As described in Section 5.7, virtual memory uses a page table to track the mapping of virtual addresses to physical addresses. This exercise shows how this table must be updated as addresses are accessed. The following data constitutes a stream of virtual addresses as seen on a system. Assume 4 KiB pages, a 4-entry fully associative TLB, and true LRU replacement. If pages must be brought in from disk, increment the next largest page number. [15] <§5.7> Repeat 5.11.1, but this time use 16 KiB pages instead of 4 KiB pages. What would be some of the advantages of having a larger page size? What are some of the disadvantages?

Answer:

SECTION LEARNING OBJECTIVES

By the end of this section, you will be able to do the following:

Distinguish between static friction and kinetic friction

Solve problems involving inclined planes

Section Key Terms

kinetic friction static friction

Static Friction and Kinetic Friction

Recall from the previous chapter that friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice.

There are different types of friction—kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects.

Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going.

Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them.

60 degrees east or west would be the two optimum radar look directions.

The optimum radar look directions for enhancing lineaments that trend 60 degrees west would be 60 degrees east and 60 degrees west.

This is because radar look directions are generally chosen to be perpendicular to the direction of the lineaments, as this allows the radar to "see" the lineaments more clearly and provide better resolution.

In this case, a look direction of 60 degrees east or west would be perpendicular to the lineaments, which would provide the best possible radar imaging of the lineaments in the exploration area.

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No, the resistor is not within its tolerance rating.To determine if the resistor is within its tolerance rating, we need to decode the resistor color bands.

The color bands represent the resistance value and tolerance. In this case, the color bands are brown (1), black (0), red (100), and gold (±5%). Using the resistor color code, we can calculate the resistance value as 10 * 100 = 1000 ohms (1 kΩ). The tolerance band indicates that the resistor's actual resistance may vary by ±5%. Therefore, the tolerance range for this resistor would be 950 ohms to 1050 ohms. However, since the voltmeter measures a voltage drop of 6.5 V, we can conclude that the resistor is operating outside its tolerance range.

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Answer:

The intake and exhaust ports

Explanation:

The intake manifold function is to channel air through the passages in the cylinder head and into the combustion chamber.

(hope this help) :)

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

so 14.55

Option a is correct

Channel width will decrease as you move from headwaters to the mouth.

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The missing term in the presented paragraph is environmental.

In recent years the effects of climate change have been strongly manifested around the world causing:

Therefore, many companies have begun to investigate productive alternatives to reduce their environmental impact. In the case of the car industry, companies have deepened their research on other sources of energy such as:

One of them is Toyota, the Japanese brand that has developed cars that work hybrid, that is, with gasoline and electric power. This will reduce the emissions of CO2 and other harmful gases into the atmosphere.

Therefore, it is possible to infer that the missing term in the paragraph is Environmental.

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The distance below the top of the cliff that the two balls cross paths is 7.53 meters.

Given the following data:

Scientific data:

To determine how far (distance) below the top of the cliff that the two balls cross paths, we would apply the third equation of motion.

Mathematically, the third equation of motion is given by this formula:

\(V^2 = U^2 +2aS\)

Where:

Substituting the parameters into the formula, we have;

\(V^2 = 0^2 +2(9.8) \times 30\\\\V^2 = 588\\\\V=\sqrt{588}\)

V = 24.25 m/s.

Note: The final velocity of the first ball becomes the initial velocity of the second ball.

The time at which the two balls meet is calculated as:

\(Time = \frac{S}{U} \\\\Time = \frac{30}{24.25}\)

Time = 1.24 seconds.

The position of the ball when it is dropped from the cliff is calculated as:

\(y_1 = h-\frac{1}{2} at^2\\\\y_1 = 30-\frac{1}{2} \times 9.8 \times 1.24^2\\\\y_1 = 30-7.53\\\\y_1=22.47\;meters\)

Lastly, the distance below the top of the cliff is calculated as:

\(Distance = 30-22.47\)

Distance = 7.53 meters.

