Tüm alıştırma soruları

5215 soru

Soru 1701Soru →

Arrange the following points in a fluid system in order of their gauge hydrostatic pressure, from the GREATEST pressure to the LEAST pressure. Assume standard gravity applies equally to all systems.

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Cevap

The correct order from greatest to least pressure is: the 1515-meter oil pipe, the layered vat, the 1111-meter lake, the 99-meter brine tank, and finally the 1414-meter conical vessel.
By calculating the equivalent water depth (Depth ×\times Specific Gravity) for each scenario, we find the values are 12.012.0 (pipe), 11.511.5 (layered vat), 11.011.0 (lake), 10.810.8 (brine tank), and 9.89.8 (conical vessel). Ranking these from highest to lowest provides the correct sequence.

Adım Adım Çözüm

1
Recall the formula for hydrostatic pressure.
P=ρ⋅g⋅hP = \rho \cdot g \cdot h, where ρ\rho is density, gg is gravity, and hh is depth.
This establishes that pressure depends solely on fluid density and depth, not on container shape or total volume.
2
Determine a standard unit for comparison across all items.
Use 'equivalent meters of water' by multiplying the depth of each fluid by its Specific Gravity (SG).
This simplifies calculations and provides a single proportional metric to compare.
3
Calculate the equivalent pressure for the single-fluid containers.
Pipe: 15×0.8=12.015 \times 0.8 = 12.0. Lake: 11×1.0=11.011 \times 1.0 = 11.0. Brine: 9×1.2=10.89 \times 1.2 = 10.8. Conical vessel: 14×0.7=9.814 \times 0.7 = 9.8.
Applying the formula directly to the given depth and Specific Gravity.
4
Calculate the equivalent pressure for the layered fluids in the vat.
Top layer (water): 5×1.0=5.05 \times 1.0 = 5.0. Bottom layer (glycerin): 5×1.3=6.55 \times 1.3 = 6.5. Total equivalent depth = 5.0+6.5=11.55.0 + 6.5 = 11.5.
The total pressure at the bottom of layered fluids is the sum of the hydrostatic pressures of each individual layer.
5
Rank the calculated equivalent water depths from greatest to least.
12.0>11.5>11.0>10.8>9.812.0 > 11.5 > 11.0 > 10.8 > 9.8.
This corresponds to the final required ordering.

Anahtar Kavram

Hydrostatic pressure is determined solely by fluid density and depth, completely independent of the container's shape, width, or total fluid volume.

Alternatif Yöntem

Instead of calculating equivalent water depth, you can calculate exact pressure if you assign g=9.8 m/s2g = 9.8 \text{ m/s}^2 and density of water as 1000 kg/m31000 \text{ kg/m}^3. For example, the pipe pressure is 0.8×1000×9.8×15=117,600 Pa0.8 \times 1000 \times 9.8 \times 15 = 117,600 \text{ Pa}. However, comparing the relative (h×SG)(h \times \text{SG}) values is much faster for mental arithmetic.
Tahmini Süre:2m 0s
Soru 1702Soru →

A mechanic uses a hydraulic floor jack to lift a heavy vehicle transmission. The jack's pump (input) piston has an area of 3 in23 \text{ in}^2, and the lifting (output) piston has an area of 15 in215 \text{ in}^2. If the mechanic applies a downward force of 60 lbs60 \text{ lbs} on the input piston, what is the resulting upward force exerted by the lifting piston?

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Cevap: 300 lbs300 \text{ lbs}

Cevap

The lifting piston exerts an upward force of 300 lbs300 \text{ lbs}.
According to Pascal's Principle, the fluid pressure is equal throughout the system. The output piston has an area 55 times larger than the input piston (15/3=515 / 3 = 5). Therefore, to maintain the same pressure, it will output a force 55 times greater than the input force (60×5=30060 \times 5 = 300).

Adım Adım Çözüm

1
Determine the area ratio between the output and input pistons.
Area Ratio=15 in23 in2=5\text{Area Ratio} = \frac{15 \text{ in}^2}{3 \text{ in}^2} = 5
This ratio represents the mechanical advantage of the hydraulic system.
2
Multiply the input force by the mechanical advantage to find the output force.
Output Force=60 lbs×5=300 lbs\text{Output Force} = 60 \text{ lbs} \times 5 = 300 \text{ lbs}
Because fluid pressure (P=FAP = \frac{F}{A}) is transmitted equally, a piston with 55 times the area will output 55 times the force.

Anahtar Kavram

Pascal's Principle states that pressure is transmitted equally throughout a confined fluid, allowing hydraulic systems to multiply force proportionally to the ratio of their piston areas.
Soru 1703Soru →

A worker is using a wheelbarrow, which acts as a second-class lever, to transport materials. The wheel's axle serves as the fulcrum. The empty wheelbarrow has a downward weight of 200 N200 \text{ N}, with its center of gravity located 0.5 m0.5 \text{ m} horizontally from the axle. A load of bricks weighing 800 N800 \text{ N} is placed in the tray so that its center of gravity is 0.4 m0.4 \text{ m} horizontally from the axle. The worker lifts upward at the ends of the handles, which are located 1.2 m1.2 \text{ m} horizontally from the axle.

Assuming the wheelbarrow is in a horizontal position, what is the minimum upward force required to lift the rear supports off the ground?

