Fluid Mechanics

118 soru

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A sealed industrial holding tank contains water with a pocket of compressed air trapped above the water surface. The total depth of the water in the tank is $8.0\text{ m}$ . A pressure gauge located at the absolute bottom of the tank reads a gauge pressure of $130\text{ kPa}$ . A secondary discharge valve is located on the side of the tank, exactly $5.0\text{ m}$ above the tank's bottom. Assuming the density of water is $1,000\text{ kg/m}^3$ and the acceleration due to gravity is $g = 10\text{ m/s}^2$ , what is the gauge pressure, in kilopascals (kPa), at the secondary discharge valve?

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Cevap: 80

Cevap

The pressure at any point in a static fluid is equal to the pressure at a lower point minus the hydrostatic pressure of the fluid column between them. Since the valve is

5.0\text{ m}

above the bottom, the pressure there is less than the bottom pressure by

\rho g \Delta h = 1,000 \times 10 \times 5.0 = 50\text{ kPa}

. Subtracting this from the bottom pressure (

130\text{ kPa}

) gives the correct pressure of

80\text{ kPa}

. Alternatively, calculating the trapped air pressure (

130 - 80 = 50\text{ kPa}

) and adding the

3.0\text{ m}

water column above the valve (

30\text{ kPa}

) yields the same correct result.

Adım Adım Çözüm

Determine the vertical distance between the pressure gauge at the bottom and the secondary valve.

\Delta h = 5.0\text{ m}

Hydrostatic pressure changes linearly with vertical height within a continuous fluid.

Calculate the hydrostatic pressure difference exerted by this

5.0\text{ m}

column of water.

\Delta P = \rho g \Delta h = 1,000 \times 10 \times 5.0 = 50,000\text{ Pa} = 50\text{ kPa}

This represents the amount of pressure lost as you move upward from the bottom to the valve.

Subtract the pressure difference from the gauge pressure at the bottom of the tank.

130\text{ kPa} - 50\text{ kPa} = 80\text{ kPa}

The pressure at the valve must be lower than the pressure at the bottom by exactly the weight of the water column between them.

Anahtar Kavram

Hydrostatic pressure varies directly with depth. The pressure difference between any two points in a continuous static fluid depends only on the vertical distance between them, regardless of surface air pressure.

Soru 62Soru →

Water flows at a steady rate through a horizontal pipe that consists of four distinct sections with different diameters. According to Bernoulli's principle and the continuity equation, arrange the pipe sections in order of their static pressure, from the lowest static pressure to the highest static pressure.

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Cevap

The correct sequence from lowest to highest static pressure is: Section W (1-inch), Section X (2-inch), Section Y (3-inch), Section Z (4-inch).

According to the continuity equation, a fluid travels faster through a narrower section of pipe. According to Bernoulli's principle, an increase in fluid velocity results in a decrease in static pressure. Therefore, the pipe sections with the smallest diameters will have the highest velocities and the lowest static pressures, while the sections with the largest diameters will have the lowest velocities and the highest static pressures.

Adım Adım Çözüm

Determine the relationship between the pipe's diameter and the fluid's velocity.

The continuity equation indicates that fluid must travel faster through narrower sections to maintain a steady flow rate. Therefore, the smallest diameter has the highest velocity, and the largest diameter has the lowest velocity.

Because the volume of fluid passing through any section per second must remain constant.

Determine the relationship between fluid velocity and static pressure.

Bernoulli's principle states that an increase in fluid velocity occurs simultaneously with a decrease in static pressure.

The total energy in a steady, horizontal fluid flow is constant; as kinetic energy (speed) increases, pressure potential energy must decrease.

Rank the sections based on these relationships.

Since smallest diameter = highest velocity = lowest pressure, the order from lowest to highest pressure corresponds exactly to the order from smallest to largest diameter.

This allows us to arrange the sections directly by their diameter sizes: 1-inch, 2-inch, 3-inch, and 4-inch.

Anahtar Kavram

Fluid Dynamics and Bernoulli's Principle

Soru 63Soru →

A marine salvage team is recovering two solid objects resting on the ocean floor: a 100-pound lead weight and a 100-pound concrete block. Lead has a significantly higher density than concrete. While both objects are completely submerged, which of the following statements is correct regarding the buoyant forces acting on them?

Both objects experience the same buoyant force because they have identical weights.

The concrete block experiences a greater buoyant force because it displaces a larger volume of water.

