Tüm alıştırma soruları

Soru 301Soru →

A technician uses a standard steel rule to measure a piece of flat bar stock that is $5\frac{5}{8}\text{ inches}$ long. If a piece measuring $\frac{3}{8}\text{ inch}$ is cut off from one end, what is the remaining length of the stock in inches, expressed as a decimal?

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Cevap: 5.25

Cevap

The remaining length of the stock is

5.25\text{ inches}

.

Subtracting

\frac{3}{8}\text{ inch}

from the original length of

5\frac{5}{8}\text{ inches}

yields

5\frac{2}{8}\text{ inches}

. Simplifying the fraction gives

5\frac{1}{4}\text{ inches}

, which converts to

5.25\text{ inches}

in decimal notation.

Adım Adım Çözüm

1

Subtract the cut length from the initial measured length.

5\frac{5}{8}\text{ in} - \frac{3}{8}\text{ in} = 5\frac{2}{8}\text{ in}

Cutting off material reduces the total length, requiring subtraction of the two values.

2

Simplify the fractional measurement and convert to decimal form.

5\frac{2}{8}\text{ in} = 5\frac{1}{4}\text{ in} = 5.25\text{ in}

Reducing

\frac{2}{8}

to

\frac{1}{4}

allows direct decimal conversion as

0.25

added to the whole number

5

.

Anahtar Kavram

Linear measurement arithmetic and fractional reduction on standard shop rules.

Alternatif Yöntem

Convert the mixed fractions directly to decimals before subtracting:

5\frac{5}{8} = 5.625

and

\frac{3}{8} = 0.375

. Then calculate

5.625 - 0.375 = 5.25\text{ inches}

.

Tahmini Süre:45s

Soru 302Soru →

During the operation of a pneumatic nailer, its internal firing reservoir holds a compressed air volume of $20 \text{ in}^3$ at an absolute pressure of $120 \text{ psi}$ . When the trigger is pulled, the air expands into the driver chamber at a constant temperature until the absolute pressure drops to $30 \text{ psi}$ . Based on Boyle's Law, what is the new total volume of the expanded air in cubic inches?

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Cevap: 80

Cevap

80

According to Boyle's Law, pressure and volume are inversely proportional at a constant temperature (

P_1 \times V_1 = P_2 \times V_2

). The pressure drops from 120 psi to 30 psi, which is a factor of 4 decrease. Because the relationship is inversely proportional, the volume must increase by a factor of 4. Therefore, the initial volume of 20 cubic inches is multiplied by 4, resulting in a final volume of 80 cubic inches.

Adım Adım Çözüm

1

Identify the initial and final states of the gas.

Initial pressure

P_1 = 120 \text{ psi}

, Initial volume

V_1 = 20 \text{ in}^3

. Final pressure

P_2 = 30 \text{ psi}

.

Establishing the known parameters is required before applying the appropriate gas law.

2

Apply Boyle's Law formula.

P_1 \times V_1 = P_2 \times V_2

Since the temperature is constant, Boyle's Law relates the pressure and volume of the trapped air.

3

Substitute values and solve for the unknown final volume (

V_2

).

120 \times 20 = 30 \times V_2 \rightarrow 2400 = 30 \times V_2 \rightarrow V_2 = 80

Isolating

V_2

provides the final expanded volume of the air.

Anahtar Kavram

Boyle's Law states that the absolute pressure of a confined gas is inversely proportional to its volume, provided the temperature remains constant.

Alternatif Yöntem

Use proportional reasoning: Observe that the final pressure (30 psi) is exactly one-quarter (1/4) of the initial pressure (120 psi). Because pressure and volume are inversely proportional under constant temperature, the final volume must be four times (4x) the initial volume. 4 × 20 = 80.

Tahmini Süre:1m 0s

Soru 303Soru →

A carpenter uses a claw hammer to extract a stubborn nail from a wooden board. The hammer acts as a first-class lever. The distance from the nail to the pivot point (fulcrum) on the hammer's head is $2\text{ inches}$ . The carpenter applies an effort force perpendicular to the handle at a distance of $16\text{ inches}$ from the pivot point. If a resistance force of $280\text{ pounds}$ is required to extract the nail, how many pounds of effort force must the carpenter apply?

