Fluid Mechanics

118 soru

Soru 41Soru →

When a steady flow of fluid passes through a horizontal pipe that suddenly narrows in diameter, the fluid's velocity increases in the constricted section, resulting in a higher static pressure against the pipe walls.

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Cevap: False

Cevap

False
The statement is false because Bernoulli's principle dictates that as a fluid's velocity increases in a constricted horizontal pipe, its static pressure decreases, creating a low-pressure zone rather than a high-pressure one.

Adım Adım Çözüm

1
Analyze the change in fluid velocity through the constriction.
Based on the continuity equation (A1V1=A2V2A_1V_1 = A_2V_2), the fluid's velocity increases as it enters the narrower section of the pipe.
The same volume of fluid must pass through a smaller area in the same amount of time, requiring it to speed up.
2
Determine the effect of increased velocity on static pressure.
Static pressure decreases.
Bernoulli's principle states that for a horizontal flow, an increase in the speed of the fluid occurs simultaneously with a decrease in static pressure.
3
Evaluate the statement's claim against the established physical principles.
The statement claims static pressure increases, which contradicts the actual outcome (a decrease in static pressure).
The inverse relationship between fluid velocity and static pressure confirms the statement is false.

Anahtar Kavram

Bernoulli's principle and the continuity equation for incompressible fluids.
Tahmini Süre:45s
Soru 42Soru →

A mechanic uses a hydraulic floor jack to lift the front axle of a delivery truck. The jack's output piston has a cross-sectional area of 25 in225 \text{ in}^2 and must support a load of 5,000 lbs5,000 \text{ lbs}. If the input piston has a cross-sectional area of 0.5 in20.5 \text{ in}^2, what is the minimum input force the mechanic must apply to lift the truck? (Assume a frictionless, ideal system)

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Cevap: 100 lbs100 \text{ lbs}

Cevap

100 lbs100 \text{ lbs}
In a closed hydraulic system, pressure is uniform. The pressure created by the load on the output piston is its force divided by its area (5,000/25=200 psi5,000 / 25 = 200 \text{ psi}). To produce this same pressure at the input piston, you multiply the pressure by the input piston's area (200×0.5200 \times 0.5), which equals an input force of 100 lbs100 \text{ lbs}.

Adım Adım Çözüm

1
Calculate the fluid pressure required at the output piston using the formula P = F/A.
5,000 lbs/25 in2=200 psi5,000 \text{ lbs} / 25 \text{ in}^2 = 200 \text{ psi}
Pascal's Principle states that pressure applied to a confined fluid is transmitted equally in all directions.
2
Apply the calculated pressure to the input piston to find the required input force using F = P × A.
200 psi×0.5 in2=100 lbs200 \text{ psi} \times 0.5 \text{ in}^2 = 100 \text{ lbs}
The input piston must generate the identical system pressure to balance and lift the load.

Anahtar Kavram

Pascal's Principle (Hydraulic Mechanical Advantage)
Soru 43Soru →

A small metal anchor weighs 120 N120 \text{ N} in the air. When the anchor is completely submerged in water, the tension on its lifting cable is measured to be 85 N85 \text{ N}. What is the magnitude of the buoyant force acting on the submerged anchor?

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Cevap: 35

Cevap

35
According to Archimedes' principle, the upward buoyant force exerted on a submerged object reduces its apparent weight. The buoyant force is equal to the difference between its true weight in the air and its apparent weight in the fluid. Subtracting the apparent weight (85 N) from the true weight (120 N) gives a buoyant force of 35 N.

Adım Adım Çözüm

1
Identify the actual weight and the apparent weight.
Actual weight = 120 N, Apparent weight = 85 N
These values represent the downward force of gravity and the net downward force when submerged, respectively.
2
Calculate the buoyant force.
120 N - 85 N = 35 N
The buoyant force is the upward force that accounts for the reduction in weight. It is found by subtracting the apparent weight from the actual weight.

Anahtar Kavram

Apparent Weight and Buoyant Force
Soru 44Soru →

An industrial cardboard baler utilizes a closed hydraulic system to compress recyclable materials. The system features an input pump piston with a cross-sectional area of 3 in23 \text{ in}^2 and a large output driving piston with a cross-sectional area of 45 in245 \text{ in}^2. If the input piston is pushed downward for a single stroke distance of 15 inches15 \text{ inches}, how many inches will the output piston move? (Assume no fluid is lost and the fluid is incompressible.)

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Cevap: 1

Cevap

1
In a hydraulic system, the volume of fluid displaced must be conserved. The output piston has 15 times the cross-sectional area of the input piston (45÷3=1545 \div 3 = 15). Because it is 15 times larger, it will only travel 1/15th of the distance to displace the identical volume of fluid. Therefore, an input distance of 15 inches results in an output distance of exactly 1 inch.

