QUESTION 16 Which of these is the in HTML code for centering all of the content in the browser window? (Identify the correct code by number and then select it in the answers below) Web Site Home Page 9 O a. 5 O b.7 O c. 1 O d. 9

Answer:

\(I = 45mA\)

Explanation:

Given

\(I = 90mA\) --- Current

Required

Determine the new current when resistance is doubled

Using \(V = IR\)

Initially, we have:

\(V = 90mA * R\)

When resistance is doubled and voltage remains unaltered, we have:

\(V = I* 2R\)

2R represents the new resistance and I represents the new current

Equate both values of V

\(90mA * R = I* 2R\)

Make I the subject

\(I = \frac{90mA * R}{2R}\)

\(I = \frac{90mA }{2}\)

\(I = 45mA\)

The new current is 45milliAmps

Answer:

The estimated batching production is 210 cy/hr

Explanation:

Given;

capacity of the batching chamber of the plant, Q = 12.5 cy

average batching cycle time, t = 3 min

plant efficiency, n = 84 %

The estimated batching production is calculated as efficiency of the machine multiplied by the production rate in hour. This is given by the solution below;

Batching production = (nQ )/ t

\(Batching \ production = 0.84(\frac{12.5 \ cy}{3 \ min} *\frac{60 \ min}{1 \ hr})\\\\ Batching \ production = 210 \ cy/hr\)

Therefore, the estimated batching production is 210 cy/hr

Association rules use the apriori algorithm to find frequently associated attributes in a data set.

Attributes are characteristics or qualities that describe an entity. They are used to define and identify an object. Attributes can be physical, such as color, size, shape, or material composition, or they can be abstract, such as a person's age or a company's customer service rating. Attributes can also be used to classify objects into groups. For example, if a person is asked to classify a set of cars, they may use attributes such as make, model, and year to categorize them. Attributes are an important part of data modeling, which is used to create databases and organize data. Attributes are used to define the structure of a database, determine relationships between entities, and set the conditions for data retrieval. Attributes also help to make data easier to understand, analyze, and query.

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Technical drawings employ both an engineering scale and an architect's scale to measure and display items and structures accurately. However, the measurement increments between them vary.

ĵ Numbers on architect scales move incrementally from right to left as well as left to right. The scale those numbers represent is indicated by a whole number or fraction to the left or right of the number line. ĵ Numbers on engineer scales move progressively from left to right.

When designing a structure, architects take the needs and specifications of the client into account. In accordance with the architect's plans, engineers develop the building's plumbing, electrical, and structural systems.

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The following that is not one of the steps in the risk management process is risk tracing.

Risk management involves identifying, analyzing, and responding to risk factors that are part of an organization's life. Effective risk management means trying to control future outcomes as much as possible by being proactive rather than reactive. Effective risk management therefore has the potential to reduce both the likelihood of a risk occurring and its potential impact.

Responses to risks typically take one of the following forms:

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Unprotected sides and edges are where there is no stair rail system in place at a height of 4 inches or higher.

Unprotected sides and edges are regions along stairs, platforms, or elevated surfaces that do not have a protective barrier in situ, such as a stair rail system. Individuals may step off the edge or lose their balance and fall in these situations, posing a possible fall hazard.

The 4-inch barrier was chosen with the notion that a fall from this height or above could result in serious damage. A stair rail system, which includes handrails and guardrails, helps to provide a physical barrier and support for those using the stairs or accessing high locations, lowering the risk of slipping down the unprotected sides or edges.

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The initial pressure in the cylinder of 10 bar of argon with \(C_p\) = (5/2)·R

and the added gas at pressure of 50 bar gives the following values.