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Cevap: 35

Cevap

35 pounds

In a first-class lever, the mechanical advantage is determined by the ratio of the effort arm to the resistance arm. The hammer provides a mechanical advantage of

16 / 2 = 8

. Therefore, the effort force required is the load divided by 8, which is

280 / 8 = 35\text{ pounds}

. Using the fundamental torque balance equation (

F_1 \times d_1 = F_2 \times d_2

),

280 \times 2 = 560

, and

560 / 16 = 35

.

Adım Adım Çözüm

1

State the torque balance equation for a lever.

F_E \times d_E = F_R \times d_R

For the lever to overcome the resistance, the torque applied by the effort must equal the torque exerted by the load.

2

Substitute the given values into the equation.

F_E \times 16 = 280 \times 2

The effort distance (

d_E

) is

16\text{ inches}

, the resistance force (

F_R

) is

280\text{ pounds}

, and the resistance distance (

d_R

) is

2\text{ inches}

.

3

Calculate the total torque of the resistance.

280 \times 2 = 560\text{ inch-pounds}

This represents the total rotational force required to loosen the nail.

4

Solve for the effort force (

F_E

).

F_E = 560 / 16 = 35\text{ pounds}

Dividing the total required torque by the longer effort arm gives the smaller effort force needed.

Anahtar Kavram

Torque Balance and Mechanical Advantage

Soru 304Soru →

A marine engineer is filling a sealed buoyancy tank with a specialized flotation fluid. The system requires a fluid with a specific gravity of $0.85$ to provide the correct lift. If the internal volume of the tank is $0.4 \text{ m}^3$ , what must the total mass of the fluid be? (Assume the density of water is $1000 \text{ kg/m}^3$ )

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Cevap: 340

Cevap

The correct mass is

340 \text{ kg}

.

The correct answer is found by first converting the unitless specific gravity into an actual density by multiplying it by the density of water (

0.85 \times 1000 = 850 \text{ kg/m}^3

). The total mass is then calculated by multiplying this density by the given volume (

850 \times 0.4 = 340 \text{ kg}

).

Adım Adım Çözüm

1

Calculate the density of the fluid using its specific gravity.

Density =

850 \text{ kg/m}^3

Specific gravity is defined as the ratio of a fluid's density to the density of water (

1000 \text{ kg/m}^3

). Therefore, density equals

0.85 \times 1000

.

2

Calculate the mass of the fluid using its density and volume.

Mass =

340 \text{ kg}

Mass is the product of density and volume.

850 \text{ kg/m}^3 \times 0.4 \text{ m}^3 = 340 \text{ kg}

.

Anahtar Kavram

The relationship between specific gravity, density, and volume to determine the mass of a fluid.

Soru 305Soru →

An industrial oven used for curing aerospace composites is set to a precise temperature of $680^\circ\text{F}$ . To program a secondary backup sensor that operates exclusively on the Celsius scale, what numerical value must the technician enter?

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Cevap: 360

Cevap

360

The correct equivalent temperature is found by first subtracting 32 from the Fahrenheit value to adjust for the different zero points, resulting in

648

. Multiplying this difference by

\frac{5}{9}

accounts for the different degree increments between the scales, producing an exact value of

360^\circ\text{C}

.

Adım Adım Çözüm

1

Set up the conversion equation.

C = \frac{5}{9}(680 - 32)

This is the standard formula to convert degrees Fahrenheit into degrees Celsius.

2

Perform the subtraction inside the parentheses.

680 - 32 = 648

According to the order of operations, the offset between the two temperature scales must be removed first.

3

Multiply the difference by the fraction

\frac{5}{9}

.

648 \div 9 = 72

, then

72 \times 5 = 360

Multiplying by

\frac{5}{9}

scales the Fahrenheit degree magnitude to match the Celsius degree magnitude.

Anahtar Kavram

Temperature Scale Conversion

Soru 306Soru →

A vertical standpipe in a building's fire suppression system is completely filled with fresh water. The top of the water column is exactly $25\text{ meters}$ above a closed test valve on the ground floor. What is the hydrostatic pressure exerted by the water directly on this valve?

(Assume the density of water is $1000\text{ kg/m}^3$ and the acceleration due to gravity is $10\text{ m/s}^2$ . Provide your answer in kilopascals, $\text{kPa}$ .)

