ASVAB Mechanical Comprehension (MC) Practice Test 2026

ASVAB Mechanical Comprehension (MC) Practice Test (2026) covers ASVAB Mechanical Comprehension (MC) practice test in a four-choice format with a “Show Answer” toggle for review. Each item includes the correct answer, a concise explanation, and official citations where applicable so you can verify what the test is targeting. Use it to practice forces, machines, fluids, heat, and everyday mechanical principles.

FAQ

What should I know for ASVAB Mechanical Comprehension (MC) on the ASVAB?

Focus on the recurring concepts that appear in many forms, not one-off trivia. Learn the key terms, practice mixed sets, and use explanations to build a quick recognition habit so you don’t overthink simple items.

Where can I take an ASVAB Mechanical Comprehension (MC) practice test?

Use a practice set that matches real wording and keeps you answering before you peek. After each question, open “Show Answer,” compare your reasoning to the explanation, and retake missed concepts later to confirm retention.

What should I know for ASVAB Mechanical Comprehension (MC) on the ASVAB (part 2)?

What should an ASVAB Mechanical Comprehension (MC) study guide cover?

A solid guide focuses on the core terms, common situations, and definitions that show up repeatedly. Pair reading with practice runs, and use explanations to learn why wrong options are wrong, not just which option is right.

What should an ASVAB Mechanical Comprehension (MC) study guide cover (part 2)?

ASVAB Mechanical Comprehension MC Practice Test

Q001. A 4 ft pry bar is used as a first-class lever. The fulcrum is 1 ft from the load. If you push down on the opposite end, what is the mechanical advantage (ideal, ignoring friction)?

Correct Answer: C
Explanation: Mechanical advantage for a lever equals effort arm ÷ load arm. Effort arm is 3 ft (from fulcrum to effort), load arm is 1 ft, so MA = 3 ÷ 1 = 3.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q002. A hydraulic jack has an input piston area of 2 cm² and an output piston area of 50 cm². If the input force is 120 N, what ideal output force is produced?

Correct Answer: C
Explanation: Pressure is transmitted equally in a closed fluid. Output force scales with area: 120 N × (50 ÷ 2) = 120 × 25 = 3,000 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (hydraulics).

Q003. Two identical gears mesh. Gear A rotates clockwise at 60 rpm. Gear B rotates:

Correct Answer: B
Explanation: Meshed gears spin in opposite directions. Identical gears have the same speed ratio, so Gear B is 60 rpm counterclockwise.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear direction and speed ratio.

Q004. A block is pulled at constant speed across a level floor with a horizontal force of 40 lb. What does that imply about the kinetic friction force on the block?

Correct Answer: C
Explanation: Constant speed means net force is zero. The friction must balance the applied 40 lb force in the opposite direction.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Newton’s first law (equilibrium of forces).

Q005. A rope passes over a frictionless fixed pulley to lift a 200 lb load. What ideal effort force is required to lift it steadily?

Correct Answer: B
Explanation: A fixed pulley changes direction of force but does not multiply it. Ideal mechanical advantage is 1, so effort equals load: 200 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage (fixed pulley).

Q006. A 10 lb object is at rest on a 30° incline. Ignoring friction, what component of the object’s weight acts parallel to the incline pulling it downhill?

Correct Answer: A
Explanation: Parallel component is W·sin(θ). With W = 10 lb and θ = 30°, sin(30°) = 0.5, so 10 × 0.5 = 5 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: resolving forces on an incline.

Q007. A steel rod and an aluminum rod have the same length at room temperature. When heated equally, which statement is most accurate?

Correct Answer: B
Explanation: Thermal expansion depends on coefficient of linear expansion. Aluminum’s coefficient is larger than steel’s, so it expands more for the same temperature rise.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: thermal expansion of solids.

Q008. A 6 kg box is accelerated at 2 m/s² on a frictionless surface. What net force is required?

Correct Answer: C
Explanation: Newton’s second law: F = m·a = 6 × 2 = 12 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Newton’s second law.

Q009. A wheel and axle has a wheel radius of 12 inches and an axle radius of 3 inches. What is the ideal mechanical advantage?

Correct Answer: C
Explanation: Ideal MA for a wheel and axle is wheel radius ÷ axle radius. 12 ÷ 3 = 4.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: wheel-and-axle mechanical advantage.

Q010. A liquid is flowing through a pipe that narrows to half the original diameter. If the flow is steady and the liquid is incompressible, what happens to the fluid speed in the narrow section?

Correct Answer: D
Explanation: For incompressible flow, continuity applies: A₁v₁ = A₂v₂. Area scales with diameter squared, so halving diameter makes area 1/4, requiring speed to be 4×.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: continuity equation (fluid flow).

Q011. A bolt is tightened with a wrench. Which is most directly increased by using a longer wrench handle with the same applied force?

Correct Answer: B
Explanation: Torque equals force × perpendicular distance from pivot. A longer handle increases the moment arm, increasing torque for the same force.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque and moment arm.

Q012. A car’s brakes convert kinetic energy mostly into:

Correct Answer: C
Explanation: Friction in braking converts motion (kinetic energy) into heat (thermal energy) in pads and rotors.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: energy conversion (friction).

Q013. A 20 lb load is lifted with a single movable pulley (ideal, frictionless). What effort force is required to lift the load steadily?

Correct Answer: B
Explanation: A single movable pulley has ideal mechanical advantage 2 because two rope segments support the load. Effort = 20 ÷ 2 = 10 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage (movable pulley).

Q014. Two springs are connected in parallel. Compared with one spring, the combined spring constant is:

Correct Answer: C
Explanation: Parallel springs share the load and their spring constants add: k_total = k₁ + k₂, so two identical springs give 2k.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring constants in parallel.

Q015. A 1,500 lb car rests on four identical tires. If the weight is evenly distributed, approximately how much normal force acts on each tire?

Correct Answer: A
Explanation: Even distribution means each tire supports one-fourth the weight: 1,500 ÷ 4 = 375 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: normal force and load distribution.

Q016. A metal object floats in mercury but sinks in water. Which statement best explains this?

Correct Answer: B
Explanation: An object floats if it is less dense than the fluid. Floating in mercury but sinking in water means its density lies between the two fluids.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: buoyancy and density.

Q017. A fan blade spinning faster does work on the air mainly by increasing the air’s:

Correct Answer: C
Explanation: The fan accelerates air particles, increasing their speed and therefore kinetic energy.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work and kinetic energy.

Q018. A beam is supported at both ends and a load is placed exactly in the center. In an ideal symmetric case, the reactions at the supports are:

Correct Answer: B
Explanation: With a centered load and symmetric supports, the load is shared equally: each support carries half the load (idealized).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: static equilibrium (balanced supports).

Q019. A 12 V battery supplies 3 A to a resistive load. What is the resistance of the load?

Correct Answer: B
Explanation: Ohm’s law: R = V ÷ I = 12 ÷ 3 = 4 Ω.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Ohm’s law (basic electrical relationships).

Q020. In an internal combustion engine, which stroke compresses the air-fuel mixture?

Correct Answer: B
Explanation: The compression stroke moves the piston upward with valves closed, compressing the mixture before ignition.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: four-stroke engine cycle.

Q021. A machine uses 500 J of input energy to lift a load, but only 400 J becomes useful output. What is the efficiency?

Correct Answer: B
Explanation: Efficiency = useful output ÷ input = 400 ÷ 500 = 0.8 = 80%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: machine efficiency.

Q022. A car rounds a curve of fixed radius at a higher speed. What happens to the required centripetal force?