Adım Adım Çözüm

1
Identify the relationship for fluid displacement in a closed hydraulic system.
A1×d1=A2×d2A_1 \times d_1 = A_2 \times d_2
Because liquids are incompressible, the volume of fluid pushed out by the input piston (A1×d1A_1 \times d_1) must equal the volume of fluid pushing up the output piston (A2×d2A_2 \times d_2).
2
Substitute the given values for the input piston (A1=3A_1 = 3, d1=15d_1 = 15) and output piston (A2=45A_2 = 45).
3×15=45×d23 \times 15 = 45 \times d_2
This sets up the equation with all known variables to solve for the unknown distance (d2d_2).
3
Calculate the total volume displaced by the input piston.
45=45×d245 = 45 \times d_2
Multiplying the input area by its travel distance gives a total displaced volume of 45 cubic inches.
4
Solve for the output distance (d2d_2) by dividing the total volume by the output piston's area.
d2=1d_2 = 1
Dividing 45 cubic inches by 45 square inches yields a travel distance of 1 inch.

Anahtar Kavram

Conservation of fluid volume and the distance-force tradeoff in hydraulics
Soru 45Soru →

An industrial pneumatic press system utilizes trapped gas to generate force. The gas is initially at an absolute pressure of 30 psi30 \text{ psi} and occupies a volume of 200 in3200 \text{ in}^3. During the operation cycle, a mechanical ram pushes down, reducing the gas volume to 50 in350 \text{ in}^3 while the temperature remains constant. If the face of the piston has a cross-sectional area of 5 in25 \text{ in}^2, what is the total upward force, in pounds, exerted by the compressed gas on the piston at this new volume?

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Cevap: 600

Cevap

600
Since the temperature is constant, Boyle's Law (P1V1=P2V2P_1V_1 = P_2V_2) applies. The volume is reduced from 200 in3200 \text{ in}^3 to 50 in350 \text{ in}^3, which is one-quarter of its original size. Because pressure and volume are inversely proportional for a fixed amount of gas at a constant temperature, the absolute pressure must increase by a factor of four (30 psi×4=120 psi30 \text{ psi} \times 4 = 120 \text{ psi}). The total force exerted by a pressurized fluid on a surface is the product of its pressure and the surface area (F=P×AF = P \times A). Multiplying the new pressure of 120 psi120 \text{ psi} by the piston area of 5 in25 \text{ in}^2 gives a total upward force of 600 lbs600 \text{ lbs}.

Adım Adım Çözüm

1
Calculate the final pressure of the gas after compression.
P2=120 psiP_2 = 120 \text{ psi}
According to Boyle's Law, when temperature is constant, the product of absolute pressure and volume is constant (P1V1=P2V2P_1V_1 = P_2V_2).
2
Calculate the total upward force exerted by the gas.
F=600 lbsF = 600 \text{ lbs}
Force is equal to pressure multiplied by the area over which it is applied (F=P×AF = P \times A).

Anahtar Kavram

Combining Boyle's Law for isothermal compression with the basic relationship between pressure, force, and area.
Soru 46Soru →

A hydraulic dump truck bed lift system operates using a small input pump cylinder and a larger output lifting cylinder. The pump cylinder has a cross-sectional area of 2 in22 \text{ in}^2 and the lifting cylinder has a cross-sectional area of 20 in220 \text{ in}^2. To raise the truck bed, the lifting cylinder must extend a distance of 8 inches8 \text{ inches}. Assuming an ideal closed hydraulic system, what is the total distance the pump cylinder must travel to achieve this lift?

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Cevap: 80 inches80 \text{ inches}

Cevap

The input pump cylinder must travel a total distance of 80 inches80 \text{ inches}.
Because liquids are essentially incompressible, the volume of fluid pushed by the input cylinder must equal the volume received by the output cylinder. Volume is calculated as Area multiplied by Distance (V=A×dV = A \times d). Setting the volumes equal gives 2×d1=20×82 \times d_1 = 20 \times 8. Solving this equation shows that the input cylinder must travel 80 inches80 \text{ inches} to move the larger cylinder 8 inches8 \text{ inches}.

Adım Adım Çözüm

1
Identify the given values for the hydraulic system.
A1=2 in2A_1 = 2 \text{ in}^2, A2=20 in2A_2 = 20 \text{ in}^2, and d2=8 inchesd_2 = 8 \text{ inches}.
These values are needed to apply the principle of volume conservation in an incompressible fluid system.
2
Apply the volume conservation formula for hydraulic systems.
A1×d1=A2×d2A_1 \times d_1 = A_2 \times d_2
In a closed hydraulic system, the volume of fluid displaced by the input piston must equal the volume of fluid displacing the output piston.
3
Substitute the known values into the equation and solve for the input distance (d1d_1).
2×d1=20×8⇒2×d1=160⇒d1=80 inches2 \times d_1 = 20 \times 8 \Rightarrow 2 \times d_1 = 160 \Rightarrow d_1 = 80 \text{ inches}.
Isolating d1d_1 provides the total travel distance required by the input pump to move the lifting cylinder by the required amount.

