All practice questions

Question 2481Question →

A structural engineer is evaluating two submerged sensor housings within a continuous freshwater facility. Housing X is mounted at the bottom of a narrow vertical standpipe, exactly $12\text{ meters}$ below the water's surface. Housing Y is mounted on the open floor of the main reservoir, also exactly $12\text{ meters}$ below the water's surface. Both locations share the same body of water. Which of the following statements correctly compares the hydrostatic pressure acting on the two housings?

The hydrostatic pressure is equal on both housings because they are situated at the identical vertical depth.

The hydrostatic pressure is greater on Housing Y because the expansive reservoir holds a much larger total weight of water.

The hydrostatic pressure is greater on Housing X because the narrow standpipe concentrates the fluid's downward force into a smaller area.

The hydrostatic pressure cannot be compared directly unless the depth measurements are mathematically converted into volumetric flow rates.

Show answer & explanation

Answer: The hydrostatic pressure is equal on both housings because they are situated at the identical vertical depth.

Answer

The hydrostatic pressure is equal on both housings because they are situated at the identical vertical depth.

The correct answer accurately states that hydrostatic pressure depends only on fluid depth. Because both housings are situated exactly

12\text{ meters}

below the surface of the same continuous body of water, the pressure exerted upon them is identical, regardless of the surrounding container geometry.

Step-by-Step Solution

1

Identify the factors that determine hydrostatic pressure.

Hydrostatic pressure (

P

) depends on fluid density (

\rho

), gravity (

g

), and vertical depth (

h

) according to the formula

P = \rho \cdot g \cdot h

.

This establishes the physical law governing the scenario.

2

Evaluate the fluid conditions for both housings.

Both housings are in the same continuous freshwater system, meaning the fluid density (

\rho

) is identical.

Eliminates fluid density as a variable between the two locations.

3

Compare the depth (

h

) and container shapes for both housings.

Housing X is in a narrow pipe at a

12\text{ m}

depth, and Housing Y is in a wide reservoir at a

12\text{ m}

depth. Container width and total fluid volume are not variables in the hydrostatic pressure formula.

Demonstrates that despite the drastically different container shapes, the equal depth guarantees equal pressure.

Key Concept

Hydrostatic pressure relies strictly on the vertical depth and density of the fluid, and is completely unaffected by the shape of the container or the total volume of the fluid.

Question 2482Question →

A rescue helicopter is flying to a distress signal at a constant speed of $120 \text{ miles per hour}$ . Given that $1 \text{ mile} = 5,280 \text{ feet}$ , how many feet does the helicopter travel in $45 \text{ seconds}$ ?

5,400

7,920

285,120

475,200

Show answer & explanation

Answer: 7,920

Answer

7,920

The problem requires finding the distance in feet, given a speed in miles per hour and time in seconds. First, convert the speed to feet per second. Since there are

5,280 \text{ feet}

in a mile,

120 \text{ miles per hour}

is

120 \times 5,280 = 633,600 \text{ feet per hour}

. Since there are

3,600 \text{ seconds}

in an hour, the speed is

633,600 \div 3,600 = 176 \text{ feet per second}

. Finally, multiply this rate by the time of

45 \text{ seconds}

to get the distance:

176 \times 45 = 7,920 \text{ feet}

.

Step-by-Step Solution

1

Convert the helicopter's speed from miles per hour to feet per hour.

120 \times 5,280 = 633,600 \text{ feet per hour}

To align the distance unit (miles) with the requested final unit (feet).

2

Convert the speed from feet per hour to feet per second.

633,600 \div 3,600 = 176 \text{ feet per second}

There are

60 \times 60 = 3,600

seconds in an hour, and the given elapsed time is in seconds.

3

Calculate the total distance traveled in

45 \text{ seconds}

.

176 \times 45 = 7,920 \text{ feet}

Distance equals rate multiplied by time (

D = R \times T

).

Key Concept

Unit Conversions in Motion Problems

Question 2483Question →

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 $\text{m/s}$ ?

Show answer & explanation

Answer: 23

Answer

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}

).

Step-by-Step Solution

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.

