The configuration shown in Fig. P9.29 depicts a satellite repeater with two antennas, one pointed towards the antenna of ground station 1 and the other towards the antenna of ground station 2. All antennas are parabolic dishes, antennas A1 and A4 are each 4m in diameter, antennas A2 and A3 are each 2m in diameter, and the distance between the satellite and each of the ground station is 40000km. Upon receiving the signal by its antenna A2, the satellite transponder boosts the power gain by 80dB and then retransmits the signal to A4. The system operates at 10GHz with Pt=1kW. Determine the received power Pr. Assume all antennas to be lossless.

Answer:

True

Explanation:

Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true

Answer:

The most common and well known of these documents are memos and emails, which are used in every type of business. In addition to this, technical communicators also create instructions, product guides and documentation, graphs, charts, images, videos, and other forms of content.

There are many ways to communicate

technical information,

1. Planning sheet- describe the process - this

shows steps on how to make your drill drift-

telling us how us how to make it.

2.Drawings- this tells you the dimensions, size

of component/workpiece

Inspection sheets- check the measurements

are within given tolerances, this shows us

you've made it within those tolerances.

Instructions/ images of how to make your.

Answer:

No. of transistors = \($4.1524 \times 10^{10}$\) transistors

Explanation:

Given that:

N = \($1920 \times 10^{0.163(Y-1980)}$\)

Y = 2025

N = \($1920 \times 10^{0.163(2025-1980)}$\)

N = \($4.1524 \times 10^{10}$\) transistors

Now at Y = 1980

Number of transistors N = 1920

Therefore,

\($1000 = 10^{0.163(Y-1980)}$\)

\($\log_{10} 1000=0.163(Y-1980)$\)

\($\frac{3}{0.163}=Y-1980$\)

18 ≅ 18.4 = Y - 1980

Y = 1980 + 18

   = 1998

So, to increase multiples of 1000 transistors. it takes 18 years.

New, derived traits of human evolution over the past five million years include all of the following except: elongation. The correct answer to the given question is option D .

Derived traits are new traits that have been passed down to a given organism by their ancestors. Some of the derived traits of human evolution over the past five million years include:

Encephalization: Encephalization is the evolutionary trend of brain size increasing relative to body size over geological time in a lineage. This evolutionary trend was observed in the human lineage.

Lack of body hair: One of the derived traits of human evolution is the lack of body hair. Humans have a far less dense body hair than other apes and primates. The loss of body hair facilitated an increase in sweat gland density, which helps us regulate our body temperature.

Reduced sexual dimorphism: The reduced differences between the sexes in size and appearance over time in a given population is referred to as reduced sexual dimorphism.

Elongation: Elongation is not a derived trait of human evolution over the past five million years, so it is the correct answer.

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Answer:

i think D

Explanation:

Answer:

shear strength = 2682.31 Ib/ft^2

Explanation:

major principal stress = 100 Ib / in2

minor principal stress = 20 Ib/in2

Normal stress = 3000 Ib/ft2

Determine the shear strength when direct shear test is performed

To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a

for direct shear test

use Mohr Coulomb criteria relation between normal stress and shear stress

Shear strength when normal strength is 3000 Ib/ft  = 2682.31 Ib/ft^2

attached below is the detailed solution

In mathematics, is equal to F/N, where F is the frictional force and N is the normal force.

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When determining the place value of the digits in a decimal number, we can use a place value chart. Example: the number 4342.761's 3 and 6 in terms of their place values. A decimal place value chart should be used to first write the number.

This decimal chart aids in determining the place value of a decimal number's digits. A decimal point, often known as a dot or point, is used to demarcate the parts of a number that are whole and decimal.

Therefore, The place values of each digit, including those before and after the decimal point, are shown in a decimal number chart.

