ASVAB Mechanical Comprehension (MC) Practice Test 2026

ASVAB Mechanical Comprehension (MC) Practice Test

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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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