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Cevap: 350

Cevap

The minimum upward force required to lift the wheelbarrow is 350 N.
The correct answer is derived by applying the principle of moments (torque). Because there are two separate masses acting on the lever, their torques must be calculated independently and summed. The wheelbarrow contributes 100 N⋅m100 \text{ N}\cdot\text{m} of torque and the bricks contribute 320 N⋅m320 \text{ N}\cdot\text{m}. Balancing the total 420 N⋅m420 \text{ N}\cdot\text{m} torque over a 1.2 m1.2 \text{ m} lever arm requires exactly 350 N350 \text{ N} of effort.

Adım Adım Çözüm

1
Calculate the torque from the empty wheelbarrow.
τcart=100 N⋅m\tau_{cart} = 100 \text{ N}\cdot\text{m}
The intrinsic weight of the lever system applies a downward force that must be accounted for.
2
Calculate the torque from the added load.
τload=320 N⋅m\tau_{load} = 320 \text{ N}\cdot\text{m}
The load applies an additional independent downward force at its specific distance from the fulcrum.
3
Sum the downward torques.
τtotal=420 N⋅m\tau_{total} = 420 \text{ N}\cdot\text{m}
In equilibrium, the total clockwise torque must be balanced by the total counterclockwise torque.
4
Divide the total torque by the effort distance to find the effort force.
Feffort=350 NF_{effort} = 350 \text{ N}
Torque equals force times distance. Rearranging 420=F×1.2420 = F \times 1.2 isolates the required lifting force.

Anahtar Kavram

Torque Equilibrium and Composite Loads in Second-Class Levers
Soru 1704Soru →

A standard shipping crate is measured on Earth (where the acceleration due to gravity is 10 m/s210 \text{ m/s}^2) using a mechanical balance scale. The balance scale determines that the crate has a mass of 400 kg400 \text{ kg}. The crate is then transported to a planetary outpost where the local gravitational acceleration is 4 m/s24 \text{ m/s}^2. At the outpost, the crate is hoisted using a heavy-duty spring scale. This spring scale was manufactured and calibrated on Earth to display mass in kilograms based on Earth's gravity. What numerical value will this spring scale display when weighing the crate at the outpost?

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Cevap: 160

Cevap

160
The true mass of the crate is 400 kg, which does not change based on location. On the outpost, the crate's weight (force) is 400 kg × 4 m/s² = 1600 N. The spring scale stretches based entirely on this 1600 N force. Because it was calibrated on Earth (where the acceleration due to gravity is 10 m/s²), its internal mechanism is designed to display '1 kg' for every 10 N of force it supports. Therefore, an applied force of 1600 N causes the scale to point to 1600 / 10 = 160.

Adım Adım Çözüm

1
Determine the true mass of the crate.
400 kg
Mass is the amount of matter and is invariant. A balance scale compares masses and measures true mass accurately regardless of the local gravitational field.
2
Calculate the downward force (weight) exerted by the crate on the spring scale at the outpost.
1600 N
Weight depends on local gravity. It is calculated using the formula Weight = mass × local gravity, which is 400 kg × 4 m/s².
3
Determine the calibration factor of the Earth-calibrated spring scale.
10 N per 1 kg of display
A spring scale physically stretches in response to force. Because it was calibrated to display mass on Earth, it assumes that every 10 N of applied force corresponds to 1 kg of mass.
4
Calculate the value displayed by the spring scale.
160
The scale converts the 1600 N force it physically experiences into a kilogram reading using its Earth calibration: 1600 N / 10 m/s² = 160.

Anahtar Kavram

Mass is an invariant property of matter, while weight is the localized gravitational force acting on that mass. A spring scale measures force (weight) rather than mass, meaning a spring scale calibrated in kilograms on Earth will display incorrect mass readings in different gravitational fields.
Soru 1705Soru →

A stationary jib crane consists of a horizontal metal boom attached to a vertical mast. The outer end of the boom is supported by a diagonal steel cable connected higher up on the mast. When a heavy load is suspended from the outer end of the boom, what types of internal stress are primarily experienced by the diagonal cable and the horizontal boom, respectively?

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Cevap: Tension in the cable and compression in the boom

Cevap

Tension in the cable and compression in the boom
The heavy load pulls down on the end of the boom. To support this, the diagonal cable must pull upward and inward toward the mast, which stretches the cable, resulting in tensile stress (tension). The inward horizontal component of the cable's pull pushes the boom horizontally into the mast. This pushing force squeezes the boom, resulting in compressive stress (compression).

Adım Adım Çözüm

1
Analyze the forces acting on the diagonal cable.
The downward weight of the load causes the cable to pull back against it to keep the boom from falling.
Identifying the force direction determines the stress. Pulling forces acting along the length of a material create tension.
2
Analyze the forces acting on the horizontal boom.
The diagonal cable pulls inward toward the mast. This inward horizontal force pushes the boom tightly against the mast.
Pushing or crushing forces acting along the axis of a structural member create compression.

Anahtar Kavram

Identifying tension and compression in load-bearing structural members.
Soru 1706Soru →

A vintage printing press utilizes a compound gear train to feed paper. Shaft 1 is the input and is driven at 120120 RPM in a counter-clockwise direction. Gear A (4040 teeth) is mounted on Shaft 1 and meshes directly with Gear B (2020 teeth), which is mounted on Shaft 2. Also rigidly mounted on Shaft 2 is Gear C (5050 teeth). Gear C meshes directly with Gear D (2525 teeth) mounted on Shaft 3. Based on this configuration, what is the rotational speed and direction of Shaft 3?