Key Concept

Applying the constant acceleration formula (

v_f = v_i + at

) with necessary unit conversions.

Estimated Time:1m 0s

Question 2484Question →

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?

Show answer & explanation

Answer: 50

Answer

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}

.

Step-by-Step Solution

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.

Key Concept

Ideal Mechanical Advantage (IMA) of a Wheel and Axle

Question 2485Question →

A dump truck bed pivots on a rear hinge, which acts as the fulcrum. A hydraulic cylinder provides the upward effort force and is attached to the bed $2\text{ ft}$ from the rear hinge. The center of gravity of the loaded bed is located $6\text{ ft}$ from the rear hinge. If the total weight of the loaded bed is $12,000\text{ lbs}$ , what is the minimum purely vertical force the hydraulic cylinder must apply to begin lifting the horizontally resting bed?

4,000 \text{ lbs}

12,000 \text{ lbs}

24,000 \text{ lbs}

36,000 \text{ lbs}

Show answer & explanation

Answer: $36,000 \text{ lbs}$

Answer

The minimum purely vertical force the hydraulic cylinder must apply is

36,000 \text{ lbs}

.

In this 3rd class lever, the effort arm (

2\text{ ft}

) is significantly shorter than the load arm (

6\text{ ft}

). To balance the torque created by the

12,000\text{ lbs}

load (

12,000 \times 6 = 72,000\text{ ft-lbs}

), the hydraulic cylinder must apply a proportionally larger force of

36,000\text{ lbs}

at its

2\text{ ft}

distance (

36,000 \times 2 = 72,000

).

Step-by-Step Solution

1

Identify the lever class and the pivot point.

The rear hinge is the fulcrum. The hydraulic cylinder provides the effort between the fulcrum and the load, making this a 3rd class lever.

To correctly apply the principles of torque, the locations of the effort and load relative to the fulcrum must be established.

2

Calculate the torque exerted by the loaded bed.

12,000 \text{ lbs} \times 6 \text{ ft} = 72,000 \text{ ft-lbs}

.

Torque is equal to force multiplied by the distance from the fulcrum.

3

Set up the torque balance equation for the effort force (

F_e

).

F_e \times 2 \text{ ft} = 72,000 \text{ ft-lbs}

.

To begin lifting the bed, the effort torque must be equal to the load torque.

4

Solve the equation for the effort force.

F_e = 36,000 \text{ lbs}

.

Dividing both sides by

2 \text{ ft}

isolates the required force.

Key Concept

Mechanical advantage and torque balance in a 3rd class lever

Estimated Time:1m 0s

Question 2486Question →

A maintenance crew is inspecting a closed, static water pipeline fed by a mountain reservoir. They monitor static pressure sensors located at different points along the pipeline's path.

Arrange the following sensor locations in order from the LOWEST hydrostatic pressure to the HIGHEST hydrostatic pressure.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest pressure is: the surface sensor (0 feet), the sensor in the 24-inch pipe (30 feet), the sensor in the 12-inch pipe (60 feet), and the sensor at the bend (90 feet).

Hydrostatic pressure in a continuous, static fluid is directly proportional to its vertical depth below the free surface (

P = \rho g h

). Factors like pipe diameter, pipe orientation, or the total volume of fluid in a specific section do not influence the pressure at a specific point. Therefore, ranking from lowest to highest pressure simply requires ordering the depths from shallowest to deepest: 0 feet, 30 feet, 60 feet, and finally 90 feet.

Step-by-Step Solution

1

Identify the principle governing hydrostatic pressure in a static fluid.

Hydrostatic pressure is determined by the formula

P = \rho g h

, where

h

is the vertical depth from the free surface of the fluid.

Establishing the core rule is necessary to evaluate which variables provided in the scenario actually affect the pressure.

2

Evaluate the effect of pipe diameter and shape on the pressure.

The 12-inch diameter, 24-inch diameter, and horizontal pipe bend do not affect static pressure.

According to Pascal's principle and fluid statics, pressure at a given depth is independent of the container's volume, shape, or width.

3

Rank the locations based solely on their vertical depth below the reservoir surface.

Surface (0 ft) < 30 ft < 60 ft < 90 ft.