Anahtar Kavram

In a hydraulic system, work and fluid volume are conserved. A smaller input piston must travel a proportionally greater distance to move a larger output piston a shorter distance.
Soru 47Soru →

A heavy-duty hydraulic press is designed to operate using hydraulic oil with a specific gravity of 0.85. During maintenance, a large volume of air is accidentally trapped in the sealed main pressure line. When the pump applies 2,000 psi of pressure to the system, which of the following best describes how the properties of the contents will affect the machine's operation?

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Cevap: The trapped air will compress significantly before the oil can transmit full force, causing a delayed and spongy cylinder response.

Cevap

The trapped air will compress significantly before the oil can transmit full force, causing a delayed and spongy cylinder response.
Hydraulic systems rely on the incompressibility of liquids to transmit force instantaneously and efficiently. When air (a gas) enters the system, it acts like a weak mechanical spring because gases are highly compressible. The pump's applied pressure will first crush the air pocket into a smaller volume before enough pressure builds up to move the heavy cylinder, resulting in a delayed, inefficient, and spongy mechanical action.

Adım Adım Çözüm

1
Identify the two different fluids in the closed hydraulic system.
The system contains a liquid (hydraulic oil) and a gas (air).
Different phases of matter respond completely differently to applied pressure.
2
Evaluate the compressibility of each fluid.
Liquids are nearly incompressible, meaning their volume remains virtually constant under high pressure. Gases are highly compressible, meaning their volume decreases significantly when pressure is applied.
Understanding compressibility is essential to predicting the physical behavior of the system.
3
Determine the mechanical effect on the hydraulic press.
When the 2,000 psi is applied, the work energy initially goes into compressing the trapped air pocket rather than moving the cylinder. This creates a delay and a soft, 'spongy' feel until the air is fully compressed.
Force cannot be efficiently transmitted through a fluid medium until all highly compressible elements in the line are reduced in volume.

Anahtar Kavram

Fluid Compressibility
Tahmini Süre:1m 0s
Soru 48Soru →

A deep-sea recovery team locates a lost concrete mooring block resting on the ocean floor. The solid concrete block has a physical volume of 2 m32 \text{ m}^3 and a total downward weight of 45,000 N45,000 \text{ N}. The surrounding seawater has a weight density of 10,000 N/m310,000 \text{ N/m}^3. According to Archimedes' principle, what is the magnitude of the upward buoyant force acting on the completely submerged block?

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Cevap: 20,000 N20,000 \text{ N}

Cevap

The correct answer is 20,000 N20,000 \text{ N}, which is found by multiplying the block's volume by the weight density of the seawater.
According to Archimedes' Principle, the buoyant force exerted on an object is exactly equal to the weight of the fluid that the object displaces. Because the block is completely submerged, it displaces a volume of water equal to its own volume (2 m32 \text{ m}^3). Multiplying this volume by the water's weight density (10,000 N/m310,000 \text{ N/m}^3) yields an upward buoyant force of 20,000 N20,000 \text{ N}. The actual weight of the concrete block is irrelevant to calculating the buoyant force.

Adım Adım Çözüm

1
Identify the formula for buoyant force based on Archimedes' Principle.
Buoyant Force = Volume of displaced fluid × Weight density of the fluid.
To establish that buoyancy depends exclusively on fluid properties and displaced volume, rather than the object's own weight.
2
Extract the necessary physical values from the problem description.
The displaced volume is equal to the block's volume (2 m32 \text{ m}^3), and the fluid's weight density is 10,000 N/m310,000 \text{ N/m}^3.
To separate the relevant variables needed for the calculation from the distractor information (the block's 45,000 N45,000 \text{ N} weight).
3
Calculate the magnitude of the upward buoyant force.
2 m3×10,000 N/m3=20,000 N2 \text{ m}^3 \times 10,000 \text{ N/m}^3 = 20,000 \text{ N}.
To determine the final upward thrust exerted by the seawater.

Anahtar Kavram

Archimedes' Principle states that the buoyant force on a completely submerged object equals the weight of the fluid it displaces, independent of the object's own downward weight.
Soru 49Soru →

A heavy-duty pneumatic shock absorber relies on a sealed chamber of air to dampen impacts. Before an impact, the air inside the chamber has an absolute pressure of 90 psi90 \text{ psi}, occupies a volume of 60 in360 \text{ in}^3, and is at a temperature of 300 K300 \text{ K}. During a severe impact, the piston rapidly compresses the air volume to 20 in320 \text{ in}^3, causing the internal temperature to spike to 400 K400 \text{ K}. Assuming no air escapes, what is the new absolute pressure inside the shock absorber?