Correct Answer: D
Explanation: Centripetal force is F = m·v²/r. With the same mass and radius, increasing speed increases force with v².
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force relationship.

Q023. A heat engine is redesigned so less heat is lost to the surroundings. All else equal, what happens to its efficiency?

Correct Answer: B
Explanation: Reducing waste heat means a larger fraction of input heat becomes useful work, so efficiency increases.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: energy conservation and heat-engine efficiency.

Q024. A pulley system has an ideal mechanical advantage of 4. If the load is 240 lb, what ideal effort is needed to lift it steadily?

Correct Answer: A
Explanation: Effort equals load ÷ MA. 240 ÷ 4 = 60 lb (ideal).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: mechanical advantage (pulleys).

Q025. Two different metals are connected end-to-end and heated at one end (a bimetal strip). If one metal expands more than the other, the strip will:

Correct Answer: C
Explanation: The side that expands more becomes effectively longer, so the strip bends toward the side that expands less (shorter side on the inside of the curve).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: differential thermal expansion (bimetal bending).

Q026. A 300 lb load is lifted by an ideal system with a mechanical advantage of 6. About how much input force is needed to lift it steadily?

Correct Answer: B
Explanation: Ideal effort equals load ÷ MA. 300 ÷ 6 = 50 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: mechanical advantage (ideal machines).

Q027. A ramp raises a crate 3 ft using a ramp length of 12 ft (ignore friction). What is the ideal mechanical advantage of the ramp?

Correct Answer: C
Explanation: Ideal MA for an inclined plane is length ÷ height. 12 ÷ 3 = 4.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: inclined plane mechanical advantage.

Q028. Two gears mesh: Gear A has 12 teeth and drives Gear B with 36 teeth. If Gear A turns at 90 rpm, Gear B turns at:

Correct Answer: A
Explanation: Speed is inversely proportional to teeth count. 90 × (12 ÷ 36) = 30 rpm (opposite direction).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q029. A wrench applies 25 lb of force perpendicular to a bolt at a distance of 8 inches from the center. What torque is applied (in lb·in)?

Correct Answer: C
Explanation: Torque = force × lever arm = 25 × 8 = 200 lb·in.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque calculation.

Q030. A 5 kg object moving at 4 m/s is brought to rest. How much kinetic energy is removed?

Correct Answer: C
Explanation: Kinetic energy = ½mv² = 0.5×5×(4²)=2.5×16=40 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic energy.

Q031. A motor does 900 J of work in 3 seconds. What is the power output?

Correct Answer: B
Explanation: Power = work ÷ time = 900 ÷ 3 = 300 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: power (work rate).

Q032. In an ideal hydraulic system, the input piston area is 4 in² and the output piston area is 32 in². If input force is 40 lb, output force is:

Correct Answer: C
Explanation: Force scales with area: 40 × (32 ÷ 4) = 40 × 8 = 320 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle.

Q033. A 2.0 kg object has a volume of 1.0 liter. Will it float in water (1.0 kg/L)?

Correct Answer: B
Explanation: Density = 2.0 kg ÷ 1.0 L = 2.0 kg/L, which is greater than water, so it sinks.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: density and buoyancy.

Q034. A uniform 10 ft beam weighs 40 lb and is supported at one end like a cantilever. Where does the beam’s weight act for torque calculations?

Correct Answer: C
Explanation: For a uniform beam, its weight acts at the center of mass, halfway along its length.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: center of mass (uniform objects).

Q035. A 100 lb crate requires 30 lb of horizontal force to slide at constant speed on a level floor. The kinetic friction force is:

Correct Answer: B
Explanation: Constant speed means net force is zero, so friction balances the applied 30 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: friction and equilibrium.

Q036. Two identical springs each have k = 200 N/m and are connected in series. The equivalent spring constant is:

Correct Answer: A
Explanation: For identical springs in series, k_eq = k/2. 200/2 = 100 N/m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: springs in series.

Q037. A device draws 2.5 A from a 12 V source. The electrical power used is:

Correct Answer: C
Explanation: Power P = V×I = 12×2.5 = 30 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: electrical power.

Q038. Two rods of equal length and cross-section are used as heat conductors, one copper and one wood. Which transfers heat faster?

Correct Answer: B
Explanation: Copper’s thermal conductivity is much higher, so it conducts heat faster.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: thermal conductivity.

Q039. Water pressure at a certain depth depends most directly on:

Correct Answer: C
Explanation: Hydrostatic pressure increases with depth (p = ρgh), not container shape.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: fluid pressure with depth.

Q040. A block-and-tackle has 4 supporting rope segments holding the load (ideal). If you pull 8 ft of rope, the load rises:

Correct Answer: C
Explanation: Distance tradeoff equals MA. With 4 supporting segments, MA = 4, so load rises 8/4 = 2 ft.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work conservation in pulleys.

Q041. A wheel-and-axle has a wheel radius of 10 cm and axle radius of 2 cm. Ideal mechanical advantage is:

Correct Answer: C
Explanation: MA = wheel radius ÷ axle radius = 10 ÷ 2 = 5.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: wheel-and-axle MA.

Q042. A spring with k = 300 N/m is stretched 0.20 m. The spring force is:

Correct Answer: A
Explanation: Hooke’s law: F = kx = 300×0.20 = 60 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Hooke’s law.

Q043. A fluid’s speed increases as it flows through a narrower section of pipe. In general, the static pressure in that narrow section tends to:

Correct Answer: B
Explanation: For steady flow, higher speed is commonly associated with lower static pressure (Bernoulli effect).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (qualitative).

Q044. Why do ball bearings reduce friction in rotating shafts?

Correct Answer: B
Explanation: Rolling friction is typically much less than sliding friction, so bearings reduce resistive torque.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: friction types (rolling vs sliding).

Q045. A 12 ft plank is used as a lever with the fulcrum 3 ft from the load end. Effort is applied at the far end. Ideal mechanical advantage is:

Correct Answer: C
Explanation: Effort arm is 9 ft and load arm is 3 ft, so MA = 9 ÷ 3 = 3.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q046. A machine has an ideal mechanical advantage of 10 but an actual mechanical advantage of 8. The efficiency is:

Correct Answer: A
Explanation: Efficiency ≈ AMA ÷ IMA = 8 ÷ 10 = 0.8 = 80%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: machine efficiency.

Q047. A 60 N force pushes a box 5 m in the direction of motion. How much work is done?

Correct Answer: C
Explanation: Work = F×d = 60×5 = 300 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work.

Q048. A screw has a pitch of 0.25 in per turn. If you turn it 8 turns, it advances about:

Correct Answer: C
Explanation: Advance = pitch × turns = 0.25×8 = 2 inches.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: screw motion (pitch).

Q049. A third-class lever is best described as one where the effort is located:

Correct Answer: A
Explanation: In a third-class lever, the effort is between fulcrum and load, giving speed advantage but MA < 1.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever classes.

Q050. A 2 m long wrench is used with a 100 N force applied at a 30° angle to the wrench (relative to the handle). What is the torque magnitude?

Correct Answer: B
Explanation: Only the perpendicular component produces torque: τ = rF sinθ = 2×100×sin30° = 200×0.5 = 100 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque with angle.

Q051. A chain drive uses a 10-tooth front sprocket driving a 30-tooth rear sprocket. If the front rotates at 120 rpm, the rear rotates at about:

Correct Answer: A
Explanation: Speed ratio is inverse of teeth: 120×(10/30)=40 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: sprocket/gear ratio.