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Cevap: 250

Cevap

250

The hydrostatic pressure at the bottom of the water column is found using the equation

P = \rho g h

. Substituting the given values gives

1000 \times 10 \times 25 = 250,000\text{ Pascals}

. To express this in kilopascals, you divide by 1,000, resulting in exactly

250\text{ kPa}

.

Adım Adım Çözüm

1

Identify the relevant formula for hydrostatic pressure.

P = \rho g h

Hydrostatic pressure at a given depth depends strictly on the fluid's density, gravity, and the depth of the fluid column.

2

Substitute the given values into the formula.

P = 1000\text{ kg/m}^3 \times 10\text{ m/s}^2 \times 25\text{ m}

These are the precise physical parameters provided in the scenario.

3

Calculate the total pressure in standard SI units (Pascals).

P = 250,000\text{ Pa}

Multiplying the standard SI units yields the pressure in Pascals, since

1\text{ Pa} = 1\text{ kg}/(\text{m}\cdot\text{s}^2)

.

4

Convert the pressure from Pascals to kilopascals.

250,000 / 1000 = 250\text{ kPa}

The question specifically requests the final numerical answer in kilopascals.

Anahtar Kavram

Hydrostatic pressure is directly proportional to the density of the fluid, the acceleration due to gravity, and the vertical depth of the fluid column (

P = \rho g h

), regardless of the total fluid volume.

Tahmini Süre:1m 0s

Soru 307Soru →

A mechanic presses a brake pedal with a force of $25\text{ pounds}$ . The pedal pivots on a hinge at the top. The mechanic's foot applies the force $16\text{ inches}$ from the pivot point. The brake's pushrod, which transfers the force to the master cylinder, is attached to the pedal $4\text{ inches}$ from the pivot point. Calculate the force, in pounds, applied to the pushrod.

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Cevap: 100

Cevap

100 pounds

The brake pedal functions as a second-class lever. To find the force transmitted to the pushrod, use the principle of moments where Effort Force × Effort Distance = Load Force × Load Distance. By substituting the known values, we get

25\text{ lbs} \times 16\text{ inches} = F_{\text{load}} \times 4\text{ inches}

. Multiplying

25

by

16

yields

400\text{ inch-pounds}

of torque. Dividing this torque by the load distance of

4\text{ inches}

reveals that a force of

100\text{ pounds}

is applied to the pushrod.

Adım Adım Çözüm

1

Identify the known variables from the problem.

Effort Force = 25 lbs, Effort Distance = 16 in, Load Distance = 4 in.

Organizing the known values is the first step to applying the lever equation.

2

Apply the mechanical advantage principle for levers.

Effort Force × Effort Distance = Load Force × Load Distance

In a state of equilibrium, the torque applied by the effort must equal the torque applied to the load.

3

Substitute the known values into the equation.

25 × 16 = Load Force × 4

This sets up the equation to isolate the unknown load force.

4

Calculate the effort torque.

400 = Load Force × 4

Multiplying the effort force by its distance from the pivot gives the total torque in inch-pounds.

5

Solve for the final load force.

Load Force = 100

Dividing the total torque by the load's distance from the pivot gives the force applied to the pushrod.

Anahtar Kavram

Torque and Mechanical Advantage in Second-Class Levers

Soru 308Soru →

A maintenance worker uses a $50\text{-inch}$ rigid steel bar to lift one edge of a heavy generator. The fulcrum is positioned at the far end of the bar, and the worker applies an upward force at the opposite end. If the weight of the generator applies a downward force of $320\text{ pounds}$ at a point exactly $10\text{ inches}$ from the fulcrum, how much upward effort force (in pounds) must the worker apply to maintain static equilibrium?

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Cevap: 64

Cevap

64 pounds

The problem describes a second-class lever setup where the fulcrum is at one end, the effort is at the opposite end, and the resistance (load) is located between them. According to the principle of moments (law of the lever), the system is in static equilibrium when the effort torque equals the resistance torque (

F_e \times d_e = F_r \times d_r

). The effort arm length is the total length of the bar (

50\text{ inches}

), and the resistance arm length is the distance from the pivot to the load (

10\text{ inches}

). Substituting the values gives

F_e \times 50 = 320 \times 10

, which simplifies to

50F_e = 3200

. Dividing both sides by 50 yields an effort force of

64\text{ pounds}

.