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Cevap: 360 psi360 \text{ psi}

Cevap

360 psi360 \text{ psi}
The correct answer correctly applies the Combined Gas Law. The volume is reduced to one-third of its original size, which alone would triple the pressure to 270 psi270 \text{ psi}. Simultaneously, the absolute temperature increases by a factor of 4/34/3, which further scales the pressure to 270×43=360 psi270 \times \frac{4}{3} = 360 \text{ psi}.

Adım Adım Çözüm

1
Identify the initial state (P1,V1,T1P_1, V_1, T_1) and final state (V2,T2V_2, T_2) variables from the problem.
P1=90 psiP_1 = 90 \text{ psi}, V1=60 in3V_1 = 60 \text{ in}^3, T1=300 KT_1 = 300 \text{ K}; V2=20 in3V_2 = 20 \text{ in}^3, T2=400 KT_2 = 400 \text{ K}.
Establishing known parameters is required to use the Combined Gas Law.
2
Set up the Combined Gas Law equation: P1×V1T1=P2×V2T2\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}.
90×60300=P2×20400\frac{90 \times 60}{300} = \frac{P_2 \times 20}{400}
Since both volume and temperature are changing simultaneously, the combined relationship must be used.
3
Rearrange the formula to isolate the final pressure (P2P_2).
P2=P1×(V1V2)×(T2T1)P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) \times \left(\frac{T_2}{T_1}\right)
This setup groups the variables into easy-to-calculate ratios.
4
Substitute the values and calculate the final result.
P2=90×(6020)×(400300)=90×3×43=360 psiP_2 = 90 \times \left(\frac{60}{20}\right) \times \left(\frac{400}{300}\right) = 90 \times 3 \times \frac{4}{3} = 360 \text{ psi}.
Mathematical evaluation yields the final absolute pressure inside the shock absorber.

Anahtar Kavram

Combined Gas Law and Compressibility
Soru 50Soru →

Four different objects are completely submerged in a tank of fresh water. The mass of each object and the volume of water it displaces are recorded below. Arrange the objects in order of the buoyant force exerted on them by the water, from LEAST buoyant force to GREATEST buoyant force.

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Cevap

The correct sequence from least to greatest buoyant force is: Iron weight, Lead brick, Aluminum gear, Wooden block.
According to Archimedes' principle, the buoyant force acting on an object is equal to the weight of the fluid it displaces. Because all the objects are in the same fresh water, the buoyant force is directly proportional to the displaced volume. By ordering the displaced volumes from smallest to largest (0.2 L0.2 \text{ L}, 0.5 L0.5 \text{ L}, 1.0 L1.0 \text{ L}, 2.0 L2.0 \text{ L}), we accurately order the buoyant forces from least to greatest.

Adım Adım Çözüm

1
Recall the definition of Archimedes' principle.
The buoyant force acting on a submerged object is equal to the weight of the fluid that the object displaces.
This principle establishes that when objects are submerged in the same fluid, the buoyant force depends entirely on the volume of fluid displaced, not the mass of the objects themselves.
2
Extract the relevant data for each object.
Iron = 0.2 L0.2 \text{ L}, Lead = 0.5 L0.5 \text{ L}, Aluminum = 1.0 L1.0 \text{ L}, Wood = 2.0 L2.0 \text{ L}.
The mass values provided (0.5 kg0.5 \text{ kg} to 3.0 kg3.0 \text{ kg}) are distractor information and should be ignored for calculating buoyant force.
3
Rank the objects based on their displaced volumes from least to greatest.
Iron (0.2 L0.2 \text{ L}) < Lead (0.5 L0.5 \text{ L}) < Aluminum (1.0 L1.0 \text{ L}) < Wood (2.0 L2.0 \text{ L}).
Since greater displaced volume equals greater buoyant force, this is the final correct order.

Anahtar Kavram

According to Archimedes' Principle, the buoyant force on a submerged object depends only on the density of the fluid and the volume of the fluid displaced, completely independent of the object's mass or weight.
Tahmini Süre:45s
Soru 51Soru →

Match each closed hydraulic system scenario to its calculated unknown value. Assume the fluid is incompressible, there is no friction, and conditions are ideal.

