A class design requires that a particular member variable must be accessible by any subclasses of this class, but otherwise not by classes which are not members of the same package. What should be done to achieve this?Question 18 options:A) The variable should have no special access modifier.B) The variable should be marked protected.C) The variable should be marked private.D) The variable should be marked public.E) The variable should be marked private and an accessor method provided.

There are a lot of ways employees uses in controlling exposure routes.  But when risk assessment is not be performed is not a part of the control methods.

The ways to control workplace hazards are known to be means taken to ensure safety in the workplace.

The examples are:

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Answer: Hello The required formula is attached below

answer:

A) KH =  19.2 m/day ,  Kv = 4.045 m/day

B) flow directions : ∝1 = 30°, ∝2 = 16.102°, ∝3 = 23.62, ∝4 = 55.55°,

∝5 = 1.66°

Explanation:

Given data :

Thickness of five horizontal formations ( K )  = 10 m each

Hydraulic conductivities of formations ( b )  =

b1= 20,  b2= 10,  b3= 15,  b4= 50 and b5  = 1        (m/day)

a)Determine the equivalent horizontal and vertical conductivities

kH ( equivalent  horizontal conductivity )

   = ∑ Kb / b  = ( K1b1 + K2b2 + k3b3 + k4b4 + k5 b5 ) / b1 + b2 +b3+b4+b5

input given values into the above equation

kH = 960 / 50  = 19.2 m/day

Kv ( equivalent vertical conductivity )

= ∑( b / (b/k) ) = ( b1 + b2 + b3 + b4 + b5 ) / ( b1/k1 + b2/k2 + b3/k3 + b4/k4 + b5/k5 )

input given values into equation above

Kv = 50 / 12.36 = 4.045 m/day

b) Determine the flow directions in all the formations

given that ∝1 = 30°  and Ki / kj = bi / bj

k1 /k2 = tan ∝1 / tan ∝2

20 / 10 = tan 30 / tan ∝2

∴ ∝2 = 16.102°

K2/k3 = tan 16.102 / tan ∝3

= 10 / 15 = tan 16.102 / tan ∝3

∴ ∝3 = 23.62

K3/k4 = tan ∝3 / tan ∝4

= 15 / 50 = tan 23.62 / tan ∝4

∴∝4 = 55.55°

k4 /k5 = tan ∝4 / tan∝5

= 50 / 1 = tan 55.55 / tan ∝5

∴ ∝5 = 1.66°

The term that refers to agreement as to what type of information the language is conveying is "semantic".

Semantic refers to the study of meaning in language, specifically the meaning of words, phrases, and sentences. It focuses on understanding how language conveys information and the relationship between words and their meanings.

For example, consider the sentence "The cat is on the mat." In a semantic analysis, we would examine the meanings of the words "cat," "on," and "mat" to understand the message being conveyed. We would determine that the sentence is expressing the location of the cat on the mat.

In contrast, the other options mentioned in the question have different meanings:

- Syntactic refers to the study of sentence structure and how words are combined to form grammatically correct sentences. It does not specifically focus on the meaning of the words.

- Logical refers to the study of reasoning and how statements relate to each other in a logical manner. It is concerned with the validity and soundness of arguments, rather than the specific meaning of words.

- Lexical refers to the study of words and their meanings. It examines the vocabulary of a language and how words are used to convey meaning. While lexical analysis is related to semantic analysis, it is not specifically concerned with the agreement of the type of information being conveyed.

Therefore, in this context, the correct term that refers to agreement as to what type of information the language is conveying is "semantic".

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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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Answer:

rain

Explanation:

evaoration causes clouds

clouds condense and rain

Answer:

rain

Explanation:

Answer:

winter viscosity grades

Explanation:

The “W”/winter viscosity grades describe the oil's viscosity under cold temperature engine starting conditions. There's a Low Temperature Cranking Viscosity which sets a viscosity requirement at various low temperatures to ensure that the oil isn't too thick so that the starter motor can't crank the engine over.

