which of the following code segments will move the robot to the gray square along the path indicated by the arrows?

The Access Control List (ACL) that would permit traffic from hosts only on the 192.168.8.0/22 subnet is "permit 192.168.8.0 0.0.7.255." The correct option is (d).

To break it down, the "permit" keyword is used to indicate that traffic from the specified subnet will be allowed. The IP address 192.168.8.0 represents the starting address of the subnet, and the wildcard mask 0.0.7.255 indicates the range of IP addresses in the subnet.

A wildcard mask is used in Cisco IOS devices to identify which bits in an IP address should be ignored when comparing it to another IP address. In this case, the 0s in the wildcard mask indicate that the corresponding bits in the IP address should be considered, while the 1s indicate that they should be ignored.

Therefore, this ACE will permit traffic only from hosts on the 192.168.8.0/22 subnet and deny traffic from any other sources. It is important to ensure that the correct ACL is applied to the appropriate interface to prevent unauthorized access to your network.

Therefore, the correct answer is an option (d).

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Answer:

an ability to identify, formulate, and solve engineering problems. an understanding of professional and ethical responsibility. an ability to communicate effectively. the broad education necessary to understand the impact of engineering solutions in a global, economic, environmental, and societal context

Explanation:

umm can you explain me what i have to answer

Answer:

I looked at the comments said oh h e double hockey sticks no

Explanation:

To solve this problem, we can use the following formula:

t = (m * c * ΔT) / P

where t is the time taken to warm the plasma to 90°, m is the mass of the plasma, c is the specific heat capacity of the plasma, ΔT is the change in temperature (90° - 40° = 50°), and P is the power of the oven.

We can assume that the mass and specific heat capacity of the plasma are constant.

If the oven is set at 100°, we have:

t = (m * c * ΔT) / P

t = (m * c * 50) / (100 - 40) (since P = 100 - 40 = 60)

t = (m * c * 50) / 60

t = (5m * c) / 6

If the oven is set at 140°, we have:

t = (m * c * ΔT) / P

t = (m * c * 50) / (140 - 40) (since P = 140 - 40 = 100)

t = (m * c * 50) / 100

t = (m * c) / 2

If the oven is set at 80°, we have:

t = (m * c * ΔT) / P

t = (m * c * 50) / (80 - 40) (since P = 80 - 40 = 40)

t = (m * c * 50) / 40

t = (5m * c) / 8

Therefore, it will take 5/6 times as long (or approximately 42.5 minutes) if the oven is set at 100°, half as long (or 22.5 minutes) if the oven is set at 140°, and 5/8 times as long (or approximately 28.1 minutes) if the oven is set at 80°, compared to the original time of 45 minutes when the plasma was placed in an oven at 120°.

Answer:

i) SF = 0.83

ii) 10 psi

iii) 3.16 psi

iv) 0.0083

Explanation:

Constant load = 1000 Psi

mean resistance = 1200 psi

Probability of failure = 1.9 * 10^-3

Determine

i) Safety factor

S.F =  1 / (mean load / load design ) = 1 / ( 1200 / 1000 ) = 0.83

ii) Standard deviation

1.9 X 10^-3 = e  ( -1/2 ( μ / б) ^3  /  (2.5 * 6  )

0.004756 = e^- 20000 / б^3

hence std = 10 psi

iii) Variance

= \(\sqrt{10}\) = 3.16 psi

iv) coefficient of variation

Cv = std / mean resistance

    = 10 / 1200 = 0.0083

According to the National Electrical Code (NEC), the minimum size copper grounding electrode conductor   (GEC) required for a 6000 amp service supplied by ten 750 kcmil Copper service conductors in parallel for each ungrounded serviceentrance phase conductor is 3/0 AWG.

This is determined by   NEC 250.66(A)(1),which states that the GEC must be not smaller than the largest ungrounded service entrance conductor. In this case, the largest    ungrounded service entrance conductor is 750 kcmil, so the GEC must be 3/0 AWG.

It is important to note that the   GEC must be copper. AluminumGECs are not permitted   for services over 400 amperes.