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Öğeler

Input: A=2 in2A = 2 \text{ in}^2, F=50 lbsF = 50 \text{ lbs} →\rightarrow Output: A=10 in2A = 10 \text{ in}^2 (Find Output Force)
Input: A=3 in2A = 3 \text{ in}^2, F=30 lbsF = 30 \text{ lbs} →\rightarrow Output: F=150 lbsF = 150 \text{ lbs} (Find Output Area)
Input: A=10 in2A = 10 \text{ in}^2 →\rightarrow Output: A=20 in2A = 20 \text{ in}^2, F=100 lbsF = 100 \text{ lbs} (Find Input Force)
Input: F=40 lbsF = 40 \text{ lbs}, moves 5 in5 \text{ in} →\rightarrow Output: F=200 lbsF = 200 \text{ lbs} (Find Output Distance)

Eşleşmeler

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Cevap

The correct matches are based on applying Pascal's Principle and the conservation of work. Output Force is 250 lbs, Output Area is 15 in², Input Force is 50 lbs, and Output Distance is 1 in.
Each matching pair applies either Pascal's Principle (P=F1A1=F2A2P = \frac{F_1}{A_1} = \frac{F_2}{A_2}) or the conservation of work (F1⋅d1=F2⋅d2F_1 \cdot d_1 = F_2 \cdot d_2) to solve for the missing variable in the hydraulic system.

Adım Adım Çözüm

1
Calculate the Output Force for the first scenario.
250 lbs
Using Pascal's Principle (F1/A1=F2/A2F_1/A_1 = F_2/A_2), the system pressure is 50 lbs / 2 in² = 25 psi. The output force is 25 psi * 10 in² = 250 lbs.
2
Calculate the Output Area for the second scenario.
15 in²
The system pressure is 30 lbs / 3 in² = 10 psi. Rearranging the formula for area (A=F/PA = F/P), the output area is 150 lbs / 10 psi = 15 in².
3
Calculate the Input Force for the third scenario.
50 lbs
System pressure is determined by the output side: 100 lbs / 20 in² = 5 psi. The required input force is 5 psi * 10 in² = 50 lbs.
4
Calculate the Output Distance for the fourth scenario.
1 in
By conservation of work (F1×d1=F2×d2F_1 \times d_1 = F_2 \times d_2), the input work is 40 lbs * 5 in = 200 in-lbs. Thus, the output distance is 200 in-lbs / 200 lbs = 1 in.

Anahtar Kavram

Pascal's Principle states that pressure applied to an enclosed fluid is transmitted undiminished. The work done on the input piston equals the work done by the output piston.
Soru 52Soru →

Evaluate the gauge hydrostatic pressure exerted at the specific points in the four situations described below. Rank them in order from the HIGHEST hydrostatic pressure to the LOWEST hydrostatic pressure.

Assume standard gravity (gg) applies equally to all scenarios.

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Cevap

The correct ranking from highest to lowest pressure is: the narrow pipe with oil, the scuba diver in the lake, the submersible in the ocean, and finally the municipal water reservoir.
The correct sequence is found by calculating the relative pressure factor (density multiplied by depth) for each situation. The narrow pipe with oil has a factor of 16,000, making it the highest. The scuba diver is next at 15,000. The submersible is third at 12,300, and the municipal reservoir is lowest at 10,000. The shape and total volume of the containers are irrelevant distractors.

Adım Adım Çözüm

1
Identify the relevant formula and variables for gauge hydrostatic pressure.
The formula is P=ρ⋅g⋅hP = \rho \cdot g \cdot h. Since gravity (gg) is constant across all scenarios, we only need to compare the product of density (ρ\rho) and depth (hh) to determine the relative pressures.
This establishes a direct mathematical way to compare the situations without needing to calculate the exact final pressure in Pascals.
2
Evaluate whether container shape or total fluid volume affects the pressure.
We must ignore the 'extremely narrow' pipe and the 'massive, wide' reservoir descriptions.
According to the principles of fluid mechanics, hydrostatic pressure at a specific depth depends only on depth and density, not on the total volume or the shape of the container (this is known as the hydrostatic paradox).
3
Calculate the pressure factor (ρ⋅h\rho \cdot h) for each scenario.
Submersible: 1025⋅12=12,3001025 \cdot 12 = 12,300. Narrow pipe: 800⋅20=16,000800 \cdot 20 = 16,000. Reservoir: 1000⋅10=10,0001000 \cdot 10 = 10,000. Scuba diver: 1000⋅15=15,0001000 \cdot 15 = 15,000.
These proportional values allow us to easily and accurately rank the pressures.
4
Rank the calculated values from highest to lowest.
Highest: 16,00016,000 (Narrow pipe). Second: 15,00015,000 (Scuba diver). Third: 12,30012,300 (Submersible). Lowest: 10,00010,000 (Reservoir).
This provides the final correct ordering required by the question.

Anahtar Kavram

Hydrostatic pressure at a given depth is determined solely by the fluid's density and the vertical depth (P=ρghP = \rho g h), and is completely independent of the container's shape or the total volume of fluid.
Soru 53Soru →

A hydraulic lifting mechanism on a heavy-duty forklift contains an input cylinder and an output cylinder. The output cylinder has 55 times the cross-sectional area of the input cylinder. If the operator applies an input force of 50 lbs50 \text{ lbs} to move the input piston down by 15 inches15 \text{ inches}, what is the resulting lifting force and how far does the output piston move?

