according to the practical evidence, in comparison to intermittent compression, constant compression units are primarily used for what two purposes?

Based on the code given , the Step 1  is done by enforcing maximum() and min() in ARM7 assembly while Step 2  is done by enforcing the main while circle in ARM7 assembly.

To apply the maximum() and min() functions in ARM7 assembly, we can use some instructions to optimize the law and code speed. There is one possible way to do it based on the image attached.

Therefore, To apply the main while loop, we need to check if the left over of the division between result and d isn't zero, and if so, add c to affect. We can use the SWI 0x6 instruction to gain the balance of the division.

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See text below

We will translate the following C function, which calculates the Lowest Common Multiple (1cm) between two positive numbers, into ARM7 assembly:

unsigned int lcm(unsigned int a, unsigned int b)

{

(unsigned int c, d, result; // cannot be translated)

c = max(a, b); // i.e. c-a if a >= b, otherwise c=b d = min(a, b); // same, but vice-versa

result = c;

while ((result % d) != 0) {

}

result = result + c;

}

(return(result);

// cannot be translated)

In doing so, you are asked to optimize the assembly code for execution speed. i.e., a) fewest number of cycles, taking into account pipeline emptying/refilling on branch instructions for example, and b) fewest number of instructions.

Disregard the first/last lines of the lcm() function, which cannot be translated into assembly. This exercise is broken into these two steps:

1. First, implement the two successive lines min() and max() in ARM7 assembly:

c = max(a, b); // i.e. c-a if a>=b, otherwise c=b

d = min(a, b); // same, but vice-versa

[Hint 1]: Remember that the function's arguments, a and b, will be found in registers r0 and rl respectively. Every other register is free for you to use in your code. Remember to use conditional instructions whenever possible, to optimize code density and execution speed.

2. Implement the main while loop in ARM7 assembly.

[Hint 2]: The remainder of a division (i.e., the “modulo” operation %) can be obtained by relying on the BIOS, using the ARM7 instruction: SWI 0x6 (which cannot be conditional) after placing the input in the appropriate, predefined registers (see lecture notes), which cannot be selected/changed, and reading the output from the appropriate register.

Remember to comment each line of your program to explain what your code does.



Describe the changes to the memory and the registers, after the execution of each of the following five load/store instructions in the (five-lined) program below (i.e.. these five instructions are run as a sequence from the initial memory state shown below).

We assume big endian formatting.

Initial Memory State 0x420014 DE OC 63 20 0x420010 FF AE 10 00 0x42000c 13 46

FA 08

0x420008 0x420004 0x420000

24 AB

A0

22

00 00 CO

FF

00 1C OE 3B

Initial Registers

State

r0=0x00000000, r1=0x00000000, r2=0x00420008, 13=0x00000007,

r4=0x00000001

# Start of the program

LDR

r0, [r2, #-4]

LDRB

r1, [r2, r3]

STR

r1, [2], r4 LSL #2

SWP

STMDA

r4, r0, [r2]

# End of the program

r2!, (r4, r3, r0}

The scenario described, where HC (hydrocarbons) and CO (carbon monoxide) levels are high while CO2 (carbon dioxide) and O2 (oxygen) levels are low, is typically caused by a rich mixture.

A rich mixture refers to an air-fuel mixture in which the fuel component is higher than the ideal stoichiometric ratio (the perfect balance of air and fuel for efficient combustion). When the mixture is rich, there is an excess of fuel relative to the available oxygen. As a result, incomplete combustion occurs, leading to higher levels of HC and CO emissions.

Excessive fuel can be caused by various factors such as a malfunctioning fuel injection system, a faulty oxygen sensor, a clogged air filter, or issues with the engine's electronic control system. These conditions can disrupt the proper air-fuel ratio, leading to a rich mixture and the observed increase in HC and CO emissions.

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Answer:

Engineer V provides expert consultation in one or more areas for the design, development and implementation of technical products and systems. Resolves highly complex technical issues and conducts advanced research. Being an Engineer V recommends alterations to development and design to improve quality of products and/or procedures.