Adım Adım Çözüm

1

Identify the lever class and the lengths of the moment arms.

The setup is a second-class lever. The effort arm (

d_e

) is

50\text{ inches}

and the resistance arm (

d_r

) is

10\text{ inches}

.

Correctly identifying the distances from the fulcrum to the forces is required to calculate torques.

2

Set up the torque balance equation.

F_e \times d_e = F_r \times d_r

For the lever to be in static equilibrium, the effort torque must equal the resistance torque.

3

Substitute the known values and solve for the effort force (

F_e

).

F_e \times 50 = 320 \times 10 \rightarrow 50F_e = 3200 \rightarrow F_e = 64

Isolating the unknown variable determines the exact force required.

Anahtar Kavram

The Law of the Lever (Torque Balance)

Soru 309Soru →

A heavy-duty vehicle's pneumatic braking system relies on an accumulator tank to store potential energy. Initially, a volume of $150 \text{ in}^3$ of air is compressed at an absolute pressure of $90 \text{ psi}$ inside the accumulator. When the operator engages the brakes, the air expands isothermally (at a constant temperature) into the connected brake chambers, causing the system's absolute pressure to drop to $30 \text{ psi}$ . What is the total volume that the air now occupies, in cubic inches?

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Cevap: 450

Cevap

450

According to Boyle's Law, the pressure and volume of a confined gas are inversely proportional when the temperature is held constant. The initial state is

90 \text{ psi}

and

150 \text{ in}^3

. The final pressure drops to

30 \text{ psi}

, which is one-third of the original pressure. Therefore, to balance the equation and conserve the quantity of gas, the volume must expand to three times its original size.

150 \times 3 = 450 \text{ in}^3

.

Adım Adım Çözüm

1

Identify the known variables from the scenario.

P_1 = 90 \text{ psi}

,

V_1 = 150 \text{ in}^3

,

P_2 = 30 \text{ psi}

.

Establishing the initial and final states of the gas is necessary to select the correct formula.

2

Select the appropriate gas law.

Use Boyle's Law:

P_1V_1 = P_2V_2

.

The problem states the expansion happens isothermally (constant temperature) and involves inversely proportional pressure and volume changes.

3

Substitute the values and solve for

V_2

.

90 \times 150 = 30 \times V_2

, so

V_2 = 13500 / 30 = 450 \text{ in}^3

.

Mathematical calculation is required to find the unknown final volume.

Anahtar Kavram

Boyle's Law (Pressure-Volume relationship at constant temperature)

Alternatif Yöntem

Instead of using the formula

P_1V_1 = P_2V_2

algebraically, you can use proportional reasoning. The pressure drops from

90 \text{ psi}

to

30 \text{ psi}

, which is a reduction by a factor of 3 (

90 / 30 = 3

). Because pressure and volume are inversely proportional at a constant temperature, the volume must increase by that exact same factor. Therefore, simply multiply the initial volume by 3:

150 \times 3 = 450 \text{ in}^3

.

Tahmini Süre:1m 0s

Soru 310Soru →

An open industrial mixing vat contains a liquid chemical with a specific gravity of $1.25$ . The liquid in the vat reaches a depth of $4\text{ meters}$ . Assuming the density of fresh water is $1,000\text{ kg/m}^3$ and the acceleration due to gravity is $10\text{ m/s}^2$ , what is the hydrostatic pressure exerted by the liquid at the bottom of the vat? (Enter your answer in kilopascals)

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Cevap: 50

Cevap

50

The hydrostatic pressure at the bottom of a fluid column is determined by the formula

P = \rho g h

. First, the specific gravity of

1.25

indicates the fluid is

1.25

times as dense as water, giving a density of

1,250\text{ kg/m}^3

. Multiplying this density by the acceleration due to gravity (

10\text{ m/s}^2

) and the depth (

4\text{ meters}

) yields

50,000\text{ Pascals}

. Since

1,000\text{ Pa}

equals

1\text{ kPa}

, this is equivalent to

50\text{ kPa}

.

Adım Adım Çözüm

1

Determine the density of the fluid.

Density = 1,250 kg/m³

Specific gravity is the ratio of the fluid's density to the density of water. Multiplying 1.25 by 1,000 kg/m³ gives the actual density.

2

Calculate the hydrostatic pressure in Pascals.

