Which are two additional important issues of web based UI design vs. local desktop Ul design? (Pick two) A)Load Time B)Browser compatibility C)Ad space D)Screen Size

The frequency the remote-pilot should use to monitor air traffic near an uncontrolled airport is the Common Traffic Advisory Frequency (CTAF).

What is air traffic?
Air traffic is the movement of aircrafts such as planes and helicopters in a certain area. This may include both civilian and military aircrafts. Air traffic is managed by air traffic controllers who are responsible for ensuring the safety of aircrafts by providing directions and instructions on the ground and in the air. They also inform pilots about the current weather conditions and ensure the proper separation of aircrafts from each other. Air traffic controllers must constantly monitor the positions and movements of aircrafts in their airspace and keep track of all the data related to their flight paths.

The CTAF is a frequency used by pilots and air traffic control to exchange information about aircraft operating in the vicinity of an airport.

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Answer: the bottom is a harder material than, the actual w33d inside it

Explanation:

the battery and ODC input voltages, as well as the main parameters for the power source, fireworks, relays, anti-static resistances, etc. The statements below refer to the design and operation of the power sub-system on a spacecraft.

Solar panels on these spacecraft convert solar energy into the electricity they require for propulsion. The electricity generated by the solar panels is used to recharge a battery within the spacecraft. These batteries can power the spaceship even as it is traveling away from the sun.

The engine, temperature management, power and power distribution, attitude control, telemetry command and control, transmitters/antenna, computers/on-board processing/software, and structural components are only a few of the many subsystems that make up a spaceship bus.

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Answer: true

Explanation:

About 296.5 N of anchoring force is required to keep the elbow in place. About 134.8 is the direction of the anchoring force.

The speed of an object is its magnitude (or value), which is the velocity. The item is traveling in the direction indicated by the velocity vector. Imagine a circle (or, better yet, draw one) and an object traveling along the path it defines.

Units of mass times length over time squared are used to express the strength of a force. The most used unit in metric measurements is the newton (N), which is equal to one-kilogram times one meter over one second squared.

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The velocity of point A on the rim of the gear at the instant shown is 39.8°.

Velocity is the directional speed of a moving object as an indication of its rate of change in location as perceived from a specific frame of reference and measured by a specific time standard.

Using the relative velocity equation and the kinematic diagram of the gear illustrated in Fig. a, apply the relative velocity equation to points B and C.

Va = Vc + w x r A/C

Va = 4i + (-3.11k) x ( -1.06i + 2.56j)

Va = 3.9665i + 3.2998j

Equating the i and j components yields and its direction:

[tex]tan^{-1} = \dfrac{3.2998}{3.9665} = 39.8^\circ[/tex]

Therefore, the velocity of point A on the rim of the gear at the instant shown is 39.8°

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The weight needed to hold the wall is equal to 14.9285 N.

The force equilibrium can be described as the addition of the forces about x, y, and z-axis will be equal to zero and the sum of the moment also equal to zero.

Given, the width of the wall, b = 10 m

The height of the water, d = 4 m

The height of the wall, l = 7m

The area of the wall, A = bd = 10(4) = 40 m²

The expression of the centroid of the wall,  [tex]\displaystyle \bar x = \frac{d}{2}[/tex]

[tex]\displaystyle \bar x = \frac{4}{2} = 2 m[/tex]

The hydrostatic force, F = ρgAx

F = (1000) (9.81) (40) (2)

F = 784,000 N

The expression for  the moment of inertia: [tex]\displaystyle I_C =\frac{bd^3}{12}[/tex]

[tex]I_c = 10\times 4/12 = 53.34 m^4[/tex]

The relation of the center of the pressure gate can be written as:

[tex]\displaystyle \bar h = \bar x +\frac{I_C}{A\bar x}[/tex]

[tex]\bar h = 2 + \frac{53.34}{40\times 2} = 2.667 \; m[/tex]

The moment about point: l × W - F(d - h) = 0

Substitute the values in the above equation:

7 × W - 78400 (4 - 2.667) = 0

7 W = 78400 (4 - 2.667)

W = 14.92 kN

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The principal of the loan that TMI Systems repaid was $90,000.

The formula to calculate the principal of a loan is P = L + I, where P is the principal, L is the loan amount, and I is the total interest paid over the duration of the loan.

In this case, we are given that the loan amount was 10% per year simple interest, which can be calculated by multiplying the principal by the interest rate (0.1) and the duration of the loan (5 years). Therefore, the total interest paid over the duration of the loan is P x 0.1 x 5 = $50,000.

We are also given the total amount repaid, which is the sum of the principal and the total interest paid. Therefore, we can calculate the principal by subtracting the total interest paid from the total amount repaid.

This gives us $140,000 - $50,000 = $90,000. As the principal must be a positive value, the principal of the loan is $90,000.

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The elevation of lowest point of the curve is 445.077 ft.

Height above or below the mean sea level is referred to as elevation. A map's elevation can be depicted either by labelling the precise elevations of specific points or by using contour lines, which link points of the same elevation. Topographic maps are depicted as having elevations.