The lead weight experiences a greater buoyant force because its higher density creates a stronger downward reaction.

The lead weight experiences a greater buoyant force because its concentrated mass exerts more pressure on the water.

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Cevap: The concrete block experiences a greater buoyant force because it displaces a larger volume of water.

Cevap

The concrete block experiences a greater buoyant force because it displaces a larger volume of water.

According to Archimedes' Principle, the upward buoyant force acting on a submerged object is exactly equal to the weight of the fluid it displaces. Because both objects weigh 100 pounds but concrete is less dense than lead, the concrete block must take up more space (larger volume). When completely submerged, the larger volume of the concrete block displaces a larger volume of water, resulting in a significantly greater upward buoyant force compared to the smaller, denser lead weight.

Adım Adım Çözüm

Analyze the relationship between weight, density, and volume for both objects.

Both weigh 100 pounds, but lead is denser than concrete. Therefore, the 100-pound concrete block must have a larger physical volume than the 100-pound lead weight.

Density is mass per unit volume. For objects of equal mass, the less dense material takes up more space.

Apply Archimedes' Principle to completely submerged objects.

The buoyant force is equal to the weight of the displaced fluid. The volume of displaced fluid equals the volume of the submerged object.

Archimedes' Principle states that an object fully or partially immersed in a fluid is buoyed up by a force equal to the weight of the fluid that it displaces.

Compare the displaced fluid volumes and resulting buoyant forces.

Because the concrete block has a larger volume, it displaces more water than the lead weight. Consequently, the concrete block experiences a greater upward buoyant force.

Greater displaced volume directly results in a proportionally greater buoyant force, regardless of the object's interior weight.

Anahtar Kavram

Buoyancy and Archimedes' Principle

Soru 64Soru →

Match each hydraulic jack scenario to its correctly calculated parameter based on Pascal's Principle.

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Cevap

Jack A matches to 160 lbs output force, Jack B matches to 3 inches output distance, Jack C matches to 4 in² input area, and Jack D matches to 30 psi fluid pressure.

Each hydraulic system operates on Pascal's Principle and the conservation of volume. By applying the formulas for pressure (

P = F/A

) and mechanical advantage (

MA = A_{out}/A_{in}

), the missing parameter for each jack can be calculated and uniquely matched.

Adım Adım Çözüm

Calculate the output force for Jack A.

F_{out} = 160 \text{ lbs}

Using Pascal's principle, mechanical advantage is

12/3 = 4

F_{out} = 40 \times 4

Calculate the output distance for Jack B.

d_{out} = 3 \text{ inches}

Due to volume conservation,

d_{in} \times A_{in} = d_{out} \times A_{out}

15 \times 2 = d_{out} \times 10

Calculate the input area for Jack C.

A_{in} = 4 \text{ in}^2

Since pressure is equal across the system,

P = 180 / 24 = 7.5 \text{ psi}

A_{in} = 30 / 7.5

Calculate the fluid pressure for Jack D.

P = 30 \text{ psi}

Pressure equals force over area,

P = 45 / 1.5

Anahtar Kavram

Pascal's Principle, Fluid Volume Conservation, and Mechanical Advantage

Soru 65Soru →

A common paint sprayer operates by blowing a stream of fast-moving air across the top of an open vertical tube, with the bottom of the tube submerged in a container of paint. Based on Bernoulli's principle, what is the primary reason the paint rises up the tube?

The fast-moving air creates a region of low static pressure at the top of the tube, allowing atmospheric pressure to push the paint upward.

The high-speed air creates a region of high static pressure at the top of the tube, which pulls the paint upward.

The stream of air compresses the fluid inside the vertical tube, reducing its volume and forcing it upward.

The fast-moving air decreases the weight of the displaced fluid, causing an upward buoyant force.

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Cevap: The fast-moving air creates a region of low static pressure at the top of the tube, allowing atmospheric pressure to push the paint upward.

Cevap

The fast-moving air creates a region of low static pressure at the top of the tube, allowing atmospheric pressure to push the paint upward.

According to Bernoulli's principle, an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure. The fast-moving air across the top of the tube creates a localized low-pressure area. The relatively higher atmospheric pressure pushing down on the surface of the paint in the container then forces the liquid up the tube.

Adım Adım Çözüm

Analyze the relationship between fluid speed and pressure.

The air moving quickly across the top of the tube has a higher velocity than the still air surrounding the container.

