An electronics engineer is interested in the effect on tube conductivity of five different types of coating for cathode ray tubes in a telecommunications system display device. The following conductivity data are obtained. Coating type | Conductivity 1 143 141 150 146
2 152 149 137 143
3 134 133 132 127
4 129 127 132 129
5 147 148 144 142
Is there any difference in conductivity due to coating type? Use a = 0.01. Approximate your answer up to 2 decimal places. F0 = ___

Answer:

B!

Explanation:

Answer:

B. Flying solo across the atlantic

Explanation:

history class

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 number of nodes present in the given circuit is the number of points where two or more components are connected together.

The number of branches present in the given circuit is the number of individual paths between the nodes.

To further elaborate on the topic, nodes and branches are essential components of an electrical circuit, as they are essential for the flow of current throughout the circuit. Nodes represent the points in a circuit where two or more components are connected together, while branches represent the individual paths that the current can take between the nodes.

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Answer:true

Explanation:

It is true that around 50 percent of all sales of farm commodities are from dairy products.

Dairy products or milk products, also known as lacticinia, are milk-based foods.

Cows, water buffalo, nanny goats, and ewes are the most common dairy animals. Dairy products include common grocery store foods such as yogurt, cheese, and butter in the Western world.

Milk, yogurt, cheese, lactose-free milk, and fortified soy milk and yogurt are all part of the Dairy Group. The Dairy Group excludes milk-based foods with low calcium and a high fat content. Cream cheese, sour cream, cream, and butter are all examples of this.

Dairy products account for roughly half of all farm commodity sales.

Thus, the given statement is true.

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Answer:

yes maybe

Explanation:

I have a project and everyone in the world will know about it.

Answer:

both

Explanation:

In plotting grain-size distribution curves, it is preferable to use a logarithmic scale for the grain diameter instead of an arithmetic scale.

This is because grain sizes can span several orders of magnitude, and plotting them on a logarithmic scale allows for a more accurate representation of the distribution, as well as making it easier to compare different distributions.On an arithmetic scale, the distances between points increase linearly with the diameter, meaning that the smallest and largest grain sizes can appear to be much more widely spaced than they actually are. On a logarithmic scale, the distances between points increase exponentially with the diameter, meaning that the relative distribution of grain sizes is more accurately represented.

The shapes of grain-size distributions plotted on an arithmetic scale can differ from those plotted on a logarithmic scale, as the representation of the relative proportions of different grain sizes can be altered. The shapes of distributions plotted on a logarithmic scale, however, are comparable and can be used to make meaningful comparisons between different distributions. The presence or absence of peaks, for example, and the positions of those peaks, can be used to compare distributions and make inferences about the processes that formed the sediments.

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It is right to state that Technician B is correct. A piston compresses the air in the cylinder, causing it to become very hot. The diesel is then atomized in the injectors and sprayed into the heated air as a mist. The heated air ignites the gasoline quickly, creating ignition.

As the fuel injectors drive air and fuel into the combustion chamber, compression occurs in the internal combustion cylinders.

The mixture ignites, and the piston is driven by the expansion of the burning gases in the cylinders, converting the energy from combustion into mechanical energy that propels the vehicle.

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Before joining a fitting and pipe using socket fusion, it is important to ensure that both the fitting and pipe are clean and free from any debris or contaminants. Additionally, the correct size fitting and pipe must be used to ensure a proper fit.

Socket fusion is a method of joining plastic pipes and fittings together by heating the material and then pressing the heated ends together to form a strong bond. Before the socket fusion process begins, it is important to prepare both the fitting and pipe by ensuring they are clean and free from any debris or contaminants.

This can be achieved by using a specialized cleaning tool or wiping the surfaces with a clean cloth. Additionally, it is crucial to use the correct size fitting and pipe to ensure a proper fit and prevent any leaks or issues in the future. Proper preparation of the materials is crucial to ensuring a successful socket fusion joint.

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To display the user agent in a `

` tag in the webpage, the `writeln()` method of the document object can be used. The `navigator.userAgent` property of the `window.navigator` object contains the user agent. The code below demonstrates how to do this.

To display the user agent in a `

` tag in the webpage, the following steps are required:1. Create a `

` element and assign it an id to make it easier to reference. For example, `

`2. Use the `writeln()` method of the document object to display the user agent in the `

` element. The `navigator.userAgent` property of the `window.navigator` object can be used to get the user agent. The code below demonstrates how to do this.document.writeln(`

${window.navigator.userAgent}

`);Note that the backticks (`) and ${} notation are used to interpolate the value of `window.navigator.userAgent` into the `

` element.