(a) 400.6 K

(b) 51.184 J

(c) The pressure of the cylinder when it is shipped is 37.194 bar

(a) Adiabatic compression, we have;

\(\dfrac{p_{1}}{p_{2}} = \mathbf{\left (\dfrac{T_{_{1}}}{T_{2}} \right )^{\dfrac{k}{k-1}}}\)

Where;

\(k = \dfrac{C_p}{C_v} = \mathbf{\dfrac{C_p}{C_p - R}}\)

Therefore;

\(K= \mathbf{\dfrac{\frac{5}{2} \cdot R}{\frac{5}{2} \cdot R - R} }= \dfrac{5}{3}\)

Which gives;

\(\dfrac{10}{50} = \left (\dfrac{298}{T_{2}} \right )^{\dfrac{\frac{5}{3} }{\frac{5}{3} -1}} = \left (\dfrac{298}{T_{2}} \right )^{2.5}\)

\(\dfrac{50}{10} = \mathbf{ \left (\dfrac{T_2}{298.15} \right )^{2.5}}\)

\(ln(5)= 2.5\cdot ln\left (\dfrac{T_2}{298.15} \right )\)

\(T_2 = 298 \times e^{\dfrac{ln(5)}{2.5} } \approx \mathbf{567.29\, K}\)

The temperature

Using an ideal gas model, we have;

\(\dfrac{V_{1}}{V_{2}} = \mathbf{\dfrac{P_{2}\times T_{1}}{T_{2} \times P_1}}\)

Which gives;

\(\dfrac{V_{1}}{V_{2}} = \dfrac{50\times 298}{567.29 \times 10} \approx 2.625\)

V₁ = 2.625·V₂

Volume occupied by the gas in the cylinder after the pressure increases is therefore;

\(V_2 = \dfrac{V_1}{2.625} \approx 0.381 \cdot V_1\)

Volume of gas added is therefore;

\(V =V_1 - \dfrac{V_1}{2.625} = 0.619\cdot V_1\)

Considering a molar volume of gas, we have;

0.040 × 0.619 × 2.5×8.3145×(T₃ - 298) = 0.040 × 0.381 × 2.5×8.3145×(567.29 - T₃)

Solving gives;

T₃ ≈ 400.6 K

For a molar volume of gas cylinder, the temperature just after the valve is closed is T₃ ≈ 400.6 K

(b) If the cylinder sits in the storage for a long time, and heat is conducted out, we have;

The temperature of the cylinder will revert to room temperature of 298 K

The heat transferred is therefore;

0.040 × (2.5 × 8.3145 - 8.3145) ×(400.6 - 298) ≈ 51.184

The heat transferred from the cylinder is 51.184 J

(c) The pressure of the gas in the cylinder is therefore;

\(\dfrac{P_1}{T_1} = \mathbf{\dfrac{P_2}{T_2}}\)

Which gives;

\(P_2= \mathbf{\dfrac{P_1}{T_1} \times T_2}\)

Therefore;

\(P_2= \dfrac{50}{400.6} \times 298 \approx 37.194\)

The pressure of the cylinder when it is shipped is 37.194 bar

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The algorithm provided correctly generates T using a random number generator and the exponential distribution formula. It is mathematically proven and ensures that Y is uniformly distributed over the interval (0, 1) when X is uniformly distributed over the same interval.

To generate T given a random number generator of a random variable X uniformly distributed over the interval (0,1), follow the steps below:

Algorithm to generate T

Here, λ is the rate parameter of the exponential distribution represented by T.This algorithm for generating T is correct and can be proved mathematically as follows:

Proof:

First, we will find the cumulative distribution function (CDF) of T. We know that T is exponentially distributed with parameter λ. Hence, the CDF of T is given by: F(t) = P(T ≤ t) = ∫0t λe^(-λt) dt = 1 - e^(-λt)

Now, let's find the inverse of the CDF. Since 0 < X < 1, we have: P(T ≤ t) = X

Therefore, we have:1 - e^(-λt) = X e^(-λt) = 1 - X -λt = ln(1 - X)

Thus, the inverse of the CDF is given by:T = - ln(1 - X)/λ

Now, we need to show that Y is uniformly distributed over the interval (0, 1) if X is uniformly distributed over the interval (0, 1).

Since X is uniformly distributed over the interval (0, 1), its CDF is given by: F(x) = x for 0 ≤ x ≤ 1

Let's find the CDF of Y. Since Y = F(T), we have: P(Y ≤ y) = P(F(T) ≤ y) = P(T ≤ F^-1(y))= F(F^-1(y))= y

Thus, the CDF of Y is: F(y) = y for 0 ≤ y ≤ 1

Hence, Y is uniformly distributed over the interval (0, 1) if X is uniformly distributed over the interval (0, 1).

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Answer:

Land capability is one aspect of land classification. For the determination of land capability, the usefulness of land for agriculture, forest and tourism is assessed solely on the basis of physical environmental factors.

