A combined cycle power plant has a net power output of 214 MW. Air enters the compressor of the Brayton cycle at 100 kPa and 37oC. The pressure ratio is 18, the turbine inlet temperature is 1620 K and the gases leaving the turbine are used to heat the steam of the Rankine cycle to 600oC and 6 MPa. The combustion gas leaves the heat exchanger (HRSG) at 420 K and the condenser pressure is 10 kPa. The isentropic efficiency of the compressor is 91%, the gas turbine isentropic efficiency is 93%, the steam turbine isentropic efficiency is 92% and the pump efficiency is 65%. The condenser sub-cools the water by 10oC (10oC less than the saturation temperature). Using variable specific heat analysis, determine:


a. Rate of heat input in the combustion chamber, Qin in kW.

b. Rate of heat transfer in the HRSG, in kW.

c. Power input to the pump, in kW.

d. Power input to the compressor, in kW.

e. Net power output from the Brayton cycle, in kW.

f. Net power output from the Rankine cycle, in kW.

g. Overall efficiency.

Answer:

hello a diagram attached to your question is missing attached below is the missing diagram

The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P = 70kN , let d = 2.4 m , L = 4m  

Answer :

Stress = force / area

for Bar 1 = (40.03 * 1000) / 500  = 80.06 MPa

Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa

for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa

Explanation:

Given data:

Type of steel = A-36

cross-sectional area = 500 mm^2

Calculate the average normal stress in each bar

we have to  make some assumptions

assume forces in AB, CD, EF  to be p1,p2,p3  respectively

∑ Fy = 0 ; p1 + p2 + p3 = 70kN  ----------  ( 1 )

∑ Mc = 0 ; P1 * d - p * d/2 - p3 * d = 0

where d = 2.4 hence ; p1 - p3 = 35 -------- ( 2 )

Take ; Tan∅

Tan∅  = MN / 2d = OP/d

i.e. s1 - 2s2 - s3 = 0

\(\frac{P1L}{AE} - \frac{2P2}{AE} + \frac{P3L}{AE} = 0\)

L , E and A are the same hence

P1 - 2p2 + p3 = 0 ----- ( 3 )

Next resolve the following equations

p1 = 40.03 kN,  p2 = 23.33 kN, p3 = 5.33 kN

Stress = force / area

for Bar 1 = (40.03 * 1000) / 500  = 80.06 MPa

Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa

for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa

Answer:

The resulting pressure of the gas when its volume decreases is 300 kN/m².

Explanation:

Given;

initial volume of the gas, V₁ = 80 L

number of moles of the gas, n = 0.5 moles

initial pressure of the gas, P₁ = 150 kN/m² = 150 kPa

Determine the constant temperature of the gas using ideal gas equation;

PV = nRT

where;

R is ideal gas constant = 8.315 L.kPa/K.mol

T is the constant temperature

\(T = \frac{P_1V_1}{nR} \\\\T = \frac{150.kPa \ \times \ 80 .L}{0.5 .mol \ \times \ 8.315(L.kPa/mol.K)} \\\\T = 2,886.35 \ K\)

When the gas is compressed to half of its volume;

new volume of the gas, V₂ = ¹/₂ V₁

                                             = ¹/₂ x 80L = 40 L

The new pressure, P₂ is calculated as;

\(P_2V_2 = nRT\\\\P_2 = \frac{nRT}{V_2} \\\\P_2 = \frac{0.5 \times 8.315\times 2886.35}{40} \\\\P_2 = 300 \ kPa = 300 \ kN/m^2\)

Therefore, the resulting pressure of the gas when its volume decreases is 300 kN/m².

The ratio of maximum shear stresses in the hollow and solid bars is (15/16).To calculate the ratio of torsional stiffnesses between a hollow and solid bar, and the ratio of their maximum shear stresses, we can use the concept of polar moment of inertia (J) and the torsion formula.

Ratio of Torsional Stiffnesses: The torsional stiffness of a bar is proportional to its polar moment of inertia. For a solid bar, the polar moment of inertia is given by J_solid = (π/32) * (d^4), where d is the diameter of the solid bar.

