After impact testing a sample at -100oC you realize that the fracture surface is very dull and fibrous. Is the sample behaving in a ductile of brittle manner at this temperature

Answer: This is true, including the top, because I myself have used this info

Explanation:

None of the following is a preventive measure aimed at reducing motor vehicle accidents caused by poor decision-making in young drivers: driver “buddy systems”, increased penalties for violations of driving laws, or driver’s education classes.

The most effective measures to reduce motor vehicle accidents are for young drivers to be taught the rules of the road and to develop good decision-making skills. These measures include: using defensive driving techniques, limiting distractions while driving, avoiding driving while impaired, and understanding the risks of speeding or aggressive driving. Young drivers should also be taught the importance of wearing seatbelts, checking tire pressure and vehicle maintenance, and having emergency preparedness kits in their vehicles. Educating young drivers on these safety measures is essential to reduce motor vehicle accidents.

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To determine the peak stress on the outer surface of a 50% fibre composite vaulting pole that has a diameter of about 50 mm and is being bent to a radius of curvature, R, of about 0.5 m, we can use the following formula:

σ = (M/Z) * y

where,σ = stress

M = bending moment

Z = section modulus

y = distance from the neutral axis

To find M, we can use the following formula:

M = (Em/Ef) * (ρc/ρf) * (π/4) * d^3 * (1/R)where,

Em = modulus of elasticity of matrixEf = modulus of elasticity of fibre

ρc = density of the matrixρf = density of the fibreπ = 3.14d = diameter of the pole

R = radius of curvatureNow, substituting the given values:

M = (3/90) * (0.5/1.5) * (π/4) * (0.05)^3 * (1/0.5)M

= 0.000216 Nm

To find Z, we can use the following formula:

Z = (π/32) * d^4 * (ρf/ρc) * (Ef/Em) * (1 + (3h/d)^2)

where,h = height of the pole

Now, substituting the given values:

Z = (π/32) * (0.05)^4 * (0.5/1.5) * (90/3) * (1 + (3(0.5)/0.05)^2)Z

= 0.0001586 m^3To find y, we can use the following formula:

y = (d/2) * (ρf/ρc)where,ρc = density of the matrix

ρf = density of the fibre

Now, substituting the given values:y = (0.05/2) * (0.5/1.5)y = 0.00833 m

Substituting the values of M, Z, and y in the formula for stress, we get:σ = (M/Z) * yσ = (0.000216/0.0001586) * 0.00833σ = 11.77 MPa

Therefore, the peak stress on the outer surface of a 50% fibre composite vaulting pole with a diameter of about 50 mm and being bent to a radius of curvature, R, of about 0.5 m is 11.77 MPa.

Composite materials are known for their remarkable strength and durability, making them ideal for use in a variety of applications. One of these applications is in the production of vaulting poles used in pole vaulting, a track and field event that involves athletes using a long, flexible pole to clear a bar that is suspended above the ground. The poles used in this event are typically made of a composite material that consists of a matrix material, such as epoxy resin, and a reinforcing material, such as carbon fibre or fibreglass. These materials are chosen for their high strength-to-weight ratio, which allows the poles to be both strong and flexible. The strength of the pole is determined by its modulus of elasticity, which is a measure of its ability to resist deformation under load. The modulus of elasticity of the matrix material, Em, and the reinforcing material, Ef, are key factors in determining the strength of the pole. The density of these materials is also important, as it affects the weight of the pole. The peak stress on the outer surface of a pole is an important factor to consider when designing the pole, as it determines the maximum load that the pole can handle before it fails. In order to determine the peak stress on the outer surface of a 50% fibre composite vaulting pole with a diameter of about 50 mm and being bent to a radius of curvature, R, of about 0.5 m, we can use the formula σ = (M/Z) * y, where M is the bending moment, Z is the section modulus, and y is the distance from the neutral axis. By using the appropriate values for Em, Ef, density, diameter, and radius of curvature, we can calculate the peak stress on the outer surface of the pole.