Q052. If a car’s speed doubles, the stopping distance from braking (all else equal) tends to:

Correct Answer: B
Explanation: Braking work must remove kinetic energy, which scales with v², so distance tends to scale with v².
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic energy and braking work.

Q053. A 200 lb crate is pushed up a frictionless ramp 10 ft long to a height of 2 ft. The ideal input force along the ramp is:

Correct Answer: B
Explanation: Force = load × (height/length) = 200×(2/10)=200×0.2=40 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: inclined plane (work conservation).

Q054. Which mode of heat transfer requires bulk movement of a fluid?

Correct Answer: B
Explanation: Convection transfers heat via moving fluid (air/water) carrying energy.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: heat transfer modes.

Q055. Expansion joints are used in bridges primarily to:

Correct Answer: B
Explanation: Materials expand/contract with temperature; joints allow movement to avoid stress damage.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: thermal expansion in structures.

Q056. A fluid in a container exerts pressure on the bottom because of:

Correct Answer: B
Explanation: Pressure at depth comes from the weight of fluid above: p = ρgh.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure.

Q057. A pulley arrangement supports a load with 3 rope segments (ideal). A 150 lb load would require about what effort to lift steadily?

Correct Answer: A
Explanation: With 3 supporting segments, MA = 3, so effort = 150 ÷ 3 = 50 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q058. A 30 lb force is applied to a 1.5 ft long screwdriver handle used as a wheel-and-axle with an axle radius of 0.25 in. Which best describes the effect?

Correct Answer: A
Explanation: A large handle radius compared with small axle radius provides substantial torque multiplication.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: wheel-and-axle torque multiplication.

Q059. A wedge is essentially two inclined planes back-to-back. Increasing wedge length while keeping thickness the same will generally:

Correct Answer: B
Explanation: A longer wedge has a smaller slope, increasing ideal mechanical advantage.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: wedge as inclined plane.

Q060. A 2,000 N load is lifted by a jack with efficiency 70%. If the ideal required input is 200 N, the actual input is closest to:

Correct Answer: C
Explanation: Efficiency = ideal/actual for same output. Actual ≈ ideal ÷ 0.70 = 200/0.70 ≈ 286 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency calculation.

Q061. Two identical resistors are connected in parallel. Compared with one resistor, the equivalent resistance is:

Correct Answer: A
Explanation: For two equal resistors in parallel, R_eq = R/2.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: parallel resistance.

Q062. A 6 Ω and a 3 Ω resistor are in series across 18 V. The current is:

Correct Answer: B
Explanation: Series resistance is 6+3=9 Ω. Current I = V/R = 18/9=2 A.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Ohm’s law (series circuit).

Q063. A flywheel is added to a machine. Its main benefit is that it:

Correct Answer: B
Explanation: A flywheel’s inertia stores energy and resists speed changes, smoothing motion.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational inertia.

Q064. A 20 N force is applied at the end of a 0.5 m lever, but only 60% of the force is perpendicular. The torque is:

Correct Answer: B
Explanation: Perpendicular component is 0.60×20=12 N. Torque = 12×0.5 = 6 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular component.

Q065. A solid disk and a solid ring of equal mass and radius roll down the same incline without slipping. Which reaches the bottom first?

Correct Answer: A
Explanation: The disk has smaller rotational inertia than a ring, so more energy goes to translational speed, arriving first.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational inertia and rolling motion.

Q066. A 400 lb load is supported by a hydraulic lift with output piston area 20 in². What is the fluid pressure (psi) in the system (ideal)?

Correct Answer: B
Explanation: Pressure equals force divided by area: 400 ÷ 20 = 20 psi.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pressure (force/area).

Q067. A 600 lb load rests on a base plate area of 30 in². The pressure on the surface is about:

Correct Answer: B
Explanation: Pressure = 600 ÷ 30 = 20 psi.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pressure (force/area).

Q068. A motor delivers 746 W of power. This is approximately:

Correct Answer: B
Explanation: 1 horsepower is defined as about 746 watts.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: power unit conversion.

Q069. An idler gear is inserted between two meshing gears. Compared to a direct mesh, the idler gear causes the driven gear to rotate:

Correct Answer: A
Explanation: An odd number of gear meshes reverses direction; adding an idler makes two meshes, so driver and driven rotate the same direction.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear direction with idler.

Q070. A wheel has radius 0.30 m and rotates at 10 rad/s. The linear speed at the rim is:

Correct Answer: A
Explanation: Linear speed v = rω = 0.30×10 = 3 m/s.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: angular to linear speed.

Q071. A 2,000 N force acts for 0.05 s on an object. The impulse delivered is:

Correct Answer: B
Explanation: Impulse equals force times time: 2,000×0.05 = 100 N·s.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse-momentum.

Q072. A steel cable is loaded in tension. “Yield strength” refers to the stress at which the cable:

Correct Answer: B
Explanation: Yield strength is where deformation transitions from elastic to plastic (permanent).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress–strain (material behavior).

Q073. In bending a beam, the “neutral axis” is the region where the material experiences:

Correct Answer: C
Explanation: At the neutral axis, bending stress changes sign and is zero.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: beam bending basics.

Q074. A 10 lb object displaces 0.12 ft³ of water (62.4 lb/ft³). The buoyant force is closest to:

Correct Answer: B
Explanation: Buoyant force equals weight of displaced fluid: 62.4×0.12 ≈ 7.49 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle.

Q075. A fluid is pushed through a pipe that narrows so cross-sectional area is reduced to one-half. For steady incompressible flow, speed in the narrow section becomes:

Correct Answer: C
Explanation: Continuity: A₁v₁ = A₂v₂. If area halves, speed doubles.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: continuity equation.

Q076. A refrigerator removes heat from its interior primarily by causing a fluid to:

Correct Answer: B
Explanation: Evaporation of refrigerant absorbs heat from the inside (latent heat).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: phase change and heat transfer.

Q077. A block-and-tackle has 5 supporting rope segments (ideal). A 250 lb load requires about what effort?

Correct Answer: B
Explanation: Effort = 250 ÷ 5 = 50 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q078. A 120 lb crate is on a ramp at 20° with coefficient of kinetic friction 0.20. Which is closest to the friction force while sliding?

Correct Answer: B
Explanation: Normal force ≈ W cos20° ≈ 120×0.94 ≈ 113 lb. Friction = μN ≈ 0.20×113 ≈ 23 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: friction on an incline.

Q079. A 10 kg cart is pulled with 30 N on a level surface. If friction is 10 N, the acceleration is:

Correct Answer: B
Explanation: Net force = 30−10=20 N. a = F/m = 20/10 = 2 m/s².
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Newton’s second law with friction.

Q080. A car moves around a curve of radius r. If speed triples, the required centripetal acceleration becomes:

Correct Answer: C
Explanation: Centripetal acceleration a = v²/r. Tripling v makes a 9 times larger.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal acceleration.

Q081. A 2 kg object slides down a frictionless 5 m high hill. Its speed at the bottom is closest to:

Correct Answer: B
Explanation: Potential energy mgh becomes kinetic: v = √(2gh)=√(2×9.8×5)=√98≈9.9 m/s.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: energy conservation.

Q082. A bolt requires 60 lb·ft of torque to loosen. Using a 2 ft wrench, the minimum perpendicular force needed is:

Correct Answer: B
Explanation: τ = F r ⇒ F = τ/r = 60/2 = 30 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque and lever arm.

