Once the external shape and internal features of a part are represented by a combination of lines, further information is provided by ____.

Answer:

b

Explanation:

Option A. The techniques of the cybersecurity discipline include encryption, access control, physical security, training, threat analysis, and authentication.

Encryption is the process of encoding information so that only authorized parties can access it. Access control is a process used to limit access to resources to only those individuals who have the necessary credentials. Physical security involves the use of physical barriers, such as locks and fences, to protect assets.

Training provides individuals with the skills and knowledge to maintain the security of the system. Threat analysis involves evaluating risks and threats to the system and identifying countermeasures. Authentication is the process of verifying an individual's identity to ensure they are who they say they are.

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Answer: Cleaning of mechanical parts is necessary to remove contaminants, and to avoid clogging of wastes which could restrict the functioning of the machine.

Explanation:

There are different agents used for cleaning different machine instruments to prevent their corrosion and experience proper cleaning.

Backing plates must be dry cleaned using a cotton cloth to remove the dirt, dust or any other dry contaminant.

Struts can be wet cleaned by applying alcoholic solvent.

Levers can be cleaned using a mineral spirit.

Metallic plates can be cleaned using water based solution or water.

Answer:

0.49

0.51

Explanation:

Probability that bulb is defective :

Let :

b1 = box 1 ; b2 = box 2 ; b3 = box 3

d = defective

P(defective bulb) = (p(b1) * (d|b1)) + (p(b2) * p(d|b2)) + (p(b3) * p(d|b3))

P(defective bulb) = (1/3 * 10/20) + (1/3 * 7/15) + (1/3 * 5/10))

P(defective bulb) = 10/60 + 7/45 + 5/30

P(defective bulb) = 1/6 + 7/45 + 1/6 = 0.4888

= 0.49

P(bulb is good) = 1 - P(defective bulb) = 1 - 0.49 = 0.51

The area of the gutter is therefore equal to the area of the rectangle plus the area of the triangles.

A = rectangle + 2 * triangle

= 10h + 2 * (1/2 hw)

Using trig work out h and w.

h = 10 sin ?

w = 10 cos ?

A = 10h + hw

= 100sin? + 100sin?cos?

Differentiate the expression (either use the product rule for sin? cos?, or write 100sin?cos? as 50sin2? using the double angle formula)

A = 100sin? + 50sin2?

dA/dt = 100cos? + 100cos2?

For A to be a max or min the derivative must equal zero

100cos? + 100cos2? = 0

cos? + cos2? = 0

Use the double angle formula to write cos2? as 2cos^2 ? - 1

cos? + 2cos^2 ? - 1 = 0

2cos^2 ? + cos? - 1 = 0

This quadratic factorizes:

(2cos? - 1)(cos? + 1) = 0

This means that

cos ? = 1/2 or cos ? = 1

giving:

? = pi/3 or ? = pi

Which means that to achieve the maximum area ? should be pi/3 = 60 degrees

You can check that this is the max by substituting in these values of theta and the two end points (theta = 0, theta = pi), and seeing that pi/3 gives the maximum.

The maximum area is therefore

A = 100sin? + 100sin?cos?

= 100sin(pi/3) + 100sin(pi/3)cos(pi/3)

= 100 sqrt(3) / 2 + 100 sqrt(3)/2 * 1/2

= 100 * 3/2 * sqrt(3) / 2

= 3 * 25 * sqrt(3)

= 75 sqrt(3)

= 129.9...

= 130 cm^3

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The original amounts of the liquids A and B based on the information will be 1260kg and 872kg.

An equation is the statement that illustrates that the variables given. In this case, two or more components are taken into consideration to describe the scenario. It is vital to note that an equation is a mathematical statement which is made up of two expressions that are connected by an equal sign.

Based on the information given, let us call the liquids illustrated as A and B, while the weight is in kg.

The first equation is the amount of solids and this will be represented thus:

0.05A + 0.11B + 18.5 = 0.0825(2150) = 177.375

0.05A + 0.11B = 158.875

5A + 11B = 15887.5

The second equation will be represented thus:

A + B + 18.5 = 2150

Collect like terms

A + B = 2150 - 18.5

A + B = 2131.5

Multiplying the equation will be:

6B = 5230

Divide through by 6

6B / 6 = 5230 / 6

B = 872 kg

Therefore, A will be 1260 kg.