For a hollow bar with an inner diameter of d/2, the polar moment of inertia is given by J_hollow = (π/32) * (d^4 - (d/2)^4).

The ratio of torsional stiffnesses (K) between the hollow and solid bars is then: K = J_hollow / J_solid = [(d^4 - (d/2)^4)] / d^4 = (15/16)

Ratio of Maximum Shear Stresses: The maximum shear stress (τ_max) in a bar under torsion is given by the formula τ_max = (T * r) / J, where T is the applied torque and r is the radial distance from the center of the bar.

Since both the hollow and solid bars are subjected to the same torque, the ratio of their maximum shear stresses will be the same as the ratio of their polar moments of inertia:

τ_max_hollow / τ_max_solid = J_hollow / J_solid = (15/16)

Therefore, the ratio of maximum shear stresses in the hollow and solid bars is (15/16).

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Answer:

Explanation:

From the information given:

original diameter \(d_o\) = 10 mm

final diameter \(d_f =\) 7.5 mm

Cold work tensile strength of brass = 380 MPa

Recall that;

\(\text {The percentage CW }= \dfrac{\pi (\dfrac{d_o}{2})^2 - \pi(\dfrac{d_f}{2})^2 }{\pi(\dfrac{d_o}{2})^2} \times 100\)

\(\implies \dfrac{\pi (\dfrac{10}{2})^2 - \pi(\dfrac{7.5}{2})^2 }{\pi(\dfrac{10}{2})^2} \times 100\)

\(\implies43.87\% \ CW\)

→ At 43.87% CW, Brass has a tensile strength of around 550 MPa, which is greater than 380 MPa.

→ At 43.87% CW, the ductility is less than 5% EL, As a result, the conditions aren't met.

To achieve 15% EL, 28% CW is allowed at most

i.e

The lower bound cold work = 15%

The upper cold work = 28%

The average = \(\dfrac{15+28}{2}\) = 21.5 CW

Now, after the first drawing, let the final diameter be \(d_o^'\); Then:

\(4.5\% \ CW = \dfrac{\pi (\dfrac{d_o^'}{2})^2 - (\dfrac{7.5}{2})^2}{\pi (\dfrac{d_o^'}{2})^2}\times 100\)

By solving:

\(d_o^'} = 8.46 mm\)

To meet all of the criteria raised by the question, we must first draw a wire with a diameter of 8.46 mm and then 21.5 percent CW on it.

Answer:

1. Yes, they are all necessary.

2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.

Both Tech A and Tech B are correct in their statements about planetary gear sets, but they are describing different scenarios.

Tech A is correct in saying that to obtain a direct drive or a 1:1 gear ratio in a planetary gear set, the transmission locks together any two members of the set. This can be achieved by locking either the sun gear, the planet carrier, or the ring gear, depending on the specific configuration and desired gear ratio.

Tech B is correct in saying that to obtain reverse in a planetary gear set, the ring gear is attached to the output shaft (the driven component) while the planet carrier is held stationary. By keeping the planet carrier fixed, the rotation of the sun gear is transferred to the ring gear in the opposite direction, resulting in a reverse gear ratio.

In summary, both Tech A and Tech B are providing accurate information about the operation of a planetary gear set. Tech A explains the concept of obtaining a direct drive or a 1:1 gear ratio by locking two members together, while Tech B describes how reverse can be achieved by holding the planet carrier stationary and attaching the ring gear to the output shaft.

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Title: Energy Efficient Air Conditioning Delivery Systems: Pros and Cons of Selecting Based on SEER Rating

Introduction:

Air conditioning systems play a vital role in keeping our homes and workplaces comfortable, but they can also consume a significant amount of energy. As a result, it is important to consider the energy efficiency of these systems when selecting the best option for your needs. In this paper, we will research the most energy-efficient air conditioning delivery systems and analyze the pros and cons of selecting an energy-efficient system based on its SEER rating. We will also discuss green technologies and provide case studies and data to support our research.