Hence the capability denotes

Answer:

B is correct I believe

Answer:

They both have the same efficiency.

Explanation:

The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.

Therefore in comparison, the efficiency is the same for both.

Answer:

Email etiquette is important

Explanation:

It is important to understand how to use correct email etiquette because it helps you communicate more clearly. It also makes you seem a bit more professional too. For example depending in who you're emailing like say you're emailing your teacher for help then here's how it'd go:

Dear(teacher name, capitalize, never use first name unless they allow it)

Hello (teacher name), my name is (first and last name) from your (number class) and I was wondering if you could please help me out with (situation, be clear on what you need help with otherwise it won't get through to them)? If you could that would be greatly appreciated!

Sincerely,

(your name first and last)

1. Foreign key constraints among the relations:
- In the Enrolled relation, the StudentID should be a foreign key referencing the Students relation (specifically the ID attribute) to ensure that only valid students can be enrolled in courses.
- Also in the Enrolled relation, the CourseID should be a foreign key referencing the Courses relation (specifically the ID attribute) to ensure that students can only enroll in valid courses.
- In the Teaches relation, the FacultyID should be a foreign key referencing the Faculty relation (specifically the ID attribute) to ensure that only valid faculty members can teach courses.
- Also in the Teaches relation, the CourseID should be a foreign key referencing the Courses relation (specifically the ID attribute) to ensure that faculty members can only teach valid courses.
- In the Meets relation, the CourseID should be a foreign key referencing the Courses relation (specifically the ID attribute) to ensure that only valid courses have meeting information.
- Also in the Meets relation, the RoomID should be a foreign key referencing the Rooms relation (specifically the ID attribute) to ensure that courses can only meet in valid rooms.

2. An example of a plausible constraint involving one or more of these relations that is not a primary key or foreign key constraint:
- A check constraint can be added to the Enrolled relation to ensure that the enrollment date is within the valid time period for a specific course. This constraint would involve checking the enrollment date attribute in the Enrolled relation against the course start and end dates in the Courses relation.

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a) Binary representation of 1033.24 in IEEE 754 single precision: 00111111110100010010001101000001 (hex: 3F 42 43 61)

b) Binary representation of 71.37 in IEEE 754 double precision: 0100000001011101100110011001100110011001100110011001100110011010 (hex: 40 9D B3 33 33 33 33 33)

c) The corresponding signed decimal number for the 32-bit number 10101010 11100000 00000000 00000000 is -1,489,800.

a) For the number 1033.24, we need to convert it to binary in IEEE 754 single precision. The integer part of 1033 is 10000011001 in binary. To represent the fractional part 0.24, we need to multiply it by 2 repeatedly until the fractional part becomes 0. The binary representation of the fractional part is 0.01. Combining the integer and fractional parts, we get 10000011001.01. In IEEE 754 single precision, the sign bit is 0 for positive numbers. The exponent is 127 + number of bits required to represent the integer part and fractional part (11 in this case). The significand is the binary representation of the integer and fractional part without the leading 1. Putting all these together, we get the binary representation as 00111111110100010010001101000001. Converting this to hex, we have 3F 42 43 61.

b) For the number 71.37, the integer part is 1000111 and the fractional part is 0.37. Converting the fractional part to binary, we get 0.010111. Combining the integer and fractional parts, we have 1000111.010111. In IEEE 754 double precision, the sign bit is 0 for positive numbers. The exponent is 1023 + number of bits required to represent the integer part and fractional part (11 + 52 = 63 in this case). The significand is the binary representation of the integer and fractional part without the leading 1. Putting all these together, we get the binary representation as 0100000001011101100110011001100110011001100110011001100110011010. In hex format, it is 40 9D B3 33 33 33 33 33.

c) The given 32-bit number 10101010 11100000 00000000 00000000 in IEEE 754 representation is a signed number. The first bit represents the sign, where 0 indicates a positive number and 1 indicates a negative number. In this case, the first bit is 1, indicating a negative number. The remaining bits are used to represent the magnitude of the number. Converting the remaining bits to decimal, we have 10101010 11100000 00000000 00000000, which is equal to 2,109,153,024. Since the sign bit is 1, the corresponding signed decimal number is -2,109,153,024.

