if a pump remove 1.65 gal/min of water from a tank, how long will it take to empty the tank if it contains 7425 lb of water? (specific weight of water is 62.4 lb/ft3)

Answer:

A

Explanation:

A problem with Cat 5 cable must be fixed by a technician. When the cable, which is attached to a network jack, is moved around, it breaks. The technician must use the RJ45 connector and Crimper to resolve the issue.

Ethernet cables, also known as category 5 (5e) or 6) are used to support computer networks. In addition to standard computer data, it is able to transmit video and telephone signals. Hosted VoIP typically only requires a cat 5 connection. 10/100 Mbps Ethernet, also known as Fast Ethernet, was introduced by at5 Ethernet over distances of up to 100 meters. Even though CAT5 cable is still used in some older deployments, it is now considered out of date and has been replaced by Cat5e.

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Phillips Head Screwdriver can be used to loosen or tighten crosshead screws. These instruments are further explained in details below.

A crosshead screw is simple metal machine that can be used to fasten one object to another. They have an X–shaped slot at the head of the screw where the screwdriver is inserted.

The Philips head screwdriver is a perfect instrument that can be used to loosen or tighten crosshead screws when fastening an object onto another one.

The Philips head screwdriver should be the same as the width of slotted screw head in order to fit onto the head of the screw.

Therefore, Phillips Head Screwdriver can be used to loosen or tighten crosshead screws.

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The surface area to volume (S/V) ratio (the three dimensional extrapolation of the perimeter to area ratio) is an important factor determining heat loss and gain. The greater the surface area the more the heat gain/ loss through it. So small S/V ratios imply minimum heat gain and minimum heat loss.

To minimize the losses and gains through the fabric of a building a compact shape is desirable. The most compact orthogonal building would then be a cube. This configuration, however, may place a large portion of the floor area far from perimeter daylighting. Contrary to this, a building massing that optimizes daylighting and ventilation would be elongated so that more of the building area is closer to the perimeter. While this may appear to compromise the thermal performance of the building, the electrical load and cooling load savings achieved by a well-designed daylighting system will more than compensate for the increased fabric losses.

In hot dry climates S/V ratio should be as low as possible as this would minimize heat gain. In cold-dry climates also S/V ratios should be as low as possible to minimize heat losses. In warm-humid climates the prime concern is creating airy spaces. This might not necessarily minimize the S/V ratio. Further, the materials of construction should be such that they do not store heat.

The surface area can be important for that because for a package design the people need to know like the sizes and measurements and other things and surface area includes some of that stuff so it can be important.

Deep foundations are used when the surface soils are not capable of supporting the loads of a structure. Two common types of deep foundations are replacement and displacement foundations.

1. Replacement foundations:

Replacement foundations are created by removing the soil beneath a structure and replacing it with concrete or other materials. This creates a solid, stable foundation that can support the weight of the structure.

Illustration: The process of creating a replacement foundation begins by excavating the soil beneath the structure, leaving a large hole. The hole is then filled with concrete, which is allowed to cure and harden. The concrete provides a solid base for the structure to rest on, ensuring that it is stable and secure.

2. Displacement foundations:

Displacement foundations work by pushing soil aside as they are installed, creating a cavity in the ground that is then filled with concrete. These foundations are often used when soil conditions are too difficult to excavate or when a more efficient installation method is desired.

Illustration: The process of creating a displacement foundation involves driving a steel pile or tube into the ground. The pile is typically filled with concrete as it is being driven, creating a cavity in the soil. Once the pile has reached the desired depth, it is left in place, and the cavity is filled with concrete. This creates a solid, stable foundation that can support the weight of the structure.

In summary, both replacement and displacement deep foundations are effective methods for creating stable foundations for structures when the soil conditions are not suitable for shallow foundations. The choice of foundation type depends on the specific site conditions, load requirements, and project budget.

When driving, it is recommended to place your hands on the steering wheel at the "9 and 3" or "8 and 4" positions for optimal control and safety.