Q083. A bicycle uses a 48-tooth chainring driving a 16-tooth rear cog. If the rider turns pedals at 60 rpm, the rear wheel sprocket rotates at:

Correct Answer: C
Explanation: Speed ratio = 48/16 = 3. Driven cog rotates 3×: 60×3=180 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: sprocket ratio.

Q084. A hydraulic lift has input area 3 in² and output area 24 in². If the input piston moves 8 inches, the output piston moves (ideal):

Correct Answer: A
Explanation: Volume conserved: A_in d_in = A_out d_out. d_out = 3×8 / 24 = 1 inch.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydraulic displacement (volume conservation).

Q085. A 50 lb box is lifted vertically 6 ft. The increase in gravitational potential energy is:

Correct Answer: C
Explanation: Potential energy increase = weight × height = 50×6 = 300 ft·lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity.

Q086. A 1,000 kg car moving 15 m/s is stopped by brakes doing work. The work done by brakes is closest to:

Correct Answer: C
Explanation: Work removed equals kinetic energy: ½mv² = 0.5×1000×225 = 112,500 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic energy and work.

Q087. A spring is compressed 0.10 m and stores 12 J of energy. The spring constant k is closest to:

Correct Answer: C
Explanation: Energy U = ½kx² ⇒ k = 2U/x² = 24 / 0.01 = 2,400 N/m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy.

Q088. A beam is in static equilibrium. Which condition must be true in addition to net force being zero?

Correct Answer: A
Explanation: For no rotation, the sum of torques about any point must be zero.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: static equilibrium (torque balance).

Q089. A 1,600 lb vehicle weight is evenly distributed on four tires. If each tire’s contact patch is 20 in², the average pressure under each tire is about:

Correct Answer: B
Explanation: Each tire supports 400 lb. Pressure = 400 ÷ 20 = 20 psi.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pressure (force/area).

Q090. Which loading type tends to cause layers of material to slide past each other?

Correct Answer: C
Explanation: Shear stress acts parallel to a surface, promoting sliding between layers.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: types of stress.

Q091. A first-class lever has the fulcrum between effort and load. If the fulcrum is moved closer to the load, the lever’s mechanical advantage:

Correct Answer: B
Explanation: Moving fulcrum toward the load increases effort arm relative to load arm, increasing MA.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever moment arms.

Q092. An ideal pulley system has a mechanical advantage of 8. If you want to lift a load 3 ft, you must pull approximately:

Correct Answer: C
Explanation: In an ideal system, input distance = MA × output distance. 8×3 = 24 ft.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work conservation in machines.

Q093. A machine outputs 520 J of useful work with 65% efficiency. The required input energy is closest to:

Correct Answer: C
Explanation: Efficiency = output/input ⇒ input = 520/0.65 ≈ 800 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (energy).

Q094. A screw jack has a handle radius of 10 inches and a screw pitch of 0.25 inch per turn (ideal). Its ideal mechanical advantage is closest to:

Correct Answer: C
Explanation: IMA ≈ (2πr)/pitch = (2π×10)/0.25 ≈ (62.8)/0.25 ≈ 251.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: screw mechanical advantage.

Q095. A block-and-tackle lifts a load 2 ft while you pull 16 ft of rope (ideal). The mechanical advantage is:

Correct Answer: C
Explanation: MA = input distance ÷ output distance = 16 ÷ 2 = 8.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: mechanical advantage from distance ratio.

Q096. Three gears are in a line: A meshes with B, and B meshes with C. If A rotates clockwise, C rotates:

Correct Answer: B
Explanation: Each mesh reverses direction. Two reversals (A→B, B→C) make C rotate opposite A.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear direction (multiple meshes).

Q097. A single movable pulley attached to a load is combined with a fixed pulley above it (ideal). The primary benefit of the fixed pulley is to:

Correct Answer: B
Explanation: The movable pulley provides MA; the fixed pulley mainly redirects the effort direction.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley function (fixed vs movable).

Q098. A copper rod and a stainless-steel rod have the same dimensions. Which will generally conduct heat faster?

Correct Answer: B
Explanation: Copper has higher thermal conductivity than stainless steel, so heat flows faster.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: thermal conductivity (materials).

Q099. A 200 lb crate on a level floor does not move until the horizontal pull reaches 70 lb. This 70 lb represents the:

Correct Answer: B
Explanation: Static friction adjusts up to a maximum; the threshold pull equals the maximum static friction.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: static friction threshold.

Q100. An ideal hydraulic press has input area 2 cm² and output area 50 cm². If the output piston rises 1 cm, the input piston must move about:

Correct Answer: C
Explanation: Volume conserved: A_in d_in = A_out d_out ⇒ d_in = (50×1)/2 = 25 cm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydraulic displacement (volume conservation).

Q101. A 8 kg toolbox is lifted vertically 3 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 8×9.8×3 ≈ 235.2 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q102. A 9 kg toolbox is lifted vertically 4 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 9×9.8×4 ≈ 352.8 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q103. A 10 kg toolbox is lifted vertically 5 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 10×9.8×5 ≈ 490.0 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q104. A 11 kg toolbox is lifted vertically 3 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 11×9.8×3 ≈ 323.4 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q105. A 12 kg toolbox is lifted vertically 4 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 12×9.8×4 ≈ 470.4 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q106. A 13 kg toolbox is lifted vertically 5 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 13×9.8×5 ≈ 637.0 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q107. A 14 kg toolbox is lifted vertically 3 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 14×9.8×3 ≈ 411.6 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q108. A 15 kg toolbox is lifted vertically 4 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 15×9.8×4 ≈ 588.0 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q109. A 16 kg toolbox is lifted vertically 5 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 16×9.8×5 ≈ 784.0 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q110. A 17 kg toolbox is lifted vertically 3 m at constant speed. Approximately how much work is done against gravity?

Correct Answer: B
Explanation: Work against gravity equals W = mgh = 17×9.8×3 ≈ 499.8 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: work against gravity (mgh).

Q111. A pry bar acts as a first-class lever. The effort arm is 18 in and the load arm is 3 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 18/3 ≈ 6.00.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q112. A pry bar acts as a first-class lever. The effort arm is 17 in and the load arm is 4 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 17/4 ≈ 4.25.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q113. A pry bar acts as a first-class lever. The effort arm is 16 in and the load arm is 5 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 16/5 ≈ 3.20.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q114. A pry bar acts as a first-class lever. The effort arm is 15 in and the load arm is 6 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 15/6 ≈ 2.50.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q115. A pry bar acts as a first-class lever. The effort arm is 14 in and the load arm is 3 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 14/3 ≈ 4.67.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q116. A pry bar acts as a first-class lever. The effort arm is 13 in and the load arm is 4 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 13/4 ≈ 3.25.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q117. A pry bar acts as a first-class lever. The effort arm is 12 in and the load arm is 5 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 12/5 ≈ 2.40.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q118. A pry bar acts as a first-class lever. The effort arm is 11 in and the load arm is 6 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 11/6 ≈ 1.83.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q119. A pry bar acts as a first-class lever. The effort arm is 10 in and the load arm is 3 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 10/3 ≈ 3.33.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q120. A pry bar acts as a first-class lever. The effort arm is 9 in and the load arm is 4 in. What is the ideal mechanical advantage?

Correct Answer: B
Explanation: Ideal MA = effort arm ÷ load arm = 9/4 ≈ 2.25.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: lever mechanical advantage (moment arms).