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Answer:

the range of K can be said to be :  -3.59 < K< 0.35

Explanation:

The transfer function of a typical tape-drive system is given by;

\(KG(s) = \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]}\)

calculating the characteristics equation; we have:

1 + KG(s) = 0

\(1+ \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]} = 0\)

\({s[s+0.5)(s+1)(s^2+0.4s+4)]} +{K(s+4)}= 0\)

\(s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ 2s+K(s+4) = 0\)

\(s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ (2+K)s+ 4K = 0\)

We can compute a Simulation Table for the Routh–Hurwitz stability criterion Table as  follows:

\(S^5\)             1                     5.1                          2+ K

\(S^4\)            1.9                   6.2                           4K

\(S^3\)             1.83            \(\dfrac{1.9 (2+K)-4K}{1.9}\)          0

\(S^2\)        \(\dfrac{11.34-1.9(X)}{1.83}\)       4K                         0

S          \(\dfrac{XY-7.32 \ K}{Y}\)        0                            0

\(\dfrac{1.9 (2+K)-4K}{1.9} = X\)

\(\dfrac{11.34-1.9(X)}{1.83}= Y\)

We need to understand that in a given stable system; all the elements in the first column is usually greater than zero

So;

11.34  - 1.9(X) > 0

\(11.34 - 1.9(\dfrac{3.8+1.9K-4K}{1.9}) > 0\)

\(11.34 - (3.8 - 2.1K)>0\)

7.54 +2.1 K > 0

2.1 K > - 7.54

K > - 7.54/2.1

K > - 3.59

Also

4K >0

K > 0/4

K > 0

Similarly;

XY - 7.32 K > 0

\((\dfrac{3.8+1.9K-4K}{1.9})[11.34 - 1.9(\dfrac{3.8+1.9K-4K}{1.83}) > 7.32 \ K]\)

0.54(2.1K+7.54)>7.32 K

11.45 K < 4.07

K < 4.07/11.45

K < 0.35

Thus the range of K can be said to be :  -3.59 < K< 0.35

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

Then  you go to find the system requirements are the D. None of these as none of it dictates about the  find the system requirements hen needed.

User necessities describe what the person ought to do . System necessities describe how will the person acquire person necessities while interacting with the device plus nonpurposeful necessities such as "the device ought to cope with one hundred thousand customers on the identical time". So: User requirement : The person ought to see their check results.Where is device necessities.

For packaged products, device necessities are regularly imprinted on the packaging. For downloadable products, the device necessities are regularly indicated at the download page. System necessities may be widely categorized as purposeful necessities, records necessities, pleasant necessities and constraints.

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The quality of an asphalt mixture is determined by factors such as the type and gradation of aggregate, the quality of the asphalt binder, the proportion of each component, and the method of mixing and compaction.

A quality asphalt mixture is composed of several components that are carefully selected and proportioned to meet specific performance requirements. The key components of a quality asphalt mixture include: Aggregate: The aggregate is the primary structural component of the asphalt mixture and is composed of a combination of coarse and fine particles of crushed rock, gravel, and sand. The aggregate provides strength and stability to the pavement structure. Asphalt binder: The asphalt binder is a sticky, black, viscous liquid that holds the aggregate together and provides flexibility and durability to the pavement. The binder is typically made from crude oil and may be modified with polymers to improve its performance. Mineral fillers: Mineral fillers such as limestone dust, hydrated lime, or Portland cement are often added to the asphalt mixture to improve its strength, stability, and resistance to cracking. Additives: Additives such as fibers, rubber, or recycled asphalt pavement (RAP) may be added to the asphalt mixture to improve its performance and reduce costs. A well-designed and properly constructed asphalt mixture can provide a durable, long-lasting pavement that can withstand heavy traffic loads and extreme weather conditions.

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Answer:

whatsthe question choices

Answer:

i think its called leggings thats wut my shop teacher told me

Explanation:

The same as when rounding to three decimal places, we also round a number to three significant figures.

For three digits, we start counting with the first non-zero digit. Next, we round the final digit. Any empty spaces to the right of the decimal point are filled with zeros. As a result, when 1,500 is written without a decimal point, the two trailing zeros are not important; instead, the number has two significant figures. The fact that 1,500.00 has a decimal point, however, makes all six digits important. Significant figures allow us to demonstrate a number's accuracy. A number's integrity is compromised if it is used to express something that is outside of its real range of measurement.

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Coal-burning power plants emit pollutants such as sulfur dioxide, nitrogen oxides, and particulate matter. These pollutants can cause local air pollution, which can affect human health and the environment.

What is power plant?
A power plant is a type of industrial building used to produce electricity using raw materials. To meet society's electrical needs, the majority of power plants employ one or more generators that transform mechanical energy into electrical energy. Solar power plants are an exception, as they produce this electricity without the use of a turbine by using photovoltaic cells.

Air pollution from coal-burning power plants can travel through the atmosphere and deposit on other surfaces, including vegetation and water bodies. This deposition of pollutants can lead to acid rain, which can damage forests and aquatic ecosystems.

