles in \(CH_3NH_2 = 0.200\ M \times 0.100\ L = 0.0200\ moles\)

Calculating the Moles in \(HCl = 0.0200 \ moles\)

Calculating the volume of \(HCl\):

\(\to \text{Molarity} = \frac{ \text{moles}}{\text{volume \ (L)}} \\\\\to \text{Volume} = \frac{\text{moles}}{\text{molarity}}\\\\\)

                \(= \frac{0.0200 \ moles}{ 0.100\ M}\\\\= 0.200 \ L\\\\= 200 \ mL\\\\\)

Calculating the reaction among the acid and base:

\(CH_3NH_2 + HCl \longrightarrow CH_3NH_3^{+} + Cl^-\)

\(0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200\)

Therefore the conjugate acid of the bases exists at the standard solution.

Then we must calculate the new molar mass of \(CH_3NH_3^+\).

Total volume\(= 100 + 200 = 300\ mL = 0.300\ L\)

\([CH_3NH_3^+] = \frac{0.0200\ mole}{ 0.300\ L}= 0.0667\ M\)

Using the ICE table

\(CH_3NH_3^+ + H_2O \longrightarrow CH_3NH_2 + H_3O^+\)

\(I \ \ \ \ \ \ \ \ \ \ 0.0667 \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ 0\\\\C\ \ \ \ \ \ \ \ -x\ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ +x\\\\E \ \ \ \ \ \ \ \ \ \ \ \ 0.0667-x \ \ \ \ \ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ \ \ \ \+x\\\\\to Ka = \frac{[CH_3NH_2] [H_3O^+] }{[CH_3NH_3^+]}\)

Calculating \(K_a\) from \(K_b\)

\(\to K_a \times K_b = 1\times 10^{-14}\\\\\to K_a = \frac{1\times 10^{-14}}{4.4\times 10^{-4}} = 2.27\times 10^{-11}\\\\\)

                           \(= 2.27\times 10^{-11} \\\\= x\times \frac{x}{(0.0667-x)}\)

The x in the 0.0667-x can be ignored since the Ka value is just too small and it also does not follow the five percent criteria.

\(\to 2.27 \times 10^{-11} \times 0.0667 = x_2\\\\\to x_2 = 1.515\times 10^{-12}\\\\\to x = 1.23\times 10^{-6}\ M\\\\\to [H_3O^+] = x = 1.23\times 10^{-6}\ M\\\\\)

We have the formula to calculate pH.

\(\to pH = - \log [H_3O^+] = - \log 1.23\times 10^{-6}\ M= 5.91\)

The pH at the equivalent point for this titration is "5.91".

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The ratio of the induced EMF in the loop CDBC to the induced EMF in the loop CADC can be calculated as follows:

ecdbc/ecadc = -dΦ_cdbc/dt / (-dΦ_cadc/dt) = dΦ_cadc/dt / dΦ_cdbc/dt

Let's dive deeper into the details below

The induced EMF is the voltage generated by a changing magnetic field in a coil of wire. In a loop, the induced EMF is proportional to the rate of change of the magnetic flux that is threading the loop. Therefore, in a loop, the induced EMF can be calculated as:

induced EMF = -dΦ/dt, where Φ is the magnetic flux threading the loop.

We can assume that both loops are parallel to the surface and therefore perpendicular to the magnetic field. This means that the magnetic flux threading each loop is proportional to the area of the loop, as follows:

Φ_cadc = B A_cadc and Φ_cdbc = B A_cdbc

Therefore, the ratio of the induced EMF in the loop CDBC to the induced EMF in the loop CADC can be calculated as follows:

ecdbc/ecadc = dΦ_cadc/dt / dΦ_cdbc/dt = (B A_cadc)/dt / (B A_cdbc)/dt = A_cadc / A_cdbc

The answer is the ratio of the areas of the loops.

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Answer: true

Explanation: I haves

A sidecut from the vacuum distillation column known as heavy vacuum gas oil is often subjected to hydrocracking, a second catalytic cracking operation.

The long chain hydrocarbon molecules in the oil are divided into smaller ones by a catalyst when it is combined with hydrogen and passes over it. The temperature and the presence of catalysts have a significant impact on the pace of cracking and the hydrocracking's final products. Hydrogen is utilized to break the C-S and C-N bonds in Hydrotreatment, a similar cracking process (previously discussed in Section 4.3); in Hydrocracking, the desired break is in the C-C bonds, and the main products are typically jet fuel and diesel. Therefore, hydrocracking is applied where there is a high demand for diesel fuel.