When driving, it is recommended to place your hands on the steering wheel at the "9 and 3" or "8 and 4" positions. These positions refer to the clock face, with the left hand at the 9 or 8 o'clock position and the right hand at the 3 or 4 o'clock position.

This hand placement allows for optimal control and maneuverability of the vehicle. Placing your hands in these positions also helps to minimize the risk of injury in the event of a deploying airbag. It is important to maintain a comfortable grip on the wheel while keeping your thumbs outside of the steering wheel spokes to prevent potential hand injuries.

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Drilling mud does all of the following EXCEPT keeps oil underground.

Drilling mud is a specialized fluid used in drilling operations to facilitate the drilling process and maintain the stability of the wellbore.

One of the functions of drilling mud is to cool the drill bit by circulating it through the bit and carrying away heat generated during drilling. This helps prevent the bit from overheating and prolongs its lifespan.

Drilling mud acts as a barrier, preventing the gas from entering the wellbore and causing blowouts. It helps control the pressure and allows for safe drilling operations.

Drilling mud has several important functions in drilling operations, including cooling the bit, allowing gas to escape, and flushing cuttings out of the hole. However, it is not intended to keep oil underground.

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Answer:

Efficiency is defined as any performance that uses the fewest number of inputs to produce the greatest number of outputs. Simply put, you're efficient if you get more out of less.

Explanation:

All of the following statements with respect to manual transmission are true EXCEPT releasing the clutch pedal closes the contacts of the clutch safety switch.

Manual automobiles are useful due to the fact they normally value much less and are extra fuel-efficient. You can also see decreased upkeep prices and coverage prices. They additionally provide you higher management of the car for the reason that driver, now no longer the car, is in fee of converting gears.

With the pedal totally released, there's complete touch between the engine and the driveshaft, through the take hold of the plate, because of this the engine can observe energy immediately to the driveshaft. However, it's miles feasible to have the take hold of plate partly engaged, permitting the take hold of to slip.

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Option A holds the correct answer. Because the statement given in option A truly reflects the radial load and axial load.

When the load pressure is perpendicular to the axis of the shaft, a radial load is applied and when the load pressure is parallel to the axis of the shaft, an axial load is applied. In other words, the radial load is applied when the pressure from the load is 'perpendicular to the axis of the shaft'. In contrast, the axial load is applied when the pressure from the load is 'parallel to the axis of the shaft'.

However, the rest of the statements are not correct in the context of radial loads and axial loads.

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An airplane flying across the sky experience drag force determined by the factors including the speed of flight, coefficient of skin friction and the reference surface area

The boundary layer thickness is approximately 0.233 cm

The total skin friction drag, is approximately 265 N

Reason:

First part:

Given parameters are;

Chord length, L = 3 m

Velocity of the plane, V = 200 m/s

Density of the air, ρ = 1.225 kg/m³

Viscosity of the air, μ = 1.81 × 10⁻⁵ kg/(m·s)

The Reynolds number is given as follows;

\(R_{eL} = \dfrac{\rho \times V \times L}{\mu}\)

Therefore;

\(R_{eL} = \dfrac{1.255 \times 200 \times 3}{1.81 \times 10^{-5}} = 4.16022099448 \times 10^7 \approx 4.16 \times 10^7\)

Boundary layer thickness, \(\delta_L\), for laminar flow, is given as follows;

\(\dfrac{ \delta_L }{L}=\dfrac{5.0}{\sqrt{R_{eL} } }\)

\({ \delta_L }=\dfrac{5.0 \times L}{\sqrt{R_{eL} } }\)

Which gives;

\({ \delta_L }=\dfrac{5.0 \times 3}{\sqrt{4.16 \times 10^{7}} } \approx 2.33 \times 10^{-3 }\)

The boundary layer thickness, \(\delta_L\) ≈ 2.33 × 10⁻³ m = 0.233 cm

Second Part

The total skin friction is given as follows;

\(Dynamic \ pressure, q = \dfrac{1}{2} \cdot \rho \cdot V^2\)

Therefore;