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The 90% confidence interval for the true average yield point of the modified bar is approximately 8428.5 lb to 8475.5 lb. To increase the confidence level to 92%, the interval becomes 8426.97 lb to 8477.03 lb.

(a) To compute the 90% confidence interval (CI) for the true average yield point of the modified bar, we can use the formula:

CI = sample mean ± \($$(critical value \times $ standard deviation / \sqrt{(sample size)})\)

Find the critical value corresponding to a 90% confidence level. This can be obtained from the standard normal distribution table or using a statistical calculator. For a 90% confidence level, the critical value is approximately 1.645.

Calculate the standard deviation (σ) of the yield point. The given standard deviation (σ) is 100.

Determine the sample size (n) which is 49.

Now, we can calculate the confidence interval:

CI = 8452 ± \((1.645 \times 100 / \sqrt{(49)})\)

CI = 8452 ± \((1.645 \times 100 / 7)\)

CI = 8452 ± 23.55

CI ≈ (8428.5, 8475.5) lb

Therefore, the 90% confidence interval for the true average yield point of the modified bar is approximately (8428.5 lb, 8475.5 lb).

(b) To modify the interval to obtain a confidence level of 92%, we need to find the new critical value.

Find the critical value corresponding to a 92% confidence level. This can be obtained from the standard normal distribution table or using a statistical calculator. For a 92% confidence level, the critical value is approximately 1.751.

Now, we can calculate the modified confidence interval:

CI = 8452 ± \((1.751 \times 100 / 7)\)

CI = 8452 ± 25.03

CI ≈ (8426.97, 8477.03) lb

Therefore, the modified confidence interval for the true average yield point of the modified bar, with a confidence level of 92%, is approximately (8426.97 lb, 8477.03 lb).

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Answer:

PF= .54

Explanation:

Power Factor equals working/real power (W) over apparent power (VA). 1.0 PF is an efficient equipment. PF= 22/(120*.34)

The power factor of the ballast is 0.54 which is the ratio of working power to apparent power.

The power factor is a measure of energy efficiency. It is typically expressed as a percentage, with a lower percentage indicating inefficient power usage.

The power factor (PF) is the ratio of working power (in kW) to apparent power (in kilovolt amperes) (kVA).

Given that a 120-volt fluorescent ballast has an input current of 0.34 ampere and an input power rating of 22 watts.

PF = (True power)/(Apparent power)

PF = W/VA

Here W = 22 watts, V = 120-volt, and A = 0.34 ampere

PF = 22 / (120 × 0.34)

PF = 22 / 40.8

PF = 0.5392

PF = 0.54

Thus, the power factor of the ballast is 0.54.

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Answer: Part tolerancing  

Answer:

You could send each person one half of the coordinates of the secret location, such as "110th street" to one person and "2nd avenue" to the other person. This way, they would need to work together to share their information and determine the exact location of the intersection.

This approach ensures that neither person can find the location on their own, as they only have half of the information needed to determine the intersection. Additionally, sharing the coordinates separately adds an extra layer of security to the meeting location as it would be difficult for anyone to determine the meeting location without both pieces of information.

However, it's important to ensure that each person understands the instructions clearly, so they know to work together to determine the secret location. It's also important to choose a location that is not well-known, so the possibility of someone stumbling upon the meeting location by chance is reduced.

Explanation:

Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%  

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( \(\frac{M1}{100}\) )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

The change in the diameter of the wood would be 5%

the new diameter would be 10.5 inches

Wood pole diameter = 10 inches

Moisture content = 5%

Fiber saturation point = 30 %

\(\frac{1}{5} (30-5)\)

= 25/5

= 5%

The percentage change in the diameter of the wood would be 5%

b. This wood is going to rise up instead of shrinking. This is due to the fact that the moisture content that it has has gone up by 55%

c. The new diameter that this wood would have

diameter = 10

moisture = 5%

D = D+D(m)

= 10 + 10(5%)

= 10.5 inches

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The carriage handwheel is used to manually position and/or hand feed the carriage in the longitudinal or Z axis.

IamSugarBee

Answer:

Explanation:

Power is the product of volts and amps:

  P = VI = (240 V)(34 A) = 8160 W = 8.16 kW

Energy is the product of power and time:

  (8.16 kW)(2 h) = 16.32 kWh

The resolved shear stress for the (010) [100] slip system is 186 MPa.