In addition, coal-burning power plants emit greenhouse gases such as carbon dioxide and methane, which contribute to climate change. Climate change can cause shifts in weather patterns, rising sea levels, and an increase in extreme weather events. These changes can affect human health, agriculture, and ecosystems across the globe.

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The magnitude of the velocity vector represents the object's current speed.The velocity vector points in the same general direction as the motion of the object.

The vector quantity velocity (v), denoted by the equation v = s/t, quantifies displacement (or change in position, s), over change in time (t).

A hanging mass of 25 kg will exert a force of:

25 * 9.81 = 254.25 Newtons on the string

Thus, during rotation, the force exerted should remain less than this. Force on an object in circular motion is given by:

F = (m * v^2) / r

F = mv2/r = 3.0(v2)/0.8 = (25)9.8 = 245

v = sqrt(245 x 0.8 / 3.0) = 8.1 m/s

v is from 0 to 8.1 m/s

Thus, the magnitude of the velocity should not exceed 8.1  meters per second.

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Answer:

The answer is "\(\bold{1.32 \times 10^{-3} \ cm}\)".

Explanation:

All of the atoms in a BCC crystalline structure are contained in the 8-corner unit cell.  

Each corner connects the atom to a single cell \(\frac{1}{8}\)  

Therefore, the unit cell number of atoms:  

\(= 8 \times \frac{1}{8}+ 1 \\\\= 1+1 \\\\= 2 \ atom\)

\(The mass unit cell = \frac{ \text{Number of atoms} \times \text{atomic weight}} {Avagadro number}\\\\= \frac{2 \times 50.9}{6.023 \times 10^{23}}  \\\\= 1.69 \times 10^{-22} \ g\\\\Area Of the atom= \frac{4r}{\sqrt{3}}\\\\   5.96 = \frac{1.69 \times 10^{-22}}{volume}\\\\volume= 2.835 \times 10^{25}\\\\v=d^3\\\\v= (\frac{4r}{\sqrt{3}})^3\\\\\to 2.835 \times 10^{-23} \times (\sqrt{3})^3 = 4^3 r^3\)  

\(\to \sqrt[3]{\frac{{2.835 \times 10^{-23} \times (\sqrt{3})^3}}{4^3}} =r\\\\\to r= 1.32 \times 10^{-3} \ cm\)

It is recommended that the switched portion of an outlet be mounted towards the top.

The reasoning behind this recommendation is that it will allow easier access to the switch while also reducing the risk of accidental contact with the live portion of the outlet. However, there may be some situations where it is more practical to mount the switched portion towards the bottom.

For example, if the outlet is being installed in a location where accessibility is more important than safety, such as in a garage or workshop, it may be preferable to mount the switch at the bottom to make it easier to reach from a standing or seated position. In general, though, mounting the switched portion of an outlet towards the top is the safer and more practical choice in most situations.

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Medium grains are those that are between 1/64 to 1/16 mm in size. The correct answer is option D.

Medium grains, in the context of particle size classification, refer to particles that fall within a specific size range. The size range for medium grains is between 1/64 to 1/16 mm.

To understand this size range, it helps to know that particle sizes are often measured in terms of fractions or decimal equivalents of an inch. In this case, the fractions 1/64 and 1/16 represent specific divisions of an inch.

1/64 inch is a smaller fraction than 1/16 inch. It means that the size of the particles falling within the medium grain range is larger than particles in the fine grain range (which would be smaller than 1/64 inch) but smaller than particles in the coarse grain range (which would be larger than 1/16 inch).

So, when it is stated that medium grains are between 1/64 to 1/16 mm in size, it means that the particles within this range have sizes larger than 1/64 inch but smaller than 1/16 inch.

It's important to note that particle size classification can vary depending on the specific industry or application. Different classification systems might use different size ranges and units of measurement.

Therefore option D is correct.

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Answer:

0.35

Explanation:

We resolve the component of the weight of the car along and perpendicular to the grade. We have mgsinФ and mgcosФ where Ф = angle of grade.

Now, the frictional force f = μN = μmgcosФ where μ = coefficient of friction

So, the net force along the grade is F = mgsinФ - μmgcosФ.