Energy-Efficient Air Conditioning Delivery Systems:

There are several types of air conditioning delivery systems available in the market, each with its unique features and level of energy efficiency. The following are some of the most energy-efficient air conditioning delivery systems:

Ductless Mini-Split Systems:

Ductless mini-split systems are an excellent option for homes without existing ductwork. They consist of an outdoor unit that connects to one or more indoor units and allow for zoning to save energy. They are highly energy-efficient and can achieve a SEER rating of up to 30.

Central Air Conditioning Systems:

Central air conditioning systems are the most commonly used air conditioning delivery systems in residential and commercial buildings. They consist of a compressor unit that pumps refrigerant to a network of ducts, which deliver cool air to different rooms. Central air conditioning systems can achieve a SEER rating of up to 26, depending on the model.

Geothermal Heat Pumps:

Geothermal heat pumps use the earth's natural heat to provide heating and cooling. They are highly energy-efficient and can achieve a SEER rating of up to 40. However, they are also more expensive to install than traditional air conditioning systems.

Pros and Cons of Selecting an Energy-Efficient System Based on SEER Rating:

SEER stands for Seasonal Energy Efficiency Ratio and is a measure of the cooling output of an air conditioning system divided by the amount of energy it consumes over a season. The higher the SEER rating, the more energy-efficient the system is. Here are the pros and cons of selecting an energy-efficient system based on its SEER rating:

Lastly, Pros:

Energy Savings: By selecting a system with a high SEER rating, you can save a significant amount of energy and money on your electricity bills.

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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

\(\triangle h=4.935m\)

The calculated torque is 91.55 KN-mm. Yes, the radius of the pulley has an effect on the belt tensions.

For a given frictional torque at the axle, the tension difference between the pull side and the load side is smaller because a larger radius produces a greater moment.

Each pulley's radius is 50 mm;

R1 = R2 = 50 mm.

R1 = 50 mm for the driver's radius

50mm for the radius of the driven,

1800 for the angle of contact (in radians), and

0.3 for the coefficient of friction. T1 = 3KN of permissible stresses Always, T1 is bigger than T2.

Utilizing the formula T1/T2 = e After adding together all the values,

T2 = 1.169KN and 3/T2 = e(0.3)().

Because both pulleys' radiuses are the same, they both experience the same amount of torque (T1–T2). R1

= (T1–T2) (T1–T2). R2

When we add up everything,

(3 - 1.169) X 50

= 91.55 KN-mm.

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Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is  that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa  against the claim Ha: u< 3.1 MPa

The alternate hypothesis is  that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68  falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

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.

The rectangular survey description for Dodger Stadium can be computed using the tools used in rectangular surveying.

Rectangular surveying is a land surveying method that divides land into square-shaped areas or townships. It makes use of the township and range system, which is a grid-like system that is used to identify locations and boundaries of land.The first step in computing the rectangular survey description for Dodger Stadium is to gather input data, including the latitude and longitude of the stadium, and the meridian and baseline that are used to define the township and range lines. The meridian and baseline are typically set by the government or a surveying authority.Next, the required distances must be calculated. This includes the distance from the baseline to the stadium, and the distance from the meridian to the stadium. These distances are used to determine the township and range lines that intersect at the location of the stadium. The township and range lines are numbered based on their distance from the baseline and meridian, respectively.Once the township and range lines are determined, the stadium can be located within a specific section. Sections are 1-mile square areas that are numbered within each township and range. The section number is determined by counting the number of sections between the township line and the stadium, and the number of sections between the range line and the stadium.Finally, the location of the stadium within the section can be determined by dividing the section into smaller portions. This is typically done using a system of fractions, where the section is divided into halves, quarters, and so on. The location of the stadium is then described in terms of the fraction that it is located within. For example, if the stadium is located in the southeast quarter of the section, it would be described as being located in section 27, township 1 north, range 1 west, southeast quarter.

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Answer:

I can help but I need to know what it looking for

A computer, programme, piece of hardware, or piece of software that depends on another programme, piece of hardware, or piece of software to access a service offered by a server is referred to as a client.