To determine the resolved shear stress for the (010) [100] slip system when a tensile stress of 352 MPa is applied along the [110] direction of a Po single crystal, one can use the Schmid's law of slip.

Schmid's law states that the resolved shear stress on a particular slip system must exceed a critical value for the slip system to be activated.

Mathematically, the resolved shear stress (τ) on a slip system is given by:

τ = σsin Φcos θ

where σ is the applied tensile stress, Φ is the angle between the slip direction and the tensile axis, and θ is the angle between the slip plane and the tensile axis.

For the (010) [100] slip system, Φ = 45° and θ = 35.3°.

Therefore,τ = 352 × sin 45 × cos 35.3τ = 186 MPa

Therefore, the resolved shear stress for the (010) [100] slip system is 186 MPa.

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Answer:

999999999999999999999

Answer:

no, the pure fission bomb is a hydrogen bomb

Answer: a pure fission bomb is not a nuclear

bomb

Explanation: fusion is the combining of hydrogen particles, whereas fission is the splitting of atoms.

Answer:

the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement is 47.8 in

Explanation:

Given that;

Weight of car W = 2600 lb

power = 80 hp = 44000 lb ft/s

Engine rpm = 3296

gear reduction ratio e = 10

drivetrain efficiency n = 95% = 0.95

wheel radius R = 16 in  = 1.3333 ft

Length of wheel base L = 95 in =

coefficient of road adhesion u = 0.60

height of center of gravity above pavement  h = 22 in

we know that;

Coefficient of rolling resistance frl = 0.01 for good wet pavement

distance of center of gravity behind the front axle lf = ?

Maximum tractive effort (Fmax) =  (uW / L) (lf - frl h) / (1 - uh / L)

First we calculate our Fmax to help us find lf

Power = Torque × 2π × Engine rpm / 60 )

44000 = Torque ( 2π×3296 / 60)

Torque = 127.5 lb ft  

so

Fmax = Torque × e × n / R

so we substitute in our values

Fmax = 127.5 × 10 × 0.95 / 1.333

Fmax = 908.66 lb

Now we input all our values into the initial formula

(Fmax) =  (uW / L) (lf - frl h) / (1 - uh / L)

908.66 =  [(0.6×2600/95) (lf - 0.01×22)] / [1 - 0.6×22) / 95]

908.66 = (16.42( lf - 0.22)) / 0.86

781.4476 = (16.42( lf - 0.22))

47.59 = lf - 0.22

lf = 47.59 + 0.22

lf = 47.8 in

Therefore the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement is 47.8 in

A variable increases or declines at a pace proportional to its present value in a relationship described by an exponential function.

The formula for exponential functions is y = a(b)x, where a and b are constants and x is the independent variable. For the specified function; P = (0.99987) ᵃ And is the height in metres. By adding the height to the equation, the altitude at 5000 feet can be computed. But first, we must convert from feet to metres. 1m Equals 12 feet. 5000 feet Equals x metres. x=1524 metres. putting the values in the equation as substitutes; P = (0.99987) ¹⁵²⁴. The provided exponential function yields 0.82atm as the pressure at 500 feet. 0.82 atmospheres are P. In a connection modelled by an exponential function, a variable grows or shrinks according to its present value.  

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a) Truth Table is: Truth table for 3-input combinational circuit with one outputS.No.InputsOutputA B C 1 0 0 0 0 2 0 0 1 0 3 0 1 0 0 4 0 1 1 1 5 1 0 0 0 6 1 0 1 1 7 1 1 0 1 8 1 1 1 1 b)K-map for the output is:

In the above K-Map, we are getting two groups of 4 '1's. So, SOP form of the output is:So, the SOP form of output is, c) Circuit diagram using minimum number of logic gates based on the simplified Boolean expression can be given as:Here, the simplified Boolean expression is given as:O = A'B'C + A'BC' + AB'C' + ABCSo, the minimum number of logic gates required to obtain the output is three.