This scenario directly sets up an application of Bernoulli's principle.

Determine the pressure difference.

Due to its high velocity, the air stream creates an area of low static pressure directly above the open tube.

Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in its static pressure.

Evaluate the effect on the fluid in the container.

The standard atmospheric pressure resting on the surface of the paint in the container is greater than the low pressure at the top of the tube, thus forcing the paint upward into the stream.

Fluids naturally move from areas of higher pressure to areas of lower pressure.

Anahtar Kavram

Bernoulli's principle establishes an inverse relationship between a fluid's velocity and its static pressure.

Tahmini Süre:45s

Soru 66Soru →

A rigid, sealed cylindrical pontoon is pushed completely underwater in a testing pool, where it displaces $200 \text{ liters}$ of water. The empty pontoon weighs $400 \text{ N}$ . A technician then pulls the pontoon out, places $150 \text{ N}$ of solid lead weights inside it, seals it, and forces it completely back under the water. Which of the following best describes the change in the upward buoyant force acting on the completely submerged pontoon?

It remains exactly the same because the exterior volume of the pontoon displacing the water has not changed.

It increases by

150 \text{ N}

to directly counteract the additional weight of the lead inside.

It decreases because the overall density of the pontoon assembly has become significantly greater.

It increases due to the heavier pontoon creating a stronger downward force, which produces a larger upward reaction from the water.

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Cevap: It remains exactly the same because the exterior volume of the pontoon displacing the water has not changed.

Cevap

The upward buoyant force remains exactly the same because the rigid exterior volume displacing the water does not change.

Archimedes' principle states that the buoyant force on a completely submerged object is equal to the weight of the fluid it displaces. Because the pontoon is rigid, its exterior volume does not change when lead weights are added inside. Since it displaces the exact same volume of water when forced completely underwater in both scenarios, the upward buoyant force acting on it remains perfectly constant, regardless of its increased internal weight.

Adım Adım Çözüm

Identify the factors that determine buoyant force according to Archimedes' Principle.

Buoyant force depends only on the density of the fluid and the volume of the fluid displaced by the object (

F_b = \rho \cdot V \cdot g

This establishes the fundamental physics rule needed to solve the problem.

Analyze what changes between the first and second scenario.

The total weight of the pontoon increases from

400 \text{ N}

550 \text{ N}

To see if any variables in the buoyant force equation have been affected.

Determine if the volume of displaced fluid changes.

Because the pontoon is rigid, its exterior dimensions do not expand or shrink. Since it is completely submerged in both cases, the displaced volume remains exactly

200 \text{ liters}

To calculate the final impact on the buoyant force.

Evaluate the final state of the buoyant force.

Since both the fluid (water) and the displaced volume (

200 \text{ liters}

) remain constant, the buoyant force remains identical.

To arrive at the final conceptual conclusion.

Anahtar Kavram

Archimedes' Principle states that buoyant force is equal to the weight of the fluid displaced, meaning it depends on the object's volume, not its mass.

Soru 67Soru →

A municipal water treatment facility utilizes two open-air settling basins with flat bottoms. Basin Alpha is a wide rectangular pool with a base area of $50\text{ m}^2$ , currently holding $150\text{ m}^3$ of water. Basin Beta is a narrow cylindrical well with a base area of $5\text{ m}^2$ , currently holding $30\text{ m}^3$ of water. What is the ratio of the gauge hydrostatic pressure exerted on the bottom of Basin Alpha to the gauge hydrostatic pressure exerted on the bottom of Basin Beta?

1:2

5:1

10:1

2:1

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Cevap: $1:2$

Cevap

The ratio is

1:2

, because gauge hydrostatic pressure depends exclusively on the vertical depth of the fluid, which is

3\text{ m}

in Basin Alpha and

6\text{ m}

in Basin Beta.

Hydrostatic pressure depends exclusively on the depth of the fluid, not the total volume or the base area. The depth of the water in Basin Alpha is

3\text{ m}

(

150\text{ m}^3 / 50\text{ m}^2

). The depth in Basin Beta is

6\text{ m}

(

30\text{ m}^3 / 5\text{ m}^2

). The ratio of these depths is

3:6

, which simplifies to

1:2

Adım Adım Çözüm

Calculate the depth of water in Basin Alpha.

h_{Alpha} = \frac{150\text{ m}^3}{50\text{ m}^2} = 3\text{ m}

Hydrostatic pressure is determined by fluid depth. Depth is found by dividing the total volume by the base area.