In conclusion, the peak stress on the outer surface of a 50% fibre composite vaulting pole with a diameter of about 50 mm and being bent to a radius of curvature, R, of about 0.5 m, is 11.77 MPa. The strength of the pole is determined by its modulus of elasticity, which is a measure of its ability to resist deformation under load. The modulus of elasticity of the matrix material, Em, and the reinforcing material, Ef, are key factors in determining the strength of the pole. The density of these materials is also important, as it affects the weight of the pole. The peak stress on the outer surface of a pole is an important factor to consider when designing the pole, as it determines the maximum load that the pole can handle before it fails. Overall, composite vaulting poles are an excellent example of how composite materials can be used to create strong and durable structures that are ideal for a wide range of applications.

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Answer:

  three-phase

Explanation:

2-phase, 3-phase, and 6-phase systems are in use. The most common of these is 3-phase.

A modern industrial plant most likely uses a 3-phase system.

__

Additional comment

Serious research began on 6-phase power transmission systems in 1973. They have been shown to increase power transfer, and to reduce corona, magnetic fields, and audio noise compared to 3-phase systems. However, they are not in widespread industrial use.

Fiscal policy is the use of government spending and taxation to influence the economy. It involves decisions made by the government regarding how much money should be spent on public goods and services, such as education, healthcare, and infrastructure, and how much revenue should be collected through taxes.

Fiscal policy is often used to promote economic growth and stability, such as during a recession when the government may increase spending to stimulate the economy. However, it can also be used to combat inflation by reducing government spending and increasing taxes.

Fiscal policy is defined as the use of government revenue collection (taxation) and expenditure (spending) to influence the economy. It is a tool utilized by the government to achieve economic objectives, such as promoting growth, managing inflation, reducing unemployment, and maintaining a stable balance of payments. Fiscal policy can be adjusted through changes in tax rates, government spending, and public debt management, depending on the economic goals and prevailing conditions.

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To determine the moment at O due to the combined effects of the equilibrium 300-lb part P, the 120-lb weight of member OA (mass center at G1), and the 50-lb weight of member AB (mass center at G2).

we first need to calculate the distances of the centers of gravity of each member from point O. Let d1 be the distance of G1 from O, and d2 be the distance of G2 from O.Using the parallel axis theorem, we can calculate d1 and d2 as follows:d1 = L1/2 * sin(120°) + L1/3 * cos(120°) = L1/6 * (-√3 + 2) = -0.2887 ftd2 = L1 * sin(120°) + L2/2 = L1/2 * √3 + L2/2 = 2.732 ftNext, we can calculate the moment due to each member about O:Moment due to P = 300 * d1 * cos()Moment due to OA = 120 * d1 * cos()Moment due to AB = 50 * d2 * cos()The angle at A is locked at 120°, so we can plot the moment as a function of the angle as follows:MO() = 300 * d1 * cos() + 120 * d1 * cos() + 50 * d2 * cos()To find the maximum value of MO, we can take the derivative of MO with respect to and set it equal to 0:d/d() [300 * d1 * cos() + 120 * d1 * cos() + 50 * d2 * cos()] = -300 * d1 * sin() - 120 * d1 * sin() + 50 * d2 * sin() = 0Simplifying and solving for , we get:sin() = (50 * d2) / (300 * d1 + 120 * d1)= 0.518Taking the inverse sine, we get:() = 31.18°Therefore, the maximum moment occurs at an angle of 31.18°, and its value can be calculated using the expression for MO() above.

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The following statement is true. The American wire gage numbers specify the size of round wire in terms of its diameter and cross-sectional area.

The American Wire Gauge (AWG) is a standardized system used in the United States to specify the diameter of electrical conductors such as wires and cables. The AWG number assigned to a wire indicates its cross-sectional area, which in turn determines the wire's current-carrying capacity and other electrical properties.