Since vertical depth is the only changing variable that affects the pressure, the ranking strictly follows the depth from shallowest to deepest.

Key Concept

Hydrostatic pressure in a continuous fluid depends exclusively on fluid density, gravity, and vertical depth, remaining entirely independent of the container's volume or shape.

Question 2487Question →

An industrial conveyor belt moves continuously at a steady speed of $44 \text{ feet per minute}$ . Given that $1 \text{ mile} = 5,280 \text{ feet}$ , how many miles does a package travel if it remains on the belt for exactly $10 \text{ hours}$ ?

Show answer & explanation

Answer: 5; 5 miles; five; five miles

Answer

5

To find the total distance traveled in miles, unit conversions must be applied to align the given measurements. The speed is given in feet per minute, but the time is in hours and the requested distance is in miles. First, multiply the speed by 60 to convert it to feet per hour (

44 \times 60 = 2,640 \text{ feet per hour}

). Next, convert this speed into miles per hour by dividing by the number of feet in a mile (

2,640 \div 5,280 = 0.5 \text{ miles per hour}

). Finally, multiply this speed by the total time (

0.5 \text{ mph} \times 10 \text{ hours}

) to find that the package travels a total distance of 5 miles.

Step-by-Step Solution

1

Convert the speed from feet per minute to feet per hour.

44 \text{ feet/minute} \times 60 \text{ minutes/hour} = 2,640 \text{ feet/hour}

The travel time is given in hours, so the speed needs to reflect distance per hour to maintain consistent units of time.

2

Convert the speed from feet per hour to miles per hour.

2,640 \text{ feet/hour} \div 5,280 \text{ feet/mile} = 0.5 \text{ miles/hour}

The final requested distance is in miles, so converting the rate into miles per hour simplifies the final distance calculation.

3

Calculate the total distance traveled.

0.5 \text{ miles/hour} \times 10 \text{ hours} = 5 \text{ miles}

Multiplying the constant rate in miles per hour by the total time in hours yields the total distance in miles.

Key Concept

Unit Conversions in Motion Problems

Alternative Method

Instead of converting the speed to miles per hour first, you can convert the total time into minutes:

10 \text{ hours} \times 60 \text{ minutes/hour} = 600 \text{ minutes}

. Then, find the total distance in feet:

600 \text{ minutes} \times 44 \text{ feet/minute} = 26,400 \text{ feet}

. Finally, divide by

5,280

to convert to miles:

26,400 \div 5,280 = 5 \text{ miles}

.

Estimated Time:1m 30s

Question 2488Question →

A driver operates a heavy-duty steering wheel connected directly to a steering column (axle). The steering wheel has a radius of $16 \text{ inches}$ , and the steering column has a diameter of $4 \text{ inches}$ . If the driver applies an effort force of $20 \text{ pounds}$ to the outer rim of the steering wheel, what is the resulting output force exerted by the steering column? (Assume a perfectly efficient system.)

2.5 \text{ pounds}

5 \text{ pounds}

80 \text{ pounds}

160 \text{ pounds}

Show answer & explanation

Answer: $160 \text{ pounds}$

Answer

160 \text{ pounds}

The mechanical advantage of a wheel and axle when effort is applied to the wheel is the radius of the wheel divided by the radius of the axle. The wheel's radius is

16 \text{ inches}

. The axle's diameter is

4 \text{ inches}

, meaning its radius is

2 \text{ inches}

. The mechanical advantage is

16 / 2 = 8

. Multiplying the

20 \text{ pound}

input force by the mechanical advantage of

8

gives a total output force of

160 \text{ pounds}

.

Step-by-Step Solution

1

Determine the radius of the steering column (axle).

The diameter is

4 \text{ inches}

, so the radius is

2 \text{ inches}

.

Mechanical advantage must be calculated by comparing equivalent dimensions, such as the radii of both components.

2

Calculate the Ideal Mechanical Advantage (IMA) of the wheel and axle.

\text{IMA} = \frac{\text{Radius of Wheel}}{\text{Radius of Axle}} = \frac{16}{2} = 8

.