Calculate rate of change of the curve as below:

[tex]$$\begin{aligned}r & =\frac{g_2-g_1}{L} \\& =\frac{2.5-(-4.0)}{\left(\frac{400}{100}\right)} \\& =1.625 \%\end{aligned}$$[/tex]

Calculate distance from PC to the lowest point as below:

[tex]$$\begin{aligned}X & =\frac{-g_1}{r} \\& =\frac{-(-4.0 \%)}{1.625 \%} \\& =2.462 \mathrm{ft}\end{aligned}$$[/tex]

Calculate the elevation of the lowest point of the curve as below:

[tex]$$\begin{aligned}Y & =Y_{P C}+g_1 X+\frac{r}{2} X^2 \\& =450\mathrm{ft}+(-4.0 \times 2.462)+\frac{1.625}{2}(2.462)^2 \\& =450\mathrm{ft}-9.848 \mathrm{ft}+4.925 \mathrm{ft} \\& =\mathbf{445.077} \mathrm{ft}\end{aligned}$$[/tex]

Thus, the elevation of lowest point of the curve is 445.077ft.

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Answer:

0.00337339236

Answer:

Hey there! 16 divided by 4743 is approximately equal to 0.003366. This means that if you take 16 and divide it by 4743, the result will be a number that is very close to 0.003366. This division problem can be written as the fraction 16/4743, or as the decimal 0.003366.

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Some common mistakes to avoid when solving division problems include forgetting to carry the remainder, failing to check for divisibility, and mixing up the order of the dividend and the divisor. It is also important to pay attention to the units of measurement and to ensure that the units in the dividend and the divisor are compatible.

Some strategies for solving division problems with large numbers include using place value to break down the numbers into smaller units, using long division or a calculator to perform the division, and using estimation to check the reasonableness of the answer. It can also be helpful to use mental math techniques, such as rounding, to simplify the division problem. Other strategies might include using division to find the prime factorization of a number or to identify common factors or multiples.

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Hello! In this question, I will answer the first part of the question, in which we will determine the x, y, and z components for the reaction at joint A in the ball-and-socket.
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Explanation:

To better understand what the question entails, we will start off by drawing the free-body diagram to understand the direction of our components. The image is attached below. This will help us solve for our components.

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Solve:

To start, we will first solve for the weight of the sign, we will use the formula:

[tex]W=mg[/tex]

Whereas:

Plug in our values into the formula and solve:

[tex]W = 90\cdot9.81=882.9 N[/tex]

Since we now know our weight value, we can solve for our force in the cable BC, expressed as a vector. We will use the formula:

[tex]\bold{T}_{BC}=T_{BC}(\frac{\bold{r}_{BC}}{r_{BC}})[/tex]

Where:

Plug in our values into the formula and solve:

[tex]\bold{T}_{BC}=T_{BC}\bigg(\cfrac{\text{[i - 2j+ 2k] ft}}{\sqrt{(1)^2+(-2)^2+(2)^2}}}\bigg)\\\\\bold{T}_{BC}=T_{BC}\bigg(\cfrac{\text{[i - 2j+ 2k] ft}}{3}}}\bigg)\\\\\text{Simplify}\\\\\bold{T}_{BC}=\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k[/tex]

Now, let us solve for our force in the cable BD, also expressed as a vector. Use the formula:

[tex]\bold{T}_{BD}=T_{BD}(\frac{\bold{r}_{BD}}{r_{BD}})[/tex]

Use the same steps from solving for our vector force of cable BC. Plug in the values and solve:

[tex]\bold{T}_{BD}=T_{BD}\bigg(\cfrac{\text{[-2i - 2j+ k] ft}}{\sqrt{(-2)^2+(-2)^2+(1)^2}}}\bigg)\\\\\bold{T}_{BD}=T_{BD}\bigg(\cfrac{\text{[-2i - 2j+ k] ft}}{3}}}\bigg)\\\\\text{Simplify}\\\\\bold{T}_{BD}=-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j+\frac{1}{3} T_{BD}k[/tex]

We now need to find the moment at A at equilibrium (0). This is known as:

[tex]\sum M_{a}=0 \Rightarrow\bold{r}_{B}\times(\bold{T}_{BC}+\bold{T}_{BD}+\bold{W})=0[/tex]

Whereas:

Plug in values into the equation:

[tex]2\text{j}\times\big[\big(\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k\big)+\big(-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j+\frac{1}{3} T_{BD}k\big)\big]\\+\text{j}\times-882.9\text{k}=0[/tex]

Find the moment about the z-axis to zero. This is known as:

[tex]\sum M_{z}=0\\\\-\frac{2}{3}T_{BC}+\frac{4}{3}T_{BD}=0\\\\\text{Simplify}\\\\T_{BC}=2T_{BD}[/tex]

Now, find the moment about the x-axis to zero. This is known as:

[tex]\sum M_{x}=0\\\\\frac{4}{3}T_{BC}+\frac{2}{3}T_{BD}-882.9=0\\\\\text{We know }T_{BC}=2T_{BD}\text{, plug in }2T_{BD}\text{ in }T_{BC}\\\\\frac{4}{3}\times2T_{BD}+\frac{2}{3}T_{BD}-882.9=0\text{ Solve for }T_{BD}\\\\\frac{10}{3}T_{BD}=882.9\\\\T_{BD}=264.87\text{ N}[/tex]

Now, let us calculate the tension in wire BC.