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The process in question is referred to as the iterative design process. Engineers utilize this method by completing and repeating a sequence of steps in order to continually improve and refine their designs as they work towards achieving the project goal. This approach allows for flexibility and adaptability in the design process, as engineers can make adjustments and modifications based on feedback and testing, ultimately leading to a more successful outcome.

Answer:

Steel and wood

Explanation:

For a material to resist stress and vibration, it must have high ductility, which is the ability to undergo large deformations and tension. Modern buildings are often constructed with structural steel, a component that comes in a variety of shapes and allows buildings to bend without breaking.

Answer:

transmitter hope thus helped!

Explanation:

Raceway is the answer

"A raceway is an enclosed conduit that forms a physical pathway for electrical wiring."

a. the upper 3-dB frequency (f) of the discrete MOSFET common-source amplifier is approximately 808.65 MHz. b. the midband gain (AM) of the discrete MOSFET common-source amplifier is approximately -3.88.

(a) To find the overall midband gain (AM) of the discrete MOSFET common-source amplifier, we can use the following formula:

AM = -gm * (R || r) * (Rd || RL)

where gm is the transconductance of the MOSFET, R is the resistance connected to the drain, r is the output resistance of the MOSFET, Rd is the internal resistance of the voltage source, and RL is the load resistance.

Given:

gm = 4 mA/V

R = 2 MΩ

r = 100 kΩ

Rd = 500 kΩ

RL = 10 kΩ

We can calculate the parallel combination of resistances (R || r) as follows:

(R || r) = (R * r) / (R + r)

Substituting the given values:

(R || r) = (2 MΩ * 100 kΩ) / (2 MΩ + 100 kΩ) = 98.039 kΩ

Next, we calculate the parallel combination of resistances (Rd || RL) as follows:

(Rd || RL) = (Rd * RL) / (Rd + RL)

Substituting the given values:

(Rd || RL) = (500 kΩ * 10 kΩ) / (500 kΩ + 10 kΩ) = 9.901 kΩ

Now, we can calculate the overall midband gain:

AM = -gm * (R || r) * (Rd || RL)

  = -4 mA/V * 98.039 kΩ * 9.901 kΩ

  ≈ -3.88

Therefore, the overall midband gain (AM) of the discrete MOSFET common-source amplifier is approximately -3.88.

(b) The upper 3-dB frequency (f) can be calculated using the formula:

f = 1 / (2π * (R || r) * C)

where (R || r) is the parallel combination of resistances and C is the capacitance.

Given:

(R || r) = 98.039 kΩ

C = 2 pF

Substituting the given values:

f = 1 / (2π * 98.039 kΩ * 2 pF)

  ≈ 808.65 MHz

Therefore, the upper 3-dB frequency (f) of the discrete MOSFET common-source amplifier is approximately 808.65 MHz.

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The given problem is concerned with a company that has to produce goods for two quarters' period. The company has a regular time capacity and forecast demand, which is given in the table below per month.

We are required to determine the overtime capacity, overtime cost, backorder cost, inventory-holding cost, and the beginning inventory. Given data: Demand | Regular time capacity50 | 50Overtime capacity | 50% of regular time capacity × 1.5= 50% of 50 × 1.5= 25 × 1.5= 37.5 units per month Overtime cost.

Therefore, the overtime cost and backorder cost are zero. Inventory-holding cost= 50 units × R5 per unit= R250Therefore, the company needs to work at 100% of the regular time capacity and 74% of the overtime capacity to meet the forecast demand of 50 units per month for two quarters.

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Both technicians provide accurate explanations for different types of noises, but they are addressing different potential causes.

Both Technician A and Technician B provide plausible explanations for different types of noises in a vehicle.

Technician A is correct in stating that thumps and grinding noises can be caused by bad thrust washers. Thrust washers are responsible for controlling axial movement and preventing metal-to-metal contact in rotating components, such as the transmission. If the thrust washers are worn or damaged, it can result in abnormal noises.

Technician B is also correct in stating that noise related to engine speed and occurring in all gear ranges, including park and neutral, can be caused by a bad planetary gearset. The planetary gearset is a key component in an automatic transmission, and if it becomes worn or damaged, it can create noise during operation.

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Answer:

Due to risk of damaging of brain.