Q121. An ideal block-and-tackle supports a 120 lb load with 2 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (2), so effort = 120/2 ≈ 60.0 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q122. An ideal block-and-tackle supports a 130 lb load with 3 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (3), so effort = 130/3 ≈ 43.3 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q123. An ideal block-and-tackle supports a 140 lb load with 4 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (4), so effort = 140/4 ≈ 35.0 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q124. An ideal block-and-tackle supports a 150 lb load with 5 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (5), so effort = 150/5 ≈ 30.0 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q125. An ideal block-and-tackle supports a 160 lb load with 6 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (6), so effort = 160/6 ≈ 26.7 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q126. An ideal block-and-tackle supports a 170 lb load with 2 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (2), so effort = 170/2 ≈ 85.0 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q127. An ideal block-and-tackle supports a 180 lb load with 3 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (3), so effort = 180/3 ≈ 60.0 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q128. An ideal block-and-tackle supports a 190 lb load with 4 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (4), so effort = 190/4 ≈ 47.5 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q129. An ideal block-and-tackle supports a 200 lb load with 5 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (5), so effort = 200/5 ≈ 40.0 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q130. An ideal block-and-tackle supports a 210 lb load with 6 rope segments. About what steady effort force is required?

Correct Answer: A
Explanation: Ideal MA equals supporting segments (6), so effort = 210/6 ≈ 35.0 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: pulley mechanical advantage.

Q131. In an ideal hydraulic press, the input piston area is 2 cm² and the output piston area is 20 cm². If the input force is 80 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 80×(20/2) ≈ 800.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q132. In an ideal hydraulic press, the input piston area is 3 cm² and the output piston area is 25 cm². If the input force is 90 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 90×(25/3) ≈ 750.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q133. In an ideal hydraulic press, the input piston area is 4 cm² and the output piston area is 30 cm². If the input force is 100 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 100×(30/4) ≈ 750.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q134. In an ideal hydraulic press, the input piston area is 5 cm² and the output piston area is 35 cm². If the input force is 80 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 80×(35/5) ≈ 560.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q135. In an ideal hydraulic press, the input piston area is 2 cm² and the output piston area is 40 cm². If the input force is 90 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 90×(40/2) ≈ 1800.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q136. In an ideal hydraulic press, the input piston area is 3 cm² and the output piston area is 45 cm². If the input force is 100 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 100×(45/3) ≈ 1500.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q137. In an ideal hydraulic press, the input piston area is 4 cm² and the output piston area is 50 cm². If the input force is 80 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 80×(50/4) ≈ 1000.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q138. In an ideal hydraulic press, the input piston area is 5 cm² and the output piston area is 55 cm². If the input force is 90 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 90×(55/5) ≈ 990.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q139. In an ideal hydraulic press, the input piston area is 2 cm² and the output piston area is 60 cm². If the input force is 100 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 100×(60/2) ≈ 3000.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q140. In an ideal hydraulic press, the input piston area is 3 cm² and the output piston area is 65 cm². If the input force is 80 N, the output force is closest to:

Correct Answer: A
Explanation: F_out = F_in × (A_out/A_in) = 80×(65/3) ≈ 1733.3 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Pascal’s principle (force-area scaling).

Q141. A 150 lb crate slides on a ramp at 15° with coefficient of kinetic friction 0.15. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 150·cos(15°) ≈ 144.9 lb, so F_f ≈ 0.15×144.9 ≈ 21.7 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q142. A 155 lb crate slides on a ramp at 20° with coefficient of kinetic friction 0.16. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 155·cos(20°) ≈ 145.7 lb, so F_f ≈ 0.16×145.7 ≈ 23.3 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q143. A 160 lb crate slides on a ramp at 25° with coefficient of kinetic friction 0.17. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 160·cos(25°) ≈ 145.0 lb, so F_f ≈ 0.17×145.0 ≈ 24.7 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q144. A 165 lb crate slides on a ramp at 30° with coefficient of kinetic friction 0.18. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 165·cos(30°) ≈ 142.9 lb, so F_f ≈ 0.18×142.9 ≈ 25.7 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q145. A 170 lb crate slides on a ramp at 35° with coefficient of kinetic friction 0.19. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 170·cos(35°) ≈ 139.3 lb, so F_f ≈ 0.19×139.3 ≈ 26.5 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q146. A 175 lb crate slides on a ramp at 40° with coefficient of kinetic friction 0.15. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 175·cos(40°) ≈ 134.1 lb, so F_f ≈ 0.15×134.1 ≈ 20.1 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q147. A 180 lb crate slides on a ramp at 15° with coefficient of kinetic friction 0.16. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 180·cos(15°) ≈ 173.9 lb, so F_f ≈ 0.16×173.9 ≈ 27.8 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q148. A 185 lb crate slides on a ramp at 20° with coefficient of kinetic friction 0.17. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 185·cos(20°) ≈ 173.8 lb, so F_f ≈ 0.17×173.8 ≈ 29.6 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q149. A 190 lb crate slides on a ramp at 25° with coefficient of kinetic friction 0.18. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 190·cos(25°) ≈ 172.2 lb, so F_f ≈ 0.18×172.2 ≈ 31.0 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q150. A 195 lb crate slides on a ramp at 30° with coefficient of kinetic friction 0.19. Approximately how large is the friction force while sliding?

Correct Answer: A
Explanation: F_f = μ_kN with N = Wcosθ. N ≈ 195·cos(30°) ≈ 168.9 lb, so F_f ≈ 0.19×168.9 ≈ 32.1 lb.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: kinetic friction on an incline (μN).

Q151. A gear with 12 teeth drives a gear with 36 teeth. If the driving gear turns at 60 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 60×(12/36) ≈ 20.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q152. A gear with 16 teeth drives a gear with 38 teeth. If the driving gear turns at 70 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 70×(16/38) ≈ 29.5 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q153. A gear with 20 teeth drives a gear with 40 teeth. If the driving gear turns at 80 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 80×(20/40) ≈ 40.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q154. A gear with 24 teeth drives a gear with 42 teeth. If the driving gear turns at 90 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 90×(24/42) ≈ 51.4 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q155. A gear with 28 teeth drives a gear with 44 teeth. If the driving gear turns at 100 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 100×(28/44) ≈ 63.6 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q156. A gear with 12 teeth drives a gear with 46 teeth. If the driving gear turns at 110 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 110×(12/46) ≈ 28.7 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q157. A gear with 16 teeth drives a gear with 48 teeth. If the driving gear turns at 120 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 120×(16/48) ≈ 40.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q158. A gear with 20 teeth drives a gear with 50 teeth. If the driving gear turns at 130 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 130×(20/50) ≈ 52.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q159. A gear with 24 teeth drives a gear with 52 teeth. If the driving gear turns at 140 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 140×(24/52) ≈ 64.6 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q160. A gear with 28 teeth drives a gear with 54 teeth. If the driving gear turns at 150 rpm, the driven gear turns at about:

Correct Answer: A
Explanation: rpm_driven = rpm_driver × (teeth_driver/teeth_driven) = 150×(28/54) ≈ 77.8 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear ratio (teeth and speed).