For instance, web clients interact with web servers to retrieve and display pages.

The online or player video games are retrieved by mail clients from mail servers. The term "client" can apply to both the individuals who utilise client software as well as the computers or other gear that they use to run it.

Once connected to a VPN server,

In general, yes, as your IP address will be changed to that of the VPN service you're connected to rather than your direct address, which is where other websites and servers will see you as coming from.

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Answer:

In the case of the production Line, we know that,

No of gasoline cans = 5

probability that 1st can is out of tolerance = 0.03

probability that 2nd can is out of tolerance = 0.03

.

.

probability that the 5th can is out of tolerance = 0.03

Therefore,

probability of 1st can out of tolerance + probability of 1st can not out of tolerance = 1

Probability of 1st can not out of tolerance = 1 -- 0.03 = 0.97

probability of 2nd can not out of tolerance = 0.97

.

.

probability of 5th can not out of tolerance = 0.97

Question A:

Probability that they are all out of tolerance

= P(1st can out of tolerance) * P(2nd can out of tolerance) * P(3rd can out of tolerance) * P(4th can out of tolerance) * P(5th can out of tolerance)  

= (0.03 ) * (0.03) * (0.03) * (0.03) * (0.03) =  2.43 E⁻⁸   (2.43 ˣ 10⁻⁸)

Question B:

Probability that exactly two are out of tolerance

= P(1st can is out of tolerance) * P(2nd can is out of tolerance) * P(3rd can is not out of tolerance) * P(4th can is not out of tolerance) * P(5th can is not out of tolerance)

= (0.03) * (0.03) * (0.97) * (0.97) * (0.97) = 0.0008214057

Explanation:

Objects permit data hiding, which means they can prevent functions outside the class from changing some or all of the data. Therefore, the answer to your question is "a) changing".

Explanation:

In object-oriented programming (OOP), data hiding is a concept that allows objects (instances of a class) to encapsulate and protect their internal data from being accessed or modified directly by external functions or code. Data hiding is achieved by using access modifiers, such as private or protected, to restrict the visibility and access of class members (i.e., variables, properties, and methods) from outside the class.

Functions outside the class can still access the data, but they cannot modify it without using specific methods provided by the class. This feature is crucial for protecting the integrity and consistency of the data in object-oriented programming.

When data is hidden using access modifiers, it means that external functions or code outside the class cannot directly modify the hidden data. This includes changing the value of the data, as mentioned in the question. Therefore, the correct answer is "a) changing." External functions or code can still access and read the value of the data if the access modifier allows for it, but they cannot modify it directly.

It's worth noting that data hiding is an important principle of OOP as it promotes encapsulation and abstraction, which helps in achieving better code organization, maintainability, and security. By hiding the internal data of objects, classes can have better control over how the data is accessed and modified, preventing unintended changes that may lead to bugs or security vulnerabilities.

Therefore, Objects permit data hiding, which means they can prevent functions outside the class from changing some or all of the data.

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Answer:

2)

a)  to the right of the dipole    E_total = kq [1 / (r + a)² - 1 / r²]

b)To the left of the dipole      E_total = - k q [1 / r² - 1 / (r + a)²]

c) at a point between the dipole, that is -a <x <a  

      E_total = kq [1 / x² + 1 / (2a-x)²]

d) on the vertical line at the midpoint of the dipole (x = 0)

E_toal = 2 kq 1 / (a ​​+ y)² cos θ

Explanation:

2) they ask us for the electric field in different positions between the dipole and a point of interest. Using the principle of superposition.

This principle states that we can analyze the field created by each charge separately and add its value and this will be the field at that point

Let's analyze each point separately.