[tex]T_{BC}=2T_{BD} = 2(264.87) = 529.74\text{ N}[/tex]

Let us calculate the reaction at A by solving for the equilibrium force equation. The formula is:

[tex]\sum F=0\\\\\bold{F}_A+\bold{T}_{BC}+\bold{T}_{BD}+\bold{W}=0[/tex]

Plug in our known information into the equation and simplify.

[tex]\big[(A_xi+A_yj+A_zk)+(\frac{1}{3}T_{BC}{\text{i}}-\frac{2}{3}T_{BC}j+\frac{2}{3} T_{BC}k)+(-\frac{2}{3}T_{BD}{\text{i}}-\frac{2}{3}T_{BD}j\\+\frac{1}{3} T_{BD}k)-981\text{k}\big]=0\\\\\text{Separate by component}\\\\\big[(A_x+\frac{1}{3}T_{BC}-\frac{2}{3}T_{BD})\text{\bold{i}}+(A_y-\frac{2}{3}T_{BC}-\frac{2}{3}T_{BD})\text{\bold{j}}+(A_z+\frac{2}{3}T_{BC}+\frac{1}{3}T_{BD}\\-981)\text{\bold{k}}\big]=0\\\\\text{Plug in known values}\\[/tex]

[tex]\big[(A_x+\frac{1}{3}\times529.74-\frac{2}{3}\times264.87)\text{\bold{i}}+(A_y-\frac{2}{3}\times529.74-\frac{2}{3}\times264.87)\text{\bold{j}}+(A_z+\frac{2}{3}\times529.74+\frac{1}{3}\times264.87-882.9)\text{\bold{k}}\big]=0[/tex]

Now, we can calculate our reactions.

Calculate the reaction at A in the x-direction.

[tex]\sum F_x=0\\\\A_x+\frac{1}{3}\times529.74-\frac{2}{3}\times264.87=0\\\\\boxed{A_x=0\text{ N}}[/tex]

Calculate the reaction at A in the y-direction.

[tex]\sum F_y=0\\\\A_y-\frac{2}{3}\times529.74-\frac{2}{3}\times264.87=0\\\\A_y-529.74=0\\\\\boxed{A_y=529.74\text{ N}}[/tex]

Calculate the reaction at A in the z-direction.

[tex]\sum F_z=0\\\\A_z+\frac{2}{3}\times529.74+\frac{1}{3}\times264.87-981=0\\\\A_z-441.45=0\\\\\boxed{A_z=441.45\text{ N}}[/tex]

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Answer:

[tex]\boxed{\text{A=} < \text{0, 529.74, 441.45} > }[/tex]

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The mass of sugar cane being fed to the process with this flowrate is 0.0014 kg/s.

Flowrate is the rate at which a fluid or gas passes through a given space or container over a period of time. It is usually expressed in terms of volume per unit of time, such as liters per second or gallons per minute. Flowrate is an important factor in many engineering and scientific applications, such as fluid dynamics, hydroelectric power generation, and chemical processing. It is also used to measure the flow of liquids and gases through pipes, valves, and other components of a system.

The mass of sugar cane being fed to the process can be calculated using the following equation:
Mass (kg/s) = Flowrate (m^3/s) * Density of the mixture (kg/m^3)
The flowrate can be calculated using the equation given:
q(m^3/s) = 0.0307 m^3/hr atm^1/2 √ Pmanometer
Therefore, the flowrate is:
q(m^3/s) = 0.0307 m^3/hr * (atm^1/2 * (6.3 in/12 in/ft)^1/2)
q(m^3/s) = 0.0015 m^3/s
The density of the mixture can be calculated using the following equation:
Density of the mixture (kg/m^3) = Mass of sugar (kg) / Volume of the mixture (m^3)
The mass of sugar can be calculated using the following equation:
Mass of sugar (kg) = Mass of the mixture (kg) * (Mass fraction of sugar (kg/kg) / Mass fraction of the mixture (kg/kg))
The mass of the mixture can be calculated using the following equation:
Mass of the mixture (kg) = Volume of the mixture (m^3) * Density of the mixture (kg/m^3)
The volume of the mixture can be calculated using the following equation:
Volume of the mixture (m^3) = Flowrate (m^3/s)
The mass fraction of sugar can be calculated using the following equation:
Mass fraction of sugar (kg/kg) = Mass of Sugar (kg) / Mass of the mixture (kg)
The mass fraction of the mixture can be calculated using the following equation:
Mass fraction of the mixture (kg/kg) = Mass of the mixture (kg) / Mass of the mixture (kg)
Substituting the values into the equation, we get:
Mass (kg/s) = 0.0015 m^3/s * (0.91 kg/m^3)
Mass (kg/s) = 0.0014 kg/s
Therefore, the mass of sugar cane being fed to the process with this flowrate is 0.0014 kg/s.

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