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Answer:

The thickness of the walls of each hollow lump of "iron ore" is 2.2 cm

Explanation:

Here we have that the density of solid gold = 19.3 g/cm³

Density of real iron ore = 5.15 g/cm³

Diameter of sphere of gold = 4 cm

Therefore, volume of sphere = 4/3·π·r³ = 4/3×π×2³ = 33.5 cm³

Mass of equivalent iron = Density of iron × Volume of iron = 5.15 × 33.5

Mass of equivalent iron = 172.6 cm³

∴ Mass of gold per lump = Mass of equivalent iron = 172.6 cm³

Volume of gold per lump = Mass of gold per lump/(Density of the gold)

Volume of gold per lump = 172.6/19.3 = 8.94 cm³

Since the gold is formed into hollow spheres, we have;

Let the radius of the hollow sphere = a

Therefore;

Total volume of the hollow gold sphere = Volume of gold per lump - void sphere of radius, a

Therefore;

\(33.5 = 8.94 - \frac{4}{3} \times \pi \times a^3\)

\(\frac{4}{3} \times \pi \times a^3 = 33.5 - 8.94\)

\(a^3 = \frac{24.6}{\frac{3}{4} \pi } = 5.9\)

a = ∛5.9 = 1.8

The thickness of the walls of each hollow lump of "iron ore" = r - a = 4 - 1.8 = 2.2 cm.

Answer:

Fast Fourier ( FFT ) algorithms speed up computation of Fourier coefficients by simply reducing the the computing time of a traditional direct form  Fourier series. it  achieves this by breaking complex DFTS into smaller DFTS to reduce its complexity and in turn reduce its computing time

Explanation:

Fast Fourier ( FFT ) algorithms speed up computation of Fourier coefficients by simply reducing the the computing time of a traditional direct form  Fourier series. it  achieves this by breaking complex DFTS into smaller DFTS to reduce its complexity and in turn reduce its computing time. an example of such FFT is Cooley-Tukey algorithm

Answer:

Explanation:

Your 3-B Annual Credit Reports & Scores with Enhanced Credit Report Monitoring.

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The credit building options that I think  is the easiest method that you can see yourself doing is Secured credit cards.

Secured credit cards are known to be a type of card that are said to be backed by a specific refundable security deposit.

This kind of deposit is known to limit or lower the lender's risk and thus makes it better for people who do not have credit or when they do not  qualify for that type of credit. This credit option is easier to get when compared to unsecured cards, as it does not need a lot of requirement.

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Answer:

See calculation below

Explanation:

Given:

W = 30 kN = 30x10³ N

d = 75 mm

p = 6 mm

D = 300 mm

μ = tan Φ = 0.2

1. Force required at the rim of handwheel

Let P₁ = Force required at the rim of handwheel

Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm

Mean diameter of screw:    *d = \(\frac{do + dc}{2}\) = (75 + 69) / 2 = 72 mm

and

tan α = p / πd  =  6 / (π x 72)  =  0.0265

∴ Torque required to overcome friction at he threads is  T = P x d/2

T = W tan (α + Ф) d/2

T =  \(W(\frac{tan \alpha + tan \theta}{1 - tan \alpha + tan \theta } ) * \frac{d}{2}\)

T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2

T = 245,400 N-mm

We know that the torque required at the rim of handwheel (T)

245,400 = P1 x D/2 = P1 x (300/2) = 150 P1

P1 = 245,400 / 150

P1 = 1636 N

2. Maximum compressive stress in the screw

                         30x10³

Qc = W / Ac = -------------- = 8.02 N/mm²

                      π/4 * 69²

Qc = 8.02 MPa

Bearing pressure on the threads (we know that number of threads in contact with the nut)

n = height of nut / pitch of threads = 150 / 6 = 25 threads

thickness of threads, t = p/2 = 6/2 = 3 mm

bearing pressure on the threads = Pb = W / (π d t n)

Pb = 30 x 10³ / (π * 72 * 3 * 25)

Pb = 1.77 N/mm²

Max shear stress on the threads = τ = 16 T / (π dc³)

τ = (16 * 245,400) / ( π * 69³ )

τ = 3.8 M/mm²

*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72

∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))

τmax = 5.5 Mpa

3. efficiency of the straightener

To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm

∴Efficiency of the straightener is η =  To / T = 28,620 / 245,400

η = 0.116 or 11.6%

By assuming the coefficient of friction for the collar as 0.2. efficiency of straightner is 11.6%.