When effort is applied to the larger wheel, the system acts as a force multiplier based on the ratio of the radii.

3

Calculate the output force exerted by the steering column.

\text{Output Force} = \text{Effort Force} \times \text{IMA} = 20 \times 8 = 160 \text{ pounds}

.

The system multiplies the driver's input force by the mechanical advantage to turn the steering mechanism.

Key Concept

Ideal Mechanical Advantage of a Wheel and Axle

Estimated Time:1m 0s

Question 2489Question →

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.

Show answer & explanation

Answer: 175

Answer

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}

.

Step-by-Step Solution

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.

Key Concept

Torque Equilibrium and Mechanical Advantage in Second-Class Levers

Estimated Time:1m 0s

Question 2490Question →

A maglev transit pod enters a straight acceleration zone with an initial velocity of $20 \text{ m/s}$ . The pod's linear induction motors apply a constant acceleration of $5 \text{ m/s}^2$ for exactly $0.1 \text{ minutes}$ . What is the pod's final velocity at the end of this time period?

20.5 \text{ m/s}

30 \text{ m/s}

50 \text{ m/s}

70 \text{ m/s}

Show answer & explanation

Answer: $50 \text{ m/s}$

Answer

The correct answer is

50 \text{ m/s}

.

The final velocity is found by first converting the given time to standard units (

0.1 \text{ minutes} = 6 \text{ seconds}

). Next, determine the total velocity gained by multiplying acceleration by time (

5 \text{ m/s}^2 \times 6 \text{ s} = 30 \text{ m/s}

). Finally, add this gained velocity to the initial velocity of

20 \text{ m/s}

to get a final velocity of

50 \text{ m/s}

.

Step-by-Step Solution

1

Convert the given time from minutes to seconds.

0.1 \text{ minutes} \times 60 \text{ seconds/minute} = 6 \text{ seconds}

Acceleration is given in

\text{m/s}^2

, so time must be in seconds to match units.

2

Calculate the total change in velocity (

\Delta v

) during the acceleration period.

\Delta v = a \times t = 5 \text{ m/s}^2 \times 6 \text{ s} = 30 \text{ m/s}

This determines how much speed the pod gained.

3

Add the change in velocity to the initial velocity to find the final velocity.

v_f = v_i + \Delta v = 20 \text{ m/s} + 30 \text{ m/s} = 50 \text{ m/s}

The final velocity is the starting velocity plus whatever velocity was gained.

Key Concept

Applying kinematic equations for constant acceleration (

v_f = v_i + at

) with required unit conversions.

Question 2491Question →

A hydroelectric dam holds back a massive artificial lake. Along the submerged face of the dam, there is a small diagnostic sensor port and a massive turbine water intake. Both the sensor port and the center of the turbine intake are located exactly $45\text{ meters}$ below the surface of the lake. The diameter of the turbine intake is twenty times larger than the diagnostic port. When the water in the lake is perfectly still, how does the hydrostatic pressure at the diagnostic port compare to the pressure at the center of the turbine intake?

The pressure is identical at both locations because they are positioned at the exact same depth.

The pressure is twenty times greater at the turbine intake because its larger diameter captures more force from the water.

The pressure is significantly higher at the diagnostic port because the fluid is compressed into a narrower opening.

The pressure at both locations is determined by the total mass of the reservoir, so the depth of the specific openings is not the primary factor.

Show answer & explanation

Answer: The pressure is identical at both locations because they are positioned at the exact same depth.

Answer

The pressure is identical at both locations because they are positioned at the exact same depth.

Hydrostatic pressure in a static fluid is calculated using the formula

P = \rho g h

(Pressure = density × gravity × depth). Because both the diagnostic port and the center of the turbine intake are located exactly

45\text{ meters}

below the surface, they experience the exact same water pressure. The size, diameter, or area of the opening affects the total force exerted on that opening, but it does not change the pressure, which is a measure of force per unit area.

Step-by-Step Solution

1

Identify the factors that determine hydrostatic pressure in a static fluid.

Hydrostatic pressure (

P

) is determined by the formula

P = \rho g h

, where

\rho

is fluid density,

g

is gravity, and

h

is depth.