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Answer: the shear strength at a depth of 12 ft is 1034.9015 lb/ft²

Explanation:

Given that;

Weight of soil r = 118 lb/ft³

stress parameter C = 250 lb/ft²

φ total = 29°

depth Z = 12 ft

The shear strength on a horizontal plane at a depth of 12ft

ζ = C + δtanφ

where δ = normal stress

normal stress δ = r × z = 118 × 12 = 1416

so

ζ = C + δtanφ

ζ = 250 + 1416(tan29°)

ζ = 250 + 1416(tan29°)

ζ = 250 + 784.9016

ζ = 1034.9015 lb/ft²

Therefore the shear strength at a depth of 12 ft is 1034.9015 lb/ft²

The Library of Muyinga's building construction was related to the product water cycle. Thus, option B is correct.

It is also known as the hydrological cycle and is basically all about water evaporating from liquid masses such as lakes, seas, oceans, and rivers, being water vapor in the air, and then falling back down in the form of rain or snow or similar things.

Sometimes they fall nearby, sometimes it gets carried to other regions due to winds.The cycle in hich water is changed and distributed to other areas is called water cycle.

Therefore, The Library of Muyinga's building construction was related to the product water cycle. Thus, option B is correct.

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The resistance of a parallel branch affects the current flowing through it in an opposite manner.

The sum of the power used by each resistor makes up the total power used in a parallel circuit. The voltage drop across each bulb is 6.0 V when the light bulbs are linked in parallel, and they all glow as the current passing through them combines to generate the current flowing through the battery. Each component of a series circuit receives the same amount of current. The circuit would be open and no current would flow if it were to be interrupted at any point. Therefore, the remaining bulbs would not be able to shine if one of them burns out.

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The current in the incandescent bulb is approximately 0.625 Amperes (A).

An incandescent light bulb, incandescent lamp or incandescent light globe is an electric light with a wire filament heated until it glows. The filament is enclosed in a glass bulb with a vacuum or inert gas to protect the filament from oxidation.

The amount of current in an incandescent bulb rated at 75W and 120V can be calculated using Ohm's Law, which states that current (I) is equal to power (P) divided by voltage (V).

Using this formula, the current can be calculated as:

I = P / V

I = 75W / 120V

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Solution :

\($P_1 = 120 \ psia$\)

\($P_2 = 20 \ psia$\)

Using the data table for refrigerant-134a at P = 120 psia

\($h_1=h_f=40.8365 \ Btu/lbm$\)

\($u_1=u_f=40.5485 \ Btu/lbm$\)

\($T_{sat}=87.745^\circ F$\)

∴ \($h_2=h_1=40.8365 \ Btu/lbm$\)

For pressure, P = 20 psia

\($h_{2f} = 11.445 \ Btu/lbm$\)

\($h_{2g} = 102.73 \ Btu/lbm$\)

\($u_{2f} = 11.401 \ Btu/lbm$\)

\($u_{2g} = 94.3 \ Btu/lbm$\)

\($T_2=T_{sat}=-2.43^\circ F$\)

Change in temperature, \($\Delta T = T_2-T_1$\)

                                         \($\Delta T = -2.43-87.745$\)

                                           \($\Delta T=-90.175^\circ F$\)

Now we find the quality,

\($h_2=h_f+x_2(h_g-h_f)$\)

\($40.8365=11.445+x_2(91.282)$\)

\($x_2=0.32198$\)

The final energy,

\($u_2=u_f+x_2.u_{fg}$\)

   \($=11.401+0.32198(82.898)$\)

   \($=38.09297 \ Btu/lbm$\)

Change in internal energy  

\($\Delta u= u_2-u_1$\)

   = 38.09297-40.5485

  = -2.4556        

An abundance of condensation nuclei from combustion products is the main reason why fog is usually prevalent in industrial areas.