As the AWG number increases, the diameter of the wire decreases, and vice versa. For example, a wire with a higher AWG number (such as 24) has a smaller diameter and lower current-carrying capacity than a wire with a lower AWG number (such as 12).

In summary, the AWG number of a wire specifies its diameter and cross-sectional area, which are important factors in determining its electrical properties.

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The distance between the two fluid levels of the manometer, when using mercury as the fluid, is approximately 0.59 meters.

To determine the distance between the two fluid levels in the manometer, we need to consider the pressure difference between the gas tank and the atmosphere. If the reading on the pressure gauge is 80 kPa, we can assume that the pressure inside the gas tank is also 80 kPa.

In a manometer, the pressure difference between the two fluid levels is equal to the difference in height of the fluid columns. The fluid used in this case is mercury. To calculate the distance, we need to know the density (rho) of mercury. The density of mercury is approximately 13,600 kg/m^3.

Using the formula P = ρgh, where P is the pressure difference, ρ is the density, g is the acceleration due to gravity, and h is the height difference:

80 kPa = (13,600 kg/m^3) * (9.8 m/s^2) * h

Simplifying the equation, we find:

h = (80,000 N/m^2) / (13,600 kg/m^3 * 9.8 m/s^2)

Calculating the result, we get:

h ≈ 0.59 meters

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Answer:

The mechanism that is typically used to bring a carbureted engine to a fast idle is a throttle cable or linkage. This allows the driver to adjust the throttle and increase the engine speed to a designated fast idle setting. This is often used when the engine is cold to help it warm up more quickly, or when the vehicle is stopped and the engine needs to be kept running at a higher speed.

The surprising realization scientists are making about bacteria and what danino hopes to eventually be able to do using bacteria are the Corallococcus exercitus.

Antonie van Leeuwenhoek used single-lens microscopes, which he made, to make the primary observations of micro organism and protozoa. His full-size studies at the boom of small animals inclusive of fleas, mussels, and eels helped disprove the concept of spontaneous era of life.

Researchers found a brand new species of alga in vital New York and named it Gormaniella terricola, with the genus named after poet Amanda Gorman.

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Answer:

i dob't know

Explanation:

Answer:

  A. Oil changes

Explanation:

It depends on the car and its usage and environment. Usually oil is supposed to be changed every few months, more often if the car is driven a lot. Coolant changes may be indicated as seasons change, so will generally occur less frequently than oil changes.

Tire and brake replacement depend on usage and driving habits. Some owners may never have to replace either one, if they trade their car every year or two. Folks who drive with their foot on the brake pedal may have to replace brakes relatively often.

The most frequent task is generally oil changes.

Answer:

A smudge was analogous to acid mine drainage because the appearance of the two elements is similar.

Analogy refers to the act of comparing things in order to promote more understanding of them. Smudge refers to smearing of things in a messy way and the appearance that this word connoted is similar to what is obtainable in an acid mine drainage that often drains and gets stored beneath the rivers.

The appearance of two analogies has an appearance that readers can relate to. When making analogies, it is vital to only use elements that are similar in some ways and make the understanding of the other clearer.

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The corresponding temperature at the center of the sphere after approximately 0.309 seconds is approximately 198.83°C.

To determine the time required for a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy, we can use the concept of thermal diffusion through a sphere. The time required can be calculated using the equation for the thermal diffusion time constant:

τ = (ρ * c * r^2) / (4 * k)

where:

τ is the thermal diffusion time constant,

ρ is the density of the sphere material (in this case, aluminum) = 2700 kg/m^3,

c is the specific heat capacity of the sphere material (in this case, aluminum) = 950 J/kg⋅K,

r is the radius of the sphere (diameter/2) = 75 mm / 2 = 37.5 mm = 0.0375 m,

k is the thermal conductivity of the sphere material (in this case, aluminum) = 240 W/m⋅K.