Pressure = 50,000 Pa

Hydrostatic pressure is the product of density, gravity, and depth (1,250 × 10 × 4 = 50,000).

3

Convert the pressure to kilopascals (kPa).

50 kPa

The question asks for the answer in kilopascals, so divide the Pascal value by 1,000.

Anahtar Kavram

Hydrostatic Pressure

Soru 311Soru →

A machinist operates a manual drill press. The feed lever (acting as the wheel) has an effective length of $8 \text{ inches}$ from the center of rotation. It turns a pinion gear (acting as the axle) with a radius of $1.6 \text{ inches}$ . If the machinist applies a tangential pull of $25 \text{ lbs}$ to the end of the feed lever, what downward force is applied by the pinion gear to the drill quill? (Assume $100\%$ efficiency).

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Cevap: 125

Cevap

The downward force applied by the pinion gear is 125 lbs.

In a wheel and axle system where effort is applied to the larger wheel, the mechanism multiplies the input force. The Ideal Mechanical Advantage (IMA) is found by dividing the effort radius (lever length of 8 inches) by the load radius (pinion gear radius of 1.6 inches). With an IMA of 5, the input force of 25 lbs is multiplied by 5 to yield 125 lbs of output force.

Adım Adım Çözüm

1

Identify the given dimensions and force.

Wheel radius (lever) = 8 inches, Axle radius (pinion) = 1.6 inches, Input Force = 25 lbs.

Identifying known variables is the first step in solving a mechanics problem.

2

Calculate the Ideal Mechanical Advantage (IMA).

IMA = Wheel Radius / Axle Radius = 8 / 1.6 = 5.

The IMA determines how much the wheel and axle mechanism multiplies the input force.

3

Calculate the output force.

Output Force = Input Force × IMA = 25 × 5 = 125 lbs.

In a 100% efficient system, the output force is the product of the input force and the mechanical advantage.

Anahtar Kavram

Ideal Mechanical Advantage of a wheel and axle, calculated as the ratio of their radii.

Soru 312Soru →

A construction worker uses a rigid pry bar to lift a concrete slab that weighs $400\text{ lbs}$ . The fulcrum is placed exactly $1\text{ foot}$ away from the point where the pry bar contacts the slab. If the worker applies a downward force exactly $5\text{ feet}$ away from the fulcrum, how many pounds of effort force must be applied to just balance the slab?

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Cevap: 80

Cevap

80 lbs

To balance the lever, the torque on both sides of the fulcrum must be equal. The load torque is generated by the slab:

400\text{ lbs} \times 1\text{ ft} = 400\text{ ft-lbs}

. To create an equal effort torque on the other side at a distance of

5\text{ feet}

, an effort force of

80\text{ lbs}

is required because

80\text{ lbs} \times 5\text{ ft} = 400\text{ ft-lbs}

.

Adım Adım Çözüm

1

Identify the known variables given in the problem statement.

Load Force (

F_L

) =

400\text{ lbs}

, Load Distance (

d_L

) =

1\text{ ft}

, Effort Distance (

d_E

) =

5\text{ ft}

.

These values are required to calculate the unknown effort force using the law of the lever.

2

Apply the law of the lever (torque balance equation).

F_E \times d_E = F_L \times d_L

A lever is in static equilibrium when the input torque exactly equals the output torque.

3

Substitute the known values and solve for the effort force (

F_E

).

F_E \times 5 = 400 \times 1 \rightarrow 5F_E = 400 \rightarrow F_E = 80

Isolating

F_E

provides the exact pounds of force needed to balance the concrete slab.

Anahtar Kavram

Torque Balance and the Law of the Lever

Tahmini Süre:45s

Soru 313Soru →

An automotive technician operates a hydraulic floor jack by pressing down on its handle. The handle pivots on a fixed pin at one end. The technician applies a steady downward force of $30\text{ lbs}$ at the opposite end of the handle. The total length of the handle from the pivot pin to the point of effort is $24\text{ inches}$ . The handle pushes down on a pump piston that is located $1.5\text{ inches}$ from the pivot pin. What is the total force, in pounds, applied to the pump piston? (Assume the handle is horizontal and ignore friction.)