This establishes the physical rule governing the scenario, showing that only depth matters when the fluid is consistent.

2

Compare the depth of the two locations provided in the scenario.

Both the diagnostic port and the turbine intake center are located at a depth of

45\text{ meters}

.

Since the

h

value is identical for both locations, the pressure calculated from

P = \rho g h

will be equal.

3

Evaluate the effect of the opening sizes on the hydrostatic pressure.

The diameter of the openings has no effect on the hydrostatic pressure, though it does affect the total force (

F = P \times A

).

It separates the concept of pressure (force per unit area) from the concept of total force, confirming that the pressure remains identical.

Key Concept

Hydrostatic pressure in a liquid depends only on the depth of the liquid and its density, and is entirely independent of the shape or size of the container or opening.

Question 2492Question →

Match each real-world mechanism to its correct functional 'wheel' and 'axle' components based on how it operates.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Manual Screwdriver matches 'Wheel = Outer wide grip; Axle = Narrow metal shaft'. Bicycle Rear Wheel Assembly matches 'Wheel = Outer tire and rim; Axle = Central toothed gear'. Hand-cranked Water Well matches 'Wheel = Circular path traced by the handle; Axle = Central drum'. Ceiling Fan matches 'Wheel = Extended rotating blades; Axle = Central motorized spindle'.

Each mechanism is correctly categorized by identifying the component with the larger rotational diameter as the wheel and the smaller central component as the axle, accurately reflecting how they transmit rotational force and speed.

Step-by-Step Solution

1

Define the physical characteristics of a wheel and axle system.

The 'wheel' is always the component or path with the larger radius, and the 'axle' is the rigidly attached component with the smaller radius.

Establishing the definitions allows us to look past the physical shape of the object and focus on its mechanical function.

2

Analyze tools designed to multiply force (Screwdriver and Windlass).

In these tools, effort is applied to the larger radius (the wide grip or the crank's path) to turn the smaller radius (the shaft or the drum) with greater force.

Identifying where the effort is applied helps determine which part acts as the wheel.

3

Analyze mechanisms designed to multiply speed or distance (Bicycle and Ceiling Fan).

In these systems, effort is applied to the smaller central radius (the sprocket or the motor spindle), which forces the larger radius (the tire or the blades) to spin faster and cover more distance.

This confirms the axle is driving the wheel in these specific applications.

Key Concept

In a wheel and axle simple machine, the 'wheel' is the component with the larger diameter, and the 'axle' is the connected component with the smaller diameter. A wheel does not have to be a solid disk; it can be a crank path or fan blades.

Question 2493Question →

An interstate bus travels from a central terminal to a destination city $150$ miles away and immediately returns along the same route. The bus averages a fuel efficiency of $12$ miles per gallon. If the bus starts the journey with exactly $60$ gallons of fuel, how many gallons of fuel will remain in the tank at the end of the round trip?

25

35

47.5

57.5

Show answer & explanation

Answer: 35

Answer

There will be 35 gallons of fuel remaining in the tank.

The correct answer is derived by first finding the total round-trip distance, which is

150 \times 2 = 300

miles. Dividing this distance by the fuel efficiency of

12

miles per gallon gives the total fuel consumed:

300 \div 12 = 25

gallons. Finally, subtracting the consumed fuel from the initial

60

gallons yields the remaining amount:

60 - 25 = 35

gallons.

Step-by-Step Solution

1

Calculate the total distance for the round trip.

150 miles * 2 = 300 miles

The bus travels to the destination and returns, meaning the one-way distance must be doubled.

2

Determine the total fuel consumed during the entire trip.

300 miles / 12 miles per gallon = 25 gallons

Dividing the total distance by the fuel efficiency gives the total amount of fuel required.

3

Calculate the remaining fuel in the tank.

60 gallons - 25 gallons = 35 gallons

Subtracting the consumed fuel from the initial fuel amount provides the remaining balance.

Key Concept

Calculating remaining fuel by finding the total round-trip distance, dividing by fuel efficiency to find consumed fuel, and subtracting from initial capacity.