The best practice that Raven should follow is to use DC2 or DC3 as the Domain Naming Master of the following is a best practice that Raven should follow. The correct option is A.

The management of the addition or deletion of domains from the forest is the responsibility of the Domain Naming Master. For redundancy and fault tolerance, it is advised to split the FSMO roles among several domain controllers.

Since DC1 already has a copy of the global catalog, it is advantageous to choose a different domain controller (DC2 or DC3) as the Domain Naming Master to disperse the workload and guarantee high availability. This ensures that the forest's operations may continue even if one domain controller goes offline and prevents the creation of a single point of failure.

Thus, the ideal selection is option A.

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Small, flat-bottom boats, such as duck hunting boats, are prone to capsizing or tipping over. This is because they have a shallow draft and a flat bottom, which makes them unstable in rough or choppy water. In addition, they are often small and lightweight, which can make them more susceptible to wind and waves.

These boats are also prone to taking on water due to their low freeboard (the distance from the waterline to the deck), which can make them vulnerable to flooding in heavy rain or rough seas. It is important to properly distribute weight and to avoid overloading these boats, as excess weight can also increase the risk of capsizing.

For these reasons, it is important to exercise caution when operating small, flat-bottom boats and to always wear a personal flotation device (PFD) or life jacket while on the water. Additionally, it is recommended to take a boater safety course to learn proper boating techniques and safety procedures.

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Small, flat-bottom boats like duck hunting boats are prone to tipping over or capsizing due to their high center of gravity and lack of stability. The concept of buoyancy and stability plays a key role in understanding why these boats are prone to tipping over.

Small, flat-bottom boats, such as duck hunting boats, are prone to tipping over or capsizing if they encounter rough water or strong waves. This is due to their center of gravity being high and their lack of stability.

To understand why these boats are prone to tipping over, it is important to consider the concept of buoyancy and stability. When a boat is floating on the water, it experiences an upward force called buoyancy that is equal to the weight of the water displaced by the boat. This buoyant force helps to keep the boat afloat.

However, if the boat is not designed with a low center of gravity or with a wider base, it can become unstable and prone to tipping over. This is especially true when the boat encounters turbulent water or when there is a shift in weight distribution, such as when a person stands up or moves around in the boat.

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In this scenario, vehicles in three lanes are traveling at different speeds with a constant spacing between them.

By collecting spot speed data for 30 minutes as vehicles cross a specific observation point, we can calculate the time-mean speed and space-mean speed for the traffic stream.

To calculate the time-mean speed, we need to determine the average speed of all vehicles over the given time period. Since the speeds in each lane are constant, the time-mean speed will be the average of the speeds in each lane. In this case, lane 1 has vehicles traveling at 30 mph, lane 2 at 45 mph, and lane 3 at 60 mph. Thus, the time-mean speed can be calculated by taking the average of these speeds: (30 + 45 + 60) / 3 = 45 mph.

To calculate the space-mean speed, we need to consider the average speed of vehicles at a specific location along the roadway. In this scenario, the spacing between vehicles in each lane is constant at 0.5 mile. Therefore, the space-mean speed can be calculated by dividing the total distance traveled by all vehicles in the given time period by the total time taken. Since the vehicles maintain a constant speed, the total distance traveled will be the same as the time-mean speed. Hence, the space-mean speed in this case will also be 45 mph.

In summary, for the given traffic stream with different speeds in each lane and a constant spacing between vehicles, the time-mean speed and space-mean speed are both equal to 45 mph.

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Answer:

10.80 mm

Explanation:

Given that the reading of the fixed scale is : 10.5 mm

The thimble scale specifies : 0.30 mm

The micrometer reading will be : 10.5 + 0.30 =10.80 mm

The function ExpPower(x, n) returns the value of x raised to the power of 3n.