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Cevap: 480480 RPM, counter-clockwise

Cevap

The output shaft rotates at 480480 RPM in a counter-clockwise direction.
In a compound gear train, mechanical advantage and speed ratios are multiplied across stages. In the first stage (Gears A to B), a 4040-tooth gear drives a 2020-tooth gear, doubling the speed to 240240 RPM and reversing the direction to clockwise. Both gears on Shaft 2 rotate together. In the second stage (Gears C to D), a 5050-tooth gear drives a 2525-tooth gear, doubling the speed again to 480480 RPM and reversing the direction back to counter-clockwise.

Adım Adım Çözüm

1
Calculate the rotational speed of Shaft 2.
Speed = 120×(40/20)=240120 \times (40 / 20) = 240 RPM.
The speed ratio between meshing gears is inversely proportional to their tooth count. A smaller driven gear rotates faster than the driving gear.
2
Determine the rotational direction of Shaft 2.
Clockwise.
When two external gears mesh, they rotate in opposite directions. Since Shaft 1 is counter-clockwise, Shaft 2 must be clockwise. Because Gears B and C are on the same shaft, Gear C also rotates at 240240 RPM clockwise.
3
Calculate the rotational speed of Shaft 3.
Speed = 240×(50/25)=480240 \times (50 / 25) = 480 RPM.
Applying the speed ratio principle again from Gear C (driving) to Gear D (driven).
4
Determine the rotational direction of Shaft 3.
Counter-clockwise.
Gear C (clockwise) meshes with Gear D, forcing Gear D to rotate in the opposite direction.

Anahtar Kavram

Compound Gear Trains and Speed Ratios
Soru 1707Soru →

Match each mechanical operational scenario to its resulting acceleration using Newton's Second Law (F=maF = ma).

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Öğeler

A 1,200 kg1,200 \text{ kg} aerospace capsule is tested in a track, propelled by a net horizontal force of 3,600 N3,600 \text{ N}.
A 1,500 kg1,500 \text{ kg} forklift experiences a net force of 2,250 N2,250 \text{ N} while accelerating on a warehouse floor.
A 2,500 kg2,500 \text{ kg} bridge section is positioned by cranes applying a net lateral force of 1,250 N1,250 \text{ N}.
A 4,000 kg4,000 \text{ kg} submarine ballast drops through the ocean, experiencing a net downward force of 16,000 N16,000 \text{ N}.

Eşleşmeler

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Cevap

The correct pairings are determined by dividing the net force by the mass for each scenario: the aerospace capsule accelerates at 3.0 m/s23.0 \text{ m/s}^2, the forklift accelerates at 1.5 m/s21.5 \text{ m/s}^2, the bridge section accelerates at 0.5 m/s20.5 \text{ m/s}^2, and the submarine ballast accelerates at 4.0 m/s24.0 \text{ m/s}^2.
Newton's Second Law dictates that acceleration is directly proportional to net force and inversely proportional to mass (a=F/ma = F/m). By dividing the net force by the mass in each of the four scenarios, the correct acceleration values are directly calculated and paired.

Adım Adım Çözüm

1
Identify the known variables provided in each scenario.
Each scenario provides the net force (FF) in Newtons and the object's mass (mm) in kilograms.
Newton's Second Law requires both net force and mass to determine an object's resulting acceleration.
2
Rearrange Newton's Second Law (F=maF = ma) to solve for the missing variable.
The formula becomes a=F/ma = F/m.
Because the right items represent acceleration, the equation must be isolated for acceleration (aa) before computing.
3
Calculate the specific acceleration for each mass and force pairing.
Capsule: 3,600/1,200=3.0 m/s23,600 / 1,200 = 3.0 \text{ m/s}^2. Forklift: 2,250/1,500=1.5 m/s22,250 / 1,500 = 1.5 \text{ m/s}^2. Bridge section: 1,250/2,500=0.5 m/s21,250 / 2,500 = 0.5 \text{ m/s}^2. Ballast: 16,000/4,000=4.0 m/s216,000 / 4,000 = 4.0 \text{ m/s}^2.
Dividing the specific force by the given mass maps each scenario to exactly one value in the right column.

Anahtar Kavram

Solving for acceleration using Newton's Second Law (a=F/ma = F/m)
Soru 1708Soru →

A manual boat winch is used to pull a small boat onto a trailer. The crank handle travels in a circle with a radius of 15 inches15 \text{ inches}, turning a central spool (the axle) with a radius of 3 inches3 \text{ inches}.

What is the Ideal Mechanical Advantage (IMA) of this winch?

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Cevap: 5

Cevap

The Ideal Mechanical Advantage is 5.
The Ideal Mechanical Advantage of a wheel and axle is calculated by dividing the radius of the wheel by the radius of the axle. Here, the wheel's radius is 15 inches15 \text{ inches} and the axle's radius is 3 inches3 \text{ inches}, which gives 15÷3=515 \div 3 = 5.