Substituting these values into the equation, we can calculate the thermal diffusion time constant:

τ = (2700 kg/m^3 * 950 J/kg⋅K * (0.0375 m)^2) / (4 * 240 W/m⋅K)

Now we can solve for τ:

τ ≈ 0.309 seconds

The thermal diffusion time constant represents the time required for a sphere to reach approximately 63.2% of the maximum possible thermal energy. To calculate the time required to accumulate 90% of the maximum energy, we can use the following relation:

t = τ * ln((90% - 63.2%) / (100% - 63.2%))

Substituting the values into the equation:

t = 0.309 seconds * ln((90% - 63.2%) / (100% - 63.2%))

t ≈ 0.309 seconds * ln(0.271 / 0.368)

t ≈ 0.309 seconds * ln(0.736)

t ≈ 0.309 seconds * (-0.305)

t ≈ -0.094 seconds

The negative value obtained implies that 90% of the maximum possible thermal energy cannot be accumulated within a sphere near the inlet of the system. This suggests that the system may need additional time or adjustments to reach the desired energy level.

To calculate the corresponding temperature at the center of the sphere, we can use the concept of one-dimensional transient heat conduction through a sphere. The equation for this scenario is:

T = Ti + (Tg,i - Ti) * (1 - exp(-t / τ))

where:

T is the temperature at the center of the sphere at time t,

Ti is the initial temperature of the spheres = 25°C,

Tg,i is the gas temperature at the inlet of the system = 300°C,

t is the time,

τ is the thermal diffusion time constant calculated earlier.

Let's calculate the corresponding temperature at the center of the sphere when t = 0.309 seconds

T = 25°C + (300°C - 25°C) * (1 - exp(-0.309 / 0.309))

T ≈ 25°C + 275°C * (1 - exp(-1))

T ≈ 25°C + 275°C * (1 - 0.3679)

T ≈ 25°C + 275°C * 0.6321

T ≈ 25°C + 173.8275°C

T ≈ 198.8275°C

Therefore, the corresponding temperature at the center of the sphere after approximately 0.309 seconds is approximately 198.83°C.

Now, let's consider the advantage of using copper instead

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Answer:

Explanation:

The process that Piaget called "adding another example to an existing scheme" is called assimilation. Assimilation refers to the cognitive process of incorporating new information or experiences into existing mental schemas or frameworks. In this case, Tammy assimilates the new example of a golf ball into her existing "ball" scheme, expanding her understanding of what falls under the category of a ball.

Answer:

I think it is D.service manual

Explanation:

Because it is the vehicle identification

To determine whether the marketing campaign will add value to the business, the marketing agency should apply the Present Net Value formula.

A calculation known as the Present Net Value brings all of an investment's future cash flows to the present. The company must identify the revenue totally attributed to the marketing campaign in order to evaluate the campaign marketing strategy as a prospective investment. By doing this, the company will be able to determine the annual inflows (sales) that must be deducted from the outflows (marketing expenses). Every year's net value is discounted at a certain rate, and if the Present Net Value is more than 0, the marketing plan is anticipated to add value to the company. To determine whether the marketing campaign will add value to the business, the marketing agency should apply the Present Net Value formula.

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Answer:

Q = [ mCp ( ΔT) ] \(_{cooling water }\)

(ΔT)\(_{cooling water}\) and  Q  is given

\(m_{cooling water}\)  = \(\frac{Q}{Cp[ T_{out} - T_{in} ] }\)

next the rate of condensation of the steam

Q = [ m\(h_{fg}\) ]\(_{steam}\)

  \(m_{steam} = \frac{Q}{h_{fg} }\)

Total resistance of the condenser is

R = \(\frac{Q}{change in T_{cooling water } }\)

Explanation:

How will the rate of condensation of the steam and the mass flow rate of the cooling water can be determined

Q = [ mCp ( ΔT) ] \(_{cooling water }\)

(ΔT)\(_{cooling water}\) and  Q  is given

\(m_{cooling water}\)  = \(\frac{Q}{Cp[ T_{out} - T_{in} ] }\)

next the rate of condensation of the steam

Q = [ m\(h_{fg}\) ]\(_{steam}\)

  \(m_{steam} = \frac{Q}{h_{fg} }\)

Total resistance of the condenser is

R = \(\frac{Q}{change in T_{cooling water } }\)

Thus, the maximum length of the specimen before elastic deformation is 0.00427 meters or 4.27 mm.