The given function, ExpPower(x, n), calculates the value of x raised to the power of 3n. It takes two inputs: x, a real number, and n, a non-negative integer. The function follows a recursive approach to compute the result.

If the value of n is 0, indicating the base case, the function returns a placeholder value such as "?" (as specified in the question) or an appropriate value for an undefined result. Otherwise, the function recursively calls itself with n-1 as the new exponent, assigning the result to the variable y. Finally, the function returns y raised to the power of 3, achieved by multiplying y by itself twice.

By repeatedly squaring the result of the previous calculation, the function efficiently computes x3n, utilizing the properties of exponentiation.

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Answer:

The meole fraction of acetone in the outlet liquid is \(x_1 = 0.0072\)

Explanation:

1.

The schematic diagram to represent this process is shown in the diagram attached below:

2.

the mole fraction of acetone in the outlet liquid is determined as follows:

solute from Basis Gas flow rate \(G_s = 10(1-0.012) =9.88 kmol/hr\)

Let the entering mole be :\(y_1 = 1.2\) % = 0.012

\(y_1 =(\dfrac{y_1}{1-y_1})\)

\(y_1 =(\dfrac{0.012}{1-0.012})\)

\(y_1 =0.012\)

Let the outlet gas concentration be \(y_2\) = 0.1% = 0.001

\(y_2 = 0.001\)

Thus; the mole fraction of acetone in the outlet liquid is:

\(G_s y_1 + L_s x_2 = y_2 L_y + L_s x_1\)

\(9.88(0.012-0.001)=15*x_1\)

\(9.88(0.011) = 15x_1\)

\(x_1 = \dfrac{0.10868}{15}\)

\(x_1 = 0.0072\)

The mole fraction of acetone in the outlet liquid is \(x_1 = 0.0072\)

Answer:

11.5-10/10= 0.15x100 =15%

Explanation:

The moisture content of the board is 13.04% as per the given data.

Simply said, a product's moisture content is its water content. It affects a substance's weight, density, viscosity, conductivity, and other physical characteristics.

The original weight of the test sample was 11.5 oz, and the weight of the sample after oven-drying was 10 oz.

The amount of moisture lost during the oven-drying process is equal to the original weight of the sample minus the weight of the sample after drying, or:

11.5 oz - 10 oz = 1.5 oz

To calculate the moisture content, you then divide the amount of moisture lost by the original weight of the sample and multiply by 100 to express the result as a percentage:

(1.5 oz / 11.5 oz) x 100% = 13.04%

Thus, the moisture content of the board is 13.04%.

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The quartiles divide a set of observations into four portions, each representing 25% of the observations, together with the minimum and maximum values of the data set. The interquartile range, a measurement of variation around the median, is calculated using quartiles.

Statistical file: {2, -6, 5}

Quartile Q1: -6

Quartile Q2: 2

Quartile Q3: 5.

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There are many video games that break Newton's laws, particularly the law of inertia, which states that an object at rest will stay at rest and an object in motion will stay in motion.

At a constant velocity unless acted upon by an external force. One example is the game "Mario Kart," where players can drift around corners without losing speed or momentum, which defies the law of inertia. Additionally, many action games feature characters who can jump or move at incredible speeds without experiencing the effects of acceleration or mass, which goes against Newton's laws. Finally, in some games, players may encounter unrealistic physics, such as objects that move or behave in ways that don't match the laws of action and reaction, which state that every action has an equal and opposite reaction. Overall, while video games may not always be scientifically accurate, they can still be fun and entertaining to play.

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It would need 32 acres of land to produce 1,000 megawatts of power.

Answer:

The plan is to make your design on AutoCAD and then make technical drawings. Print those technical drawings out with a scale of 1:1 and use those to trace over wood or metal.