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Cevap: 250 lbs250 \text{ lbs} of force and 3 inches3 \text{ inches} of movement

Cevap

The output cylinder produces 250 lbs250 \text{ lbs} of lifting force and moves 3 inches3 \text{ inches}.
The hydraulic system provides a mechanical advantage equal to the ratio of the piston areas. Since the output cylinder has 55 times the area, the output force is 55 times the input force (50×5=250 lbs50 \times 5 = 250 \text{ lbs}). However, to conserve energy and fluid volume, the output piston only moves 1/51/5 of the distance of the input piston (15/5=3 inches15 / 5 = 3 \text{ inches}).

Adım Adım Çözüm

1
Determine the mechanical advantage of the hydraulic system.
Mechanical Advantage = 55.
The output cylinder has 55 times the cross-sectional area of the input cylinder.
2
Calculate the output lifting force.
F2=50×5=250 lbsF_2 = 50 \times 5 = 250 \text{ lbs}.
According to Pascal's Principle, force is directly proportional to the area. The larger piston exerts 55 times the force.
3
Calculate the distance the output piston moves.
d2=15/5=3 inchesd_2 = 15 / 5 = 3 \text{ inches}.
Because the volume of incompressible fluid displaced remains constant, the distance moved is inversely proportional to the area.

Anahtar Kavram

Pascal's Principle and Conservation of Energy in Hydraulics
Tahmini Süre:1m 0s
Soru 54Soru →

A solid steel engine block with a true weight of 2,000 N2,000 \text{ N} is completely submerged in a freshwater lake. The block displaces a volume of water that weighs 250 N250 \text{ N}. According to Archimedes' principle, what is the magnitude of the upward buoyant force acting on the engine block?

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Cevap: 250 N250 \text{ N}

Cevap

250 N250 \text{ N}
According to Archimedes' principle, the upward buoyant force on an object submerged in a fluid is exactly equal to the weight of the fluid it displaces. Since the engine block displaces water weighing 250 N250 \text{ N}, the buoyant force acting upon it is exactly 250 N250 \text{ N}.

Adım Adım Çözüm

1
Identify the values provided in the scenario.
The true weight of the block is 2,000 N2,000 \text{ N} and the weight of the displaced water is 250 N250 \text{ N}.
Extracting given values clarifies which numbers represent which physical properties.
2
Apply Archimedes' principle to determine the buoyant force.
Archimedes' principle states that the upward buoyant force is equal to the weight of the displaced fluid.
This principle directly links the volume of fluid displaced to the upward force exerted by the fluid.
3
Select the value that represents the weight of the displaced fluid.
The weight of the displaced water is 250 N250 \text{ N}, therefore the buoyant force is 250 N250 \text{ N}.
The object's own weight does not determine the buoyant force; only the displaced fluid's weight matters.

Anahtar Kavram

Archimedes' Principle and Buoyant Force
Soru 55Soru →

An industrial processing facility uses two large vertical storage tanks that are open to the atmosphere. Tank A contains two distinct layers of non-mixing liquids: a 5-meter5\text{-meter} top layer of oil (density = 800 kg/m3800\text{ kg/m}^3) and a 3-meter3\text{-meter} bottom layer of water (density = 1000 kg/m31000\text{ kg/m}^3). Tank B, which has exactly twice the cross-sectional area of Tank A, contains pure water to a total depth of 8 meters8\text{ meters}. Which of the following statements correctly compares the gauge hydrostatic pressure acting on the interior bottom surfaces of the two tanks?

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Cevap: The pressure at the bottom of Tank B is greater because the entire 8-meter8\text{-meter} fluid column consists of the denser liquid.

Cevap

The pressure at the bottom of Tank B is greater because its entire fluid column is composed of the denser water, yielding a higher hydrostatic pressure than Tank A's mixed column.
The correct answer accurately calculates that Tank B has a higher bottom pressure because it contains a full 8-meter8\text{-meter} column of the denser water (1000 kg/m31000\text{ kg/m}^3), while Tank A replaces 5 meters5\text{ meters} of that column with lighter oil (800 kg/m3800\text{ kg/m}^3). It also correctly implies that hydrostatic pressure depends only on fluid density and depth, not on the tank's cross-sectional area.

