For this problem, you are going to implement a method that processes an ArrayList that contains MyCircles. Here is the complete MyCircle class that we will assume:
public class MyCircle { private int radius, centerX, centerY;
public MyCircle (int inRadius, int inx, int inY) { radius inRadius; centery = inY;
centerX = inX;
}
public int getRadius() { return radius; }
public int getX() { return centerX; }
public int getY() { return centery; }
public double getArea() { return Math.PI * radius * radius; }
}

Answer:

not understand language

Answer:

1211 N.

Explanation:

Okay, we are given the following data or parameters or information in the question above;

=> "aluminum (E = 70 GPa) tube of length 8-m.

=> "The aluminum is used as a simply supported column carrying a 1.2 kN axial load"

The axial load to provide a factor of safety of 2 against buckling =[ (22/7)^2 × 224348 × 1000 × 70] ÷ (8 × 1000)^2 × 2.

The axial load to provide a factor of safety of 2 against buckling = 1211 N.

The government established the Federal Deposit Insurance Corporation in order to regulate the stability and public confidence in the nation's financial system.

The fundamental objective of FDIC is to examine and administer financial institutions for standard safety, soundness, and protection of the consumer. It also constructs other large and complex financial institutions with the intention of delivering the solution.

According to the context of this question, FDIC has generally served as an integral domain of the financial system of the nation for more than 50 years. It is typically established by the Banking Act of 1933 at the depth of the most severe banking crisis in the nation's history.

Therefore, the government established the Federal Deposit Insurance Corporation in order to regulate the stability and public confidence in the nation's financial system.

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Answer:

12 mins

Explanation:

The summation of the forces in vertical direction

= Fb - Fd - w = 0 ∴ Fd = Fb - w  ----- ( 1 )

Fb ( buoyant force ) = Pair * g * Vballoon ------- ( 2 )

Pair = air density , Vballoon = volume of balloon

Vballoon = \(\frac{\pi D^3}{6}\) ,  where D = 4  ∴  Vballoon = 33.51 ft^3

g = 32.2 ft/s^2

From property tables

Pair = 2.33 * 10^-3 slug/ft^3

μ ( dynamic viscosity ) = 3.8 * 10^-7 slug/ft.s

Insert values into equation 2

Fb = ( 2.33 * 10^-3 ) * ( 32.2 ) *( 33.51 )  = 2.514 Ib

∴ Fd = 2.514 - 2.5 = 0.014 Ib ( equation 1 )

Assuming that flow is Laminar  and  RE < 1

Re = (Pair * vd) / μair   -------- ( 3 )

where:  Pair = 2.33 * 10^-3 slug/ft^3 ,  vd = ( 987 * 4 ) ft^2/s ,   μair = 3.8 * 10^-7 slug/ft.s

Insert values into equation 3

Re = 2.4 * 10^7 (  this means that the assumption above is wrong )

Hence we will use drag force law

Assume Cd = 0.5

Express  Fd using the relation below

Fd = 1/2* Cd * Pair * AV^2

therefore V = 1.39 ft/s

Recalculate  Reynold's number using  v = 1.39 ft/s

Re = 34091

from the figure Cd ≈ 0.5 at Re = 34091  

Finally calculate the rise time ( time taken to reach an altitude of 1000 ft )

t = h/v

  = 1000 / 1.39 = 719 seconds ≈ 12 mins

The maximum velocity based on the information given will be 980 feet per second.

Maximum velocity is defined as the greatest speed an entity can possibly reach in a certain system or situation. This maximum rate of motion is dependent upon multiple factors such as the mass, applied force, and resistance encountered by the object, plus even the environment it is traversing.

When considering classical mechanics, the maximum velocity of an article is calculated via the given equation v = √2*E/m, with v representing its velocity, E standing for its energy, and m signifying its mass.

Based on the information, maximum velocity can be found by using TR , TA vs V curve. The intersection point gives Vmax = 980ft/s (approx)

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The color of white light bends the most when it is refracted by a prism is green after the violet.

The shorter the wavelength of the mild, the greater its miles refracted. As a result, crimson mild is refracted the least, and violet mild is refracted the most - inflicting the colored mild to unfold out to shape a spectrum.

The hues of white mild withinside the order of lowering wavelength are crimson orange-yellow inexperienced blue indigo violet. Fro the listing we are given withinside the alternatives inexperienced has the shortest wavelength than others. Therefore inexperienced will bend the most.