Calculate the depth of water in Basin Beta.

h_{Beta} = \frac{30\text{ m}^3}{5\text{ m}^2} = 6\text{ m}

The exact same volumetric relationship applies to find the vertical depth in the second basin.

Determine the ratio of their hydrostatic pressures.

Ratio =

3\text{ m} : 6\text{ m} = 1:2

Because both basins contain the same fluid (water) under the same gravitational force, the ratio of their bottom pressures is exactly equal to the ratio of their depths (

P = \rho g h

Anahtar Kavram

Hydrostatic pressure is completely independent of a container's shape, base area, or total fluid volume; it depends exclusively on fluid density, gravity, and vertical depth.

Tahmini Süre:1m 30s

Soru 68Soru →

In a mechanical workshop, two identical syringes are sealed at the tip. One syringe is filled with atmospheric gas, and the other is filled with brake fluid. When a mechanic pushes down hard on both plungers, what physical difference between the two substances becomes obvious?

The liquid volume decreases greatly, while the gas volume stays nearly constant.

The volumes of both the gas and the liquid decrease by identical proportions.

The gas volume decreases greatly, while the liquid volume stays nearly constant.

The volumes of both substances remain completely unchanged regardless of the force.

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Cevap: The gas volume decreases greatly, while the liquid volume stays nearly constant.

Cevap

The gas volume decreases greatly, while the liquid volume stays nearly constant.

Gases have widely spaced molecules and are highly compressible, meaning their volume decreases visibly under pressure. Liquids have closely packed molecules and are virtually incompressible, so their volume remains almost entirely constant even when force is applied.

Adım Adım Çözüm

Identify the states of matter for the substances in the syringes.

The atmospheric gas is in a gaseous state, and the brake fluid is in a liquid state.

Determining the state of matter is the first step in predicting physical behavior under pressure.

Recall the compressibility properties of gases versus liquids.

Gases are highly compressible because their molecules are far apart. Liquids are generally considered incompressible because their molecules are closely packed.

This fundamental principle dictates how each fluid phase reacts to an applied force.

Apply this principle to the scenario of pushing the plungers.

The force easily compresses the gas, allowing its plunger to move down significantly. The force cannot significantly compress the liquid, so its plunger barely moves.

The physical difference in molecular spacing results in a highly visible difference in volume change.

Anahtar Kavram

Compressibility differences between fluid phases

Soru 69Soru →

Arrange the following events in the correct chronological order for the extension stroke of a basic single-acting pneumatic cylinder:

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Cevap

The correct sequence is: 1) The control valve is opened, 2) Air flows into the cylinder, 3) Pressure builds against the piston, 4) The piston rod extends.

The chronological sequence follows the fundamental transfer of energy in a pneumatic system. It begins with control (valve opening), followed by fluid transmission (air flowing into the cylinder), followed by energy accumulation (pressure building against the piston due to gas compressibility), and ends with mechanical work (piston extension).

Adım Adım Çözüm

Identify the initiating control action

The control valve is opened.

A pneumatic system requires a signal or mechanical actuation to release energy from the compressor or tank.

Trace the path of the working fluid

Air enters the cylinder chamber.

Air moves from a region of higher pressure (supply) to lower pressure (cylinder) through the open valve.

Determine the physical effect inside the actuator

Pressure builds against the piston.

As air fills the confined space of the cylinder, it exerts uniform pressure on the piston face.

Identify the resulting mechanical output

The piston extends.

Once the force of the air pressure is greater than the opposing load and friction, the piston moves outward.

Anahtar Kavram

Pneumatic System Operation and Force Generation

Soru 70Soru →

Match each physical property of a fluid to its correct definition.

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Cevap

Density matches 'The mass of a fluid per unit of its volume.' Specific Gravity matches 'The ratio of a fluid's density compared to the density of pure water.' Viscosity matches 'A fluid's internal resistance to flowing.' Compressibility matches 'The measure of a fluid's volume decrease when placed under pressure.'

Each fluid property has a distinct operational definition in mechanical systems: density is mass over volume, specific gravity compares that density to water, viscosity measures flow resistance, and compressibility measures volume change under pressure.

Adım Adım Çözüm

Identify the definition of density.

Density is mapped to mass per unit volume.

This is a fundamental physical property of all matter that determines its weight for a given size.

Identify the definition of specific gravity.

Specific gravity is mapped to the ratio of a fluid's density to water.