In order to calculate the maximum length of the specimen before deformation, we need to use the formula for elastic deformation:

δ = FL / AE

where:
- δ is the elongation (in meters)
- F is the force applied (in Newtons)
- L is the original length of the specimen (in meters)
- A is the original cross-sectional area (in square meters)
- E is the elastic modulus (in Pascals)

We can rearrange this formula to solve for the maximum length before deformation:

L_max = δ * AE / F

First, let's convert the original cross-sectional diameter to area:

A = πr^2 = π(3.9/2)^2 = 11.89 x 10^-6 m^2

Now we can plug in the values given in the problem:

- F = 1630 N
- E = 104 GPa = 104 x 10^9 Pa
- A = 11.89 x 10^-6 m^2
- δ = 0.44 mm = 0.44 x 10^-3 m

L_max = (0.44 x 10^-3) * (104 x 10^9) * (11.89 x 10^-6) / 1630

L_max = 0.00427 m

Therefore, the maximum length of the specimen before deformation is 0.00427 meters or 4.27 mm.

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Answer:

When a loss of charge exists, There is not a sufficient force to throttle the value open to it's correct position.

Answer:

Scientists were surprised to find that Ötzi the Iceman's stomach contained remnants of red deer meat, as it had been previously assumed that his diet consisted primarily of plants. This finding provided new insights into the dietary habits of Copper Age Europeans and challenged previous assumptions about their reliance on agriculture. Additionally, the presence of the red deer meat in Ötzi's stomach suggested that he had eaten a last meal that was high in fat and protein, which may have provided him with the energy needed for his fateful journey through the Alps.

Answer:

Ötzi the Iceman was a well-preserved mummy found in the Italian Alps in 1991. One of the most interesting things about his remains was the discovery of his last meal. Analysis of his stomach contents showed that he had recently eaten a meal of ibex meat, a type of wild goat that lived in the region, and a type of ancient bread made from einkorn wheat. What surprised scientists was the presence of the ibex's stomach contents in Ötzi's stomach. They discovered that Ötzi had eaten the ibex meat raw, with no signs of cooking. This was surprising because it was previously believed that prehistoric humans only ate meat that had been cooked over fire. The discovery of the raw meat in Ötzi's stomach suggested that prehistoric humans may have eaten raw meat more often than previously thought and that cooking food may not have been as prevalent in ancient societies as previously believed.

When Courtney compiles the code it will be produced a programming model.

Machining learning is a branch of artificial intelligence that consists of educating the machine so that it performs functions following a reasoning similar to human thinking, so that it can make its own decisions. That is, machine learning seeks to automate the system.

It is possible to educate the machine through machine learning algorithms. In a nutshell, machine learning algorithms are ordered sequences of rules, commands, and instructions. By combining the algorithms with data, they arrive at a certain result.

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The calculated voltage that would be applied to the Control Panel is 20.8 volts.

Since the panel K is 120/208V, 3Ø, 4-W, we know that it has a high leg or wild leg that supplies 208 volts to phase-to-neutral loads and 240 volts to phase-to-phase loads.

To obtain 230 volts, which is required for the control panel, we need to step down the voltage using a transformer.

A 120/240V-12/24V Group I transformer can be used to step down the voltage from 208V to 24V. Since this is a step-down transformer, the voltage across the primary winding will be greater than the voltage across the secondary winding.