Adım Adım Çözüm

1
Identify the formula for gauge hydrostatic pressure.
The hydrostatic pressure PP at a depth hh is calculated using P=ρghP = \rho gh, where ρ\rho is the fluid density and gg is the acceleration due to gravity.
This formula is needed to find the pressure at the bottom of each tank.
2
Calculate the relative pressure at the bottom of Tank A.
For multiple layers, sum the individual pressures: PA=(ρoil×g×hoil)+(ρwater×g×hwater)=(800×g×5)+(1000×g×3)=4000g+3000g=7000g PaP_A = (\rho_{oil} \times g \times h_{oil}) + (\rho_{water} \times g \times h_{water}) = (800 \times g \times 5) + (1000 \times g \times 3) = 4000g + 3000g = 7000g\text{ Pa}.
The total bottom pressure in a multi-fluid system is the combined pressure of all stacked fluid layers.
3
Calculate the relative pressure at the bottom of Tank B.
Tank B contains only water: PB=ρwater×g×htotal=1000×g×8=8000g PaP_B = \rho_{water} \times g \times h_{total} = 1000 \times g \times 8 = 8000g\text{ Pa}.
This calculation provides the value needed to compare Tank B against Tank A.
4
Compare the pressures and analyze the effect of tank width.
Since 8000g>7000g8000g > 7000g, Tank B has the higher bottom pressure. The larger cross-sectional area of Tank B is irrelevant to the calculation.
Hydrostatic pressure depends solely on depth and density, meaning a wider container does not artificially increase pressure.

Anahtar Kavram

Hydrostatic pressure depends strictly on fluid density, gravity, and depth (P=ρghP = \rho gh), and is independent of the container's shape, base area, or total volume of fluid.
Soru 56Soru →

A barber chair operates using a closed hydraulic lifting system to raise a customer. The input pump piston has a cross-sectional area of 4 in24 \text{ in}^2, and the output lifting piston has a cross-sectional area of 60 in260 \text{ in}^2. If the barber pushes the input piston down a distance of 7.5 inches7.5 \text{ inches}, how many inches does the output piston move upward?

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Cevap: 0.5

Cevap

0.5
In a hydraulic system, the mechanical advantage that multiplies the output force must correspondingly divide the output distance. The output piston has an area 15 times larger than the input piston (60/4=1560 / 4 = 15). Therefore, while it can lift 15 times the force, it will only travel 1/15th1/15\text{th} of the input distance (7.5/15=0.5 inches7.5 / 15 = 0.5 \text{ inches}).

Adım Adım Çözüm

1
Identify the relevant principle for hydraulic piston travel distances.
The volume of fluid displaced by the input piston equals the volume gained by the output piston: A1×d1=A2×d2A_1 \times d_1 = A_2 \times d_2.
Because hydraulic fluid is essentially incompressible, the total volume in the closed system remains constant.
2
Set up the equation using the given values.
4×7.5=60×d24 \times 7.5 = 60 \times d_2
Input area (A1A_1) is 4, input distance (d1d_1) is 7.5, and output area (A2A_2) is 60.
3
Calculate the total fluid volume displaced.
4×7.5=30 in34 \times 7.5 = 30 \text{ in}^3
Multiplying area by distance gives the volume of fluid moved.
4
Solve for the output piston's upward distance (d2d_2).
30/60=0.5 inches30 / 60 = 0.5 \text{ inches}
Dividing the displaced volume by the output piston's larger area yields the smaller travel distance.

Anahtar Kavram

Conservation of volume in hydraulic systems dictates that distance and area are inversely proportional (Mechanical Advantage trades distance for force).
Soru 57Soru →

Evaluate whether the following statement is true or false:

In a mechanical carburetor, outside air flows through a narrow central constriction. As the air accelerates and its velocity increases through this narrow section, the static pressure of the air increases, forcing fuel to mix into the airstream.

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Cevap: False

Cevap

False
The statement is false because Bernoulli's principle establishes an inverse relationship between a fluid's velocity and its static pressure. As the air accelerates through the carburetor's narrow constriction, its static pressure drops, creating a low-pressure area (a partial vacuum) that siphons fuel into the airstream, rather than an area of increased pressure.

Adım Adım Çözüm

1
Recall Bernoulli's principle regarding fluid velocity and static pressure.
Bernoulli's principle states that as the velocity of a fluid increases, its static pressure decreases.
Establishing the fundamental physical law governing fluids moving through a constriction.
2
Evaluate the claim made in the statement.
The statement incorrectly claims that an increase in air velocity causes the static pressure to increase.
Comparing the scenario's claim directly against the established physics principle to determine its validity.

Anahtar Kavram

Inverse relationship between fluid velocity and static pressure (Bernoulli's Principle)
Soru 58Soru →

In an aircraft's hydraulic braking system, the master cylinder (input) connects to a brake caliper cylinder (output). The cross-sectional area of the caliper cylinder is 55 times the area of the master cylinder. If the master cylinder piston is pushed with 40 lbs40 \text{ lbs} of force and travels 5 inches5 \text{ inches}, what is the resulting force and travel distance of the caliper cylinder piston?