\(q = \dfrac{1}{2} \times 1.225 \times 200^2 = 24,500\)

The dynamic pressure, q = 24,500 N/m²

Skin friction drag coefficient, \(C_D\), is given as follows;

\(C_D = \dfrac{1.328}{\sqrt{R_{eL} } }\)

Therefore;

\(C_D = \dfrac{1.328}{\sqrt{4.16 \times 10^7 } } \approx 2.06 \times 10^{-4}\)

Skin friction drag, \(D_f\), is given as follows;

\(D_f\) = q × \(C_D\) × A

Where;

A = The reference area

∴ \(D_f\) = 24,500 N/m² × 2.06 × 10⁻⁴ × 3 m × 17.5 m = 264.9675 N ≈ 265 N

The total skin friction drag, \(D_f\) ≈ 265 N

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Wood is a composite type of CMC. What is Wood. Wood is a natural composite that is formed from cellulose fibers embedded in a matrix of lignin.

It is a natural material that has been used for thousands of years to make structures, furniture, and other useful items. What are Composites. Composite materials are made up of two or more materials with distinct and they can be made by combining any type of material. The goal of combining different materials is to create a new material with better overall properties than any of the individual components. Wood is made up of cellulose fibers embedded in a matrix of lignin. Wood, which is a natural composite, is an example of a composite made up of organic materials. CMC (Ceramic matrix composites) and MMC (Metal matrix composites) are the other two composite types.
Wood is a natural composite material and does not fall under the categories of (a) CMC (Ceramic Matrix Composite), (b) MMC (Metal Matrix Composite), or (c) PMC (Polymer Matrix Composite).

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Answer:

Energy Saved = 6.93 x 10⁹ Btu

Cost Saved = $ 30145.5

Explanation:

The energy generated by each boiler can be given by the following formula:

\(Annual\ Energy = (Heat\ In)(Combustion\ Efficiency)(Operating\ Hours)\)

Now, the energy saved by the increase of efficiency through tuning will be the difference between the energy produced before and after tuning:

\(Energy\ Saved = (Heat\ In)(Efficiency\ After\ Tune - Efficiency\ Before\ Tune)(Hours)\)\(Energy\ Saved = (5.5\ x\ 3\ x\ 10^{6}\ Btu/h)(0.8-0.7)(4200\ h)\)

Energy Saved = 6.93 x 10⁹ Btu

Now, for the saved cost:

\(Cost\ Saved = (Energy\ Saved)(Unit\ Cost)\\Cost\ Saved = (6.93\ x\ 10^{9}\ Btu)(\$4.35/10^{6}Btu)\\\)

Cost Saved = $ 30145.5

Answer:

hey. its a big question. solved from *c hegg

Explanation:

Answer:

Resistance of copper = 1.54 * 10^18 Ohms

Explanation:

Given the following data;

Length of copper, L = 3 kilometers to meters = 3 * 1000 = 3000 m

Resistivity, P = 1.7 * 10^8 Ωm

Diameter = 0.65 millimeters to meters = 0.65/1000 = 0.00065 m

\( Radius, r = \frac {diameter}{2} \)

\( Radius = \frac {0.00065}{2} \)

Radius = 0.000325 m

To find the resistance;

Mathematically, resistance is given by the formula;

\( Resistance = P \frac {L}{A} \)

Where;

First of all, we would find the cross-sectional area of copper.

Area of circle = πr²

Substituting into the equation, we have;

Area  = 3.142 * (0.000325)²

Area = 3.142 * 1.05625 × 10^-7

Area = 3.32 × 10^-7 m²

Now, to find the resistance of copper;

\( Resistance = 1.7 * 10^{8} \frac {3000}{3.32 * 10^{-7}} \)

\( Resistance = 1.7 * 10^{8} * 903614.46 \)

Resistance = 1.54 * 10^18 Ohms

Here's a for loop that populates an array in C++ with NUM_GUESSES integers:```
#include
using namespace std;
int main() {
 const int NUM_GUESSES = 3;
 int userGuesses[NUM_GUESSES];
 int i = 0;
 for (i = 0; i < NUM_GUESSES; ++i)