Cevabı ve açıklamayı göster

Cevap: 480

Cevap

480

The jack handle operates as a lever with the fulcrum at one end. Because the effort is applied much further from the fulcrum than the load, the lever multiplies the applied force. The mechanical advantage is the ratio of the effort arm to the load arm (

24 / 1.5 = 16

). Multiplying the

30\text{ lbs}

of effort by a mechanical advantage of 16 yields a

480\text{ lbs}

force on the piston.

Adım Adım Çözüm

1

Identify the lever components.

Fulcrum = pivot pin; Effort Force (E) = 30 lbs; Effort Arm (d_E) = 24 inches; Load Arm (d_L) = 1.5 inches.

To apply the torque balance equation, all distances must be measured from the fulcrum.

2

Set up the torque balance equation.

E \times d_E = L \times d_L

In a static lever system, the torque generated by the effort equals the torque generated by the load.

3

Substitute the given values.

30 \times 24 = L \times 1.5

To prepare the equation for solving the unknown load force (L).

4

Solve for the load force.

720 = 1.5L \implies L = 480\text{ lbs}

Dividing the effort torque by the load arm distance yields the resulting force applied to the piston.

Anahtar Kavram

Torque Balance and Mechanical Advantage

Soru 314Soru →

A technician uses a standard vernier caliper with 0.001-inch resolution to measure the outside diameter of a transmission output shaft. On the main beam, the zero mark on the vernier plate has moved past the 2-inch mark, past the major graduation 3 (representing 0.300 inch), and past two 0.025-inch subdivision lines. On the 25-division vernier plate, the line marked 18 aligns precisely with a graduation line on the main beam. What is the total measured diameter of the shaft in inches?

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Cevap: 2.368

Cevap

The total measured diameter of the transmission output shaft is 2.368 inches.

To read a standard inch vernier caliper, sum the whole inches (

2.000\text{ in}

), the exposed tenths on the main beam (

0.300\text{ in}

), the exposed

0.025\text{-inch}

subdivisions (

2 \times 0.025\text{ in} = 0.050\text{ in}

), and the aligning vernier division (

18 \times 0.001\text{ in} = 0.018\text{ in}

). Adding these together gives

2.000 + 0.300 + 0.050 + 0.018 = 2.368\text{ inches}

.

Adım Adım Çözüm

1

Determine the whole inch and tenths-of-an-inch value from the main beam

2.300 inches

The vernier zero index mark has passed the 2-inch mark and the numbered 3 division, representing 2.300 inches.

2

Add the value of the exposed 0.025-inch beam subdivisions

0.050 inch (running total: 2.350 inches)

Two subdivision marks are visible past the 3, and each mark represents 0.025 inch (2 × 0.025 in = 0.050 in).

3

Add the vernier plate reading based on the aligned graduation mark

0.018 inch (final total: 2.368 inches)

The 18th mark on the vernier plate aligns with a beam line, representing 18 × 0.001 in = 0.018 inch.

Anahtar Kavram

Reading a 0.001-inch vernier caliper by combining the main beam whole inches, tenths, 0.025-inch subdivisions, and the aligning vernier plate division.

Alternatif Yöntem

Convert each scale value into thousandths of an inch before summing: 2000 thousandths (whole inches) + 300 thousandths (major mark 3) + 50 thousandths (two 0.025-inch increments) + 18 thousandths (vernier alignment) = 2368 thousandths, which equals 2.368 inches.

Tahmini Süre:1m 0s

Soru 315Soru →

A fabricator is using a pair of layout dividers to step off and mark $5$ equally spaced hole centers in a straight line on a metal workpiece. The total distance from the center of the first hole to the center of the fifth hole is $14\text{ inches}$ . What is the required divider setting, in inches, between consecutive hole centers?

Cevabı ve açıklamayı göster

Cevap: 3.5

Cevap

The divider points must be set to

3.5\text{ inches}

.

The correct answer is

3.5\text{ inches}

because

5

equally spaced hole centers define

4

equal intervals along the line (

5 - 1 = 4

). Dividing the

14\text{-inch}

span by

4

gives exactly

3.5\text{ inches}

per interval.

Adım Adım Çözüm

1

Determine the number of equal intervals between the hole centers.

There are

4

intervals between

5

collinear points (

5 - 1 = 4

).

When stepping off positions with dividers between defined endpoints, the number of intervals is always one less than the number of points.