Wheels are circular frames or disks that are mounted on machines or vehicles and are designed to rotate around an axis.

Given:

W = 30 kN = 30x10³ N

d = 75 mm

p = 6 mm

D = 300 mm

μ = tan Φ = 0.2

1. Force required at the rim of handwheel :

Let P₁ = Force required at the rim of handwheel

Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm

Mean diameter of screw:    *d =  = (75 + 69) / 2 = 72 mm and

tan α = p / πd  =  6 / (π x 72)  =  0.0265

Torque required to overcome friction at he threads is  T = P x d/2

T = W tan (α + Ф) d/2

T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2

T = 245,400 N-mm

We know that the torque required at the rim of handwheel (T)

245,400 = P1 x D/2 = P1 x (300/2) = 150 P1

P1 = 245,400 / 150

P1 = 1636 N

2. Maximum compressive stress in the screw :

30x10³

Qc = W / Ac = -------------- = 8.02 N/mm²

π/4 * 69²

Qc = 8.02 MPa

Bearing pressure on the threads (we know that number of threads in contact with the nut)

n = height of nut / pitch of threads = 150 / 6 = 25 threads

thickness of threads, t = p/2 = 6/2 = 3 mm

bearing pressure on the threads = Pb = W / (π d t n)

Pb = 30 x 10³ / (π * 72 * 3 * 25)

Pb = 1.77 N/mm²

Max shear stress on the threads = τ = 16 T / (π dc³)

τ = (16 * 245,400) / ( π * 69³ )

τ = 3.8 M/mm²

*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72

max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))

τmax = 5.5 Mpa

Therefore, efficiency of the straightener :

To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm

Efficiency of the straightener is η =  To / T = 28,620 / 245,400

η = 0.116 or 11.6%

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The letters A to H on the blank, 'A' being the first step, and so on is as follows:

8-F, 9-E, 10-B, 11-C, 12-G, 13.D 14.A

In mathematics, a sequence is an ordered list of objects where repetitions are allowed. It has members and is comparable to a set (also called elements, or terms). The number of elements determines the length of the sequence (possibly infinite). In contrast to a set, the same elements can appear multiple times in a sequence at different positions, and unlike a set, the order matters.

Formally, a sequence can be defined as a function from natural numbers (the positions of the sequence's elements) to the elements at each of those positions. You can think of an indexed family, which is a function from any index set, as a generalisation of the concept of a sequence.

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Software Development and Client Needs

In Incremental method of software development customers who do not have a basic idea of the development process are being carried along on like other methods that will relegate them to the background until a product is ready.

With this model and structure in place, when softwares/ products are built from several stages e.g prototype, testing, and when new features are added customers are always carried along with their valuable feedback and suggested greatly considered to achieve the customers satisfactions

This model will work well for the customers/clients who does not have a clear idea on the systems needed for their operations.

In summary the incremental model combines features from the waterfall and prototyping model.

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The order the objective lenses listed from the smallest to largest working distance that results when they are used is:

Oil immersion, high power, and low power.

The explanation for the same is as follows:
1. Oil immersion lens: This lens has the smallest working distance, as it is designed for maximum resolution and requires immersion oil to bridge the gap between the lens and the specimen
2. High power lens: This lens has a larger working distance than the oil immersion lens but smaller than the low power lens. It is commonly used for detailed observation of specimens
3. Low power lens: This lens has the largest working distance among the three listed, and it is typically used for initial scanning and observation of specimens.

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In surface water, DDT will form a bond with water-borne particles, settle, and eventually be deposited in the sediment. DDT is absorbed by fish and other aquatic life.

DDT and its metabolites also enter the atmosphere through the volatilization of residues in soil and surface water, the majority of which are the result of previous use. The ground will be covered in these substances.

DDT remains in soil for a very long time. The majority of the DDT in soil will decompose within two to fifteen years. Microscopic plants and animals, sunlight, and/or evaporation into the atmosphere break down DDT in soil or surface water.

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