{
   cin >> userGuesses[i];
 }
 for (i = 0; i < NUM_GUESSES; ++i)

{
   cout << userGuesses[i] << " ";
 }
 return 0;
}
```This program will ask the user to enter three integers and store them in the array `userGuesses`. The `for` loop runs `NUM_GUESSES` times (in this case, 3 times), and each time it prompts the user to enter an integer using `cin`, and stores it in the array at the current index (`userGuesses[i]`). Finally, it loops through the array again and prints out each integer that was entered by the user, separated by spaces.The output will look like this if the user enters 9, 5, and 2:```952```

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Answer:0.00234

Explanation:

The Moody chart is a graphical representation used to determine the friction factor in fluid dynamics for laminar and turbulent flow in pipes.

The Moody chart uses the Reynolds number (a dimensionless quantity that describes the flow regime of the fluid) and the relative roughness of the pipe (the ratio of the pipe's roughness to its diameter) as inputs. The chart itself consists of multiple curves representing different levels of relative roughness, with the friction factor on the y-axis and the Reynolds number on the x-axis. For laminar flow (Reynolds number less than 2000), the friction factor can be calculated directly using the formula f = 64/Re. For turbulent flow, one locates the Reynolds number and the relative roughness on the chart, follows these values until they intersect, and reads the corresponding friction factor from the y-axis.

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The decomposition of Liquid A follows a first-order kinetics. It means that the rate of reaction is proportional to the concentration of A present at any given time.

The rate constant for the reaction is k. The formula for the rate of a first-order reaction is given as follows:r = k[A]where, r is the rate of reaction, and [A] is the concentration of A at any given time.The time taken for the conversion of 50% of A is given as 5 minutes. The concentration of A remaining is 50% of the initial concentration. The rate of reaction at this point is:r = k[0.5 A0]where, A0 is the initial concentration of A.

Since the reaction follows a first-order kinetics, the rate constant k will remain constant throughout the reaction.To calculate the time taken for the conversion of 75% of A, we can use the following equation:ln ([A]t/[A]0) = -ktwhere, [A]t is the concentration of A remaining after time t, and [A]0 is the initial concentration of A. We know that [A]t = 0.25[A]0.Substituting these values, we get:ln (0.25) = -k(t2 - t1)where, t1 = 5 minutes (time taken for 50% conversion), and t2 is the time taken for 75% conversion.Solving for t2, we get:t2 = t1 + (1/k) ln(0.25)Substituting the value of k from the rate equation, we get:t2 = 5 + (1/k) ln(0.25 [A]0)Therefore, we need to know the value of the rate constant k to calculate the time taken for 75% conversion.

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Correct question is;

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at −30°C by rejecting its waste heat to cooling water that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The refrigerant enters the condenser at 1.2 MPa and 65°C and leaves at 42°C. The inlet state of the compressor is 60 kPa and −34°C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the mass flow rate of the refrigerant.

Answer:

A) Quality = 0.48

B) Mass flow rate; m' = 0.0455 kg/s

Explanation:

A) From the refrigerant R-144 table I attached,

At P=60kpa and interpolating at - 34°C,we obtain enthalpy;

h1 = 230.03 Kj/kg

Also at P= 1.2MPa which is 1200kpa and interpolating at 65°C,we obtain enthalpy ;

h2 = 295.16 Kj/Kg

Also at P= 1.2MPa which is 1200kpa and interpolating at 42°C,we obtain enthalpy ;

h3 = 111.23 Kj/Kg

h4 is equal to h3 and thus h4 = 111.23 Kj/kg

We want to find the refrigerant quality at the evaporation inlet which is state 4 and P= 60 Kpa.

Thus, from the table attached, we see that hf = 3.84 at that pressure and hg = 227.8

Now, to find the quality of the refrigerant, we'll use the formula,

x4 = (h4 - hg) /(hf - hg)

Where x4 is the quality of the refrigerant. Thus;

x4 = (111.23 - 3.84)/(227.8 - 3.84) = 0.48

B) The mass flow rate of the refrigerant can be determined by applying a 1st law energy balance across the condenser. Thus, the water properties can be obtained by using a saturated liquid at the given temperatures;

So using the first table in the image i attached; interpolating at 18°C; hw1 = hf = 75.54 kJ/kg