Explanation:

True . The names of all registers updated during a procedure can be listed using the USES operator and the PROC directive.

The value that the ESP register references to is copied from the stack before the POP instruction is executed, at which point the register is incremented. The instruction ((n XOR m) XOR m) generates PUSH and POP in the 8085 microprocessor for any two integers n and m: When the PUSH instruction is carried out by the 8085 microprocessor, the stack pointer register is decremented by two, and when the POP instruction is carried out, the stack pointer is increased by two.

Mathematical calculations are carried out using arithmetic operators. A value can be assigned to a property or variable using assignment operators. Numeric, date, system, time, and tex are all acceptable assignment operators.

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Answer:

SpaceX

Explanation:

SpaceX successfully launches 60 more Starlink satellites as it continues towards 2020 service debut. SpaceX has launched another big batch of Starlink satellites, the low Earth orbit spacecraft that will provide connectivity for its globe-spanning high-bandwidth broadband internet network.

Vehicles driving in the opposite direction on a multi-lane highway with opposite lanes separated by elevated barrier

Vehicles going in the opposite direction on a multi-lane highway with opposite lanes separated by a raised barrier do not required to stop for school buses. This is so because students face extremely little risk. Since there is a raised barrier between the lanes, it is assumed that kids won't try to cross them that way. Therefore, if opposing traffic does not stop for school buses, it is unlikely that anyone will be wounded. In fact, if they cease, it is more likely that they may damage someone.

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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)

The ignition coil is the crucial part in an ignition system that increases the voltage.

It is essentially a transformer, capable of boosting the low voltage of a battery to the high voltage necessary for generating a spark at the plug - generally ranging between 20,000 and 50,000 volts.

The construction of the coil entails two coils of wire encircling an iron core, connected by either some breaker points or an electronic switch. As electricity passes through the primary coil, a magnetic field arises, inducing a high voltage in the secondary coil resulting in the generation of a spark.

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Answer:

B

Explanation:

Answer:

b. American Society of Mechanical Engineers

Explanation:

The "American Society of Mechanical Engineers" (ASME) is an organization that ensures the development of engineering fields. It is an accreditation organization that ensures parties will comply to the ASME Boiler and Pressure Vessel Code or BPVC.

The BPVC is a standard being followed by ASME in order to regulate the different pressure vessels and valves. Such standard prevents boiler explosion incidents.

Given Capacitance C = 8 μF = 8 × 10⁻⁶ F Voltage, V = 400 V Resistance, R = 0.1 MΩ = 0.1 × 10⁶ ΩNow, we have to calculate the time taken by the capacitor to develop a p.d. of 300 V.T = RC ln(1 + Vc/V).

Where R is the resistance  C is the capacitance V is the voltage of the supply Vc is the final voltage across the capacitor ln is the natural logarithm T is the time So, let's put the given values in the above formula. T = RC ln(1 + V c/V)T = 0.1 × 10⁶ × 8 × 10⁻⁶ ln(1 + 300/400)T = 0.8 ln(1.75)T = 0.8 × 0.5596T = 0.4477 seconds.

It takes 0.4477 seconds to charge the capacitor to a potential difference of 300 V. Next, we need to find the fraction of final energy that is stored in the capacitor. The energy stored in the capacitor is given as: Energy stored = (1/2) CV²Where C is capacitance and V is the voltage across the capacitor. Using the above formula.

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Answer:

It is known as (PCS)

Explanation:

That sounds a lot like:

(aka. a stylized picture of an object representing a word, syllable, or sound, as found in ancient Egyptian and other writing systems.)

Answer:

up up down down

Explanation:

left right left right b a select start

Answer:

Can't see, clearly photo then you can get your answer quickly

Explanation:

Answer:

Water enters and leaves the 6-cm diameter pipe bend in Fig. P3.45 at an average velocity of 8.5 m/s. The horizontal force to support the bend against momentum change is 300 N. Find (a) the angle ϕ; and (b) the vertical force on the bend.

a) The angle \(\phi\) is 62.06°.

b) The vertical force on the bend is 180.47 N.