Q161. A force of 120 N is applied at the end of a 0.4 m wrench at an angle of 20° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.4×120×sin(20°) ≈ 16.4 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q162. A force of 135 N is applied at the end of a 0.5 m wrench at an angle of 30° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.5×135×sin(30°) ≈ 33.7 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q163. A force of 150 N is applied at the end of a 0.6 m wrench at an angle of 40° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.6×150×sin(40°) ≈ 57.9 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q164. A force of 165 N is applied at the end of a 0.7 m wrench at an angle of 50° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.7×165×sin(50°) ≈ 88.5 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q165. A force of 180 N is applied at the end of a 0.4 m wrench at an angle of 60° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.4×180×sin(60°) ≈ 62.4 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q166. A force of 195 N is applied at the end of a 0.5 m wrench at an angle of 20° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.5×195×sin(20°) ≈ 33.3 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q167. A force of 210 N is applied at the end of a 0.6 m wrench at an angle of 30° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.6×210×sin(30°) ≈ 63.0 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q168. A force of 225 N is applied at the end of a 0.7 m wrench at an angle of 40° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.7×225×sin(40°) ≈ 101.2 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q169. A force of 240 N is applied at the end of a 0.4 m wrench at an angle of 50° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.4×240×sin(50°) ≈ 73.5 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q170. A force of 255 N is applied at the end of a 0.5 m wrench at an angle of 60° to the wrench handle. What is the torque magnitude (approximately)?

Correct Answer: A
Explanation: τ = rF sinθ = 0.5×255×sin(60°) ≈ 110.4 N·m.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: torque from perpendicular force component.

Q171. A 900 kg vehicle travels at 10 m/s around a curve of radius 40 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 900×10²/40 ≈ 2250 N ≈ 2.2 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q172. A 950 kg vehicle travels at 11 m/s around a curve of radius 45 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 950×11²/45 ≈ 2554 N ≈ 2.6 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q173. A 1000 kg vehicle travels at 12 m/s around a curve of radius 50 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 1000×12²/50 ≈ 2880 N ≈ 2.9 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q174. A 1050 kg vehicle travels at 13 m/s around a curve of radius 55 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 1050×13²/55 ≈ 3226 N ≈ 3.2 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q175. A 1100 kg vehicle travels at 14 m/s around a curve of radius 40 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 1100×14²/40 ≈ 5390 N ≈ 5.4 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q176. A 1150 kg vehicle travels at 15 m/s around a curve of radius 45 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 1150×15²/45 ≈ 5750 N ≈ 5.8 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q177. A 1200 kg vehicle travels at 16 m/s around a curve of radius 50 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 1200×16²/50 ≈ 6144 N ≈ 6.1 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q178. A 1250 kg vehicle travels at 17 m/s around a curve of radius 55 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 1250×17²/55 ≈ 6568 N ≈ 6.6 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q179. A 1300 kg vehicle travels at 18 m/s around a curve of radius 40 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 1300×18²/40 ≈ 10530 N ≈ 10.5 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q180. A 1350 kg vehicle travels at 19 m/s around a curve of radius 45 m. What centripetal force is required (approximately)?

Correct Answer: A
Explanation: F = mv²/r = 1350×19²/45 ≈ 10830 N ≈ 10.8 kN.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: centripetal force (mv²/r).

Q181. A steel pin carries a tensile load of 4000 N. Its cross-sectional area is 50 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 50 mm² = 0.000050 m². σ ≈ 4000/0.000050 ≈ 80.0 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q182. A steel pin carries a tensile load of 4500 N. Its cross-sectional area is 57 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 57 mm² = 0.000057 m². σ ≈ 4500/0.000057 ≈ 78.9 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q183. A steel pin carries a tensile load of 5000 N. Its cross-sectional area is 64 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 64 mm² = 0.000064 m². σ ≈ 5000/0.000064 ≈ 78.1 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q184. A steel pin carries a tensile load of 5500 N. Its cross-sectional area is 71 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 71 mm² = 0.000071 m². σ ≈ 5500/0.000071 ≈ 77.5 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q185. A steel pin carries a tensile load of 6000 N. Its cross-sectional area is 78 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 78 mm² = 0.000078 m². σ ≈ 6000/0.000078 ≈ 76.9 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q186. A steel pin carries a tensile load of 6500 N. Its cross-sectional area is 85 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 85 mm² = 0.000085 m². σ ≈ 6500/0.000085 ≈ 76.5 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q187. A steel pin carries a tensile load of 7000 N. Its cross-sectional area is 92 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 92 mm² = 0.000092 m². σ ≈ 7000/0.000092 ≈ 76.1 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q188. A steel pin carries a tensile load of 7500 N. Its cross-sectional area is 99 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 99 mm² = 0.000099 m². σ ≈ 7500/0.000099 ≈ 75.8 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q189. A steel pin carries a tensile load of 8000 N. Its cross-sectional area is 106 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 106 mm² = 0.000106 m². σ ≈ 8000/0.000106 ≈ 75.5 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q190. A steel pin carries a tensile load of 8500 N. Its cross-sectional area is 113 mm². The tensile stress is closest to:

Correct Answer: A
Explanation: σ = F/A. A = 113 mm² = 0.000113 m². σ ≈ 8500/0.000113 ≈ 75.2 MPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: stress = force/area (tension).

Q191. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 0.80 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×0.80 ≈ 7.8 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q192. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 1.15 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×1.15 ≈ 11.3 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q193. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 1.50 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×1.50 ≈ 14.7 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q194. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 1.85 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×1.85 ≈ 18.1 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q195. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 2.20 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×2.20 ≈ 21.6 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q196. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 2.55 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×2.55 ≈ 25.0 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q197. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 2.90 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×2.90 ≈ 28.4 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q198. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 3.25 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×3.25 ≈ 31.9 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q199. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 3.60 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×3.60 ≈ 35.3 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q200. Approximately what gauge pressure (above atmospheric) does water exert at a depth of 3.95 m?

Correct Answer: A
Explanation: p = ρgh = 1000×9.8×3.95 ≈ 38.7 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: hydrostatic pressure (ρgh).

Q201. A 0.6 kg ball moving at 8 m/s is brought to rest in 0.04 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 8→0, so F_avg = m v / Δt = 0.6×8/0.04 ≈ 120.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q202. A 0.7 kg ball moving at 9 m/s is brought to rest in 0.05 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 9→0, so F_avg = m v / Δt = 0.7×9/0.05 ≈ 126.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q203. A 0.8 kg ball moving at 10 m/s is brought to rest in 0.06 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 10→0, so F_avg = m v / Δt = 0.8×10/0.06 ≈ 133.3 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q204. A 0.9 kg ball moving at 11 m/s is brought to rest in 0.07 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 11→0, so F_avg = m v / Δt = 0.9×11/0.07 ≈ 141.4 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q205. A 1.0 kg ball moving at 12 m/s is brought to rest in 0.04 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 12→0, so F_avg = m v / Δt = 1.0×12/0.04 ≈ 300.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q206. A 1.1 kg ball moving at 13 m/s is brought to rest in 0.05 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 13→0, so F_avg = m v / Δt = 1.1×13/0.05 ≈ 286.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q207. A 1.2 kg ball moving at 14 m/s is brought to rest in 0.06 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 14→0, so F_avg = m v / Δt = 1.2×14/0.06 ≈ 280.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q208. A 1.3 kg ball moving at 15 m/s is brought to rest in 0.07 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 15→0, so F_avg = m v / Δt = 1.3×15/0.07 ≈ 278.6 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q209. A 1.4 kg ball moving at 16 m/s is brought to rest in 0.04 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 16→0, so F_avg = m v / Δt = 1.4×16/0.04 ≈ 560.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q210. A 1.5 kg ball moving at 17 m/s is brought to rest in 0.05 s. What is the average stopping force (magnitude)?

Correct Answer: A
Explanation: Impulse equals change in momentum: F_avg·Δt = mΔv. Here Δv = 17→0, so F_avg = m v / Δt = 1.5×17/0.05 ≈ 510.0 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: impulse–momentum (FΔt = mΔv).