It acts as a unitless multiplier comparing a fluid to a standard baseline.

Identify the definition of viscosity.

Viscosity is mapped to internal resistance to flow.

It dictates how easily fluids move through mechanical components like pipes and engines.

Identify the definition of compressibility.

Compressibility is mapped to how much volume decreases under pressure.

This property separates liquids (nearly incompressible) from gases (highly compressible) in hydraulic and pneumatic systems.

Anahtar Kavram

Foundational fluid properties determine how fluids behave and function in mechanical, hydraulic, and pneumatic applications.

Tahmini Süre:45s

Soru 71Soru →

Four solid machine parts are completely submerged in a deep vat of hydraulic fluid. Their respective masses and physical volumes are listed below. Arrange the machine parts in order from the LEAST buoyant force experienced to the GREATEST buoyant force experienced.

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Cevap

The correct sequence from least to greatest buoyant force is based entirely on the objects' volumes: Machine Part 3, Machine Part 1, Machine Part 4, and Machine Part 2.

Archimedes' Principle states that the buoyant force on an object equals the weight of the fluid it displaces. Because all the machine parts are entirely submerged in the exact same hydraulic fluid, the buoyant force is determined exclusively by their physical volumes. Therefore, ordering them from least to greatest buoyant force simply requires ordering them from smallest to largest volume:

4 \text{ L}

6 \text{ L}

8 \text{ L}

, and

12 \text{ L}

Adım Adım Çözüm

Identify the relevant variable for determining buoyant force on fully submerged objects.

Buoyant force depends solely on the volume of displaced fluid, not the mass of the object.

According to Archimedes' Principle, the buoyant force equals the weight of the displaced fluid. Since all parts are submerged in the same fluid, displaced volume is the only variable.

Compare the volumes of all four machine parts.

The volumes are: Part 3 (

4 \text{ L}

), Part 1 (

6 \text{ L}

), Part 4 (

8 \text{ L}

), and Part 2 (

12 \text{ L}

Extracting the volume metric from each item allows for direct comparison.

Arrange the parts in ascending order based on their volume.

The correct order is Part 3, Part 1, Part 4, Part 2.

A smaller physical volume displaces less fluid, resulting in a proportionally smaller upward buoyant force.

Anahtar Kavram

For fully submerged objects, the buoyant force is directly proportional to the object's volume (which dictates the volume of fluid displaced), completely independent of the object's mass.

Soru 72Soru →

A mechanic pushes down on the handle of a manual air pump, sealing the valve and compressing the trapped air. If the volume of the air chamber is reduced to exactly one-half of its original size while the temperature remains constant, what happens to the air pressure inside the pump?

The pressure doubles.

The pressure is reduced by half.

The pressure remains exactly the same.

The pressure quadruples.

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Cevap: The pressure doubles.

Cevap

The pressure doubles.

According to Boyle's Law, the pressure of a given mass of an ideal gas is inversely proportional to its volume when the temperature remains constant. Because the mechanic forces the same amount of air into half the space, the air molecules strike the walls of the pump chamber twice as often, causing the pressure to double.

Adım Adım Çözüm

Identify the relationship between gas volume and pressure at a constant temperature.

The relationship is inversely proportional, described by Boyle's Law (

P_1V_1 = P_2V_2

Understanding this core gas law is necessary to determine the outcome of volume changes in a sealed pneumatic system.

Apply the given volume change to the inverse relationship.

Since the volume is multiplied by

1/2

, the pressure must be multiplied by the reciprocal, which is

2

To keep the product of pressure and volume constant, halving one variable strictly requires doubling the other.

Anahtar Kavram

Boyle's Law and Gas Compressibility

Tahmini Süre:45s

Soru 73Soru →

A narrow vertical pipe and a massive, wide cylindrical storage tank are both placed on a flat horizontal surface and filled with fresh water. The water level in both containers reaches an exact depth of 15 feet. Based on the principles of fluid mechanics, how does the hydrostatic pressure at the bottom of the narrow pipe compare to the pressure at the bottom of the wide tank?

The pressure is greater in the tank because it holds a much larger total weight of water.

The pressure is exactly the same at the bottom of both containers.

The pressure is greater in the pipe because the water is concentrated over a smaller bottom area.

The pressure is greater in the tank because the large volume of water compresses the fluid at the bottom.

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Cevap: The pressure is exactly the same at the bottom of both containers.