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Answer:

1) The power developed by the engine is 14705.7739 kW

2) The thermal efficiency is approximately 61.5%

Explanation:

The given parameters are;

P₁ = 95 kPa

T₁ = 22°C

V₁ = 3.17 liters

The cutoff ratio = 2.5

Displacement volume = 3 liters

The number of times the cycle is executed per minute = 1000 times per minute

We have;

The displacement volume = V₁ - V₂ = 3 l

V₁ = 3.17 l

V₂ = 3 - 3.17 = 0.17 l

Compression ratio = V₁/V₂ = 3.17/0.17 ≈ 18.65

P₂/P₁ = P₂/(95 kPa) =  (V₁/V₂)^(k) = 18.65^1.4

P₂ = (95×18.65^(1.4)) ≈ 5710.5 kPa

T₂/T₁ = (V₁/V₂)^(k - 1)

T₂/(295 K)= (18.65)^(1.4 - 1)

T₂ = 295 * (18.65)^(1.4 - 1) = 950.81 K

The cutoff ratio = V₃/V₂ = 2.5

T₃ = T₂ × V₃/V₂  = 2.5 * 950.81 K = 2377.025 K

\(Q_{in}\) = \(C_p\)×(T₃ - T₂) = 1.006 × (2377.025 - 950.81) = 1,434.77 kJ/kg

T₄ = T₃ × (V₃/V₄)^(k-1) =

Therefore,

\(T_4 = T_3 \times \left (\dfrac{r_c}{r} \right )^{k - 1} = 2377.025 \times \left( \dfrac{2.5}{18.65} \right )^{1.4 - 1} \approx 1064 \ K\)

T₄ ≈ 1064 K

\(Q_{out}\) = \(-C_v \times (T_4 - T_1)\)

\(C_v = C_p/k = 1.006/1.4 \approx 0.7186 \ kJ/kg\)

∴ \(Q_{out}\) = 0.7186×(1064 - 295) = 552.6034 kJ/kg

1) The net work = \(Q_{in}\) - \(Q_{out}\) = 1,434.77 kJ/kg - 552.6034 kJ/kg ≈ 882.17 kJ/kg

The number of cycle per minute = 1000 rpm

The number of cycle per minute = 1000 rpm/60 = 16.67 cycles per second

The power developed by the engine = The number of cycles per second × The net work of the engine

Therefore;

The power developed by the engine = 16.67 cycles/second  × 882.17 kJ/kg

The power developed by the engine = 14705.7739 kW

2) Efficiency, \(\eta _{th}\), is given as follows;

\(\eta _{th} = \dfrac{Q_{in}-Q_{out}}{Q_{in}} \times 100 = 1 - \dfrac{Q_{out}}{Q_{in}} \times 100= 1 - \dfrac{552.6034}{1434.77}\times 100 \approx 61.5\%\)

Therefore, the thermal efficiency ≈ 61.5%.

Answer:

It's equalization of the system

Explanation: If there was a leak it would likely leak all the time even if the car was turned off. Plus, a system leak bad enough to hear would drain the system of refrigerant very quickly and would no longer cool.

One type of infrastructure support that is not currently under consideration to keep sewer systems in cities from experiencing overflows is the implementation of green infrastructure. Green infrastructure refers to the use of natural systems such as permeable pavement, rain gardens, and green roofs to manage stormwater runoff. These natural systems help to absorb and filter rainwater, reducing the amount of water that enters the sewer system.

However, despite their effectiveness in preventing overflows, green infrastructure is often not considered due to various reasons.

One reason is the lack of awareness and understanding of the benefits of green infrastructure among city planners and policymakers. Many cities still rely heavily on traditional gray infrastructure, such as large pipes and storage tanks, to manage stormwater. This outdated approach may be seen as more reliable and familiar, even though it can be costly and less environmentally friendly.

Another reason is the perception that green infrastructure is not suitable for urban environments. There is a misconception that these natural systems require a significant amount of space, which may be limited in densely populated cities.

However, there are innovative approaches to implementing green infrastructure in urban areas, such as using green roofs on buildings or retrofitting existing infrastructure.

Additionally, the upfront costs of implementing green infrastructure can be a deterrent for some cities. While green infrastructure may be more cost-effective in the long run due to lower maintenance and operation costs, the initial investment required can be higher compared to traditional gray infrastructure.

This financial barrier may prevent cities from considering green infrastructure as an option.

Green infrastructure is not currently under consideration as a type of infrastructure support to prevent sewer system overflows in cities. This can be attributed to factors such as lack of awareness and understanding, misconceptions about its suitability for urban environments, and upfront costs.

However, promoting the benefits and exploring innovative approaches can help overcome these barriers and encourage the adoption of green infrastructure to mitigate overflows in sewer systems.

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Answer:

Explanation:

i) In the attached truth table, TRUE means the respective key owner is present and their keys are inserted at the same time. The "Opens" column is TRUE when two owners are present, not including the case where the only two owners present are B and C.