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The displacement of end D with respect to end A is \(-0.488 \times 10^{-3}\;m\)

Given the following data:

To determine the displacement of end D with respect to end A:

First of all, we would do a sectional cut of the circular rods at point A and then determine the sum of forces acting on it:

\(\sum F=0\\\\ 10000+N_1=0\\\\N_1 = -10000\;N\)

Mathematically, the displacement of a circular rod is given by this formula:

\(d = \frac{NL}{AE}\)

Where:

Substituting the given parameters into the formula, we have;

\(d_1 = \frac{-10000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_1 = \frac{-4350}{0.0003142 \times 68.9 \times 10^9}\\\\d_1 =-0.201 \times 10^{-3}\;meter\)

For section cut BC:

\(\sum F=0\\\\ 10000-20000+N_2=0\\\\N_2 = 10000\;N\)

\(radius = outer\;radius^2 - inner\;radius^2\\\\radius = 0.04^2 -0.03^2=0.0007\;m\)

\(d_2 = \frac{10000 \times 0.435}{\frac{3.142 \times 0.0007^2}{4} \times 68.9 \times 10^9} \\\\d_2 = \frac{4350}{0.00054985 \times 68.9 \times 10^9}\\\\d_2 =0.115 \times 10^{-3}\;meter\)

For section cut D:

\(\sum F=0\\\\ -20000-N_3=0\\\\N_3 =- 20000\;N\)

\(d_3 = \frac{-20000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_3 = \frac{-8700}{0.0003142 \times 68.9 \times 10^9}\\\\d_3 =-0.402 \times 10^{-3}\;meter\)

For the total displacement:

The total displacement is equal to the displacement of end D with respect to end A.

\(d_T = d_1 +d_2+d_3\\\\d_T = -0.201 \times 10^{-3} + 0.115 \times 10^{-3}-0.402 \times 10^{-3}\\\\d_T = -0.488 \times 10^{-3}\;m\)

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Answer:

Explanation:

This is Answer....

To write a computer program that determines how many grades are between 0 and 19, between 20 and 39, between 40 and 59, between 60 and 79, and between 80 and 100 and display the results using the command fprintf we will follow the steps given below:

Step 1: Create a list of exam scores.

Step 2: Initialize 5 variables to 0 for each grade range

Step 3: Create a for loop that will iterate through each score in the list.

Step 4: For each score, use conditional statements to check which grade range it falls into and increment the corresponding variable for that range by 1.