The weight of the water and the reaction force must be considered to determine the vertical force exerted on the bend. This analysis provides valuable insights into the behavior and structural requirements of the pipe bend in relation to the fluid flow, ensuring its proper functioning and structural integrity.

The objective is to determine two key parameters: the angle of deflection QQ of the pipe bend and the vertical force acting on it. By applying principles of fluid mechanics and equilibrium, the solution requires calculating the change in momentum of the water passing through the bend and equating it to the known horizontal force required to support the bend.

Given that,

The diameter (d)= 6 cm

The average velocity (v)=8.5 m/s

Consider the rate of change of momentum along the x-direction.

\(F_x=mass \times (\frac{Intial \ velocity-Final \ Velocity}{time})\)

\(F_x=m'[(V-(-Vcos\phi)]\)

\(F_x=\rho AV^2[1+cos\phi]\)

\(300=100\times(\frac{\pi}{4}(0.06)^2)\times(8.5)^2[(1+cos\phi)]\)

\([(1+cos\phi)]=1.4685\)

\(cos\phi=0.4685\)

\(\phi=cos^{-1}(0.4685)\)

\(\phi=62.06^{^\circ}\)

(b) Calculate the vertical force on the bend.

Consider the rate of change of momentum along the y-direction.

\(F_y=mass \times (\frac{Intial \ velocity-Final \ Velocity}{time})\)

\(F_y=m'[(0-(V-Vsin\phi)]\)

\(F_y=\rho AVsin\phi\)

\(F_y=(1000)\times(\frac{\pi}{4}(0.06)^2)\times(8.5)^2\times sin62.06^{\circ}\)

\(F_y=180.47 \ N\)

Therefore,

a) The angle \(\phi\) is 62.06°.

b) The vertical force on the bend is 180.47 N.

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Answer:

True

Explanation:

Hai

Your answer will be A.

If you lower the Air Pressure your Object will Float Down ward. The Air Pressure allows it to Fly.

The pressure field created by faster air movement over an airfoil is; A:  higher

When the air hits the front of the wing, the air will flow in a steeper curve upward, than the bottom wing flow which will lead to the creation of a vacuum on top of the wing that pulls more air towards the top of the wing.

Finally, this air above does the same thing but it will move faster as a result of the vacuum pulling it in, and as such the vacuum now lifts the wing. Thus, Faster air movement over an airfoil creates a higher pressure field.

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Answer:

That's not true, if your house is burning down, leave. Let the firefighters do their jobs.

Explanation:

The maximum permissible variation between the two bearing indicators on a dual VOR system when checking one VOR against the other is ±4 degrees.

This variation is allowed to ensure that both VORs are working properly and are providing accurate and consistent readings. In a dual VOR system, two VOR receivers are installed in the aircraft, and the pilot can switch between them to check their accuracy. When checking one VOR against the other, the pilot must ensure that the bearings are consistent and within the permissible variation of ±4 degrees.

The permissible variation of ±4 degrees allows for small errors that may be caused by factors such as aircraft heading, magnetic deviation, and atmospheric conditions. If the variation between the two bearings is more than ±4 degrees, it may indicate that one of the VORs is not working properly and needs to be checked or serviced. In summary, the maximum permissible variation between the two bearing indicators on a dual VOR system when checking one VOR against the other is ±4 degrees.

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The type of type of building construction can include materials with no fire-resistance ratings in limited quantities is type 2.

Although many buildings appear to be similar at first appearance, their cost and durability—particularly in an emergency—are affected by the underlying materials. Building codes categorize all structures into Types 1 through 5, and each Type discloses important details like fire-resistance.

Some contemporary structures are now stronger and less expensive to construct. Engineered wood and synthetic plastics, which burn readily, cause quick collapses and present significant risks for firefighters.