Question 2494Question →

Calculate the acceleration for each of the following pieces of equipment. Arrange the scenarios in order from the greatest acceleration to the least acceleration.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from greatest to least acceleration is: Emergency response boat (

2.5 \text{ m/s}^2

), Heavy towing winch (

1.5 \text{ m/s}^2

), Runway snowplow (

1.2 \text{ m/s}^2

), and Construction crane (

0.5 \text{ m/s}^2

).

To determine the correct order, we must calculate the magnitude of acceleration for each scenario using the formula

a = \frac{v_f - v_i}{t}

. The boat has the highest acceleration (

2.5 \text{ m/s}^2

), followed by the winch (

1.5 \text{ m/s}^2

), the snowplow (

1.2 \text{ m/s}^2

), and finally the crane (

0.5 \text{ m/s}^2

). Arranging them from greatest to least yields this precise sequence.

Step-by-Step Solution

1

Calculate the acceleration of the emergency response boat.

The boat accelerates at

2.5 \text{ m/s}^2

.

Using the formula

a = \frac{v_f - v_i}{t}

, we get

a = \frac{25.0 - 5.0}{8.0} = 2.5 \text{ m/s}^2

.

2

Calculate the acceleration of the heavy towing winch.

The winch accelerates at

1.5 \text{ m/s}^2

.

Using the formula

a = \frac{v_f - v_i}{t}

, we get

a = \frac{4.0 - 1.0}{2.0} = 1.5 \text{ m/s}^2

.

3

Calculate the acceleration of the runway snowplow.

The snowplow accelerates at

1.2 \text{ m/s}^2

.

From a complete stop,

v_i = 0

.

a = \frac{12.0 - 0}{10.0} = 1.2 \text{ m/s}^2

.

4

Calculate the acceleration of the construction crane.

The crane accelerates at

0.5 \text{ m/s}^2

.

From rest,

v_i = 0

.

a = \frac{3.0 - 0}{6.0} = 0.5 \text{ m/s}^2

.

5

Order the calculated acceleration values from greatest to least.

Boat (

2.5

) > Winch (

1.5

) > Snowplow (

1.2

) > Crane (

0.5

).

The prompt requires arranging the scenarios in descending order based on their calculated magnitudes of acceleration.

Key Concept

Acceleration is defined as the rate at which an object changes its velocity over a specific interval of time.

Question 2495Question →

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)?

Show answer & explanation

Answer: 26

Answer

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}

.

Step-by-Step Solution

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.

Key Concept

Hydrostatic Pressure Calculation and Unit Conversion

Question 2496Question →

A city planner is reviewing a design where two straight, parallel avenues are intersected by a diagonal rail line. The measure of one obtuse angle at the first intersection is $(12x - 20)^\circ$ , and the measure of its corresponding angle at the second intersection is $(8x + 40)^\circ$ .

What is the measure of the acute angle formed by the intersection of the rail line and the avenues?

76^\circ

80^\circ

20^\circ

64^\circ

Show answer & explanation

Answer: $20^\circ$

Answer

The measure of the acute angle is

20^\circ

.

Because the avenues are parallel, their corresponding angles are equal. Setting the expressions equal gives

12x - 20 = 8x + 40

. Subtracting

8x

and adding

20

to both sides results in

4x = 60

, which simplifies to

x = 15

. Substituting

15

back into either expression yields the obtuse angle measure:

12(15) - 20 = 160^\circ

. The question asks for the adjacent acute angle, which forms a linear pair with the obtuse angle. Subtracting from

180^\circ

gives

180^\circ - 160^\circ = 20^\circ

.

Step-by-Step Solution

1

Set the expressions for the two corresponding angles equal to each other.

12x - 20 = 8x + 40

When two parallel lines are intersected by a transversal, corresponding angles are equal in measure.

2

Solve the algebraic equation for

x

.

4x = 60 \Rightarrow x = 15

Subtracting

8x

from both sides and adding

20

to both sides isolates the variable

x

.

3

Substitute

x = 15

back into either expression to find the measure of the obtuse angle.

12(15) - 20 = 180 - 20 = 160^\circ

The expressions represent the obtuse corresponding angles, so plugging in the value of

x

gives the actual angle measure.