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As the temperature of the material increases, the potential energy of the molecules increases. Thermal expansion occurs due to changes in temperature, and interatomic distances increase as potential energy increases.

Thermal expansion is used in a variety of applications such as rail buckling, engine coolant, mercury thermometers, joint expansion, and others.

It is to be noted that an application of the concept of liquid expansion in everyday life concerns liquid thermometers. As the heat rises, the mercury or alcohol in the thermometer tube moves in only one direction. As the heat decreases, the liquid moves back smoothly.

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This implies that the ratio of the inner and outer radii (r/rj) = 1. Hence, the use of the arithmetic mean area of the cylindrical wall with r = rj will not introduce more than 1% error in heat transfer calculation. However, the use of the arithmetic mean area always overestimates the heat transfer rate.

Given that the arithmetic mean area of the wall can be used to calculate the rate of heat transfer through a cylindrical wall of small thickness. We are required to determine the ratio of the inner and outer radii (r/rj) of the cylindrical wall for which the use of the arithmetic mean area does not introduce more than 1% error in heat transfer calculation.

The expression for the rate of heat transfer through a cylindrical wall of thickness 'dx' is given by dQ/dt = (2πL/kA) (T₁ − T₂), where 'L' is the length of the cylinder, 'k' is the thermal conductivity of the wall material, and 'A' is the area for heat transfer and is given by the arithmetic mean area of the wall as A = π(r² - rj²).

Let us assume that 'a' is the maximum allowable error, so we can express the acceptable limits of the area as (1 − a) A ≤ Am ≤ (1 + a) A, or A − aA ≤ Am ≤ A + aA, and π(r² − rj²) − aπ(r² − rj²) ≤ Am ≤ π(r² − rj²) + aπ(r² − rj²).

Since (r/rj) > 1, assume (r/rj) = α. The permissible range for the arithmetic mean area can be expressed as (1 − a) π(r² − rj²) ≤ Am ≤ (1 + a) π(r² − rj²), or (1 − a) π(r² − α²r²) ≤ Am ≤ (1 + a) π(r² − α²r²), or (1 − a)(1 − α²) ≤ Am/(π(r² − α²r²)) ≤ (1 + a)(1 − α²).

Since the arithmetic mean area does not introduce more than 1% error in heat transfer calculation, a = 0.01. Thus, (1 − 0.01)(1 − α²) ≤ Am/(π(r² − α²r²)) ≤ (1 + 0.01)(1 − α²). Therefore, (1 − α²) = 0.99(1 − α²), or 0.01(1 − α²) = 0.01, or (1 − α²) = 1. Therefore, α = 0.

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B) transistor. Typically a constant voltage CP rectifier will NOT have transistor.

A constant voltage CP (controlled potential) rectifier is a type of rectifier used in electrochemical processes. It typically consists of a transformer, a rectifying element (such as diodes), voltage taps, and other components. However, it does not typically include a transistor, as transistors are not commonly used in CP rectifiers. Transistors are electronic devices used for amplification and switching of electrical signals, and they are not necessary for the operation of a typical constant voltage CP rectifier, which primarily functions to provide a stable DC output voltage for electrochemical processes.

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To determine the allowable load P, we need to calculate the shear and normal stresses at each connection. The shear stress at each connection can be calculated using the equation τ = P/A, where P is the load and A is the area of the pin.

The normal stress at each connection can be calculated using the equation σ = P/A, where P is the load and A is the cross-sectional area of the link. Once the shear and normal stresses have been calculated, we can compare them to the ultimate shear and normal stresses to determine the allowable load. The allowable load will be the load that produces a shear and normal stress that is equal to or less than the ultimate shear and normal stresses, multiplied by the factor of safety.

To calculate the allowable load, we need to first calculate the shear and normal stresses at each connection. The shear stress at each connection can be calculated using the equation τ = P/A, where P is the load and A is the area of the pin. The area of the pin can be calculated using the equation A = πr2, where r is the radius of the pin. The normal stress at each connection can be calculated using the equation σ = P/A, where P is the load and A is the cross-sectional area of the link. The cross-sectional area of the link can be calculated using the equation A = bh, where b is the width of the link and h is the height of the link. Once the shear and normal stresses have been calculated, we can compare them to the ultimate shear and normal stresses to determine the allowable load. The allowable load will be the load that produces a shear and normal stress that is equal to or less than the ultimate shear and normal stresses, multiplied by the factor of safety.

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Answer:

The product and process are designed simultaneously

Explanation:

PENN

Answer:

16

Explanation:

Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

Article 430 that is found in  National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .

Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.

Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

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