2

Divide the total measured span by the number of intervals.

\frac{14\text{ inches}}{4} = 3.5\text{ inches}

Equal divider spacing requires distributing the total span across the

4

equal intervals.

Anahtar Kavram

Using layout dividers to step off equal intervals along a layout line

Tahmini Süre:1m 0s

Soru 316Soru →

A hydrostatic pressure sensor at the bottom of an upright agricultural water cistern reads $45\text{ psi}$ when the water depth is $30\text{ feet}$ . During a dry season, the water level drops to a depth of $18\text{ feet}$ . Assuming the fluid density remains constant, what is the new hydrostatic pressure reading in $\text{psi}$ at the sensor?

Cevabı ve açıklamayı göster

Cevap: 27

Cevap

27

Since hydrostatic pressure is directly proportional to depth (

P = \rho \cdot g \cdot h

), the ratio of pressure to depth remains constant. The initial ratio is

45\text{ psi} / 30\text{ feet} = 1.5\text{ psi/ft}

. At a new depth of

18\text{ feet}

, the pressure is

18\text{ feet} \times 1.5\text{ psi/ft} = 27\text{ psi}

.

Adım Adım Çözüm

1

Identify the relationship between hydrostatic pressure and fluid depth.

Hydrostatic pressure is directly proportional to the depth of the fluid.

The formula for hydrostatic pressure is

P = \rho \cdot g \cdot h

, meaning if density (

\rho

) and gravity (

g

) are constant, the pressure changes linearly with depth (

h

).

2

Determine the pressure per foot of water depth.

45\text{ psi} \div 30\text{ feet} = 1.5\text{ psi}

per foot.

This establishes the constant of proportionality for the fluid in this specific cistern.

3

Calculate the new pressure at the reduced depth.

1.5\text{ psi/foot} \times 18\text{ feet} = 27\text{ psi}

.

Multiplying the pressure per foot by the new depth yields the corresponding hydrostatic pressure.

Anahtar Kavram

Hydrostatic pressure is directly proportional to fluid depth.

Soru 317Soru →

A coastal patrol boat is cruising at an initial velocity of $5.0 \text{ m/s}$ . The captain orders the engines to increase power, causing the boat to accelerate at a constant rate of $0.2 \text{ m/s}^2$ for exactly $1.5 \text{ minutes}$ .

What is the final velocity of the boat in $\text{m/s}$ ?

Cevabı ve açıklamayı göster

Cevap: 23

Cevap

23.0

The correct answer is derived by first converting the time from minutes to seconds (

1.5 \text{ min} = 90 \text{ s}

), calculating the total change in velocity (

0.2 \times 90 = 18.0 \text{ m/s}

), and then adding that change to the initial cruising velocity (

5.0 + 18.0 = 23.0 \text{ m/s}

).

Adım Adım Çözüm

1

Extract and convert the time variable into seconds.

t = 1.5 \text{ minutes} \times 60 \text{ seconds/minute} = 90 \text{ seconds}

Acceleration is given in meters per second squared (

\text{m/s}^2

), so time must be in seconds to ensure unit consistency.

2

Calculate the total change in velocity during the acceleration phase.

\Delta v = a \times t = 0.2 \text{ m/s}^2 \times 90 \text{ s} = 18.0 \text{ m/s}

The change in velocity is the product of the constant acceleration and the duration of that acceleration.

3

Add the change in velocity to the initial velocity.

v_f = v_i + \Delta v = 5.0 \text{ m/s} + 18.0 \text{ m/s} = 23.0 \text{ m/s}

The final velocity is the starting velocity plus the amount the velocity increased.

Anahtar Kavram

Applying the constant acceleration formula (

v_f = v_i + at

) with necessary unit conversions.

Tahmini Süre:1m 0s

Soru 318Soru →

A worker uses a hand-operated winch to vertically lift a bundle of materials weighing $300 \text{ pounds}$ . The winch is driven by a crank handle that has a length of $15 \text{ inches}$ from its center of rotation. The cable lifting the load wraps around a drum with a diameter of $5 \text{ inches}$ .

Assuming a $100\%$ efficient system, what is the theoretical effort force required on the crank handle to lift the bundle?

