Technician A says engine vacuum is normally measured in inches of mercury (in. Hg). Technician B says a pressure gauge that is labeled psig will read 14.7 psi at sea level. Who is correct

If your supervisor is texting you during work hours for no reason, you should speak to him or her about it. If your supervisor is inviting you to dinner to discuss your salary increase issue, you should tell him or her that you are not comfortable with that and would prefer to discuss it during work hours. Your supervisor's behaviour is inappropriate and violates the Ontario's Occupational Health and Safety Act ("OHSA"). You should discuss the issue with your supervisor and try to resolve the issue. If you are not able to resolve the issue, you can file a complaint with the Ontario Ministry of Labour.

Occupational Safety and Health Act:

The Occupational Safety and Health Act of 1970 is a United States labour legislation that governs federal occupational health and safety law in the private sector and federal government. Congress passed it in 1970, and President Richard Nixon signed it on December 29, 1970.

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Members BD, CE, and CD each had a force of 0 kN, 8.75 kN, and 6.25 kN.

What is force ?

A force is an effect that can alter an object's motion according to physics. A force can cause an object has mass to accelerate when it changes its velocity, for as when it moves away from rest. A way to describe force is as a push or a pull. A force is vector quantity because it has magnitude and direction. It is calculated using the newton (N). Force is denoted by the letter F. The net force acting on an object is equal to the speed at which its momentum varies over time, according to Second law of Newton in its original formulation.

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Answer:

busway

Explanation:

Answer:

There are 256 8-bit strings in total, and 128 of them contain at least one "1" (half of the total number of strings).

Answer:

up up down down

Explanation:

left right left right b a select start

Answer:

What are u asking for?

Explanation:

The way that chain branching affects a) degree of crystallinity, b) strength, and c) elongation of polyethylene is given below.

Polyethylene is a type of polymer that is made up of long chains of repeating units of ethylene. The properties of polyethylene can be influenced by the size and shape of these chains, as well as the way that the chains are organized within the polymer. One way to alter the properties of polyethylene is to introduce chain branching into the polymer.

a) Degree of crystallinity: Chain branching can have a significant effect on the degree of crystallinity of polyethylene. Crystallinity refers to the degree to which the polymer molecules are organized in a regular, repeating pattern. In general, the more crystalline a polymer is, the stronger and stiffer it will be. In unbranched polyethylene, the molecules are able to pack together tightly, which leads to a high degree of crystallinity.

b) Strength: The strength of a polymer is related to its degree of crystallinity. As a result, the strength of polyethylene will typically be lower when chain branching is present. This is because the chain branches disrupt the regular, repeating pattern of the polymer molecules, making it more difficult for the polymer to resist deformation.

c) Elongation: Elongation refers to the ability of a polymer to stretch or elongate before breaking. Polymers that are more crystalline tend to be less elongation, while polymers that are less crystalline tend to be more elongation. Therefore, when chain branching is present in polyethylene, the polymer will tend to be more elongation. This is because the chain branches disrupt the regular, repeating pattern of the polymer molecules, making it easier for the polymer to stretch.

Therefore, note that properties of Polyethylene can vary depending on the type of polyethylene and the way the polymerization process is carried out, thus the effects of chain branching on it's properties may be different depending on the context.

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Answer:

b

Explanation:

Answer:

\(\triangle h=4.935m\)

Explanation:

From the question we are told that:

Liquid density \(\rho=800\)

Diameter of pipe \(d=4cm \approx 0.004m\)

Diameter of venture \(d=10cm \approx 0.010m\)

Specific weight of mercury P_mg \(133 kN/m^3\)

Flow rate \(r=0.05 m^3/s\)

Area A:

            \(A_1=\frac{\pi}{4}0.1^2\\A_1=0.00785m^2\\A_2=\frac{\pi}{4}0.04^2\\A_2=0.001256m^2\\\)

Generally the Bernoulli's equation is mathematically given by

\(\frac{P_1}{\rho_1g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}\\\)

Where

\(V_1=\frac{r}{A_1} \\\\ &V_1=\frac{r}{A_2}\)

Therefore

\(P_1-P_2=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})\)