The work done by this force moving a distance, d along the grade is

W = (mgsinФ - μmgcosФ)d

This work equals the change in kinetic energy of the car. So ΔK = 1/2m(v₂² - v₁²) = W = (mgsinФ - μmgcosФ)d

1/2m(v₂² - v₁²) = (mgsinФ - μmgcosФ)d

1/2(v₂² - v₁²) = (gsinФ - μgcosФ)d

(v₂² - v₁²)/2d = (gsinФ - μgcosФ)

dividing through by gcosФ, we have

(v₂² - v₁²)/2dgcosФ = (gsinФ/gcosФ) - μgcosФ/gcosФ

(v₂² - v₁²)/2dgcosФ = tanФ -  μ

μ = tanФ - (v₂² - v₁²)/2dgcosФ

given that tanФ = 3% = 3/100 and 1 + tan²Ф = 1/cos²Ф, cosФ = 1/(√1 + tan²Ф) = 1/(√1 + (3/100)²) = 1/(√1 + (9/10000)) = 1/(√10000 + 9/10000) = 1/√(10009/10000) = 100/√10009 = 100/100.05 = 0.9995.

Also, given that v₁ = 90 km/h = 90 × 1000/3600 m/s = 25 m/s and v₂ = 45 km/h = 45 × 1000/3600 m/s = 12.5 m/s, d = 75 m and g = 9.8 m/s².

So, substituting the values of the variables into the equation, we have

μ = tanФ - (v₂² - v₁²)/2dgcosФ

μ = 3/100 - ((12.5 m/s)² - (25 m/s)²)/(2 × 75 m × 9.8 m/s² × 0.9995)

μ = 3/100 - ((156.25 m/s)² - (625 m/s)²)/1,469.265 m²/s²

μ = 3/100 - (-468.75 m²/s²)/1,469.265 m²/s²

μ = 3/100 + 468.75 m²/s²/1,469.265 m²/s²

μ = 0.03 + 0.32

μ = 0.35

So, theoretical friction  coefficient is 0.35

To determine the highest value of Vr for which the transistor still operates in the active mode, we need to consider the transistor's voltage characteristics.

In the active mode, the transistor operates as an amplifier, and Vr should be within certain limits. One important parameter to consider is the base-emitter voltage (Vbe). When Vbe exceeds the threshold voltage, the transistor starts to enter saturation.

In the given circuit of Fig. P6.51, with a very high β (amplification factor), the base current (Ib) can be considered negligible. Therefore, the collector current (Ic) flows through the resistor Rc.

To find the highest value of Vr for active mode operation, we need to ensure that the voltage drop across Rc (Vrc) is sufficiently high to keep the transistor in active mode. This is typically achieved by ensuring that Vrc is at least a few times Vbe.

To find the value of Vr for which the transistor operates in saturation with a forced β of 3, we need to consider the transistor's saturation characteristics. In saturation mode, the transistor acts as a switch, and Vrc should be reduced to a low voltage level.

To provide a more accurate analysis and determine the specific values of Vr for active mode and saturation, the circuit diagram (Fig. P6.51) and transistor parameters (such as Vbe, β, etc.) need to be provided.

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With regard to the data rate, note that since the message is not that big, X is likely to be on the order of 1 millisecond. (Option C)

The speed of the cars   is irrelevant in this scenario.The only thing that matters isthe amount of time it takes to transmit the data.

The data rate of the wireless technology is   equivalent to fast Ethernet speed, which is 100 Mbps.The message is 19.9452909 MB, which is 19,945,290.9 bytes. The time it takes to transmitthe message   is

time = (19,945,290.9 bytes) / (100 Mbps)

= 199.4529 seconds

= 199.4529 milliseconds

The security technique takes X milliseconds to exchange the security keys and encrypt the message.

Therefore, X must be less than or equal to 199.4529 milliseconds. Since the message is not that big, X is likely to be on the order of 1 millisecond. (Option C).

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Answer:

B. Iron

Explanation:

took the test.

A computer is an electronic device that processes and stores data, performs calculations, and executes instructions to carry out various tasks. The most likely causes for the computer spontaneously shutting down within a short period of time are:

2.) The power connector for the fan was not connected to the motherboard: If the fan is not receiving power, the CPU may overheat quickly, triggering a shutdown to protect the system.

3.) The heat sink and fan were not installed correctly: Improper installation of the heat sink and fan can result in inadequate cooling, causing the CPU to overheat and leading to shutdowns for thermal protection.