__

ii) The second attachment is a Karnaugh map of the truth table. The circled terms are ...

  Opens = AB +AC

Answer:

Therefore the value of the line currents IR, IY, and IB are

\(I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A\)

Explanation:

Apply KVL for loop 1  

\(415\angle 120^{o}=\left ( 3+j4+3+j4 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 6+j8 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} \rightarrow \left ( 1 \right )\)

Apply KVL for loop 2

\(415\angle 0^{o}=\left ( 3+j4+3+j4 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 6+j8 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}-\left ( 5\angle 53.13^{o}\right )I_{1} \rightarrow \left ( 2 \right )\)

Solving the above equations,

\(415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} +\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{1}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{2}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}\\I_{2}= 71.88\angle -23.13^{o}A\)&

\(415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right ) \times 71.88\angle -23.13^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\415\angle 0^{o}=718.8\angle 30^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\\left ( 5\angle 53.13^{o} \right )I_{1}=415\angle 60^{o}\\I_{1}= 83\angle 6.87^{o}A\)

Hence,

\(I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A\)

Answer:

so where is the picture?

Answer:

there the drawing lol

If the siding is fastened to the studs through the sheathing, the International Residential Code prescribes 3/8-inch plywood for sheathing walls with a typical stud spacing of 16 inches.

The minimum plywood thickness required by the IRC is 1/2 inch if the nails only go through the sheathing and not through the studs.

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Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

   = 20.4kg · 9.81 \(\frac{N}{kg}\) · 1.50m

   = 35.316 \(\frac{N}{m}\)

   = 35.316 W

Answer:

The advantages of using a pure sine wave for your appliances and machinery are as follows: Reduces electrical noise in your machinery.

translates to no TV lines and no sound system hum.

Cooking in microwaves is quicker.

Explanation:

The smoothest signal is a sine wave, and sine waves are the basis of all functions.

Every other continuous periodic function is a basis function, which means that it can be described in terms of sines and cosines.

For instance, using the Fourier series, I can describe the fundamental Sinusoidal frequency and its multiples in terms of the triangle and square waves.

The SQL queries for the given set of tables: EMPLOYEE (EmpNo, Name, DoB, Address, Gender, Salary, DNumber) DEPARTMENT (DNumber, Dname, ManagerEmpNo, MnagerStartDate) is given below.

The code that display the age of 'male' employees is given below:

sql

SELECT Name, TIMESTAMPDIFF(YEAR, DoB, CURDATE()) AS Age

FROM EMPLOYEE

WHERE Gender = 'male';

The code that  display all employees in the Department named 'Marketing':

sql

SELECT *

FROM EMPLOYEE

WHERE DNumber IN (SELECT DNumber FROM DEPARTMENT WHERE Dname = 'Marketing');

The code that display the name of the highest salary paid 'female' employee:

sql

SELECT Name

FROM EMPLOYEE

WHERE Gender = 'female'

ORDER BY Salary DESC

LIMIT 1;

Therefore, This code shows the age of male workers and workers in the ‘Marketing’ team, and the name of the woman who earns the most, using special statements.

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When using the simulation mode, you can ask a wide range of questions that can be answered based on simulated scenarios and data.

Here are three examples:

"What would be the outcome if we increase the production capacity by 20%?"

Simulation mode can help you assess the potential impact of changing a specific parameter or variable in a simulated environment. By increasing the production capacity in the simulation, you can observe how it affects factors such as output, costs, resource utilization, and overall efficiency.

"How would different marketing strategies impact customer acquisition?"

Simulations can simulate different marketing strategies and their potential impact on customer acquisition. By analyzing simulated data and scenarios, you can assess the effectiveness of various marketing approaches, such as targeted advertising, promotions, or influencer collaborations, in acquiring new customers.

"What is the projected revenue growth if we expand into a new market segment?"

Simulations can provide insights into the potential outcomes of expanding into new market segments. By simulating different scenarios, you can analyze the market dynamics, customer demand, competition, and other factors to estimate the projected revenue growth associated with entering a new market segment.

These are just a few examples, and the range of questions that can be answered using simulation mode is extensive. The specific questions will depend on the context, goals, and parameters of the simulation model being used.