Adım Adım Çözüm

1
Identify the given dimensions of the wheel and the axle.
Rwheel=15 inchesR_{wheel} = 15 \text{ inches}, Raxle=3 inchesR_{axle} = 3 \text{ inches}
These are the required variables for the IMA equation.
2
Apply the Ideal Mechanical Advantage formula for a wheel and axle.
IMA=RwheelRaxleIMA = \frac{R_{wheel}}{R_{axle}}
IMA represents the ratio of the effort distance to the resistance distance.
3
Calculate the final value.
IMA=153=5IMA = \frac{15}{3} = 5
Dividing the wheel radius by the axle radius yields the mechanical advantage.

Anahtar Kavram

Ideal Mechanical Advantage (IMA) of a Wheel and Axle
Soru 1709Soru →

A flight line technician must load a cargo pallet with a mass of 240 kg240\text{ kg} into a transport aircraft. The aircraft's cargo deck is 1.5 meters1.5\text{ meters} above the tarmac. To assist with loading, the technician deploys a straight, 6.0-meter6.0\text{-meter} long rigid ramp equipped with frictionless rollers. Assuming the acceleration due to gravity is 10 m/s210\text{ m/s}^2, which of the following identifies the ideal effort force required to push the pallet up the ramp, and correctly compares the total work performed to lifting the pallet vertically?

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Cevap: 600 N600\text{ N}; the total work performed is the same as lifting the pallet vertically.

Cevap

The effort force is 600 N600\text{ N}, and the total work performed is exactly the same as lifting the pallet vertically.
The correct answer accurately calculates both the required force and applies the principle of work conservation. First, weight is determined by multiplying mass by gravity (240 kg×10 m/s2=2400 N240\text{ kg} \times 10\text{ m/s}^2 = 2400\text{ N}). The ideal mechanical advantage (IMA) of the ramp is its length divided by its height (6.0 m/1.5 m=46.0\text{ m} / 1.5\text{ m} = 4). The effort force is the weight divided by the IMA (2400 N/4=600 N2400\text{ N} / 4 = 600\text{ N}). Finally, while the ramp reduces the effort force, it requires moving the object over a longer distance, meaning the total work performed (600 N×6.0 m=3600 Joules600\text{ N} \times 6.0\text{ m} = 3600\text{ Joules}) is identical to lifting it straight up (2400 N×1.5 m=3600 Joules2400\text{ N} \times 1.5\text{ m} = 3600\text{ Joules}).

Adım Adım Çözüm

1
Calculate the weight (load force) of the cargo pallet.
W=240 kg×10 m/s2=2400 NW = 240\text{ kg} \times 10\text{ m/s}^2 = 2400\text{ N}
Mass must be converted to weight (force) before mechanical advantage formulas can be properly applied.
2
Determine the Ideal Mechanical Advantage (IMA) of the inclined plane.
IMA=6.0 m1.5 m=4\text{IMA} = \frac{6.0\text{ m}}{1.5\text{ m}} = 4
IMA for an inclined plane is the ratio of the effort distance (ramp length) to the load distance (vertical height).
3
Calculate the ideal effort force.
Effort Force=2400 N4=600 N\text{Effort Force} = \frac{2400\text{ N}}{4} = 600\text{ N}
The effort force is equal to the total load force divided by the mechanical advantage of the machine.
4
Evaluate the total work performed in both scenarios.
Work via ramp: 600 N×6.0 m=3600 J600\text{ N} \times 6.0\text{ m} = 3600\text{ J}. Work vertical: 2400 N×1.5 m=3600 J2400\text{ N} \times 1.5\text{ m} = 3600\text{ J}. Work is equal.
According to the principle of conservation of energy, simple machines trade increased distance for reduced effort force, leaving the total work unchanged.

Anahtar Kavram

Mechanical advantage reduces effort force by increasing distance, but total work is conserved.
Soru 1710Soru →

A construction worker needs to push a heavy wheelbarrow up to a raised foundation that is 2 m2\text{ m} above the ground. The worker has three different sturdy wooden planks available to use as a ramp. Arrange the following planks in order from the LEAST amount of effort force required to the MOST amount of effort force required.

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Cevap

The planks should be ordered from longest (10 m10\text{ m}) to shortest (4 m4\text{ m}). The 10 m10\text{ m} plank requires the least effort, the 8 m8\text{ m} plank requires a moderate amount, and the 4 m4\text{ m} plank requires the most effort.
The effort force needed to move an object up a ramp is inversely proportional to the ramp's length when the vertical height is fixed. The 10 m10\text{ m} plank provides the most gradual slope (highest mechanical advantage) and therefore requires the least effort force. As the plank length decreases to 8 m8\text{ m} and then 4 m4\text{ m}, the slope becomes steeper, reducing the mechanical advantage and requiring increasingly more effort force. Thus, ordering from least to most effort force is synonymous with ordering from longest to shortest plank.

Adım Adım Çözüm

1
Recall the formula for the Ideal Mechanical Advantage (IMA) of an inclined plane.
IMA=Length of the slopeHeight of the incline\text{IMA} = \frac{\text{Length of the slope}}{\text{Height of the incline}}
This formula helps us determine how much a ramp multiplies the input force.
2
Determine the relationship between length, IMA, and effort force.
Since the height (2 m2\text{ m}) is constant for all three planks, a longer plank results in a higher IMA. A higher IMA means less effort force is required to push the load.
Understanding this inverse relationship allows us to rank the effort force without doing complex calculations.
3
Rank the planks based on length to determine the required effort force.
Longest plank (10 m10\text{ m}) = Highest IMA = Least effort force. Shortest plank (4 m4\text{ m}) = Lowest IMA = Most effort force.
The question asks to order from LEAST effort force to MOST effort force, which corresponds to ordering from longest ramp to shortest ramp.