Q211. A flywheel has moment of inertia 0.45 kg·m² and spins at 180 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 180×2π/60 ≈ 18.8 rad/s. KE ≈ 0.5×0.45×(18.8)² ≈ 80 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q212. A flywheel has moment of inertia 0.50 kg·m² and spins at 200 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 200×2π/60 ≈ 20.9 rad/s. KE ≈ 0.5×0.50×(20.9)² ≈ 110 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q213. A flywheel has moment of inertia 0.55 kg·m² and spins at 220 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 220×2π/60 ≈ 23.0 rad/s. KE ≈ 0.5×0.55×(23.0)² ≈ 146 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q214. A flywheel has moment of inertia 0.60 kg·m² and spins at 240 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 240×2π/60 ≈ 25.1 rad/s. KE ≈ 0.5×0.60×(25.1)² ≈ 189 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q215. A flywheel has moment of inertia 0.65 kg·m² and spins at 260 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 260×2π/60 ≈ 27.2 rad/s. KE ≈ 0.5×0.65×(27.2)² ≈ 241 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q216. A flywheel has moment of inertia 0.70 kg·m² and spins at 280 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 280×2π/60 ≈ 29.3 rad/s. KE ≈ 0.5×0.70×(29.3)² ≈ 301 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q217. A flywheel has moment of inertia 0.75 kg·m² and spins at 300 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 300×2π/60 ≈ 31.4 rad/s. KE ≈ 0.5×0.75×(31.4)² ≈ 370 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q218. A flywheel has moment of inertia 0.80 kg·m² and spins at 320 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 320×2π/60 ≈ 33.5 rad/s. KE ≈ 0.5×0.80×(33.5)² ≈ 449 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q219. A flywheel has moment of inertia 0.85 kg·m² and spins at 340 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 340×2π/60 ≈ 35.6 rad/s. KE ≈ 0.5×0.85×(35.6)² ≈ 539 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q220. A flywheel has moment of inertia 0.90 kg·m² and spins at 360 rpm. Approximately how much rotational kinetic energy does it have?

Correct Answer: A
Explanation: Rotational KE = ½Iω². Convert speed: ω = 360×2π/60 ≈ 37.7 rad/s. KE ≈ 0.5×0.90×(37.7)² ≈ 640 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: rotational kinetic energy (½Iω²).

Q221. A wrench of length 0.35 m has two perpendicular forces applied: 80 N turning it clockwise and 30 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.35×(80−30) ≈ 17.5 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q222. A wrench of length 0.40 m has two perpendicular forces applied: 90 N turning it clockwise and 35 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.40×(90−35) ≈ 22.0 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q223. A wrench of length 0.45 m has two perpendicular forces applied: 100 N turning it clockwise and 40 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.45×(100−40) ≈ 27.0 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q224. A wrench of length 0.50 m has two perpendicular forces applied: 110 N turning it clockwise and 45 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.50×(110−45) ≈ 32.5 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q225. A wrench of length 0.55 m has two perpendicular forces applied: 120 N turning it clockwise and 30 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.55×(120−30) ≈ 49.5 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q226. A wrench of length 0.35 m has two perpendicular forces applied: 130 N turning it clockwise and 35 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.35×(130−35) ≈ 33.2 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q227. A wrench of length 0.40 m has two perpendicular forces applied: 140 N turning it clockwise and 40 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.40×(140−40) ≈ 40.0 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q228. A wrench of length 0.45 m has two perpendicular forces applied: 150 N turning it clockwise and 45 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.45×(150−45) ≈ 47.2 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q229. A wrench of length 0.50 m has two perpendicular forces applied: 160 N turning it clockwise and 30 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.50×(160−30) ≈ 65.0 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q230. A wrench of length 0.55 m has two perpendicular forces applied: 170 N turning it clockwise and 35 N turning it counterclockwise at the same radius. What is the net torque (clockwise positive)?

Correct Answer: A
Explanation: Torques add with sign. τ_net = r(F1 − F2) = 0.55×(170−35) ≈ 74.2 N·m (clockwise).
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: net torque (sum of torques).

Q231. Gear A (14 teeth) drives idler gear B (28 teeth), which drives gear C (21 teeth). If gear A spins at 120 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 120×(14/21) ≈ 80.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q232. Gear A (16 teeth) drives idler gear B (29 teeth), which drives gear C (24 teeth). If gear A spins at 130 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 130×(16/24) ≈ 86.7 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q233. Gear A (18 teeth) drives idler gear B (30 teeth), which drives gear C (27 teeth). If gear A spins at 140 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 140×(18/27) ≈ 93.3 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q234. Gear A (20 teeth) drives idler gear B (31 teeth), which drives gear C (30 teeth). If gear A spins at 150 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 150×(20/30) ≈ 100.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q235. Gear A (22 teeth) drives idler gear B (32 teeth), which drives gear C (21 teeth). If gear A spins at 160 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 160×(22/21) ≈ 167.6 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q236. Gear A (24 teeth) drives idler gear B (33 teeth), which drives gear C (24 teeth). If gear A spins at 170 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 170×(24/24) ≈ 170.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q237. Gear A (26 teeth) drives idler gear B (34 teeth), which drives gear C (27 teeth). If gear A spins at 180 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 180×(26/27) ≈ 173.3 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q238. Gear A (28 teeth) drives idler gear B (35 teeth), which drives gear C (30 teeth). If gear A spins at 190 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 190×(28/30) ≈ 177.3 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q239. Gear A (30 teeth) drives idler gear B (36 teeth), which drives gear C (21 teeth). If gear A spins at 200 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 200×(30/21) ≈ 285.7 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q240. Gear A (32 teeth) drives idler gear B (37 teeth), which drives gear C (24 teeth). If gear A spins at 210 rpm, gear C spins at about:

Correct Answer: A
Explanation: An idler gear changes direction but not the overall speed ratio. Overall ratio = teeth_A/teeth_C, so rpm_C = 210×(32/24) ≈ 280.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: gear trains with idlers (ratio depends on first and last gears).

Q241. A belt drive uses a 4-inch driver pulley turning at 900 rpm and a 10-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 900×(4/10) ≈ 360.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q242. A belt drive uses a 5-inch driver pulley turning at 950 rpm and a 12-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 950×(5/12) ≈ 395.8 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q243. A belt drive uses a 6-inch driver pulley turning at 1000 rpm and a 14-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 1000×(6/14) ≈ 428.6 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q244. A belt drive uses a 7-inch driver pulley turning at 1050 rpm and a 16-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 1050×(7/16) ≈ 459.4 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q245. A belt drive uses a 4-inch driver pulley turning at 1100 rpm and a 18-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 1100×(4/18) ≈ 244.4 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q246. A belt drive uses a 5-inch driver pulley turning at 1150 rpm and a 20-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 1150×(5/20) ≈ 287.5 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q247. A belt drive uses a 6-inch driver pulley turning at 1200 rpm and a 22-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 1200×(6/22) ≈ 327.3 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q248. A belt drive uses a 7-inch driver pulley turning at 1250 rpm and a 24-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 1250×(7/24) ≈ 364.6 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q249. A belt drive uses a 4-inch driver pulley turning at 1300 rpm and a 26-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 1300×(4/26) ≈ 200.0 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q250. A belt drive uses a 5-inch driver pulley turning at 1350 rpm and a 28-inch driven pulley. Ignoring slip, the driven pulley speed is closest to:

Correct Answer: A
Explanation: In a belt drive (no slip), rim speeds match: rpm₁D₁ = rpm₂D₂. So rpm₂ = rpm₁(D₁/D₂) = 1350×(5/28) ≈ 241.1 rpm.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: belt/pulley speed ratio (equal rim speed).