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The printed parameter-passing methods are passed by value.

Pass-by-value and pass-by-reference are the most widely used. While most other languages (including Java, JavaScript, Python, etc.) only employ pass-by-value, C++ supports both approaches.

Passing arguments by reference has the advantage over passing arguments by value in that changes made to arguments passed in by reference in the called function affect the calling function, whereas changes made to arguments passed in by value in the called function have no effect on the calling function.

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Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

Answer:

Both are correct.

Explanation:

Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.

The minimum diameter required for the wire is 0.755 millimeters.

Hooke's Law states that the force applied to a material is directly proportional to the strain (change in length) of the material, given a constant called the Young's modulus (Y).

F = Y × A × (ΔL / L)

Where:

F = Force applied (350 N)

Y = Young's modulus for steel (2.00 x 10¹¹ N/m²)

A = Cross-sectional area of the wire

ΔL = Change in length (0.270 cm = 0.00270 m)

L = Original length of the wire (2.08 m)

The cross-sectional area (A) of a wire with a circular cross-section can be calculated using the formula:

A = π × (d² / 4)

Where:

d = Diameter of the wire

Now, we can rearrange the equation for Hooke's Law to solve for the diameter (d):

d = √((4 × F× L) / (π× Y×ΔL))

Plugging in the given values, we have:

d = √((4 × 350 × 2.08) / (π ×2.00× 10¹¹ × 0.00270))

Calculating this equation gives us the minimum diameter (dmin) required for the wire. Let's compute it:

dmin = 0.755 mm

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A steel wire of length 2.08m with circular cross section must stretch no more than 0.270cm when a tensile (stretching) force of 350N is applied to each end of the wire. Part A What minimum diameter (dmin) is required for the wire? Express your answer in millimeters. Take Young's modulus for steel to be Y = 2.00

Answer:

Both technician A and technician B are correct.

Explanation: Vehicle manufacturers always specify the size of the tires required for a given vehiclefor optimal efficiency,this will ensure that the speedometer is accurate and the level of the vehicle is good enough to ensure the vehicle works efficiently.

It is also a known fact that an increase in a vehicle's rpm(revolution per minute) will eventually lead to increased fuel consumption which means the fuel economy of the vehicle will be reduced making the vehicle less efficient in its fuel consumption.

The bath tub vortex shown in Figure 4 has a rotational zone that extends up to a radius of r = 0. 6 meters and has a variable tangential velocity of the fluid under rotation.

This page is about the line segment. A bone is a radius; see Bone . In other contexts, the radius of a circle or sphere is any line segment that, in classical geometry, connects the object's center to its perimeter; in more modern usage, it also refers to the length of such line segments. The Latin origin of the term "radius" gives it the meanings "ray" and "the spoke of a chariot wheel." The term radius can be pluralized using radii or the standard English radii. The most common abbreviations and names for the mathematical variable radius are R and r. By extension, the diameter D and the radius R are equal. The phrase could relate to an absence of a center.

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The local isentropic stagnation conditions in front of the shock and estimate the stagnation pressure sensed by the pitot tube.

a) To evaluate the local isentropic stagnation conditions in front of the shock, we can use the isentropic relations for a perfect gas. The isentropic relations relate the properties of a gas across a shock wave. Given the altitude of 12 km and the provided pressure and temperature of the surrounding air (19.4 kPa and -56.45 °C), we can calculate the local isentropic stagnation conditions.

First, we need to convert the temperature from Celsius to Kelvin:

T = -56.45 °C + 273.15 = 216.7 K

Using the ideal gas equation, we can calculate the density of the surrounding air:

ρ = P / (R * T)

Where P is the pressure, R is the specific gas constant, and T is the temperature.

For air, the specific gas constant R is approximately 287 J/(kg·K).

ρ = 19.4 kPa / (287 J/(kg·K) * 216.7 K)

After performing the calculation, we obtain the density of the surrounding air.

Now, using the isentropic relations, we can determine the isentropic stagnation conditions ahead of the shock. These conditions can be obtained by relating the Mach number (M) and the local conditions (P, ρ, T) to the isentropic stagnation conditions (P0, ρ0, T0).

The specific heat ratio (gamma) for air is approximately 1.4.

M0 = M * √(γ * R * T0 / (2 * γ * R * T))

Where M0 is the isentropic Mach number and T0 is the isentropic stagnation temperature.

Using this equation, we can solve for T0 and calculate the isentropic stagnation temperature.