Anahtar Kavram

For an inclined plane of a given height, increasing the length of the plane increases its mechanical advantage, which proportionally decreases the effort force required to move a load up the plane.
Tahmini Süre:45s
Soru 1711Soru →

Match each basic structural component with its primary function and the main type of internal force it is designed to resist.

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Öğeler

Column
Beam
Tie
Strut

Eşleşmeler

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Cevap

Column matches with the vertical compressive member. Beam matches with the horizontal bending member. Tie matches with the tensile member. Strut matches with the compressive framework member.
Each structural component serves a distinct mechanical role based on orientation and force distribution. Columns handle vertical compression, beams handle transverse bending, ties handle axial tension, and struts handle axial compression.

Adım Adım Çözüm

1
Identify the primary orientation and force resisted by a Column.
A Column is a vertical member supporting heavy compressive loads.
Columns transfer the weight of the structure above down to the foundation through compression.
2
Identify the primary orientation and force resisted by a Beam.
A Beam is a horizontal member resisting bending forces.
Beams span open spaces and carry downward loads across their span, which induces bending.
3
Distinguish between a Tie and a Strut based on internal forces.
A Tie resists tension (pulling apart), while a Strut resists compression (pushing together).
Ties 'tie' things together to prevent separation. Struts act like rigid pillars holding things apart.

Anahtar Kavram

Basic Structural Components and the Internal Forces They Resist
Tahmini Süre:1m 0s
Soru 1712Soru →

Match each type of wedge or wedge application with its correct mechanical characteristic.

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Öğeler

Long, thin wedge
Short, thick wedge
Doorstop

Eşleşmeler

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Cevap

The correct matches are: Long, thin wedge corresponds to high mechanical advantage and less effort; Short, thick wedge corresponds to lower mechanical advantage but rapid splitting; Doorstop corresponds to securing objects using friction.
The correct pairings are based on the principle that the ideal mechanical advantage (IMA) of a wedge is its length divided by its thickness. A long, thin wedge has a high mechanical advantage and therefore requires less effort force to penetrate a material. Conversely, a short, thick wedge has a lower mechanical advantage, meaning it takes more effort force, but it successfully splits the object wider in a shorter distance. A doorstop is a specific application that uses the wedge shape to generate a strong normal force and resulting friction between the door and the floor, holding the object stationary.

Adım Adım Çözüm

1
Analyze the mechanical advantage of a long, thin wedge.
Matches with 'Provides high mechanical advantage and requires less applied effort force.'
The mechanical advantage of a wedge is its length divided by its thickness. A larger ratio means less effort is needed to drive the wedge.
2
Analyze the mechanical advantage of a short, thick wedge.
Matches with 'Provides lower mechanical advantage but splits material apart rapidly over a short distance.'
A smaller length-to-thickness ratio results in a lower mechanical advantage, requiring more force, but the increased thickness spreads the material faster.
3
Analyze the practical application of a doorstop.
Matches with 'Secures objects in place primarily by utilizing friction against a surface.'
Instead of splitting material, a doorstop is driven under a door to increase normal force and create high friction, keeping it wedged in place.

Anahtar Kavram

The mechanical advantage of a wedge is determined by the ratio of its length to its thickness, governing the trade-off between the effort force required and the distance the material is separated.
Soru 1713Soru →

A 500 kg500\text{ kg} steel block is dropped from rest into a deep vertical mineshaft. Ignoring air resistance, arrange the following one-second time intervals in order according to the distance the block travels during each specific interval, from the shortest distance to the longest distance.

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Cevap

The correct sequence is chronological: the 1st second, followed by the 2nd second, 3rd second, and 4th second.
In free fall, an object undergoes constant acceleration due to gravity. Because its downward speed is continuously increasing, the distance it travels during each successive equal time interval also increases. The block's heavy mass is a distracter; it does not change the fact that the shortest distance is covered in the first second (when it is slowest), and the longest distance is covered in the final second.

Adım Adım Çözüm

1
Analyze the relationship between gravity, acceleration, and the speed of a falling object.
Gravity causes a constant downward acceleration on the block, meaning its speed increases steadily as time passes.
Understanding constant acceleration is required to determine how distance changes over equal time increments.
2
Determine the distance traveled during each successive one-second interval.
Since the block's average speed is higher during each subsequent second, the distance it travels in each one-second interval must also be strictly greater than the previous interval.
Distance equals average speed multiplied by time. A higher average speed over the same one-second duration yields a longer distance.
3
Order the time intervals from the shortest distance to the longest distance.
The 1st second has the lowest average speed (shortest distance), followed in order by the 2nd, 3rd, and 4th seconds (longest distance).
This sequences the intervals properly according to the prompt's instructions.

Anahtar Kavram

Acceleration due to gravity causes a freely falling object's speed to continuously increase, resulting in strictly greater distances covered during each successive time interval, regardless of its mass.
Soru 1714Soru →

A specialized industrial press uses a compound lever system composed of two uniform steel bars to compress a heavy coil spring.