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Answer: Shielding gas solenoid

Explanation:

Answer:

Diameter of pipe is 0.535 ft

Explanation:

see attachment, its works out 1st half

Answer:

Explanation:

The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder        

Answer:

Part A

The thickness of the compacted soil is approximately 4.3467 × 10⁻¹ m

Part B

The weight of water to be added is approximately 19886.\(\overline{36}\) kN, the volume of the water added is approximately 2,027.77 m³

Explanation:

The parameters of the soil are;

The volume of sol the excavator excavates, \(V_T\) = 10,000 m³

The moist unit weight, W = 17.5 kN/m³

The moisture content = 10%

The area of the project, A = 20,000 m²

The required dry unit weight = 18.3 kN/m³

The required moisture content = 12.5%

Part A

Therefore, we have;

The moist unit weight = Unit weight = (\(W_s\) + \(W_w\))/\(V_T\)

The moisture content, MC = 10% = (\(W_w\)/\(W_s\)) × 100

∴ \(W_w\) = 0.1·\(W_s\)

∴ The moist unit weight = 17.5 kN/m³ = (\(W_s\) + 0.1·\(W_s\))/(10,000 m³)

1.1·\(W_s\) = 10,000 m³ × 17.5 kN/m³ = 175,000 kN

\(W_s\) = 175,000 kN/1.1 = 159,090.\(\overline{09}\) kN

For the required soil, we have;

The required dry unit weight = 18.3 kN/m³ = \(W_s\)/\(V_T\) = 159,090.\(\overline{09}\) kN/\(V_T\)

\(V_T\) = 159,090.\(\overline{09}\) kN/(18.3 kN/m³) ≈ 8,693.4923 m³

The total volume of the required soil ≈ 8,693.4923 m³

Volume \(V_T\) = Area, A × Thickness, d

∴ d =  \(V_T\)/A

d = 8,693.4923 m³/(20,000 m²) ≈ 4.3467 × 10⁻¹ m

The thickness of the compacted soil ≈ 4.3467 × 10⁻¹ m

Part A

The moisture content, MC = 12.5% = (\(W_w\)/\(W_s\)) × 100

\(W_w\) = \(W_s\) × MC/100 = 159,090.\(\overline{09}\) kN × 12.5/100 = 19886.\(\overline{36}\) kN

The weight of water to be added, \(W_w\) = 19886.\(\overline{36}\) kN

Where the density of water, ρ = 9.807 kN/m³

Therefore, we have;

The volume of water, V = \(W_w\)/ρ

∴ V = 19886.\(\overline{36}\) kN/(9.807 kN/m³) ≈ 2027.77 m³

The volume of water, V ≈ 2027.77 m³

Answer:

1.2 Newtons upwards

Explanation:

because the friction is opposite of the magnitude.

Answer:

Construction, the techniques and industry involved in the assembly and ... Early building materials were perishable, such as leaves, branches, and animal hides. ... The well-developed masonry technology of Mesopotamia was used to build large ... although its precise description is unknown; the concealed faces of stones

Explanation:

Answer:

Placement.

Explanation:

SAE is an acronym for Supervised Agricultural Experience programs and it is typically a planned, practical-based program designed to help students develop competent skills and experience in their various career choice. Basically, it enhances the academic knowledge and agricultural skills acquired by the students in the classroom.

Generally, there are four (4) main types of Supervised Agricultural Experience Programs;

1. Entrepreneurship SAE.

2. Exploratory SAE.

3. Research and Experimentation SAE.

4. Placement SAE.

A non-entrepreneurship, work-based, agricultural type of Supervised Agricultural Experience program (SAE), in which a student learns and gains skills in a paid or unpaid position describe a placement.

Some examples of organizations or businesses that provide placement SAE to students includes; lawn services, veterinary clinics, soil conservation firms, livestock farms, floral shops, grain farms etc.

Answer:

Placement

Explanation:

Answer:

True

Explanation:

Ture, the indemnity protects.

Answer: The correct choice is True

There are two countershafts in the forward section of a Roadranger transmission.

A countershaft is a machine component that is used to convert rotational force from one direction to another. The component works in a gearbox and is typically used to change gear ratios and transmit power throughout a car's powertrain.

A Roadranger transmission is a brand name for a specific gearbox that is used in heavy-duty trucks. This particular type of transmission has two countershafts in the forward section. There are two different designs for this type of transmission: ten-speed and thirteen-speed.

Both designs have two countershafts, which means that they are the same in this respect. The Roadranger transmission is used in many different types of heavy-duty trucks and is known for its reliability and durability.

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Answer:

approaching a lady or a man

Besides the properties of the material composing the fiber in a fiber-reinforced composite, the length of the fiber is also the most important factor for producing a significant improvement in strength.

Fiber-reinforced composite refers to any building material made up of two or more constituent materials with individually different physical properties. Fiber-reinforced composite has a combination of properties that create a superior material ideal for multiple construction applications. In addition, the fiber's length is also a vital property of fiber-reinforced composite which reinforces long fiber composites with improved strengths.

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