The test charge is a positive charge and in the reference frame it is at the midpoint between the two charges.

a) to the right of the dipole

The electric charge creates an outgoing field, to the right, but as it is further away the field is of less intensity

           E₊ = k q / (r + a)²

where 2a is the distance between the charges of the dipole and the field is to the right

the negative charge creates an incoming field of magnitude

           E₋ = -k q / r²

The field is to the left

therefore the total field is the sum of these two fields

           E_total = E₊ + E₋

           E_total = kq [1 / (r + a)² - 1 / r²]

we can see that the field to the right of the dipole is incoming and of magnitude more similar to the field of the negative charge as the distance increases.

b) To the left of the dipole

The result is similar to the previous one by the opposite sign, since the closest charge is the positive one

E₊ is to the left and E₋ is to the right

          E_total = - k q [1 / r² - 1 / (r + a)²]

We see that this field is also directed to the left

c) at a point between the dipole, that is -a <x <a

In this case the E₊ field points to the right and the E₋ field points to the right

                      E₊ = k q 1 / x²

                      E₋ = k q 1 / (2a-x)²

                      E_total = kq [1 / x² + 1 / (2a-x)²]

in this case the field points to the right

d) on the vertical line at the midpoint of the dipole (x = 0)

    In this case the E₊ field points in the direction of the positive charge and the test charge

    in E₋ field the ni is between the test charge and the negative charge,

the resultant of a horizontal field in zirconium on the x axis (where the negative charge is)

                      E₊ = kq 1 / (a ​​+ y) 2

                      E₋ = kp 1 / (a ​​+ y) 2

                      E_total = E₊ₓ + E_{-x}

                      E_toal = 2 kq 1 / (a ​​+ y)² cos θ

e) same as the previous part, but on the negative side

                        E_toal = 2 kq 1 / (a ​​+ y)² cos θ

When analyzing the previous answer there is no point where the field is zero

The different configurations are outlined in the attached

3) We are asked to repeat part 2 changing the negative charge for a positive one, so in this case the two charges are positive

a) to the right

in this case the two field goes to the right

           E_total = kq [1 / (r + a)² + 1 / r²]

b) to the left

            E_total = - kq [1 / (r + a)² + 1 / r²]

c) between the two charges

E₊ goes to the right

E₋ goes to the left

            E_total = kq [1 / x² - 1 / (2a-x)²]

d) between vertical line at x = 0

E₊ salient between test charge and positive charge

           E_total = 2 kq 1 / (a ​​+ y)² sin θ

In this configuration at the point between the two charges the field is zero

Answer:

This is an example of the conduction of electricity through metal. Free moving electrons on the car will conduct a electric field when a voltage is applied to the car;  in this case the transmission line, and would flow through the metal to the door handle causing electrocution.

Answer:

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then  

R

1

 in Figure 1(a) could be the resistance of the screwdriver’s shaft,  

R

2

 the resistance of its handle,  

R

3

 the person’s body resistance, and  

R

4

 the resistance of her shoes.

Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.

Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.

According to Ohm’s law, the voltage drop,  

V

, across a resistor when a current flows through it is calculated using the equation  

V

=

I

R

, where  

I

 equals the current in amps (A) and  

R

 is the resistance in ohms  

(

Ω

)

. Another way to think of this is that  

V

 is the voltage necessary to make a current  

I

 flow through a resistance  

R

.

So the voltage drop across  

R

1

 is  

V

1

=

I

R

1

, that across  

R

2

 is  

V

2

=

I

R

2

, and that across  

R

3

 is  

V

3

=

I

R

3

. The sum of these voltages equals the voltage output of the source; that is,

V

=

V

1

+

V

2

+

V

3

.

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation  

P

E

=

q

V

, where  

q

 is the electric charge and  

V

 is the voltage. Thus the energy supplied by the source is  

q

V

, while that dissipated by the resistors is

q

V

1

+

q

V

2

+

q

V

3

.

Explanation:

Answer:

  see attached

Explanation:

As you know, a full adder produces the binary value in (carry, output) that is the the number of true input bits among the (carry in, A, B) inputs.

The annotated diagram attached shows the bit values for a 4-bit full adder/subtractor.

__

Additional comment

If the numbers are considered "signed", then the top diagram will generate an "overflow" based on the difference between the carry C4 and the output O4.

The safety technique that is recommended for this situation of moving a heavy industrial printer is to Bend at the knees when lifting.