- Lever A (Top handle): is 20 inches20\text{ inches} long, weighs 10 lbs10\text{ lbs}, and pivots on a fulcrum at its left end. A vertical rigid connecting link is attached to Lever A at a point 5 inches5\text{ inches} from its fulcrum.
- Lever B (Bottom pressing bar): is 30 inches30\text{ inches} long, weighs 30 lbs30\text{ lbs}, and pivots on a fulcrum at its left end. The connecting link from Lever A attaches to the far right end of Lever B (30 inches30\text{ inches} from its fulcrum). A heavy coil spring is placed under Lever B, located 6 inches6\text{ inches} from Lever B's fulcrum.

If a worker applies a 40 lb40\text{ lb} downward force at the far right end of Lever A (20 inches20\text{ inches} from its fulcrum), what is the total downward compressive force (in lbs) exerted on the spring?

*Assume both levers are horizontal, the connecting link is perfectly vertical, and the system is in static equilibrium.*

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Cevap: 975

Cevap

975 lbs
In static equilibrium, the sum of clockwise and counterclockwise torques about any pivot point must be exactly zero. By calculating the torques step-by-step through the compound system—incorporating both the applied forces and the center of gravity of each uniform bar—we find the final compressive force transmitted to the spring is 975 lbs.

Adım Adım Çözüm

1
Calculate total clockwise torque on Lever A
900 lb-in
The worker's force (40 lbs at 20 inches) and the lever's own uniform weight (10 lbs at its 10-inch center of gravity) both create clockwise rotation around the fulcrum.
2
Determine the force transmitted through the connecting link
180 lbs
The link must balance the 900 lb-in torque at a distance of 5 inches (900 / 5 = 180 lbs). This 180 lbs force is transmitted downward to Lever B.
3
Calculate total clockwise torque on Lever B
5850 lb-in
The downward force from the link (180 lbs at 30 inches) and Lever B's uniform weight (30 lbs at its 15-inch center of gravity) create clockwise rotation around Lever B's fulcrum.
4
Calculate the final compressive force applied to the spring
975 lbs
The spring must balance Lever B's total 5850 lb-in torque at a distance of 6 inches (5850 / 6 = 975 lbs).

Anahtar Kavram

Torque Equilibrium in Compound and Uniform Levers
Soru 1715Soru →

An aerospace technician is testing an experimental probe with a weight of 50 N50\text{ N} on Earth (where g=10 m/s2g = 10\text{ m/s}^2). The probe is released from rest inside an evacuated drop tube, ensuring there is no air resistance. It falls freely for 200 milliseconds200\text{ milliseconds}. What is the probe's velocity right before impact, and how would this impact velocity change if a 100 N100\text{ N} probe were tested under the exact same conditions?

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Cevap: 2.0 m/s2.0\text{ m/s}; the impact velocity would remain exactly the same because free-fall acceleration is independent of an object's weight.

Cevap

The probe's impact velocity is 2.0 m/s2.0\text{ m/s}, and the velocity would remain exactly the same for the heavier probe.
First, the fall time must be converted to standard units (200 ms=0.2 s200\text{ ms} = 0.2\text{ s}). Using the velocity formula v=g×tv = g \times t, the speed is calculated as 10×0.2=2.0 m/s10 \times 0.2 = 2.0\text{ m/s}. Second, in the complete absence of air resistance, the acceleration due to gravity is a constant. Therefore, a heavier 100 N100\text{ N} probe would fall at the exact same rate and strike the ground with the exact same velocity.

Adım Adım Çözüm

1
Convert the free-fall time from milliseconds to seconds.
200 ms=0.2 seconds200\text{ ms} = 0.2\text{ seconds}
Standard kinematic equations require time to be in standard SI units (seconds) to match the acceleration rate (10 m/s210\text{ m/s}^2).
2
Calculate the impact velocity using the formula for an object falling from rest (v=g×tv = g \times t).
v=10 m/s2×0.2 s=2.0 m/sv = 10\text{ m/s}^2 \times 0.2\text{ s} = 2.0\text{ m/s}
This determines the exact speed of the 50 N50\text{ N} probe right before it strikes the bottom of the drop tube.
3
Evaluate how doubling the probe's weight to 100 N100\text{ N} affects the velocity.
The impact velocity remains 2.0 m/s2.0\text{ m/s}.
In an evacuated space (a vacuum), there is no air resistance. All objects accelerate due to gravity at the same constant rate, regardless of their individual masses or weights.

Anahtar Kavram

Free Fall Velocity and Mass Independence
Tahmini Süre:1m 0s
Soru 1716Soru →

An engineer on an unknown planet evaluates a solid metallic crate using two different instruments. First, the crate perfectly balances on a double-pan beam balance against standard reference masses totaling 150 kg150\text{ kg}. Second, the crate is suspended from a calibrated spring scale, which gives a reading of 600 N600\text{ N}. If the crate is transported back to Earth (where gravitational acceleration is 10 m/s210\text{ m/s}^2), what will the spring scale and the beam balance read, respectively?

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Cevap: Spring scale: 1500 N1500\text{ N}; Beam balance: 150 kg150\text{ kg}

Cevap

The spring scale will read 1500 N1500\text{ N} and the beam balance will read 150 kg150\text{ kg}.
The correct answer accurately distinguishes between mass and weight. Mass is the amount of matter and does not change regardless of location, meaning the beam balance will still read 150 kg150\text{ kg} on Earth. Weight is the force of gravity acting on that mass. A spring scale measures this downward force. On Earth, the force is W=mg=150 kg×10 m/s2=1500 NW = mg = 150\text{ kg} \times 10\text{ m/s}^2 = 1500\text{ N}.