Step 5: Use the command fprintf to display the results in the required format. Below is the program that will implement the above steps:```
scores = [31, 70, 92, 5, 47, 88, 81, 73, 51, 76, 80, 90, 55, 23, 43,98,36,87,22,61, 19,69,26,82,89,99, 71,59,49,64]
range1 = 0
range2 = 0
range3 = 0
range4 = 0
range5 = 0
for score in scores:
   if score >= 0 and score <= 19:
       range1 += 1
   elif score >= 20 and score <= 39:
       range2 += 1
   elif score >= 40 and score <= 59:
       range3 += 1
   elif score >= 60 and score <= 79:
       range4 += 1
   elif score >= 80 and score <= 100:
       range5 += 1
fprintf('Grades between 0 and 19: %d students\n', range1)
fprintf('Grades between 20 and 39: %d students\n', range2)
fprintf('Grades between 40 and 59: %d students\n', range3)
fprintf('Grades between 60 and 79: %d students\n', range4)
fprintf('Grades between 80 and 100: %d students\n', range5)

```The above program will print the number of students in each grade range in the required format.

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In contrast to mobile work zones, stationary work zones have clearly defined boundaries.

According to the 2009 edition of the Manual on Uniform Traffic Control Devices (MUTCD), a short-term stationary work zone is "daytime work that occupies a location for more than one hour within a single daylight period."

The following are typical daytime work shift-length tasks:

paving the road, fixing broken underground water lines, fixing storm water catch basins, trimming trees along the road, and parking work vehicles for a long time on the side of the road. The minimum traffic control planning and devices that are required to safely direct motorists, bicyclists, and pedestrians around workers and work vehicles during daylight hours will be discussed in this bulletin. Night operations necessitate additional equipment and planning, which is not the subject of this bulletin.

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Answer:

Explanation:

In satellite communications, downlink is the establishment of a communications link from an orbiting satellite down to one or more ground stations on Earth. Contrast with uplink.

Answer:

Shallow foundations, often called footings, are usually embedded about a metre or so into soil. ... Another common type of shallow foundation is the slab-on-grade foundation where the weight of the structure is transferred to the soil through a concrete slab placed at the surface.

Explanation:

Because I said so.

The uncompressed values from given data is s = 10 p(s) = 1/16

compressed file                                                                  uncompressed file

l=-(0.5 log 2 0.5)...= h bit/sample                                         2 bits sample

100,000 bits                                                                                     x bits

x bits ÷ 100000 bits = 2 bits/sample ÷ h bits sample

A data stream is the transmission of a series of coherent signals that have been digitally encoded to carry information. The sent symbols are often organized into a number of packets.

Data streaming is now commonplace. Any communication sent via the Internet is done so as a data stream. When speaking on a phone, the sound is transmitted as a data stream. Depending on the data format selected, Data Stream comprises several sorts of data. When determining the time of an event, attributes like the Timestamp attribute are helpful. An algorithmically encoded ID called "Subject ID" has been taken out of a cookie.

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Answer:

d

Explanation:

d

The match up are:

This is known to be a tool that is often used to aid the reaction exhaust gases as it moves away from a controlled combustion which often occurs in an engine.

Therefore, The match up are:

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Answer:

Hydrostatic force = 41168 N

Explanation:

Complete question

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water  so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Let "x" be the side length submerged in water.

Then

w(x)/base = (4+3-x)/altitude

w(x)/5 = (4+3-x)/3

w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of  x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))

HF = 658.69 N *62.5 = 41168 N

Answer:

120

Explanation:

In the radio communication system, multipath is the propagation phenomenon that causes diffusion or reflection in multiple different directions of a signal.

Multipath is a propagation mechanism that impacts the propagation of signals in radio communication. Multipath results in the transmission of data to the receiving antenna by two or more paths. Diffusion and reflection are the causes that create multiple paths for the signal to be delivered.

Diffraction occurs when a signal bends around sharp corners; while reflection occurs when a signal impinges on a smooth object. When a signal is received through more than one path because of the diffraction or reflection, it creates phase shifting and interference of the signal.

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Answer:

  0°

Explanation:

The resistor voltage has the same phase as the circuit current. There is no phase difference.

Answer:

Explanation:

Based on the information provided by these technicians, both of them are correct.

According to heating, ventilation, and air conditioning (HVAC), heat in an air-conditioning system generally flows from the hotter object to the colder object.

As the refrigerant evaporates in a small radiator-type unit (evaporator), it absorbs heat as it changes phase from liquid to gas. Also, as the heat is being absorbed by the refrigerant, the small radiator-type unit (evaporator) becomes cold.

In conclusion, we can logically deduce that both of them are correct.

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Answer:

i really font onow why tbh eot you

Answer:

It may be non-unique

Explanation:

Answer:al part of question

Explanation:voltage

Answer:

TRUE

Explanation: I think

protecting the ecosystem saves the species is true because a lot of species lives in the ecosystem and is there habitat / i think

This question is incomplete, the complete question is;

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of 1.5 mol/L, while the groundwater circulating around the pit flows fast enough that the contaminate concentration remains 0. There is initially no contaminant in the barrier material at the time of installation. The governing second order, partial differential equation for diffusion of the contaminant through the barrier is:

dC/dt = D( d²C / dz²)

where c(z,t) represent the concentration of containment of any depth into the barrier at anytime and D is the diffusion coefficient (a constant) for the containment in the barrier material.

a) write all boundary and initial conditions needed to solve this equation for C(z, t)

b) Find the steady  state solution (infinite time) for C(z)

Answer:

a) At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b) C(z) = z² - 4.15z + 1.5

Explanation:

a)

The boundary and initial conditions are as follows

At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b)

The governing second order, partial differential equation for diffusion of the contaminant through the barrier is :

(dC/dt) = D*(d²C/dz²) ..............equ(1)

For steady state, above equation becomes,

(d²C/dz²) =0

Integrating above equation,

(dC/dz) = Z + C1  { where C1 is integration constant) }

again integrating above equation,

C = z² + C1*z + C2    ...................equ(2)

applying boundary condition : at t =0, z= 0, c = 1.5 mol/L, to above equation

 C = z² + C1*z + C2

1.5 = 0 + 0*0 + c2

C2 = 1.5

applying boundary condition : at t =0, z= 0.4m, c = 0 mol/L, to equation (2) ,

0 = 0.4² + C1*0.4 +  1.5

0 = 0.16 + 0.4C1 + 1.5

0.4C1 = - 1.66

C1 = -1.66/0.4

C1 = -4.15

So, the steady state solution for C(z) is:

C(z) = z² - 4.15z + 1.5

Answer:

B....................

True, ABS enables steering while using the most forceful stopping possible.

The most common material used in manufacturing, ABS is available in many different varieties. It can be extruded for CNC machining, turned into pellets for injection moulding, and filament for additive manufacturing.

Three monomers—acrylonitrile, butadiene, and styrene—combine to form ABS. A manufactured monomer made from propylene and ammonia is called acrylonitrile.

This element helps ABS maintain its heat stability and chemical resilience. Butadiene: It is created as a by-product of steam crackers, which generate ethylene.

The best driving conditions for ABS are dry, level surfaces. It aids the driver in applying the brakes as quickly as feasible while keeping the car under control. However, as the system repeatedly releases the brake, it can lengthen total stopping distances.

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Answer:

The original seven wastes, or muda, are transportation, inventory, motion, waiting, overproduction, overprocessing and defects. ... When manufacturers are able to identify the seven wastes, they can correct and prevent further loss of time, money and other resources.

Explanation:

Answer:

transportation, inventory,motion, waiting,overproduction,overprocessing,and defects

Explanation:

they are often referred to by the acronym TIMWOOD