This refers to the process of executing a task in a safe and acceptable way that does not put the user or others at risk

Hence, we can see that The safety technique that is recommended for this situation of moving a heavy industrial printer is to Bend at the knees when lifting.

This is to ensure that the person doing the lifting does not injure himself and has the proper form when lifting.

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True. Serial data packets are used to transmit data serially, bit by bit, over a communication channel.

These packets contain the actual data that needs to be processed. The packets may also include additional information such as packet headers, error detection and correction codes, and synchronization information to ensure that the data is transmitted correctly and received accurately.

Serial data packets are commonly used in various applications, such as serial communication between computers and peripherals, digital audio and video transmission, and industrial automation systems. By using serial data packets, the data can be transmitted efficiently and reliably, with minimal transmission errors and high data transfer rates.

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Answer:

4 or D pressure

Explanation:

A hydraulic braking system transmits brake-pedal force to the wheel brakes through pressurized fluid, converting the fluid pressure into useful work of braking at the wheels.

Answer:

pressure is the correct answer...

Explanation:

mark me brainliest

Answer:

  A battery

Explanation:

There are several words used to describe such an assembly:

  pile, battery, electrolytic cell

Classifier could we have used based on data point to the data set is:  a. 1-nearest neighbor classifier.

When adding an additional data point and the decision boundary remains unchanged, which is a characteristic of the 1-nearest neighbor method. Logistic regression is a supervised learning algorithm that is used for binary classification used to analyzing from data point.

The probability of occurrence of a binary target variable based on one or more independent variables. One advantage of the nearest neighbor classifier is that it can capture complex decision boundaries and can be applied to a wide range of classification problems.

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Salinization is the processes results most directly from over-irrigation. Hence option c is correct.

Over irrigation is defined as excessive water being artificially applied to soil for agricultural production. Because it results in water logging, over irrigation promotes salt in the soil. Excess water evaporates, concentrating salt at the soil's surface.

When water is absorbed by plants or evaporates, the salt is left behind in the soil. Due to leakage from both irrigation and rainfall, recharge rates in irrigation areas can be significantly higher than in dryland areas. This could result in salinization rates that are quite high.

Thus, salinization is the processes results most directly from over-irrigation. Hence option c is correct.

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Your firm has been asked to design a new secondary clarifier for a conventional activated sludge plant.

For MLSS is 1,400 mg/L and setting velocity is 3 m/h.

Similarly calculate the gravity flux for remaining MLSS values.

Cmg/L    um/h         SF kg/m².h

1,400  3           4,2

2,200  1,85   4,07

3,000  1,21           3,63

3,700 0,76   2,812

4,500 0,45   2,025

5,200 0,28    1,456

6,500 0,13           0,845

8,200 0,089   0,7298

Under flow velocity = - Slope

Substitute - 0.135 m/h for slope.

Under flow velocity = - (-0.135)

                              = 0.135 m/h

Therefore, the diameter of the clarifier is 40m.

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Answer:

answer 2

Explanation:

To start a cold engine on a snowmobile, you should first make sure that the machine is in a well-ventilated area. Then, turn on the choke and give the engine a few pulls with the starter cord. Once the engine starts, gradually release the choke until the snowmobile is idling smoothly. It is important to not rev the engine too much while it is still cold, as this can cause damage to the engine. However the following steps must be followed:

1. Check the fuel level and ensure that there is enough fuel in the tank.

2. Turn the fuel valve on to allow fuel to flow into the carburetor.

3. Check the choke and ensure that it is in the "closed" or "on" position.

4. Turn the ignition key to the "on" position.

5. Pull the starter cord slowly until you feel some resistance, then give it a quick, firm pull to start the engine.

6. Once the engine starts, let it run for a few minutes to warm up.

7. If the engine is still cold and not running smoothly, you may need to adjust the choke or throttle to maintain a steady idle.

It's important to note that proper maintenance and storage practices can also affect the ease of starting a snowmobile engine. Regular tune-ups and keeping the machine in a warm, dry place can help prevent issues with starting a cold engine.

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