Combustion products such as smoke and particulate matter provide a large number of tiny particles that serve as nuclei for the formation of fog droplets.

The presence of these particles, combined with cool and humid weather conditions, can lead to the formation of thick fog in industrial areas.

Additionally, the urban heat island effect caused by industrial heating can trap moisture close to the ground and contribute to the formation of fog. However, atmospheric stabilization around cities is not a primary cause of fog in industrial areas.

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Answer:

a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces.

Explanation:

Turbulent flow is a type of fluid flow in which fluid will undergo irregular fluctuations. The tubes with rough surfaces have higher friction factors than the tubes with smooth surfaces. In laminar flow the effect of effect of surface roughness is negligible on friction factors.

Answer:

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.

Answer:

Explanation:

Generally speaking, flammable liquids will ignite (catch on fire) and burn easily at normal working temperatures. ... Fuels and many common products like solvents, thinners, cleaners, adhesives, paints, waxes and polishes may be flammable or combustible liquids.

The material which is considered flammable is: A. paint.

A chemical property can be defined as the property of a chemical compound (material) that can be observed and is measurable during a chemical reaction.

In Science, some examples of the chemical properties of a material (substance) include the following;

Flammability refers to the ability of a material to support combustion or burn continuously in the presence of air. An example of a material which is considered to be flammable is paint because it contains a high level of solvents.

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The total frictional head loss in the system is 25.6 m. If these were the measurements in the manometers for the Bernoulli experiment,

To calculate the total frictional head loss in the system, we need to use the Bernoulli equation which states that the total head at any point in a fluid flow system is constant. This means that the sum of the pressure head, velocity head, and elevation head at any point must be equal to the sum of these same variables at any other point in the system. In addition, we need to take into account the total frictional head loss which is the energy lost due to friction as the fluid flows through the pipes and fittings.
Using the manometer readings provided, we can calculate the pressure difference between points 1 and 3, and between points 5 and 6 as follows:
ΔP₁₋₃ = H₁ - H₃ = 256 - 159 = 97 mm
ΔP₅₋₆ = H₅ - H₆ = 0 - 131 = -131 mm (since the fluid is flowing from point 6 to point 5)
Next, we need to convert these pressure differences into velocity heads using the equation ΔP = ρgΔh, where ρ is the fluid density and g is the acceleration due to gravity. Assuming water at 20°C with a density of 1000 kg/m3, we get:
Δh₁₋₃ = ΔP₁₋₃ / (ρg) = 0.097 m
Δh₅₋₆ = ΔP₅₋₆ / (ρg) = -0.131 m
Now we can use these velocity heads along with the elevation heads to calculate the total head at points 1, 3, 5, and 6:
h₁ = H₁ + Δh₁₋₃ = 256 + 0.097 = 256.097 mm
h₃ = H₃ = 159 mm
h₅ = H₅ + Δh₅₋₆ = 0 - 0.131 = -0.131 mm
h₆ = H₆ = 131 mm
Since the total head is constant along the flow path, we can equate the total head at points 1 and 6:
h₁ + (v₁² / 2g) + z₁ + hL₁₋₆ = h₆ + (v₆² / 2g) + z₆
where v1 and v6 are the velocities at points 1 and 6 respectively, z1 and z6 are the elevations of points 1 and 6, and hL1-6 is the total frictional head loss between points 1 and 6.
Assuming that the velocity at point 6 is negligible, we can simplify the equation to:
h₁ + (v₁² / 2g) + z₁ = h₆ + z₆+ hL₁₋₆
Substituting the values we have calculated, we get:
256.097 + (v₁² / 2g) + 0 = 131 + 0 + hL₁₋₆
Simplifying further, we get:
hL₁₋₆ = 256.097 - 131 - (v₁² / 2g)
To calculate the velocity at point 1, we can use the Bernoulli equation between points 1 and 3:
h1 + (v₁² / 2g) + z1 = h3 + (v₃² / 2g) + z3 + hL₁₋₃
Assuming that the elevation difference between points 1 and 3 is negligible, we can simplify the equation to:
h1 + (v₁² / 2g) = h₃ + (v₃² / 2g) + hL₁₋₃
Substituting the values we have calculated, we get:
256.097 + (v₁² / 2g) = 159 + (v₃² / 2g) + hL₁₋₃
Solving for v1, we get:
v₁ = √[(2g / ρ) * (256.097 - 159 - hL₁₋₃)]
Substituting the values we have calculated, we get:
v₁ = 4.71 m/s
Finally, substituting this value into the equation for hL1-6, we get:
hL₁₋₆ = 256.097 - 131 - (4.71² / 2g) = 25.6 m