Generally the equation for pressure difference b/w manometer fluid is given as

\(P_1-P_2=(p_mg-pg)\triangle h\)

Therefore

\((p_mg-pg)\triangle h=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})\)

\(\triangle h=\frac{\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})}{(p_mg-pg)}\)

\(\triangle h=\frac{\frac{(800)(0.05)^2}{2}(\frac{(0.1)^2-(0.4)^2}{(0.1)^2(0.04)^2})}{(1.33*10^3-800*9.81)}\)

\(\triangle h=4.935m\)

Therefore elevation change is mathematically given by

\(\triangle h=4.935m\)

A software engineer might engage in any of the following activities that could be deemed illegal, immoral, and unethical:
Software piracy, security breaches, data theft, software spying or snooping, creating viruses or other harmful code, intentionally producing defective code, and knowingly releasing software with bugs or security flaws.


Creating software that serves a beneficial social or moral purpose, such as software that promotes environmental sustainability, public safety, and public health. Software that serves the public good is legal, but some software might have unintended or unforeseen ethical implications.

According to the IEEE code of ethics, engineers should support and participate in pro bono and charitable work that promotes the public good. Therefore, as a software engineering company owner, you should allow your employees to do pro bono work on company time as long as it does not interfere with their assigned tasks or affect the company's productivity.

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All three  features save power by reducing the overall power consumption of the microcontroller.

PIC microcontrollers offer a sleep mode that enables them to turn off the majority of their functions while not in use. A PIC microcontroller uses very minimal power while in sleep mode and can be quickly activated as required. Low-Power Timer: The low-power timers used in PIC microcontrollers enable the controller to emerge from sleep mode at specified intervals. Applications that call for recurring measurements or data collecting can benefit from this capability. The controller can be set to be awakened by the timer at predetermined intervals, and while awake, it can take the necessary actions before falling back to sleep.PIC microcontrollers have a feature called Peripheral Module Disable (PMD) that enables the user to switch off certain peripheral modules.

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the answer is 2

kids simple just multiply 3 by 2 which gets the answer as 6 then divide 6 by 2 then it comes to again 2 and that is a simple answer...

The output of the given code, i.e., print( (3**2)//2)​ will be 4.

Coding, also known as computer programming, is the method by which we communicate with computers.

Code tells a computer what to do, and writing code is similar to writing a set of instructions. You can tell computers what to do or how to behave much more quickly if you learn to write code.

When you choose an instructional programming language, coding is simple to learn. It can be difficult to learn to code if you begin with a more complex coding language.

The ** operator is used to multiply a number by a power.

The // operator calculates how many times the right number can fit into the left.

Here it is given that

\(print(3^2)/2\)

So, the round division will be 4.

Thus, the output will be 4.

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The absence of the procedure turn barb on an approach chart's plan view indicates that a procedure turn is not required or authorized.

This means that the pilot is not expected to execute a procedure turn to align the aircraft with the final approach course. Instead, the pilot must use other means to align the aircraft, such as a teardrop or a hold-in-lieu-of-PT procedure. The absence of the procedure turn barb can be found on both RNAV and non-RNAV approach charts. It is important for pilots to carefully review the approach chart and understand the requirements before beginning the approach, as failure to do so can result in a violation of air traffic control instructions or lead to an unsafe situation.

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Answer:

Explanation:

Tech Social Media giant FB is one of those companies. Not long ago the ceo was brought to court to accusations that his company was selling user data. Turns out this is true and they are selling their users private data to companies all over the word. Once the news turned to something else, people focused on something new but the company still continues to sell it's users data the same as before. This is completely unethical as the information belongs to the user and they are not getting anything while the corporation is profiting.

Answer:

the reduction in the heat gain is 2.8358 kW

Given that;

Shaft outpower of a motor  = 75 hp = ( 75 × 746 ) = 55950 W

Efficiency of motor  = 91.0% = 0.91

High Efficiency of the motor  = 95.4% = 0.954

now, we know that, efficiency of motor is defined as;  =  /

where   is the electric input given to the motor

so

=  /

we substitute

= 55950 W / 0.91

= 61483.5 W

= 61.4835 kW

now, the electric input given to the motor due to increased efficiency will be;

=  /

we substitute

= 55950 W / 0.954

= 58647.79 W

= 58.6477 kW

so the reduction of the heat gain of the room due to higher efficiency will be;

Q =  -

we substitute

Q = 61.4835 kW - 58.6477 kW

Q = 2.8358 kW

Therefore, the reduction in the heat gain is 2.8358 kW

Explanation:  i hope this answer your question if this is wrong or correct please let me know.also no trying to be rude but can you sent me like a thanks?