While other factors like a bad CPU (5) or an unsupported CPU by the BIOS (1) could potentially cause shutdowns, the power connector and heat sink installation issues are more commonly associated with sudden and frequent shutdowns. It is recommended to ensure the fan is connected and the heat sink is properly installed to resolve the overheating issue and prevent the spontaneous shutdowns.

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Explanation:

print() has a variable number of parameters. this is the answer.

hope this helps you

have a nice day

(a) The probability of drawing a black fuse from the second toolbox is 4/7.

(b) The probability of drawing a white fuse from the second toolbox is 3/7.

(c) The probability that one fuse is white is 27/49.

To calculate the probabilities, we need to consider the number of white and black fuses in each toolbox.

First, let's determine the total number of fuses in each toolbox:

First Toolbox:

Total fuses = 4 white fuses + 3 black fuses = 7 fuses

Second Toolbox:

Total fuses = 3 white fuses + 5 black fuses = 8 fuses

(a) The probability that the fuse drawn from the second toolbox is black:

The probability of drawing a black fuse from the second toolbox depends on the fuse selected from the first toolbox. There are two scenarios to consider:

Scenario 1: The fuse selected from the first toolbox is black.

In this case, the second toolbox will have 3 black fuses remaining out of the total 7 fuses.

Probability = (Number of black fuses in the second toolbox)/(Total number of fuses in the second toolbox)

Probability = 3/7

Scenario 2: The fuse selected from the first toolbox is white.

In this case, the second toolbox will have 5 black fuses out of the total 7 fuses.

Probability = (Number of black fuses in the second toolbox)/(Total number of fuses in the second toolbox)

Probability = 5/7

To calculate the overall probability, we need to consider the probability of each scenario and weigh it by the probability of selecting a fuse of that color from the first toolbox. Since the fuse from the first toolbox has an equal chance of being white or black, we need to take an average of the probabilities in both scenarios.

Probability of drawing a black fuse from the second toolbox = (Probability of scenario 1 + Probability of scenario 2)/2

Probability of drawing a black fuse from the second toolbox = (3/7 + 5/7)/2

Probability of drawing a black fuse from the second toolbox = 8/14

Probability of drawing a black fuse from the second toolbox = 4/7

Therefore, the probability of drawing a black fuse from the second toolbox is 4/7.

(b) The probability that the fuse drawn from the second toolbox is white:

Similarly, we can calculate the probability of drawing a white fuse from the second toolbox. It will be the complement of the probability of drawing a black fuse.

Probability of drawing a white fuse from the second toolbox = 1 - Probability of drawing a black fuse from the second toolbox

Probability of drawing a white fuse from the second toolbox = 1 - 4/7

Probability of drawing a white fuse from the second toolbox = 3/7

Therefore, the probability of drawing a white fuse from the second toolbox is 3/7.

(c) The probability that one is white:

To calculate the probability that one fuse is white, we need to consider both scenarios: either the fuse drawn from the first toolbox is white and the fuse drawn from the second toolbox is black, or the fuse drawn from the first toolbox is black and the fuse drawn from the second toolbox is white.

Scenario 1: Fuse drawn from the first toolbox is white, and the second toolbox has a black fuse:

Probability = (Probability of drawing a white fuse from the first toolbox) * (Probability of drawing a black fuse from the second toolbox)

Probability = (3/7) * (5/7)

Scenario 2: Fuse drawn from the first toolbox is black, and the second toolbox has a white fuse:

Probability = (Probability of drawing a black fuse from the first toolbox) * (Probability of drawing a white fuse from the second toolbox)

Probability = (4/7) * (3/7)

To calculate the overall probability, we need to sum up the probabilities of both scenarios:

Probability that one fuse is white = Probability

of scenario 1 + Probability of scenario 2

Probability that one fuse is white = (3/7) * (5/7) + (4/7) * (3/7)

Probability that one fuse is white = 15/49 + 12/49

Probability that one fuse is white = 27/49

Therefore, the probability that one fuse is white is 27/49.

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Answer:

B) Minimum voltage drop across meter

Explanation:

An ammeter is a measuring instrument device that is used to measure small currents flowing through a circuit in milli-ampere. The internal resistance of the milli ammeter is very low so that there would be a minimum voltage drop across the meter causing a minimum effect of the current in the circuit, this results to a measurement closer to the original value. If a high resistance is place, there would be large voltage drop therefore the current measured would be lesser than the original value.