Explanation:

The electrode therapy is not the most effective treatment for brain disorders because there are various other treatments which can treat the brain disorder without causing damage to the brain. electrode therapy greatly damaged the brain instead of treatment of brain disorder. Medication is the best way to treat brain disorder so that's why electrode therapy is not considered as the most effective treatment for brain disorders, and the doctors recommend an alternative treatment.

Solution :

1. PL \($= \frac{11.53 - 10.49}{10.49 - 4.5} \times 100$\)

       = 17.36 %

2. Liquid limit at 25 number of blows,

  \($LL = \frac{(26-25)\times 49.8 + (25 - 2.17) \times 47.5}{26-17}$\)

       \($= 47.75 $\) %

Since 84 % of the passes the number 200 sieve the soil is immediately fine grained and MI, OI or CI because LL = 47.75 % > 35%

Therefore, the Plasticity index, PI  \($=47.75 - 17.36 = 30.39$\) %

Equation of A-line \($= 0.73 \times [47.75 - 20]$\)

                             \($=20.25 $\) %

Thus the soil lies above the A-line, that is soil is medium compressible clay ---- CI

Answer:

Bluetooth is a slow wireless technology used to connect devices within a radius of about 30 feet. While Bluetooth technology is amazing, there are lots of bugs involved with Bluetooth devices, and there is still lots to be discovered in this area of tech.

Answer:

fggffgethjbdxvgrsbjb you are my world I see the attached resume

Answer:

a) 397.89 ohm

b) attached below

c) 0.707  as a ratio

Gain in dB = 20 log 0.707 = -3 dB

d)  0.8087

Gain in dB = 20 log \(|\frac{Vout}{Vin}|\) = -1.844 dB

Explanation:

A) Find the appropriate resistor

c = 0.01 uf

fc = 40 kHz

cut-off frequency ; fo = \(\frac{1}{2\pi RC }\)

from the above equation  R = \(\frac{1}{2\pi foC}\)  = 397.89 ohm

B) sketch of the circuit  is attached

C) The gain of the filter at the cutoff frequency

fc = 40 kHz,  

C = 0.01 uF ⇒ \(\frac{-j}{2\pi foC }\) =  -j 397.89

Vout = Vin * ( R / R- C )

Vout = Vin * ( 397.89 / (397.89 - j 397.89))

Vout = \(\frac{1}{\sqrt{2} }\)  Vin ∠45⁰

therefore gain = |\(\frac{Vout}{Vin }\)| = \(\frac{1}{\sqrt{2} }\) = 0.707  as a ratio

Gain in dB = 20 log 0.707 = -3 dB

D) Gain of filter at 55 kHz

c = 0.01 uF =  \(\frac{-J}{2\pi foC }\) =  -j 289.373 ohms

Vout = Vin * \(\frac{R}{R-C}\)  

        = Vin * ( 397.89 / ( 397.89 - j 289.373))

Gain in ratio \(|\frac{Vout}{Vin}|\) = 0.8087 ∠ 36.03⁰

therefore gain in ratio = 0.8087

Gain in dB = 20 log \(|\frac{Vout}{Vin}|\) = -1.844 dB

Answer: It is power stroke

Z≤ -4.852 ft, Maximum efficiency is η≅ 0.88 ≅ 88% is the maximum discharge possible

Solution

Given Data:-

D = 14.62in, N = 2134 rc/min, T=20°C. At T= 20°C ɣ=ρg= 62.35 lb/ft³, vapor pressure. Pv = 49.2 lb/ft².

The efficienies at each flow rate is computal by using formula

η = ρgθH / (550) (bhp)

→ As we can See the maximum efficiency point is at θ = 6ft³/s (close to 6ft³/s)

Maximum efficiency is η≅ 0.88 ≅ 88%

b) Given NPSHR = 16 ft,hg=22ft. Zactual. = 9ft  (below the sea surface)

To avoid cavitation NPSH < Pa - Pv/ρg - Z - hf

Z < Pa - Pv/ρg - hf

Z < 2116 - 49.2/62.35 - 16 - 22 [1 atm = 2116 lb/ft2]

Z≤ -4.852 ft

-> Keeping the pump 9 ft below the surface gives 4.148 ft of marign against cavitation.

Hence it is Sufficient to avoid cavitation.

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Answer:

expensive method is train