Adım Adım Çözüm

1
Determine the true mass of the crate using the beam balance reading.
Mass (mm) = 150 kg150\text{ kg}.
A double-pan beam balance compares unknown mass against known reference masses. Because local gravity acts equally on both pans, gravity cancels out, meaning the balance directly measures true mass, which is invariant.
2
Calculate the weight of the crate on Earth.
Weight on Earth (WW) = 1500 N1500\text{ N}.
Weight is the gravitational force exerted on an object, calculated using W=mgW = mg. Using Earth's gravity (g=10 m/s2g = 10\text{ m/s}^2), the weight is 150 kg×10 m/s2=1500 N150\text{ kg} \times 10\text{ m/s}^2 = 1500\text{ N}.
3
Determine what each instrument will read on Earth.
Spring scale reads 1500 N1500\text{ N}; Beam balance reads 150 kg150\text{ kg}.
The spring scale measures the downward force of weight (1500 N1500\text{ N}), while the beam balance continues to measure the invariant mass (150 kg150\text{ kg}).

Anahtar Kavram

Mass is an invariant property of an object measured by a balance, while weight is a force dependent on local gravity measured by a spring scale.

Alternatif Yöntem

You can verify the unknown planet's gravity (g=W/m=600 N/150 kg=4 m/s2g = W/m = 600\text{ N} / 150\text{ kg} = 4\text{ m/s}^2) to confirm the initial readings are consistent, then reapply the logic to Earth (g=10 m/s2g = 10\text{ m/s}^2) to find the new weight.
Tahmini Süre:1m 30s
Soru 1717Soru →

A mechanic needs to lift a heavy engine using a screw jack. They can equip the jack with one of two interchangeable threaded rods: Rod X has a coarse thread with 44 threads per inch (TPI), and Rod Y has a fine thread with 88 TPI.

Assuming the jack's handle length is identical for both setups and friction is negligible, which of the following statements accurately describes the required effort force and total work done if the mechanic uses Rod Y instead of Rod X to lift the engine by exactly 1010 inches?

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Cevap: Rod Y requires half the effort force on the handle, but the total work done to lift the engine remains the same.

Cevap

Rod Y requires half the effort force on the handle, but the total work done to lift the engine remains the same.
A screw is essentially an inclined plane wrapped around a central cylinder. Increasing the Threads Per Inch from 44 to 88 halves the pitch of the thread, making the 'incline' less steep. This doubles the mechanical advantage of the jack, cutting the required effort force on the handle in half. However, the fundamental principle of all simple machines is that they do not reduce the total amount of work required to move a load; they merely trade increased input distance (more rotations of the handle) for decreased input force. Because the engine is still being lifted exactly 1010 inches, the total output work remains identical.

Adım Adım Çözüm

1
Analyze the relationship between Threads Per Inch (TPI) and pitch.
Rod Y (88 TPI) has half the pitch of Rod X (44 TPI), meaning it advances half the linear distance per full rotation.
Pitch is the mathematical inverse of TPI (1/TPI1/\text{TPI}). It determines the steepness of the screw's thread.
2
Determine the effect of a smaller pitch on Mechanical Advantage and effort force.
Because Rod Y advances half as far per rotation, its Ideal Mechanical Advantage is doubled. Thus, it requires half the effort force to lift the same load.
A screw is an inclined plane wrapped around a cylinder. A finer thread equates to a shallower incline, which trades distance for force.
3
Evaluate the total work done by the system in both scenarios.
The total work done remains identical regardless of the rod used.
Total output work is calculated as Load Force ×\times Lift Distance. Since the engine's weight and the 1010-inch lift distance do not change, the total work is conserved.

Anahtar Kavram

Mechanical Advantage of Screws and Conservation of Work
Tahmini Süre:1m 0s
Soru 1718Soru →

Two astronauts are conducting a repair mission in a zero-gravity vacuum environment. A massive 800 kg communication satellite is drifting in a straight line at a constant velocity of 1 m/s. An astronaut maneuvers in front of the satellite and attempts to stop it by pushing against it with her hands. She finds it extremely difficult to bring the satellite to a halt.

Based on the principles of mechanics, which of the following best explains why the weightless satellite is so difficult to stop?

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Cevap: The satellite's significant mass provides a high degree of inertia, which strongly resists any changes to its constant velocity.

Cevap

The satellite's significant mass provides a high degree of inertia, which strongly resists any changes to its constant velocity.
According to Newton's First Law, inertia is the property of an object that resists any change to its velocity. The amount of inertia an object possesses is directly proportional to its mass, regardless of gravitational pull. Therefore, an 800 kg satellite requires a significant force (or a prolonged period of time) to decelerate, purely because of its high mass and the resulting inertia.

Adım Adım Çözüm

1
Analyze the environmental conditions described in the scenario.
The scenario takes place in a zero-gravity vacuum, meaning there is no weight and no air resistance acting on the satellite.
Understanding the environment helps eliminate forces like gravity and friction as contributing factors to the resistance felt by the astronaut.
2
Identify the relevant mechanical property of the satellite.
Although the satellite is weightless, it has a mass of 800 kg.
Mass is an intrinsic property of matter that remains constant regardless of gravitational pull.
3
Apply Newton's First Law of Motion to the situation.
The satellite's large mass means it has high inertia, which naturally resists any acceleration or deceleration.
Inertia is the tendency of an object to resist changes in its state of motion, and it is directly proportional to an object's mass.