Explanation:

Answer:

Make your design, make technical drawings, 3D print or trace

Explanation:

Answer:

Metamorphic manufacturing

Explanation:

I don't know what form of an answer you wanted but I hope this helps :)

Here's a Python program that takes a string and a non-negative integer n as inputs and returns a new string consisting of n copies of the given string:

def get_copies_of_string(string, n):

   if n < 0:

       return "Invalid input. Please provide a non-negative integer."

   return string * n

# Example usage

input_string = input("Enter a string: ")

input_n = int(input("Enter a non-negative integer: "))

result = get_copies_of_string(input_string, input_n)

print("Result:", result)

In this Python program, the get_copies_of_string function takes two parameters: string and n. It checks if n is a non-negative integer. If n is valid, it uses the string repetition operator * to create a new string by multiplying the given string n times. Finally, it returns the resulting string.

A Python program is a set of instructions written in Python syntax that enables a computer to perform specific tasks or solve problems based on the defined logic and algorithms.

You can run this program by providing a string and a non-negative integer when prompted. The program will then output the new string consisting of n copies of the given string.

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Answer:

Avogadro's law is a gas law which states that the total number of atoms or molecules of a gas (representing the amount of gaseous substance) is directly proportional to the volume that the gas occupies at constant temperature and pressure.

Answer:

byee byee bbbbbbbbbbbb

The size of the wrapper that TKIP places around the WEP encryption is A. 128-bit.

TKIP (Temporal Key Integrity Protocol) is a security protocol used in wireless networks to enhance the security of the outdated WEP (Wired Equivalent Privacy) encryption. TKIP introduces several improvements, including a larger key size and a more secure key management mechanism.

The size of the wrapper that TKIP places around the WEP encryption is 128-bit. This 128-bit wrapper is known as the TKIP Key Mixing function.

Here is a step-by-step explanation:

Step 1: TKIP uses a 128-bit key derived from a combination of the original WEP key and additional data, such as the MAC address and serial number of the packet.

Step 2: The TKIP Key Mixing function takes this 128-bit key and creates a 128-bit key stream.

Step 3: The key stream is then XORed (exclusive OR operation) with the plaintext data to produce the ciphertext.

Step 4: To enhance security, TKIP uses a different 128-bit key for each packet. This per-packet key is generated by combining the 128-bit key stream with a unique initialization vector (IV) for each packet.

Step 5: The resulting ciphertext, along with the IV, is transmitted over the wireless network.

By using a larger 128-bit key and per-packet keys, TKIP significantly improves the security of WEP by making it more difficult for attackers to crack the encryption. The additional data, such as the MAC address and serial number of the packet, adds uniqueness and randomness to the key generation process, making it harder for attackers to predict or exploit patterns in the encryption.

It's important to note that while TKIP provides better security than WEP, it has been largely superseded by the more robust WPA (Wi-Fi Protected Access) and WPA2 protocols, which use stronger encryption algorithms like AES (Advanced Encryption Standard).

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The sum of moments about A must be zero. The reaction moment at point A is 106 kN.m in the clockwise direction.

(a) To determine whether the structure is statically determinate or not, the first step is to count the number of unknown reactions. Here, we have three unknowns - the vertical reactions at A and C, and the reaction moment at A.Now let us make a cut in the structure at point A to 'explode' the structure. The following figure shows the exploded view of the structure. The cut divides the structure into two parts, and we have to consider the left and right parts of the structure separately in order to determine the unknown reactions. Since the right part of the structure is isolated, there are no external forces acting on it. Therefore, the sum of moments about A must be zero.

\($$-RA(2) + 6(4) + 12(2) = 0$$$$\Rightarrow RA = 18 \ kN.m$$\)

The left part of the structure can be treated as a simply supported beam carrying a uniformly distributed load of 2 kN/m between A and C. The vertical reactions at A and C can be determined by taking moments about C.

\($$- RA(2) - RC(10) + \frac{2 \times 8^{2}}{8} + \frac{2 \times 6^{2}}{8} + \frac{2 \times 4^{2}}{8} + \frac{2 \times 2^{2}}{8} = 0$$$$\Rightarrow RC = 11 \ kN$$$$\sum F_{y} = 0 \ \ \Rightarrow \ \ RA + RC = 2(2+4+6+8+10)$$$$\Rightarrow RA = 92 \ kN$$\)

Since the two values of RA do not match, the structure is not statically determinate.
(b) The reaction moment at point A can be determined by taking moments about A.