Cevap

The pressure is exactly the same at the bottom of both containers.

Hydrostatic pressure in an open container is determined entirely by the density of the fluid and its depth. Because both containers hold the same fluid (fresh water) at the exact same depth (15 feet), the pressure exerted on the bottom of each container is identical. The shape of the container and the total volume of the water do not alter the pressure at a given depth.

Adım Adım Çözüm

Identify the factors that determine hydrostatic pressure in a fluid.

Hydrostatic pressure is determined by the formula

P = \rho g h

, where

\rho

is density,

g

is gravity, and

h

is depth.

This formula establishes that pressure in a resting fluid is independent of the container's shape or total fluid volume.

Compare the fluid properties and depth in both containers.

Both containers hold fresh water (same density,

\rho

) and the water reaches exactly 15 feet in both (same depth,

h

Comparing the variables in the hydrostatic pressure formula reveals if the pressure differs.

Determine the relative pressure at the bottom of each container.

Since both the density and the depth are identical, the pressure calculated by

P = \rho g h

is exactly the same for both containers.

Container width and total water weight do not factor into the calculation for hydrostatic pressure at a specific point.

Anahtar Kavram

Hydrostatic pressure depends exclusively on the depth and density of the fluid, not on the shape of the container or the total volume of fluid it holds.

Soru 74Soru →

A scuba diver records the surrounding water pressure at four different locations in a freshwater lake. Arrange these locations in order from the LOWEST hydrostatic pressure to the HIGHEST hydrostatic pressure experienced by the diver.

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Cevap

The correct order from lowest to highest hydrostatic pressure is: Location W (1 meter), Location X (4 meters), Location Y (12 meters), and Location Z (18 meters).

Hydrostatic pressure is determined by the weight of the fluid directly above an object. In a single body of water like a lake, the density of the water is constant. Therefore, the hydrostatic pressure depends entirely on the depth (

h

). Since the question asks for the order from lowest to highest pressure, the locations must be ordered from the shallowest depth to the deepest depth: 1 meter, 4 meters, 12 meters, and finally 18 meters.

Adım Adım Çözüm

Recall the relationship between hydrostatic pressure and depth.

Hydrostatic pressure is calculated using the formula P = ρgh (where ρ is fluid density, g is gravity, and h is depth). Therefore, pressure increases as depth increases.

Understanding this direct relationship is required to rank the locations.

Identify the depth of each location provided in the scenario.

Location W is 1m, Location X is 4m, Location Y is 12m, and Location Z is 18m.

The depth is the only variable changing between the different locations.

Order the locations from the smallest depth to the greatest depth.

Smallest depth (1m) is Location W, followed by 4m (Location X), 12m (Location Y), and the greatest depth is 18m (Location Z).

Since pressure increases with depth, ordering from smallest to greatest depth directly corresponds to lowest to highest pressure.

Anahtar Kavram

In a fluid of uniform density, hydrostatic pressure increases proportionally with the depth of the fluid.

Soru 75Soru →

A technician drops two completely sealed metal canisters into a large vat of testing fluid. Canister A is completely hollow. Canister B has the exact same exterior dimensions and shape as Canister A, but is filled completely with solid lead weights. Both canisters are pushed down and held completely stationary underwater. Which of the following statements is true regarding the upward buoyant force acting on the canisters?

The buoyant forces acting on both canisters are exactly equal because they displace the same volume of fluid.

The buoyant force on Canister B is greater because its higher mass interacts more strongly with the surrounding fluid.

The buoyant force on Canister A is greater because its lower weight provides less resistance to the upward push of the fluid.

Canister B experiences no buoyant force because its total weight overcomes the lifting capacity of the fluid.

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Cevap: The buoyant forces acting on both canisters are exactly equal because they displace the same volume of fluid.

Cevap

The buoyant forces acting on both canisters are exactly equal because they displace the same volume of fluid.

According to Archimedes' principle, the buoyant force on an object is equal to the weight of the fluid it displaces. Since both canisters are perfectly sealed, completely submerged, and have identical exterior dimensions, they displace the exact same volume of testing fluid. Therefore, the fluid exerts the exact same upward force on both, despite the fact that Canister B is much heavier.

Adım Adım Çözüm

Identify the physics principle governing the situation.

Archimedes' principle states that the upward buoyant force is equal to the weight of the fluid displaced by the object.

To determine what variables influence buoyant force.

Determine the factors that affect the volume of fluid displaced.