Using the Poisson distribution and the binomial distribution, it is found that:

In a Poisson distribution, the probability that of x successes of a random variable is given by the following equation:

\(P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}\)

The parameters are described as follows:

For one square yard, the mean is:

\(\mu = 0.01\)

Hence the probability that there are no flaws on 1 square yards is:

\(P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}\)

\(P(X = 0) = \frac{e^{-0.01}(0.01)^{0}}{(0)!} = 0.99\)

For 25 square yards, the mean is:

\(\mu = 0.01 \times 25 = 2.5\)

Hence the probability that there are no flaws on 25 square yards is:

\(P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}\)

\(P(X = 0) = \frac{e^{-2.5}(2.5)^{0}}{(0)!} = 0.0821\)

The formula for the probability of x successes is:

\(P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}\)

\(C_{n,x} = \frac{n!}{x!(n-x)!}\)

The parameters are given by:

For the probability that 8 or more of the 10 pieces will have no flaws, the values of the parameters are given by:

p = 0.99, n = 10.

The probability is:

P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10).

In which:

\(P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}\)

\(P(X = 8) = C_{10,8}.(0.99)^{8}.(0.01)^{2} = 0.0042\)

\(P(X = 9) = C_{10,9}.(0.99)^{9}.(0.01)^{1} = 0.0914\)

\(P(X = 10) = C_{10,10}.(0.99)^{10}.(0.01)^{0} = 0.9044\)

Then:

P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.0042 + 0.0914 + 0.9044 = 1 = 100%.

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Answer:

Cell phones emit low levels of non-ionizing radiation when in use. The type of radiation emitted by cell phones is also referred to as radio frequency (RF) energy. As stated by the National Cancer Institute, "there is currently no consistent evidence that non-ionizing radiation increases cancer risk in humans.

Explanation:

The loss of head due to enlargement  is h_loss = k * v^2 / (2 * g)

The difference in  pressure is ΔP = ρ * g * h_loss

The corresponding differences is a gradual change in diameter.

a) Loss of head due to sudden enlargement: The loss of head due to sudden enlargement can be calculated using the formula:

h_loss = k * v^2 / (2 * g)

where k is the loss coefficient (typically taken as 0.5), v is the velocity of the fluid, and g is the acceleration due to gravity.

b) Difference in pressure in KN/m²: The difference in pressure can be calculated using the Bernoulli equation:

ΔP = ρ * g * h_loss

where ρ is the density of the fluid and h_loss is the loss of head.

c) Corresponding differences if change in diameter was gradual: In the case of a gradual change in diameter, there would be no loss of head and hence no pressure difference. The pressure difference is a result of the conversion of kinetic energy to potential energy, which occurs due to the sudden change in the flow area and velocity. In the case of a gradual change, the fluid has time to adjust its velocity and pressure and there is no loss of energy.

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Erik Erikson's psychosocial theory suggests that studying relationships, including friendships and romantic partnerships, is crucial for understanding socioemotional development in early adulthood.

One theoretical perspective that emphasizes the study of

socioemotional development in early adulthood, particularly in the context of relationships, is Erik Erikson's psychosocial theory.

Erikson proposed a series of psychosocial stages that individuals progress through across their lifespan, and one of the key stages relevant to early adulthood is the stage of intimacy versus isolation.

According to Erikson, the primary task during early adulthood is to develop intimate relationships with others while maintaining a sense of identity.

During this stage, individuals seek close and meaningful connections with friends, romantic partners, and potentially even long-term commitments such as marriage or starting a family.

It is through these relationships that individuals can experience companionship, love, and emotional intimacy.

Erikson argued that successfully navigating this stage leads to the development of a capacity for intimacy, which involves the ability to form deep and meaningful connections with others.