Similarly, we can calculate the isentropic stagnation pressure (P0) using the relation:

P0 = P * (1 + ((γ - 1) / 2) * M^2)^(γ / (γ - 1))

By substituting the known values, including the calculated density (ρ), pressure (P), and temperature (T), we can obtain the isentropic stagnation pressure sensed by the pitot tube.

b) To estimate the stagnation pressure sensed by the pitot tube, we can consider that the pitot tube measures the stagnation pressure, which is the total pressure (P0) ahead of the shock. Therefore, the calculated isentropic stagnation pressure (P0) from part a) represents the stagnation pressure sensed by the pitot tube.

By following these calculations, we can evaluate the local isentropic stagnation conditions in front of the shock and estimate the stagnation pressure sensed by the pitot tube.

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Answer:

7.5 feet

Explanation:

45 divided by 6 is 7.5

Answer:

7.5 feet

Explanation:

45 divided by 6 is 7.5

Answer:

Emails and phone calls are both common forms of communication that are used in professional and personal settings. There are several similarities between email and phone calls:

1. Both are asynchronous forms of communication: Unlike instant messaging or face-to-face conversations, both emails and phone calls allow the sender or recipient to respond at their convenience. They don't require immediate attention or an instant response.

2. Both are written forms of communication: While phone calls rely on spoken words, emails are written. As a result, both can be used to convey detailed information and allow the sender to carefully consider their words before sending.

3. Both are forms of direct communication: Emails and phone calls both allow for direct communication between two parties. This can be beneficial for discussing sensitive information or resolving issues quickly.

4. Both can be used for formal and informal communication: Emails and phone calls can be used in both personal and professional contexts. They are both flexible forms of communication that can be adapted to fit different situations.

5. Both require attention to tone and etiquette: Just like with phone calls, emails require attention to tone and proper etiquette. Both forms of communication should be approached professionally and respectfully to ensure effective communication.

In conclusion, while there are differences between emails and phone calls, there are also similarities that make them useful communication tools. Both allow for direct, asynchronous communication and can be adapted to fit different situations.

Explanation:

Fault location techniques are used in power systems for accurate pinpointing of the fault position.

This paper presents a comparative study between two fault location methods in distribution network with Distributed Generation (DG). Both methods are based on computing the impedance using fundamental voltage and current signals. The first method uses one-end information and the second uses both ends

Answer:

490000 dollars

Explanation:

Answer:

yes

Explanation:

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Answer:

hello some parts of your question is missing attached below is the missing part

answer :

A) Determine changes in the 50-mm dimensions

The changes are : 0.006mm compression  in y-direction

                               0.002 mm expansion in x and z directions

B) the stress required are evenly distributed

Explanation:

Given data :

50-mm cube of graphite fiber reinforced polymer matrix

subjected to 125-KN force in direction 2,

direction 2 is perpendicular to fiber direction ( direction 1 ) and cube is constrained against expansion in direction 3

A) Determine changes in the 50-mm dimensions

The changes are : 0.006mm compression  in y-direction

                               0.002 mm expansion in x and z directions

B) the stress required are evenly distributed

attached below is the detailed solution

Answer:

This is not a clear indication of the hot zone as the information of the radioactivity of the background is not provided clearly.

Explanation:

According to IAEA as well as NRCP, the hot area is defined on the basis of the radioactivity reading it shows instead of contrast or comparative reading from the background. The value of radiation activity which will be required to declare an area as hot zone is if it is greater than 0.1 mSv/h or \(1.5091\times 10^{29} kg^{-1} s^{-1}\).

Answer:

O The mix is uniform in color

The initial enthalpy and the entropy of the saturated water can be found out from the table of A-12E

i.e.   \($h_1= 101 \ \text{Btu/lbm}$\)

       \($s_1 = 0.22717 \text{ Btu/lbmR}$\)

Since the process mentioned above is an adiabatic compression process, the entropy will remain constant throughout the process. Therefore, we take the value of entropy and the final pressure using the table with few interpolations and also approximations to find the final enthalpy. It is given by :

       \($h_2= 116.09 \text{ Btu/lbm}$\)

So the work input from the energy balance equation :

       \($\dot{W} + \dot{m}h_1 = \dot{m}h_2$\)

          \($w=h_2 - h_1$\)

              = 116.09 - 101

              = 15.09

Therefore, \($w= 15.09 \text{ Btu/lbm}$\)

Answer:

Calcium Chloride

Brine Wash

Drierite

Explanation:

Removal of small amounts of water from a polar solvent is Calcium Chloride

Removal of visible pockets of water from an organic solvent is Brine Wash

Storage of solvents or other materials in a desiccator is Drierite