4

Subtract the obtuse angle from

180^\circ

to find the measure of the adjacent acute angle.

180^\circ - 160^\circ = 20^\circ

Angles on a straight line form a linear pair and are supplementary, meaning their sum must be

180^\circ

.

Key Concept

Corresponding angles on parallel lines are equal, and adjacent angles on a straight line are supplementary (sum to

180^\circ

).

Question 2497Question →

A mechanic uses a manual screwdriver to remove a tightly secured machine screw. The handle of the screwdriver has a diameter of $1.5 \text{ inches}$ , and the metal shaft has a diameter of $0.25 \text{ inches}$ . If the mechanic applies a rotational effort force of $12 \text{ pounds}$ to the outer edge of the handle, how much output force is exerted by the shaft onto the screw? (Assume $100\%$ efficiency.)

72 \text{ pounds}

36 \text{ pounds}

18 \text{ pounds}

2 \text{ pounds}

Show answer & explanation

Answer: $72 \text{ pounds}$

Answer

72 \text{ pounds}

In a wheel and axle system where effort is applied to the wheel (the handle), the force is multiplied by the ratio of the wheel's diameter to the axle's diameter. The ratio of

1.5 \text{ inches}

to

0.25 \text{ inches}

provides a mechanical advantage of

6

. Multiplying the

12 \text{ pounds}

of effort by

6

yields an output force of

72 \text{ pounds}

.

Step-by-Step Solution

1

Identify the components of the wheel and axle system.

The handle acts as the wheel (Diameter =

1.5 \text{ in}

), and the shaft acts as the axle (Diameter =

0.25 \text{ in}

).

Establishing which component is the wheel and which is the axle determines how force is multiplied.

2

Calculate the Ideal Mechanical Advantage (IMA).

IMA = \frac{1.5}{0.25} = 6

The IMA for a wheel and axle is found by dividing the diameter of the wheel by the diameter of the axle.

3

Calculate the total output force applied to the screw.

\text{Output Force} = \text{Input Force} \times IMA = 12 \text{ lbs} \times 6 = 72 \text{ lbs}

.

The system multiplies the input effort force by the mechanical advantage to produce the final output force.

Key Concept

Ideal Mechanical Advantage of a Wheel and Axle

Question 2498Question →

A manual water pump features a straight, rigid handle that pivots on a fixed hinge at one end. A vertical piston rod, which provides the resistance, is attached to the handle $0.20 \text{ meters}$ from the hinge. A user pushes down on the opposite end of the handle, located $1.00 \text{ meter}$ from the hinge. If the piston rod requires $400 \text{ N}$ of force to operate, what is the minimum effort force the user must apply, and what class of lever does this mechanism represent?

100 \text{ N}

; First-class lever

80 \text{ N}

; Third-class lever

80 \text{ N}

; Second-class lever

2000 \text{ N}

; Second-class lever

Show answer & explanation

Answer: $80 \text{ N}$ ; Second-class lever

Answer

80 \text{ N}

; Second-class lever

In this mechanism, the fulcrum (hinge) is at one end, the effort is applied at the opposite end, and the resistance (load) is located between them. This arrangement defines a second-class lever. The effort arm is the total distance from the fulcrum to the effort (

1.00 \text{ m}

), while the load arm is the distance from the fulcrum to the resistance (

0.20 \text{ m}

). The mechanical advantage (MA) is found by dividing the effort arm by the load arm (

1.00 / 0.20 = 5

). To find the required effort force, the load is divided by the MA (

400 \text{ N} / 5 = 80 \text{ N}

).

Step-by-Step Solution

1

Identify the fulcrum, effort, and load positions to determine the lever class.

The fulcrum is at one end (hinge), the effort is at the opposite end, and the load is located between them.

This specific physical arrangement defines a second-class lever.

2

Determine the lengths of the effort arm and load arm.

Effort arm =

1.00 \text{ m}

; Load arm =

0.20 \text{ m}

.

Lever arms are strictly measured from the fulcrum to the respective points where the force and resistance are applied.

3

Calculate the Mechanical Advantage (MA).