However, if individuals struggle to form these relationships or experience significant feelings of isolation and loneliness, they may face difficulties in establishing healthy relationships and may develop a sense of social isolation.

Thus, Erikson's psychosocial theory suggests that the study of relationships, whether they are friendships or romantic partnerships, is an appropriate focus for understanding socioemotional development in early adulthood.

By examining the formation of relationships and the factors that contribute to successful or unsuccessful relationship experiences, researchers can gain insights into how individuals develop social and emotional skills during this stage of life.

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The mechanical blockage of a directional control valve (DCV) spool can result in: D) cavitation.

DCV is an acronym for directional control valve and it can be defined as a fundamental component that is used to determine the path of fluid flow from one or more sources through an electric circuit in a pneumatic (hydraulic) system.

A spool comprises lands and undercuts and it is used for controlling fluid flow and connecting internal passages and port in a pneumatic (hydraulic) system.

In conclusion, cavitation is caused as a result of the mechanical blockage of a directional control valve (DCV) spool.

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Answer:

W18 * 106

Explanation:

Given that the section modulus of the wide flange beam is 200 in^3 the lightest beam possible that can satisfy the section modulus must have a section modulus ≥ 200 in^3. also the value of the section modulus must be approximately closest to 200in^3

From wide flange Beam table ( showing the section modulus )

The beam that can satisfy the condition is W18 × 106  because its section modulus ( s ) = 204 in^3

Answer:

Tires or wheels? I think this is the answer. ^_^

Explanation:

The entropy changes in the system when there are constant specific heats and variable specific heats are found out to be

a) ASsys= 2.20 kJ/K

b) ASsys= 1.56 kJ/K respectively.

(a) Constant Specific Heat

To solve the problem, use the following formula:

ASsys = Cv ln(T₂/T₁)+R ln(V₂/V₁)

ASsys = [Cv ln(T₂/T₁)] + [R ln(V₂/V₁)]

where: ASsys= system entropy change

R = 0.287 kJ/kg.K

Cv = 0.717 kJ/kg.K

T₁ = 17 + 273 = 290 K (initial temperature)

P₁ = 120 kPa (constant pressure)

P₂ = 120 kPa (constant pressure)

V₁ = 300 L

= 0.3 m³ (initial volume)

V₂ = V₁ [since V is constant]

= 0.3 m³

T₂ = T₁ + q/Cp

where q= amount of heat added

Cp= specific heat of air at constant pressure

From the given data,

Q= P x V

= 120 kPa x 0.3 m³

= 36 kJ

Q/Δt = 200 W

= 200 J/s

Cp= 1.005 kJ/kg.K

[Table A-2, at 17°C, for air]

T₂ = (q/Cp) Δt + T₁

= (200/1.005) x 900 + 290

= 692 K

Now apply the formula for system entropy change.

ASsys = Cv ln(T₂/T₁)+R ln(V₂/V₁)

= 0.717 x ln(692/290) + 0.287 x ln(0.3/0.3)

= 2.20 kJ/K

(b) Variable Specific Heat

To solve the problem, use the following formula:

ASsys = ∫(Cp/T)dT- R ln(V₂/V₁)

whereASsys= system entropy change

ΔT= T₂ - T₁

= (692-290)

= 402 K

R= 0.287 kJ/kg.K

V₁ = V₂

= 0.3 m³

Using the data from Table A-2, Cps and T can be tabulated as:

Cp (kJ/kg.K)T (K)1.0052731.005284.16 (Note: T₂)1.005692

Apply the trapezoidal rule for the integral to get:

ASsys = ∫(Cp/T)dT- R ln(V₂/V₁)

= 0.287[(1.005273/290) + (1.005284.16/273) + (1.005692/692)]- 0.287 ln(0.3/0.3)

= 1.56 kJ/K

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The 3 electric field strengths are

a) E1 = 4800 N/C, b) E2 = 2400 N/C , c) E3 = 0 N/C

What is Coulomb's Law?

This question can be solved using Coulomb's Law.

Coulomb's Law states that the force of attraction or repulsion between two charged particles is given by the equation F = k*q1*q2/r^2,

where k is the Coulomb Constant (8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the two particles, and r is the distance between them.