The displaced volume depends strictly on the exterior dimensions of the completely submerged object.

The formula for buoyant force is

F_B = \rho \cdot V \cdot g

, where

V

is the displaced volume.

Compare the exterior volumes of Canister A and Canister B.

The problem states both canisters have the exact same exterior dimensions and shape, meaning they displace the exact same volume of fluid.

To establish the relationship between the two displaced volumes.

Evaluate the buoyant force based on the displaced volumes.

Because the displaced volumes are equal, the upward buoyant forces exerted by the fluid must also be equal, regardless of the internal mass.

To reach the final conclusion regarding the buoyant forces.

Anahtar Kavram

Buoyancy and Archimedes' Principle

Soru 76Soru →

A large dam holds back a deep reservoir of fresh water. Three pressure sensors are installed on the submerged face of the dam at different depths: Sensor X is 10 meters below the water surface, Sensor Y is 2 meters below the water surface, and Sensor Z is 25 meters below the water surface. Arrange the sensors in order from the LOWEST hydrostatic pressure to the HIGHEST hydrostatic pressure.

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Cevap

The correct order from lowest to highest hydrostatic pressure is: Sensor Y, Sensor X, and Sensor Z.

Hydrostatic pressure is determined by the density of the fluid, gravity, and the depth of the fluid. Because the fluid (fresh water) is consistent across all sensors, the pressure depends entirely on the depth. The sensor closest to the water's surface (Sensor Y at 2 meters) experiences the least amount of downward force from the water column above it, giving it the lowest pressure. The sensor furthest down (Sensor Z at 25 meters) experiences the most force, resulting in the highest pressure.

Adım Adım Çözüm

Identify the relationship between depth and hydrostatic pressure.

Hydrostatic pressure increases as the depth of the fluid increases.

The formula for hydrostatic pressure,

P = \rho g h

, demonstrates that pressure (

P

) is directly proportional to depth (

h

Compare the vertical depths of the three sensors.

Sensor Y is at 2 meters, Sensor X is at 10 meters, and Sensor Z is at 25 meters.

Determining the relative depths is required to compare the relative pressures.

Order the sensors based on depth to find the sequence from lowest to highest pressure.

Lowest pressure is at Sensor Y, followed by Sensor X, and the highest is at Sensor Z.

Because the question asks for the order from lowest to highest pressure, the items must be arranged from the shallowest depth to the deepest depth.

Anahtar Kavram

Hydrostatic pressure in a fluid increases proportionally with depth.

Soru 77Soru →

A backup breathing canister has a volume of $50 \text{ in}^3$ and stores gas at an absolute pressure of $2000 \text{ psi}$ . During a transfer process, the gas is allowed to expand into a larger holding tank, reaching a total final volume of $250 \text{ in}^3$ . Assuming the temperature remains constant, what is the new absolute pressure of the gas in $\text{psi}$ ?

Cevabı ve açıklamayı göster

Cevap: 400

Cevap

400

According to Boyle's Law, the pressure and volume of a gas are inversely proportional when the temperature is constant (

P_1V_1 = P_2V_2

). Since the volume increased by a factor of 5 (from 50 to 250 cubic inches), the pressure must decrease by a factor of 5. Dividing the initial 2000 psi by 5 gives a final pressure of 400 psi.

Adım Adım Çözüm

Identify the initial and final states of the gas.

P_1 = 2000 \text{ psi}

V_1 = 50 \text{ in}^3

, and

V_2 = 250 \text{ in}^3

Organizing the known variables is the first step in solving any gas law problem.

Apply Boyle's Law to relate pressure and volume.

Equation to use:

P_1 V_1 = P_2 V_2

Boyle's Law states that for a fixed mass of gas at constant temperature, pressure and volume are inversely proportional.

Substitute the known values into the equation and solve for the final pressure (

P_2

2000 \times 50 = P_2 \times 250

, which simplifies to

100000 = 250 P_2

. Dividing both sides by 250 gives

P_2 = 400

This mathematical calculation yields the new absolute pressure after the gas has expanded.

Anahtar Kavram

Boyle's Law (Inverse relationship between pressure and volume)

Alternatif Yöntem

Instead of full algebraic substitution, recognize the proportionality: the volume increases by a factor of 5 (

250 / 50 = 5

). Because pressure and volume are inversely proportional, the pressure must be divided by 5. Thus,

2000 \text{ psi} / 5 = 400 \text{ psi}

Tahmini Süre:45s

Soru 78Soru →

A perfectly sealed, rigid steel air compressor tank is left outside in direct sunlight on a very hot summer day. As the temperature of the trapped air inside the tank increases, what will happen to the internal air pressure?