\text{MA} = 1.00 \text{ m} / 0.20 \text{ m} = 5

.

The MA of a lever is the ratio of the effort arm to the load arm.

4

Calculate the required effort force.

400 \text{ N} / 5 = 80 \text{ N}

.

The required effort is the load force divided by the mechanical advantage.

Key Concept

Second-Class Levers and Mechanical Advantage

Alternative Method

Instead of using the Mechanical Advantage (MA) ratio, the static equilibrium torque equation can be used directly: (Effort Force) × (Distance from Fulcrum to Effort) = (Load Force) × (Distance from Fulcrum to Load). This gives: Effort × 1.00 = 400 × 0.20, solving directly to Effort = 80 N.

Estimated Time:1m 0s

Question 2499Question →

An aircraft carrier's munitions elevator is used to transport heavy ordnance to the flight deck. The elevator starts from a complete stop and accelerates upward at a constant rate of $0.5 \text{ m/s}^2$ . How many seconds will it take for the elevator to reach its maximum operational upward velocity of $120 \text{ meters per minute}$ ?

4

240

60

1

Show answer & explanation

Answer: 4

Answer

4

To find the elapsed time, the units must first be standardized. Since acceleration is given in meters per second squared (

0.5 \text{ m/s}^2

), the target velocity must be converted from meters per minute to meters per second by dividing by 60. This gives a final velocity of

2 \text{ m/s}

. Using the formula for time (

t = \frac{\Delta v}{a}

), dividing

2 \text{ m/s}

by

0.5 \text{ m/s}^2

results in exactly 4 seconds.

Step-by-Step Solution

1

Identify the given variables and the unknown.

Initial velocity (

v_i

) =

0 \text{ m/s}

. Acceleration (

a

) =

0.5 \text{ m/s}^2

. Final velocity (

v_f

) =

120 \text{ m/min}

. Unknown is time (

t

).

Establishing the known kinematic values is the first step in solving acceleration problems.

2

Convert the final velocity into standard units (meters per second) to match the acceleration unit.

120 \text{ m/min} \div 60 \text{ s/min} = 2 \text{ m/s}

.

All units must be consistent before applying mathematical formulas.

3

Apply the acceleration formula

a = \frac{v_f - v_i}{t}

and solve for

t

.

0.5 = \frac{2 - 0}{t}

, which rearranges to

t = \frac{2}{0.5} = 4 \text{ seconds}

.

This calculates the elapsed time required to achieve the desired change in velocity.

Key Concept

Kinematic equations and unit conversion for acceleration.

Estimated Time:1m 0s

Question 2500Question →

A circular metal plate has a total area of $100\pi \text{ cm}^2$ . A smaller circular hole is drilled completely out of its exact center. If the area of the remaining metal plate is $64\pi \text{ cm}^2$ , what is the radius of the drilled hole?

6 \text{ cm}

12 \text{ cm}

18 \text{ cm}

36 \text{ cm}

Show answer & explanation

Answer: $6 \text{ cm}$

Answer

The radius of the drilled hole is

6 \text{ cm}

.

The area of the removed hole is the difference between the original total area and the remaining area, which is

100\pi - 64\pi = 36\pi \text{ cm}^2

. Using the circle area formula (

A = \pi r^2

), setting

\pi r^2 = 36\pi

yields

r^2 = 36

. Taking the square root of both sides confirms the radius is

6 \text{ cm}

.

Step-by-Step Solution

1

Find the area of the drilled hole by subtracting the remaining area from the total area.

100\pi - 64\pi = 36\pi \text{ cm}^2

.

The drilled hole accounts for the area that was removed from the total plate.

2

Set up the area formula for the drilled hole to solve for the unknown radius.

A = \pi r^2 \rightarrow 36\pi = \pi r^2

.

The area of a circle is calculated using the formula

A = \pi r^2

.

3

Solve the equation for the radius

r

.

r^2 = 36

, so

r = 6 \text{ cm}

.

Dividing both sides by

\pi

leaves

r^2 = 36

, and taking the square root of both sides gives the radius.

Key Concept

Perimeter and Area of 2D Figures

Estimated Time:1m 0s

Answer

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