Anahtar Kavram

Newton's First Law (Inertia)
Soru 1719Soru →

A driver must execute an emergency stop on a dry asphalt road. The vehicle is equipped with an anti-lock braking system (ABS) that uses threshold braking, allowing the tires to continue rotating just on the verge of locking up. If the ABS were disabled and the driver fully locked the brakes, causing the tires to skid continuously across the surface, how would the frictional resistance between the tires and the road change?

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Cevap: The resistance would decrease, because the skidding tires rely on kinetic friction, which is generally weaker than the maximum static friction maintained during threshold braking.

Cevap

The resistance would decrease, because the skidding tires rely on kinetic friction, which is generally weaker than the maximum static friction maintained during threshold braking.
As long as a tire is rolling without slipping (as is maintained during threshold braking), the exact point of the tire touching the road is momentarily stationary relative to the ground. This means it utilizes static friction. Once the brakes lock and the tire skids, the rubber slides across the surface, switching the force to kinetic friction. Because the coefficient of static friction is greater than the coefficient of kinetic friction, a sliding tire actually provides less overall stopping resistance than a tire rolling on the verge of locking.

Adım Adım Çözüm

1
Identify the type of friction present during threshold braking when the tires are still rotating.
The bottom of a rolling tire does not slide against the road, meaning the system utilizes static friction at the contact patch.
Understanding the physical state of rolling without slipping is necessary to determine the active frictional forces.
2
Identify the type of friction present during a full brake lockup (skid).
When the tires lock, they slide across the asphalt, meaning the system transitions to utilizing kinetic (sliding) friction.
Skidding fundamentally changes the interaction between the surfaces from gripping to sliding.
3
Compare the magnitudes of the relevant friction coefficients.
For standard materials like rubber on asphalt, the coefficient of static friction is greater than the coefficient of kinetic friction.
This physical rule dictates which braking method provides the most stopping force.
4
Determine the net effect on frictional resistance.
Because kinetic friction is weaker than maximum static friction, locking the brakes and skidding results in decreased frictional resistance.
Combining the physical state with the friction coefficients provides the final answer.

Anahtar Kavram

Static versus Kinetic Friction in Rolling Mechanisms
Soru 1720Soru →

A mechanical engineer is testing a hydraulic actuation system. The standard baseline configuration features an input piston with an area of 2 in22 \text{ in}^2, an output piston with an area of 20 in220 \text{ in}^2, an applied input force of 50 lbs50 \text{ lbs}, and an input stroke distance of 12 in12 \text{ in}. Match each experimental setup below to its resulting output force and output travel distance.

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Öğeler

Input area is increased to 4 in24 \text{ in}^2 (all other baseline parameters unchanged)
Output area is increased to 60 in260 \text{ in}^2 (all other baseline parameters unchanged)
Input force is increased to 100 lbs100 \text{ lbs} and input stroke is reduced to 6 in6 \text{ in} (areas unchanged)
Both input and output areas are tripled to 6 in26 \text{ in}^2 and 60 in260 \text{ in}^2 respectively

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

Each setup modifies the mechanical advantage or the input parameters, requiring independent calculations for output force (Fout=Fin×MAF_{out} = F_{in} \times \text{MA}) and output distance (dout=din/MAd_{out} = d_{in} / \text{MA}).
Correct pairings are achieved by systematically calculating the ratio of the new areas (MA), multiplying the corresponding input force by this ratio, and dividing the corresponding input stroke by this ratio. This ensures Pascal's Principle and the conservation of volume are strictly followed.

Adım Adım Çözüm

1
Calculate the Mechanical Advantage (MA) for each experimental setup.
MA is found by dividing the output area by the input area (Aout/AinA_{out} / A_{in}). Setup 1: 20/4=520/4=5; Setup 2: 60/2=3060/2=30; Setup 3: 20/2=1020/2=10; Setup 4: 60/6=1060/6=10.
MA dictates the multiplier for force and the divisor for distance.
2
Determine the resulting output force for each setup.
Multiply the applied input force by the MA. Setup 1: 50×5=250 lbs50 \times 5 = 250 \text{ lbs}; Setup 2: 50×30=1500 lbs50 \times 30 = 1500 \text{ lbs}; Setup 3: 100×10=1000 lbs100 \times 10 = 1000 \text{ lbs}; Setup 4: 50×10=500 lbs50 \times 10 = 500 \text{ lbs}.
According to Pascal's Principle, pressure is transmitted equally, so force scales proportionally with area.
3
Determine the resulting output travel distance for each setup.
Divide the input stroke distance by the MA. Setup 1: 12/5=2.4 in12 / 5 = 2.4 \text{ in}; Setup 2: 12/30=0.4 in12 / 30 = 0.4 \text{ in}; Setup 3: 6/10=0.6 in6 / 10 = 0.6 \text{ in}; Setup 4: 12/10=1.2 in12 / 10 = 1.2 \text{ in}.
Because the fluid is incompressible, work and volume are conserved; travel distance scales inversely to force.

Anahtar Kavram

In hydraulics, Mechanical Advantage (MA) equals the ratio of output to input areas. Output force is multiplied by MA, while output travel distance is divided by MA to conserve work.
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