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Cevap: 200 lbs200 \text{ lbs} of force and 1 inch1 \text{ inch} of travel

Cevap

The caliper cylinder piston will output 200 lbs200 \text{ lbs} of force and move 1 inch1 \text{ inch}.
Based on Pascal's principle and the conservation of energy, the larger output cylinder multiplies the input force by the area ratio (40×5=200 lbs40 \times 5 = 200 \text{ lbs}). Correspondingly, the output distance must be divided by that same ratio (5÷5=1 inch5 \div 5 = 1 \text{ inch}) to ensure the work input equals the work output.

Adım Adım Çözüm

1
Determine the area ratio between the input and output cylinders.
Area Ratio = 5.
The problem states the caliper cylinder has an area 5 times larger than the master cylinder.
2
Calculate the resulting force at the output cylinder.
Output Force = Input Force x Area Ratio = 40 lbs×5=200 lbs40 \text{ lbs} \times 5 = 200 \text{ lbs}.
According to Pascal's principle, pressure is transmitted equally, so the force is multiplied by the ratio of the piston areas.
3
Calculate the travel distance of the output cylinder.
Output Distance = Input Distance / Area Ratio = 5 inches/5=1 inch5 \text{ inches} / 5 = 1 \text{ inch}.
Because fluid is incompressible, the volume displaced by the input piston equals the volume displaced by the output piston. A larger area requires proportionally less travel distance.

Anahtar Kavram

In a closed hydraulic system, moving from a smaller to a larger cylinder multiplies the force by the area ratio but decreases the travel distance by that same ratio, conserving the total work done.
Soru 59Soru →

A municipal water tower supplies fresh water to a nearby town. Assuming the water is not flowing, which of the following factors primarily determines the hydrostatic pressure of the water at a ground-level faucet in a house?

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Cevap: The vertical distance from the faucet to the top of the water level in the tower

Cevap

The vertical distance from the faucet to the top of the water level in the tower.
According to the principles of fluid mechanics, hydrostatic pressure is determined solely by the fluid's density, gravity, and the vertical depth of the fluid column. Because density and gravity are constant in this scenario, the vertical distance from the water's surface to the faucet is the only factor that dictates the pressure. The total amount of water or the width of the pipes does not change this static pressure.

Adım Adım Çözüm

1
Recall the formula for hydrostatic pressure.
P=ρ⋅g⋅hP = \rho \cdot g \cdot h
This fundamental equation identifies the variables that affect pressure in a static fluid.
2
Analyze the variables in the context of the water tower.
The density of water (ρ\rho) and the acceleration due to gravity (gg) remain constant.
Since ρ\rho and gg are fixed, the pressure depends entirely on the remaining variable, hh.
3
Identify what hh represents in this physical scenario.
The variable hh represents the vertical height or depth from the fluid's surface down to the measurement point (the faucet).
Hydrostatic pressure is based solely on vertical depth, making the vertical distance from the water level to the faucet the determining factor.

Anahtar Kavram

Hydrostatic Pressure Dependency on Depth
Soru 60Soru →

During flight, an aircraft wing (airfoil) is designed so that air travels significantly faster over its curved upper surface compared to its flatter lower surface. According to Bernoulli's principle, what is the result of this difference in air velocity?

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Cevap: The static pressure above the wing becomes lower than the pressure below it, generating upward lift.

Cevap

The static pressure above the wing becomes lower than the pressure below it, generating upward lift.
According to Bernoulli's principle, as the velocity of a fluid (such as air) increases, its static pressure decreases. Because the wing is shaped to make the air travel faster over its top surface, a region of low pressure is created above the wing. The slower-moving air beneath the wing maintains a relatively higher pressure. This difference in pressure—high on the bottom, low on the top—results in a net upward force that lifts the aircraft.

Adım Adım Çözüm

1
Recall the fundamental rule of Bernoulli's principle regarding fluid flow.
In a flowing fluid, an increase in velocity is accompanied by a decrease in static pressure.
This inverse relationship is the core concept of Bernoulli's principle.
2
Apply the principle to the specific scenario of the aircraft wing described in the question.
The air moving over the top of the wing has a higher velocity than the air moving underneath. Therefore, the static pressure on the top surface must be lower than the static pressure on the bottom surface.
Mapping the physical conditions (faster top, slower bottom) to the principle reveals the pressure distribution.
3
Determine the physical consequence of this pressure difference.
The relatively higher pressure beneath the wing pushes upward toward the area of lower pressure above the wing, creating a net upward force known as lift.
Objects are naturally pushed from areas of higher pressure toward areas of lower pressure.

Anahtar Kavram

According to Bernoulli's principle, an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure. This pressure differential across surfaces like airfoils creates mechanical forces such as lift.
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