The pressure will increase.

The pressure will decrease.

The pressure will stay constant because the tank volume is fixed.

The pressure will drop as the heated air molecules move faster.

Cevabı ve açıklamayı göster

Cevap: The pressure will increase.

Cevap

The pressure will increase.

According to Gay-Lussac's Law, the pressure of a fixed mass of gas is directly proportional to its temperature when the volume remains constant. As the air inside the rigid steel tank absorbs heat from the sun, the air molecules gain kinetic energy. They move faster and collide with the internal walls of the tank more frequently and with greater force, which results in a higher internal pressure.

Adım Adım Çözüm

Identify the relationship between temperature and pressure for a trapped gas.

Gay-Lussac's Law states that pressure is directly proportional to absolute temperature when volume is constant.

The tank is described as 'rigid' and 'perfectly sealed', meaning the volume cannot change.

Apply the temperature change to the system.

The sunlight heats the tank, increasing the temperature of the air inside.

To determine the final state of the gas.

Determine the outcome based on the direct relationship.

Because temperature increases, pressure must also increase.

Higher temperature means higher kinetic energy; molecules strike the rigid walls harder and more frequently.

Anahtar Kavram

Gay-Lussac's Law (Temperature-Pressure Relationship)

Soru 79Soru →

A salvage crane is lifting a solid brass propeller from the bottom of a freshwater lake. The propeller has a true weight of $1,200 \text{ N}$ in the air. When it is completely submerged in the water, it displaces $35 \text{ liters}$ of water.

If $1 \text{ liter}$ of water weighs exactly $10 \text{ N}$ , what is the apparent weight of the propeller, in Newtons, while it is still completely underwater?

Cevabı ve açıklamayı göster

Cevap: 850

Cevap

850

The propeller experiences an upward buoyant force equal to the weight of the water it displaces. Since it displaces

35 \text{ liters}

and each liter weighs

10 \text{ N}

, the buoyant force is

350 \text{ N}

. The apparent weight is the true weight (

1,200 \text{ N}

) minus this buoyant force (

350 \text{ N}

), resulting in

850 \text{ N}

Adım Adım Çözüm

Calculate the weight of the displaced water to find the buoyant force.

Buoyant force =

350 \text{ N}

According to Archimedes' Principle, the buoyant force on an object is equal to the weight of the fluid it displaces. Here,

35 \text{ liters} \times 10 \text{ N/liter} = 350 \text{ N}

Subtract the buoyant force from the true weight to find the apparent weight.

Apparent weight =

850 \text{ N}

The apparent weight of a submerged object is its actual weight in air minus the upward buoyant force exerted by the fluid (

1,200 \text{ N} - 350 \text{ N} = 850 \text{ N}

Anahtar Kavram

Archimedes' principle states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object. The apparent weight of a submerged object is its true weight minus the buoyant force.

Tahmini Süre:1m 0s

Soru 80Soru →

Place the following operational stages of a basic reciprocating air compressor in their proper sequential order, starting with the introduction of fresh air into the system.

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

The correct operational sequence begins with the intake of ambient air, followed by the sealing of the inlet valve, then the compression of the trapped air, and finally the discharge of the pressurized air into the tank.

A basic reciprocating compressor operates in a continuous four-stage loop: it draws air in (intake), seals the chamber, squeezes the air into a smaller volume to build pressure (compression), and then pushes that pressurized air out to a tank (discharge).

Adım Adım Çözüm

Identify the starting point of the pneumatic cycle.

The cycle must begin by drawing ambient air into the empty cylinder.

An air compressor needs a fresh supply of air to perform any mechanical work.

Determine the necessary condition for compression.

The inlet valve must close to seal the chamber.

If the chamber is not sealed, upward movement of the piston would simply push the air back out instead of compressing it.

Identify the actual compression action.

The piston moves up to decrease the internal volume.

Decreasing the volume of the sealed gas increases its pressure.

Identify the final stage of the cycle.

The outlet valve opens to release the compressed air.

The system must transfer the high-pressure air into the storage tank for future use.

Anahtar Kavram

Reciprocating compressor cycle and basic application of Boyle's Law

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Fluid Mechanics

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