Q251. A spring with k = 150 N/m is compressed 0.08 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×150×(0.08)² ≈ 0.48 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q252. A spring with k = 175 N/m is compressed 0.09 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×175×(0.09)² ≈ 0.71 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q253. A spring with k = 200 N/m is compressed 0.10 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×200×(0.10)² ≈ 1.00 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q254. A spring with k = 225 N/m is compressed 0.11 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×225×(0.11)² ≈ 1.36 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q255. A spring with k = 250 N/m is compressed 0.12 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×250×(0.12)² ≈ 1.80 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q256. A spring with k = 275 N/m is compressed 0.08 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×275×(0.08)² ≈ 0.88 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q257. A spring with k = 300 N/m is compressed 0.09 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×300×(0.09)² ≈ 1.21 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q258. A spring with k = 325 N/m is compressed 0.10 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×325×(0.10)² ≈ 1.62 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q259. A spring with k = 350 N/m is compressed 0.11 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×350×(0.11)² ≈ 2.12 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q260. A spring with k = 375 N/m is compressed 0.12 m. Approximately how much elastic potential energy is stored?

Correct Answer: A
Explanation: Spring energy U = ½kx² = 0.5×375×(0.12)² ≈ 2.70 J.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: spring potential energy (½kx²).

Q261. Two resistors 4 Ω and 10 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 4+10 = 14 Ω. Current I = V/R = 24/14 ≈ 1.71 A. Power P = VI ≈ 24×1.71 ≈ 41.1 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q262. Two resistors 5 Ω and 9 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 5+9 = 14 Ω. Current I = V/R = 24/14 ≈ 1.71 A. Power P = VI ≈ 24×1.71 ≈ 41.1 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q263. Two resistors 6 Ω and 8 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 6+8 = 14 Ω. Current I = V/R = 24/14 ≈ 1.71 A. Power P = VI ≈ 24×1.71 ≈ 41.1 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q264. Two resistors 7 Ω and 7 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 7+7 = 14 Ω. Current I = V/R = 24/14 ≈ 1.71 A. Power P = VI ≈ 24×1.71 ≈ 41.1 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q265. Two resistors 8 Ω and 10 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 8+10 = 18 Ω. Current I = V/R = 24/18 ≈ 1.33 A. Power P = VI ≈ 24×1.33 ≈ 32.0 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q266. Two resistors 9 Ω and 9 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 9+9 = 18 Ω. Current I = V/R = 24/18 ≈ 1.33 A. Power P = VI ≈ 24×1.33 ≈ 32.0 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q267. Two resistors 10 Ω and 8 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 10+8 = 18 Ω. Current I = V/R = 24/18 ≈ 1.33 A. Power P = VI ≈ 24×1.33 ≈ 32.0 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q268. Two resistors 11 Ω and 7 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 11+7 = 18 Ω. Current I = V/R = 24/18 ≈ 1.33 A. Power P = VI ≈ 24×1.33 ≈ 32.0 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q269. Two resistors 12 Ω and 10 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 12+10 = 22 Ω. Current I = V/R = 24/22 ≈ 1.09 A. Power P = VI ≈ 24×1.09 ≈ 26.2 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q270. Two resistors 13 Ω and 9 Ω are connected in series across 24 V. What is the total power dissipated (approximately)?

Correct Answer: A
Explanation: Series resistance R = 13+9 = 22 Ω. Current I = V/R = 24/22 ≈ 1.09 A. Power P = VI ≈ 24×1.09 ≈ 26.2 W.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: series circuits (Ohm’s law, P = VI).

Q271. An object displaces 0.015 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.015 ≈ 147 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q272. An object displaces 0.017 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.017 ≈ 167 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q273. An object displaces 0.019 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.019 ≈ 186 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q274. An object displaces 0.021 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.021 ≈ 206 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q275. An object displaces 0.023 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.023 ≈ 225 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q276. An object displaces 0.025 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.025 ≈ 245 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q277. An object displaces 0.027 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.027 ≈ 265 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q278. An object displaces 0.029 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.029 ≈ 284 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q279. An object displaces 0.031 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.031 ≈ 304 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q280. An object displaces 0.033 m³ of water. What is the buoyant force on the object (approximately)?

Correct Answer: A
Explanation: Buoyant force equals weight of displaced fluid: F_b = ρgV = 1000×9.8×0.033 ≈ 323 N.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Archimedes’ principle (ρgV).

Q281. Water speeds up from 2.0 m/s to 4.0 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(4.0²−2.0²) ≈ 6.0 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q282. Water speeds up from 2.2 m/s to 4.3 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(4.3²−2.2²) ≈ 6.8 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q283. Water speeds up from 2.4 m/s to 4.6 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(4.6²−2.4²) ≈ 7.7 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q284. Water speeds up from 2.6 m/s to 4.9 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(4.9²−2.6²) ≈ 8.6 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q285. Water speeds up from 2.8 m/s to 5.2 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(5.2²−2.8²) ≈ 9.6 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q286. Water speeds up from 3.0 m/s to 5.5 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(5.5²−3.0²) ≈ 10.6 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q287. Water speeds up from 3.2 m/s to 4.0 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(4.0²−3.2²) ≈ 2.9 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q288. Water speeds up from 3.4 m/s to 4.3 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(4.3²−3.4²) ≈ 3.5 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q289. Water speeds up from 3.6 m/s to 4.6 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(4.6²−3.6²) ≈ 4.1 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q290. Water speeds up from 3.8 m/s to 4.9 m/s in a pipe at the same height. Ignoring losses, approximately how much does static pressure drop?

Correct Answer: A
Explanation: Bernoulli (same height): pressure drop ΔP ≈ ½ρ(v₂²−v₁²) = 0.5×1000×(4.9²−3.8²) ≈ 4.8 kPa.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: Bernoulli principle (pressure–velocity tradeoff).

Q291. A heat engine takes in 500 J of heat and produces 175 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 175/500 ≈ 0.350 → about 35%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

Q292. A heat engine takes in 550 J of heat and produces 202 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 202/550 ≈ 0.368 → about 37%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

Q293. A heat engine takes in 600 J of heat and produces 230 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 230/600 ≈ 0.383 → about 38%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

Q294. A heat engine takes in 650 J of heat and produces 227 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 227/650 ≈ 0.350 → about 35%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

Q295. A heat engine takes in 700 J of heat and produces 255 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 255/700 ≈ 0.364 → about 36%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

Q296. A heat engine takes in 750 J of heat and produces 282 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 282/750 ≈ 0.377 → about 38%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

Q297. A heat engine takes in 800 J of heat and produces 280 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 280/800 ≈ 0.350 → about 35%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

Q298. A heat engine takes in 850 J of heat and produces 308 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 308/850 ≈ 0.362 → about 36%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

Q299. A heat engine takes in 900 J of heat and produces 335 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 335/900 ≈ 0.372 → about 37%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

Q300. A heat engine takes in 950 J of heat and produces 332 J of work. What is its efficiency (closest)?

Correct Answer: A
Explanation: Efficiency η = W_out/Q_in = 332/950 ≈ 0.350 → about 35%.
Citation: ASVAB content domain — Mechanical Comprehension; underlying principle: efficiency (η = W/Q).

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