\($$M_{A} = -RC(10) + \frac{2 \times 8^{2}}{8} + \frac{2 \times 6^{2}}{8} + \frac{2 \times 4^{2}}{8} + \frac{2 \times 2^{2}}{8}$$$$\Rightarrow M_{A} = -11(10) + 4 + 9 + 4 + 1$$$$\Rightarrow M_{A} = -106 \ kN.m$$\)

Therefore, the reaction moment at point A is 106 kN.m in the clockwise direction.

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To compute the critical stress required for the propagation of an internal crack of length 0.3 mm in aluminum oxide with a specific surface energy of 0.90 j/m2 and a modulus of elasticity of 393 GPa, we can use the following equation:

Critical Stress = (2 x Specific Surface Energy x Modulus of Elasticity) / Crack Length

Plugging in our values:

Critical Stress = (2 x 0.90 j/m2 x 393 GPa) / 0.3 mm

Critical Stress = 8,760 MPa

The specific surface energy for aluminum oxide is given as 0.90 J/m2 and its modulus of elasticity is 393 GPa. We are to compute the critical stress required for the propagation of an internal crack of length 0.3 mm.

Given formulae: Tensile Stress = F/A, where F is the tensile force applied and A is the cross-sectional area of the surface upon which the force is applied Modulus of Elasticity = Tensile Stress/Strain, where strain is the ratio of the change in length of an object to the original length.

Critical Stress = σ_0 √πa/2 * Y_b

The first step is to compute the tensile stress, which is given as:

Tensile Stress = F/A

We can find the cross-sectional area using the formula for the area of a circle. This formula is given as: Area = πr^2, where r is the radius of the circle. We can find the radius of the circle as r = 0.3/2 mm = 0.15 mm = 0.00015 m. We can now compute the area of the circle as follows: Area = π(0.00015)^2 = 7.07 × 10^-8 m^2.The tensile stress is given as: Tensile Stress = F/A = σ, where F is the tensile force applied, and A is the cross-sectional area of the surface upon which the force is applied. We can rearrange the equation above to get: F = Aσ = 7.07 × 10^-8 × σThe modulus of elasticity is given as:

Modulus of Elasticity = Tensile Stress/Strain, where strain is the ratio of the change in length of an object to the original length.

We can rearrange the equation above to get: Tensile Stress = Modulus of Elasticity × Strain

Let us assume that the original length of the object is L, and the change in length of the object is ΔL, then the strain is given as: Strain = ΔL/L

The critical stress is given as: Critical Stress = σ_0 √πa/2 * Y_b . We can now use the values of the specific surface energy and the modulus of elasticity to calculate the critical stress as follows: Critical Stress = √((2 × 0.90 × 10^-6 × 393 × 10^9)/((π × 0.3 × 10^-3)/2)) = 97.2 MPa

Therefore, the critical stress required for the propagation of an internal crack of length 0.3 mm is 97.2 MPa.

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Answer:

it's not included

Explanation:

plz exact ur explain

Answer:

si amor

Explanation:

Hoiykñjdnlklbutrk

Answer:

35.92 kpsi

Explanation:

Given data:

diameter of the steel bar d = 0.875 in

Area A = πd^2/4 = π(0.875)^2/4

length L = 15.0 ft

Load P = 21.6 kip

Modulus of elesticity E = 29×10^6 Psi

Assume we are asked to determine axial stress in the bar which is given as

\(\sigma = Load, P/ Area, A\)

\(\sigma = 4P/\pi d^2\)

substitute the value

\(\sigma = \frac{4\times 21.6}{\pi \times (0.875)^2} \\=35.92\ kpsi\)