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Answer:

10.5

Explanation:

Measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial reads 22.  Diameter of the drum is 10.5.

Any straight line segment that cuts through the center of a circle and has ends that are on the circle is considered a circle's diameter in geometry. It is also known as the circle's longest chord.

The diameter of a sphere can be defined using either of the two methods. In more recent usage, the diameter can also refer to a diameter's length, or displaystyle d.

As opposed to a diameter, which refers to the line segment itself, one uses the term "diameter" in this context since all diameters of a circle or sphere have the same length, which is equal to twice the radius "displaystyle r.

Therefore, Measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial reads 22.  Diameter of the drum is 10.5.

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The sum and product of each hexadecimal number are for a D6A, 1A3F88 for b 216BB, 1438D5A for c ACD8F, B051FA2E for d E0BAA8, 92A26ABAE4

Here is how you find the sum and product of each of these pairs of hexadecimal numbers.

a) (1AE)16, (BBC)16
Sum: 1AE + BBC = D6A (in hexadecimal)
Product: 1AE * BBC = 1A3F88 (in hexadecimal)

b) (20CBA)16, (A01)16
Sum: 20CBA + A01 = 216BB (in hexadecimal)
Product: 20CBA * A01 = 1438D5A (in hexadecimal)

c) (ABCDE)16, (1111)16
Sum: ABCDE + 1111 = ACD8F (in hexadecimal)
Product: ABCDE * 1111 = B051FA2E (in hexadecimal)
d) (E0000E)16, (BAAA)16
Sum: E0000E + BAAA = E0BAA8 (in hexadecimal)
Product: E0000E * BAAA = 92A26ABAE4 (in hexadecimal)

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Answer:

b

Explanation:

you must check that the head gasket is seated correctly and is not dislodged or slightly moved in some way or youl result in oil or coolant in the cylinders and loss of compression is possible to or a vacum leak in 1 or more cylinders

Answer:

I think its A

Explanation:

Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3  = √9 = 3

P5/P6 = P7/P8  = √9 =3  

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860  

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238  

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158  

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg  

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg  

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =  

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))  

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg  

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33  

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg  

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg  

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =  

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg  

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg  

Calculate the net work done

Wnet = (Wturb)out  - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg  

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

The characteristic of a primary key states that the selected primary key must not be composed of any attribute(s) that might be considered a violation is preferably single-attribute.

What is a primary key?A primary key is a single field or a set of fields in a table that are used to identify the uniqueness of each record.

A primary key is a special kind of unique key that can't be null and must be unique throughout the table.

It is used to ensure the uniqueness of a row within a table. Therefore, it is necessary to select a primary key to identify each row of a table uniquely.

The characteristic of a primary key states that the selected primary key must not be composed of any attribute(s) that might be considered a violation is preferably single-attribute. It should be as simple as possible, but it must be unique for each record.

The value of the primary key should be in the same data type as the foreign key's value that references it.

The primary key is always indexed, and it has the unique constraint. It cannot contain null values, and it must have values that are not repeated or duplicated.

In conclusion, the characteristic of a primary key states that the selected primary key must not be composed of any attribute(s) that might be considered a violation is preferably single-attribute. The primary key is always indexed, and it has the unique constraint. It cannot contain null values, and it must have values that are not repeated or duplicated.