Type 1 constructions, which are the most fire-resistant buildings, are made of shielded steel and concrete because they can sustain high temperatures without collapsing. Type 5 constructions, on the other hand, are the least fire-resistant since they are built of flammable materials, are lightweight, and they burn out quickly. Type 2 constructions, modern structures with metal roofs and tilt-slab or reinforced masonry walls.

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My mother lives.

From the point of view of the sentence, the mother is living far away from the point of her child. She is living not very far but from the point of view of the streets. She is living in her old apartment that is at the end of the road.

Hence says lives just the corner from me.

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The two main functions of a variable attenuator are to reduce the signal level or amplitude of a signal and to maintain a precise signal-to-noise ratio (SNR). By adjusting the attenuation level, it is possible to control the level of power that is transmitted or received.

The two main functions of a variable attenuator are to reduce the power of an incoming signal and to adjust the gain of an amplifier or other device.  An attenuator is a device that is used to reduce the power of an incoming signal without distorting the waveform. One of the main functions of a variable attenuator is to reduce the power of an incoming signal. This is important in order to prevent overloading or damaging the equipment that the signal is being sent to.

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a. To find the compression of the spring needed to launch the rock horizontally at 35 mph, we can use conservation of energy. The potential energy stored in the compressed spring is equal to the kinetic energy of the rock when it leaves the spring:

1/2 k x^2 = 1/2 m v^2

where k is the spring constant, x is the compression distance, m is the mass of the rock, and v is the velocity of the rock.

Converting the velocity to meters per second:

35 mph = 15.6 m/s

Plugging in the values and solving for x:

1/2 (23,000 N/m) x^2 = 1/2 (13 kg) (15.6 m/s)^2

x = sqrt[(13 kg) (15.6 m/s)^2 / (23,000 N/m)] = 0.263 m

Therefore, the spring must be compressed by 0.263 meters.

b. To find the maximum height the rock will reach when the spring is oriented vertically, we can again use conservation of energy. The potential energy stored in the compressed spring is converted into gravitational potential energy of the rock when it leaves the spring:

1/2 k x^2 = m g h

where g is the acceleration due to gravity and h is the maximum height reached by the rock.

Plugging in the values and solving for h:

1/2 (23,000 N/m) (0.5 m)^2 = (13 kg) (9.8 m/s^2) h

h = (1/2) (23,000 N/m) (0.5 m)^2 / (13 kg) (9.8 m/s^2) = 0.605 m

Therefore, the rock will reach a height of 0.605 meters above the ground.

c. To find the compression distance when the rock is dropped onto the spring from a height of 14 meters, we need to consider both the potential energy of the rock and the energy absorbed by the spring. When the rock hits the spring, it will come to a stop, so all of its initial potential energy will be converted into potential energy stored in the compressed spring:

m g h = 1/2 k x^2

where h is the initial height of the rock and x is the compression distance of the spring.

Plugging in the values and solving for x, we get a quadratic equation:

1/2 (23,000 N/m) x^2 - (13 kg) (9.8 m/s^2) (14 m) = 0

Simplifying and solving for x using the quadratic formula:

x = sqrt[(13 kg) (9.8 m/s^2) (14 m) / (23,000 N/m)] = 0.473 m

Therefore, the spring will compress by 0.473 meters before the rock comes to a stop.

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a. To find the compression of the spring needed to launch the rock horizontally at 35 mph, we can use conservation of energy. The potential energy stored in the compressed spring is equal to the kinetic energy of the rock when it leaves the spring:

1/2 k x^2 = 1/2 m v^2

where k is the spring constant, x is the compression distance, m is the mass of the rock, and v is the velocity of the rock.

Converting the velocity to meters per second:

35 mph = 15.6 m/s

Plugging in the values and solving for x:

1/2 (23,000 N/m) x^2 = 1/2 (13 kg) (15.6 m/s)^2

x = sqrt[(13 kg) (15.6 m/s)^2 / (23,000 N/m)] = 0.263 m

Therefore, the spring must be compressed by 0.263 meters.

b. To find the maximum height the rock will reach when the spring is oriented vertically, we can again use conservation of energy. The potential energy stored in the compressed spring is converted into gravitational potential energy of the rock when it leaves the spring:

1/2 k x^2 = m g h

where g is the acceleration due to gravity and h is the maximum height reached by the rock.