Also interpolating at 26°C; hw2 = hf = 109.01 kJ/kg

Now;

(m')(h2 − h3)= (m_w)(hw2 − hw1)

m' is mass flow rate

Making m' the subject, we get;

m' = [(m_w)(hw2 − hw1)]/(h2 − h3)

m' = [(0.25 kg/s)(109.01 − 75.54) kJ/kg] /(295.13 − 111.37) kJ/kg

m' = 8.3675/183.76

m' = 0.0455 kg/s

Of the three types of hazard controls – engineering, administrative, and personal protective equipment (PPE) – PPE is the last line of defense: True.

OSHA is simply an abbreviation for occupational safety and health administration that was created under the Occupational Safety and Health Act, so as to develop work-safety standards for workers and providing precautionary measures against all workplace hazards.

Generally, there are three types of hazard controls and these include the following:

PPE is an acronym for personal protective equipment and it can be defined as a terminology that is used to denote any piece of equipment which offer protection to different parts of the body while working in a potentially hazardous environment.

Also, some examples of personal protective equipment (PPE) used to protect the different parts of the body are:

According to occupational safety and health administration (OSHA), the use of a faulty personal protective equipment (PPE) should be the last line of defense.

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Complete Question:

Of the three types of hazard controls – engineering, administrative, and personal protective equipment (PPE) – PPE is the last line of defense. True or False?

Answer:

Explanation: 2 is thy answer

The CIA triad and the Parkerian hex are the fundamental principles of information security.

Parkeriano, or Parkerian hexad: is a set of six elements of information security proposed by Donn B. Parker.

1. Confidentiality.

2. Ownership or Control.

3. Integrity.

4. Authenticity.

5. Availability.

6. Utility

The Parkerian hexagram adds three more attributes to the three classic security attributes of the CIA triangle

these are the fundamental principles of information security.

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Answer:

UUUUUUMMMM do you mean in soccer ????????????????

Explanation:

Telnet and SSH are both remote management systems that enable users to access and control remote computers over a network.

Explanation:

Telnet is a remote management system that operates on the application layer of the OSI model. It is a simple and lightweight protocol that enables users to establish a connection and interact with a remote computer using a command-line interface. However, because Telnet does not provide encryption, it is considered insecure, and data transmitted over Telnet can be intercepted and read by malicious actors.

SSH, or Secure Shell, is a more advanced remote management system that provides encryption, authentication, and data integrity. SSH encrypts all data transmitted between the client and the server, making it difficult for attackers to intercept and read sensitive information. It also provides authentication mechanisms, such as passwords and public-key cryptography, to ensure that only authorized users can access the remote computer. Additionally, SSH uses digital signatures to ensure data integrity, preventing malicious actors from altering the data in transit. As a result, SSH is considered a more secure protocol than Telnet and is widely used in modern network environments.

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Answer:

say in English plz I don't understand

Answer:

These formal systems state exactly what work is to be done, where and when. In many organisations, people doing the actual work are required to sign a permit to work document to confirm that they have understood all processes and guidelines of the system. Permits also serve as a means of communication between site workers and the organisation’s management.

However, they should not be mistaken as a replacement for robust risk assessment, which should happen simultaneously. Having a permit to work systems as part of organisational operations reduces the risks of losses or damage due to unsafe activities in a non-trivial work atmosphere.

These systems ensure that the correct jobs are entrusted to appropriate individuals and review or audit of conditions are also undertaken by experts or authorised personnel who have proven capabilities of undertaking the respective responsibilities. Permit to work systems also confirms that work is completed safely and the workplace is restored to its safe, original form for normal operations to commence.