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Answer:

true

Explanation:

If your accelerator gets stuck down, do the following:

Shift to neutral.

Apply the brakes.

Keep your eyes on the road and look for a way out.

Warn other drivers by blinking and flashing your hazard lights.

Try to drive the car safely off the road.

Turn off the ignition when you no longer need to change direction.

Answer:Kick the accelerator to dislodge it

Explanation:

Never reach down with your hand while driving, it's not safe.

Hi! The advantages of having shared fields and methods in superclasses, rather than throughout multiple extended classes, are as follows:

1. Code Reusability: By having shared fields and methods in a superclass, you can reuse the code in multiple extended classes without having to rewrite the same code in each class. This makes the code more efficient and easier to maintain.

2. Consistency: With shared fields and methods in a superclass, all extended classes will have access to the same fields and methods, ensuring consistent behavior across all subclasses.

3. Easier Maintenance: If a change needs to be made to a shared field or method, it only needs to be updated in the superclass, and the change will automatically propagate to all extended classes. This reduces the risk of errors and inconsistencies.

4. Modularity: By organizing shared fields and methods in a superclass, you are creating a more modular and organized codebase, making it easier to understand and manage.

In summary, having shared fields and methods in superclasses offers advantages such as code reusability, consistency, easier maintenance, and modularity, which can lead to a more efficient and maintainable codebase.

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Fluid balance is essential for the proper functioning of the human body, and it is maintained by the means of gaining and losing fluid. Fluid can be gained through ingestion and metabolism of food and drink and can be lost through urine, feces, sweat, and respiration.Regulating Intake and Output of FluidIntake: Thirst mechanisms in the hypothalamus, stimulation of saliva production, and swallowing are all responsible for the intake of fluids. The pituitary gland is also responsible for regulating thirst and urine concentration in the body.

Output: Urine is the primary means of fluid loss in the body, with an average of 1-2 liters of urine produced every day. Feces and sweat can also contribute to the body's fluid loss, and breathing is another vital means of fluid loss.Volume Depletion and DehydrationVolume depletion occurs when there is a decrease in the amount of fluid circulating in the body, leading to a decrease in blood volume. This can occur due to a variety of factors, including severe bleeding, vomiting, and diarrhea. Dehydration, on the other hand, occurs when there is a net loss of water from the body, leading to an imbalance in electrolyte and fluid levels. This can occur due to insufficient fluid intake, excessive sweating, or a combination of factors.Regulation of Intake and OutputFluid balance is regulated by the kidneys, which filter waste and excess fluid from the blood and produce urine. Antidiuretic hormone (ADH) and aldosterone are two hormones that regulate fluid balance by affecting the amount of water that is reabsorbed or excreted by the kidneys. When fluid intake is insufficient, the kidneys conserve water by decreasing urine output and increasing water reabsorption. Conversely, when fluid intake is high, the kidneys excrete excess water to maintain fluid balance.ConclusionIn conclusion, the human body gains and loses fluids through various means such as ingestion, metabolism, urination, sweating, and respiration. The body's fluid balance is regulated through the kidneys and hormones like ADH and aldosterone. Volume depletion and dehydration are two conditions that affect fluid balance in the body and can be caused by various factors like severe bleeding, vomiting, diarrhea, and insufficient fluid intake.

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The gelatinous membrane that stretches across the tube to form a fluid-tight seal akin to the skin of an ear drum covers the semicircular canal receptor cells, also known as hair cells, only in the center of the circular tubes in a special epithelium.

You can sense your body's position and keep your balance thanks to the stimulation of hair-like receptors that send signals to the cerebellum at the back of the brain.

Your inner ear contains three tiny, liquid-filled tubes called semicircular canals that aid in balance. The fluid in the semicircular canals sloshes around as your head moves, moving the fine hairs that line each canal.

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