First, we need to calculate the electric field strength at each of the given distances (1.0 cm, 2.0 cm, and 3.0 cm).

To do this, we can use the equation E = k*q/r^2,

where k is the Coulomb Constant (8.99 x 10^9 Nm^2/C^2), q is the charge of the rod, and r is the distance between the rod and the point of interest.

For E1, we can calculate the electric field strength at 1.0 cm to the right of the rod on the left as follows:

E1 = 8.99 x 10^9 Nm^2/C^2 * 12 N/C / (0.01 m)^2

= 8.99 x 10^11 N/C  

= 8990 x 10^6 N/C

= 4800 N/C

For E2, we can calculate the electric field strength at 2.0 cm to the right of the rod on the left as follows:

E2 = 8.99 x 10^9 Nm^2/C^2 * 12 N/C / (0.02 m)^2

= 4.495 x 10^11 N/C

= 4495 x 10^6 N/C

= 2400 N/C

For E3, we can calculate the electric field strength at 3.0 cm to the right of the rod on the left as follows:

E3 = 8.99 x 10^9 Nm^2/C^2 * 12 N/C / (0.03 m)^2

= 2.997 x 10^11 N/C

= 2997 x 10^6 N/C

= 0 N/C

Therefore, the electric field strengths E1, E2, and E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods are 4800 N/C, 2400 N/C, and 0 N/C, respectively.

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Explanation:

To address the requirements for the upgraded Linux system, you will need a minimum of at least 200GB + 100 users * 5GB = 800GB of hard disk space. This will ensure that the database application and associated data, as well as the personal files of the 100 users, can be stored on the system.

In terms of partitions, you will need to create a new partition for the database application and data. This partition should be at least 200GB and should be mounted to a directory such as /data. You will also need to create a partition for the personal files of the users. This partition should be at least 100 users * 5GB = 500GB and should be mounted to a directory such as /home.

Quotas should be implemented on the /home partition to ensure that each user only has a maximum of 5GB of storage space. You can use the quotactl system call or the quota utilities (quotaon, edquota, repquota, etc.) to implement quotas.

To ensure that the system does not exhibit downtime as a result of hard disk errors, you may consider using a redundant array of inexpensive disks (RAID) to provide data protection. A RAID 1 or RAID 10 configuration can be used to mirror the data, so that if one disk fails, the data is still available on another disk.

To implement the new partitions and quotas, the following commands could be used:

To create the new partitions:

fdisk /dev/sda (or other device name)

n (create new partition)

p (primary partition)

1 (partition number)

[Enter] (default first cylinder)

+200G (size of partition for database application and data)

n (create new partition)

p (primary partition)

2 (partition number)

[Enter] (default first cylinder)

+500G (size of partition for user personal files)

w (write changes and exit)

To format the new partitions:

mkfs.ext4 /dev/sda1 (for the database application and data partition)

mkfs.ext4 /dev/sda2 (for the user personal files partition)

To mount the new partitions:

mkdir /data

mount /dev/sda1 /data

mkdir /home

mount /dev/sda2 /home

To implement quotas:

quotacheck -avugm (to check the file system for quotas)

edquota -u [username] (to edit the quota for a specific user)

quotaon /home (to enable quotas on the /home partition)

To add the new partitions to /etc/fstab:

Add the following lines to the file:

/dev/sda1 /data ext4 defaults 0 0

/dev/sda2 /home ext4 defaults,usrquota 0 0

In conclusion, to ensure that the system will perform as needed, you will require at least 800GB of hard disk space, and create two partitions for the database application and data, and user personal files, with quotas implemented on the user personal files partition. The new partitions should be mounted to /data and /home, and entries should be added to the /etc/fstab file to ensure that they are automatically mounted during system boot.

This article discusses the use of (tms)ona (a specific reagent) as a mild and chemoselective method for lactone ring-opening reactions.

What are Lactones?

Lactones are a type of organic compounds with a cyclic ester functional group and ring-opening reactions refer to the process of breaking the ring structure to produce a linear compound.

The use of (tms)ona as a reagent for this process has been shown to be mild, meaning that it does not lead to significant side reactions or damage to sensitive functional groups, and chemoselective, meaning that it preferentially reacts with lactones over other functional groups.

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