Plugging in the values and solving for h:

1/2 (23,000 N/m) (0.5 m)^2 = (13 kg) (9.8 m/s^2) h

h = (1/2) (23,000 N/m) (0.5 m)^2 / (13 kg) (9.8 m/s^2) = 0.605 m

Therefore, the rock will reach a height of 0.605 meters above the ground.

c. To find the compression distance when the rock is dropped onto the spring from a height of 14 meters, we need to consider both the potential energy of the rock and the energy absorbed by the spring. When the rock hits the spring, it will come to a stop, so all of its initial potential energy will be converted into potential energy stored in the compressed spring:

m g h = 1/2 k x^2

where h is the initial height of the rock and x is the compression distance of the spring.

Plugging in the values and solving for x, we get a quadratic equation:

1/2 (23,000 N/m) x^2 - (13 kg) (9.8 m/s^2) (14 m) = 0

Simplifying and solving for x using the quadratic formula:

x = sqrt[(13 kg) (9.8 m/s^2) (14 m) / (23,000 N/m)] = 0.473 m

Therefore, the spring will compress by 0.473 meters before the rock comes to a stop.

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In order to draw the amplitude modulation, connect the power supply to the protoboard, onnect the fixed capacitor (220 microfarads) in parallel with the power supply to filter any noise and connect the fixed resistor (100 ohms) in series with the base of the NPN transistor.

A transmitting device that employs Amplitude Modulation (AM) is a necessary component for transferring information across radio waves.

The audio input is passed through an equipping capacitor C1, barring any Direct Current (DC) potential from the audio generator. Then, transistor Q1 intensively amplifies the incoming signal after its been regulated in a base bias configuration by R1 and R2 voltage dividers.

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The given statement "the va shear stress represents the maximum shear stress found anywhere in the structure" is false. The term "va" has no relevance to shear stress and therefore the statement cannot be true.

Let's further understand what shear stress is and how it is represented in a structure.Shear stress is the force that acts in parallel to a surface in opposite directions. It is also called tangential stress. Shear stress comes into play when two forces act parallel to each other but in opposite directions on an object. The formula to calculate shear stress is given as:T = F/AWhere T is shear stress, F is the applied force, and A is the cross-sectional area of the object.

The unit of shear stress is Pascal (Pa) in the SI system and psi (pound per square inch) in the Imperial system.However, the maximum shear stress found anywhere in the structure is represented by the symbol τ_max (tau-max). It is important to note that τ_max is not represented by the term "va" but by the symbol τ_max itself. Therefore, the given statement "the va shear stress represents the maximum shear stress found anywhere in the structure" is false.

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Answer:

yes.

Explanation:

Engineers may choose an abrasive machining process over traditional machining for several reasons. Abrasive machining offers advantages such as increased material removal rates, the ability to work with harder materials, improved surface finish, and reduced tool wear.