Cevabı ve açıklamayı göster

Cevap: 50

Cevap

50

The system operates as a wheel and axle where the crank handle is the wheel and the drum is the axle. By finding the ratio of their radii (

15 \text{ inches}

to

2.5 \text{ inches}

), we determine that the mechanical advantage is

6

. This means the effort force required is one-sixth of the total load, yielding

50 \text{ pounds}

.

Adım Adım Çözüm

1

Determine the effective radius of both the wheel and the axle.

Wheel radius =

15 \text{ inches}

. Axle radius =

5 \text{ inches} \div 2 = 2.5 \text{ inches}

.

The formula for mechanical advantage requires consistent measurements, typically the radius of each component.

2

Calculate the Ideal Mechanical Advantage (IMA) of the system.

IMA = \frac{15}{2.5} = 6

.

The IMA indicates how many times the simple machine multiplies the input effort force.

3

Calculate the theoretical effort force required.

F_{effort} = \frac{300}{6} = 50 \text{ pounds}

.

In a frictionless (

100\%

efficient) system, the effort force is equal to the load force divided by the mechanical advantage.

Anahtar Kavram

Ideal Mechanical Advantage (IMA) of a Wheel and Axle

Soru 319Soru →

An automotive brake pedal operates as a second-class lever. The pivot point (fulcrum) is located at the top of the pedal assembly. The pushrod, which connects to the brake master cylinder, is attached $4 \text{ inches}$ below the pivot. The driver applies force to the pedal pad, which is located $14 \text{ inches}$ below the pivot. If the driver applies a perpendicular force of $50 \text{ lbs}$ to the pedal pad, what is the force, in pounds, transmitted to the pushrod? Assume the system is in equilibrium.

Cevabı ve açıklamayı göster

Cevap: 175

Cevap

175

The brake pedal acts as a second-class lever in equilibrium. According to the principle of moments, the effort torque (

50 \text{ lbs} \times 14 \text{ inches} = 700 \text{ in-lbs}

) must equal the load torque. Dividing the total torque by the load arm distance (

4 \text{ inches}

) gives an output force of

175 \text{ lbs}

.

Adım Adım Çözüm

1

Identify the lever components and their distances from the fulcrum.

Effort arm =

14 \text{ inches}

, Load arm =

4 \text{ inches}

.

In torque calculations, all distances must be measured directly from the pivot point (fulcrum).

2

Apply the torque balance equation for a lever in equilibrium.

F_{effort} \times d_{effort} = F_{load} \times d_{load}

To find the unknown force, the applied torque must equal the output torque.

3

Substitute the known values into the equation and solve for the load force.

50 \times 14 = F_{load} \times 4 \rightarrow 700 = F_{load} \times 4 \rightarrow F_{load} = 175 \text{ lbs}

.

Isolating the unknown variable gives the final transmitted force.

Anahtar Kavram

Torque Equilibrium and Mechanical Advantage in Second-Class Levers

Tahmini Süre:1m 0s

Soru 320Soru →

A vertical ballast column on a maritime vessel is completely filled with fresh water. A pressure gauge is mounted precisely $60\text{ feet}$ below the open surface of the water. If the specific weight of the fresh water is $62.4\text{ lb/ft}^3$ , what is the hydrostatic gauge pressure at this depth in pounds per square inch (psi)?

Cevabı ve açıklamayı göster

Cevap: 26

Cevap

The hydrostatic pressure is 26 psi.

The correct calculation requires two phases: first, determining the pressure based on depth and specific weight, which yields

3744\text{ psf}

(

60 \times 62.4

). Second, converting area units from square feet to square inches by dividing by 144, resulting in exactly

26\text{ psi}

.

Adım Adım Çözüm

1

Calculate the hydrostatic pressure in pounds per square foot (psf).

62.4\text{ lb/ft}^3 \times 60\text{ ft} = 3744\text{ lb/ft}^2

(psf)

Hydrostatic pressure is the product of the fluid's specific weight and the depth of the fluid column.

2

Convert the pressure from psf to pounds per square inch (psi).

3744\text{ psf} \div 144\text{ in}^2/\text{ft}^2 = 26\text{ psi}

There are

144\text{ square inches}

in

1\text{ square foot}

(

12\text{ inches} \times 12\text{ inches}

). Dividing the psf value by 144 converts the units appropriately.

Anahtar Kavram

Hydrostatic Pressure Calculation and Unit Conversion

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