The transformer turns ratio is calculated as follows:

Turns ratio = primary voltage / secondary voltage

For the given transformer, the turns ratio is:

Turns ratio = 240V / 24V = 10

Since the input voltage is exactly 208 volts, the voltage across the primary winding of the transformer will also be 208 volts. Therefore, the voltage across the secondary winding can be calculated as follows:

Secondary voltage = Primary voltage / Turns ratio

Secondary voltage = 208V / 10 = 20.8V

Thus, the calculated voltage that would be applied to the Control Panel is 20.8 volts.

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Answer:

Both zinc plating and galvanizing is an application of zinc plating. The big difference is thickness: zinc plating is normally 0.2 mils thick. Hot dip galvanizing might be 1.0 mil thick – you get over 5 times the protection with galvanizing. ... After 20 years outdoors a galvanized product will not show signs of rust.

Explanation:

hope this helps! a brainiest is always appreciated

The missing angle of the stated expression for i1(t) is 376t - π/2.

The phasor for i1(t) is represented by:

I1 = L1 * jω

L1 measures the phasor amplitude, ω denoting angular frequency whereas j represents an imaginary unit.

Upon substituting the values provided, we obtain:

I1 = (2.8e^(-jπ/3)) * j(376)

I1 = (-2.8/2) * j * e^(-jπ/3) * 752

I1 = -1.4j * e^(-jπ/3) * 752

I1 = 1204.16 e^(-jπ/3 + jπ/2)

I1 = 1204.16 e^(-jπ/6)

Using Euler's formula to convert this phasor onto a corresponding sinusoid:

i1(t) = Re(I1 * e^(jωt))

i1(t) = Re(1204.16 e^(-jπ/6) * e^(j376t))

i1(t) = Re(1204.16 e^(j(376t - π/6)))

i1(t) = 1204.16 cos(376t - π/6)

After comparing this expression with the given one for i1(t), it can be inferred that:

phi/3 = π/6

phi = π/2

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Answer:Referencing is a standardised method of acknowledging the sources of information and ideas you have used.

Explanation:Depending on the way in which they record sources, scholarly reference styles can be divided into three main categories: documentary notes styles, parenthetical (or author-date) styles, and numbered styles. Within each category there are several, slightly different reference styles.

Referencing allows you to acknowledge the contribution of other writers and researchers in your work. Any university assignments that draw on the ideas, words or research of other writers must contain citations. Referencing is also a way to give credit to the writers from whom you have borrowed words and ideas.

Answer:

I am pretty confident it is the Employer!

Explanation:

They have the responsibility to provide a safe workplace that is free from serious hazards, according to the General Duty Clause of the OSH Act (OSHA Standards)

I hope this helped you!! :D

Answer:

Employers

Explanation:

OSHA requires employers to: Provide working conditions that are free of known dangers. Keep floors in work areas in a clean and, so far as possible, a dry condition. Select and provide required personal protective equipment at no cost to workers.

how am I going to do this, I have a friend that might be able to help I will check

The given statement "draft is not necessary in die casting due to the shrink and ejector pins" is false.

Die casting is a manufacturing process in which molten metal is injected into a steel die under high pressure. This produces a part that is accurate, smooth, and often has a very thin wall. Die casting is the most efficient way to produce complex, accurate metal parts with fine detail, and it is frequently used in the automotive and aerospace industries, among others.

To make die casting possible, a slight taper, known as a draft angle, is added to the walls of the die. This makes it easier to eject the finished piece and prevents it from becoming stuck inside the die. The need for a draft angle in die casting. Die casting requires the use of a draft angle on all vertical faces.

The angle is typically 1-3 degrees for aluminum alloys and 3-5 degrees for zinc alloys. The inclusion of draft angles on castings reduces the overall cost of the process. Because the part can be easily removed from the mold, less metal is used in the part, and the manufacturing process is faster.

Furthermore, draft angles can aid in the flow of molten metal, allowing for better filling of cavities and minimizing porosity in the finished product.

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