Additionally, abrasive machining processes can be more cost-effective and suitable for complex geometries. However, it is important to consider factors such as the required precision, the specific material being worked on, and the desired final product characteristics when deciding between abrasive and traditional machining methods. Engineers often opt for abrasive machining processes over traditional machining methods due to various benefits. One significant advantage is the higher material removal rate offered by abrasive machining. In processes like grinding or lapping, abrasive particles are used to remove material from the workpiece, allowing for faster stock removal compared to traditional machining techniques like milling or turning. This increased efficiency can be particularly advantageous when working on large quantities or time-sensitive projects. Another key factor is the ability of abrasive machining to handle harder materials. Traditional machining methods can struggle with materials that are extremely hard or have high toughness, leading to excessive tool wear and reduced efficiency. Abrasive machining, on the other hand, employs abrasive particles that are capable of cutting through or wearing away even the toughest materials. This makes it a suitable choice for working on materials such as ceramics, hardened steel, or superalloys. Furthermore, abrasive machining processes often yield improved surface finishes. By carefully selecting the size and type of abrasive particles, engineers can achieve fine surface finishes with minimal surface roughness. This is especially crucial in applications where precision and aesthetics are important, such as in the production of optical lenses or critical components for aerospace industries. Cost-effectiveness is another aspect that influences the decision to choose abrasive machining. While the initial setup costs for abrasive machining equipment can be higher, the long-term operational costs can be significantly lower due to reduced tool wear and longer tool life. Moreover, abrasive machining processes can be more adaptable to complex geometries, making them suitable for intricate shapes or parts with non-traditional designs that would be challenging to produce using traditional methods. However, it is crucial for engineers to consider certain factors before selecting abrasive machining. The required precision of the final product, for instance, may play a role in the decision-making process. Traditional machining methods can offer higher accuracy and tighter tolerances, which may be necessary in applications where dimensional precision is critical. The specific material being worked on also needs to be taken into account. Some materials may react poorly to abrasive processes, such as those prone to heat damage or with low thermal conductivity. In such cases, traditional machining techniques that generate less heat, such as milling or turning, might be more suitable. Lastly, the desired characteristics of the final product should be considered. Traditional machining can create smoother, more precise edges and contours, which may be essential for certain applications. On the other hand, abrasive machining processes can introduce compressive residual stresses, potentially improving the fatigue life and durability of the workpiece. Ultimately, the choice between abrasive and traditional machining methods depends on a comprehensive evaluation of the project requirements, including considerations of material properties, cost-effectiveness, surface finish, and the desired final product characteristics.

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Approximately 1 solar heating panel is required to heat the water from 20ºC to 40ºC.

To estimate the number of 4'x8' solar heating panels required to heat water at a home from 20ºC to 40ºC, we need to consider the energy requirements, efficiency, solar irradiation, and the heat capacity of water.

- Daily water usage: 125 gallons

- Efficiency (η): 0.7

- Direct Normal Irradiation (DNI): 7 kW-hr/m2

- Heat capacity of water: 4200 J/(kg ºC)

First, we need to convert the daily water usage from gallons to kilograms. Since 1 gallon is approximately 3.78541 kilograms, the daily water usage is approximately 471.9275 kg.

Next, we calculate the energy required to heat the water from 20ºC to 40ºC using the formula:

Energy = Mass of water * Specific heat capacity * Temperature change

Temperature change = (40ºC - 20ºC) = 20ºC

Energy = 471.9275 kg * 4200 J/(kg ºC) * 20ºC = 19,773,090 J

Now, we need to calculate the energy received from the solar panels. The total energy received can be obtained by multiplying the DNI by the area of the solar panels and the efficiency.

Area of a 4'x8' panel = 4 ft * 8 ft = 32 ft2

Converting to square meters: 32 ft2 * 0.092903 m2/ft2 = 2.97256 m2

Total energy received = DNI * Area of panels * Efficiency

Total energy received = 7 kW-hr/m2 * 2.97256 m2 * 0.7 * 3600 kJ/kWh * 1000 J/kJ = 65,647,040 J

Finally, we can calculate the number of panels required by dividing the energy required by the energy received per panel:

Number of panels = Energy required / Total energy received

Number of panels = 19,773,090 J / 65,647,040 J = 0.301

Please note that this calculation is an estimation based on the given data and assumptions. Other factors such as system losses, temperature variations, and specific panel efficiency may affect the actual number of panels required in a real-world scenario.

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A warning sign that indicates a divided highway is ending is typically a "Divided Highway Ends" sign.

This sign is designed to alert drivers that the divided section of the road is coming to an end and the road will transition back into a two-way traffic pattern. The sign usually consists of a rectangular shape with a black symbol on a yellow background.

The symbol represents a divided highway with two solid lines merging into one. It serves as a visual signal for drivers to be prepared for changes in traffic flow and adjust their driving accordingly.

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