Physics
Practice Test 8 for Physics: real questions and explanations from the Varsity Tutors practice-test pool.
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Two objects are separated by the same distance r. The gravitational force is Fgrav=Gm1m2/r2 and the electric force is Felec=kq1q2/r2. If the distance between the objects is doubled to 2r, how do Fgrav and Felec change, and what happens to the ratio Felec/Fgrav?
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Two objects are separated by the same distance r. The gravitational force is Fgrav=Gm1m2/r2 and the electric force is Felec=kq1q2/r2. If the distance between the objects is doubled to 2r, how do Fgrav and Felec change, and what happens to the ratio Felec/Fgrav?
Explanation: This question tests understanding of how gravitational and electric forces compare in their dependence on distance. Both gravitational force (F = Gm₁m₂/r²) and electric force (F = k|q₁q₂|/r²) follow inverse square laws, decreasing with the square of the distance between objects, but they differ dramatically in strength: the Coulomb constant k = 9.0 × 10⁹ N·m²/C² is about 10²⁰ times larger than the gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg², making electric forces intrinsically much stronger than gravitational forces for comparable numerical values; additionally, gravity is always attractive while electric force can be attractive or repulsive, and gravity depends on mass while electric force depends on charge. The ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of distance r (since both forces have r² in denominator which cancels in the ratio), meaning the relative strength depends only on the charge-to-mass ratios of the objects—this is why electric forces dominate at atomic scales while gravity dominates at cosmic scales. Choice A is correct because it properly applies both force formulas and compares the resulting magnitudes after doubling distance, showing both become 1/4 as large and ratio unchanged. Choice B suggests that the distance dependence is different for the two forces (claiming one decreases faster with distance than the other), when actually both follow inverse square laws (F ∝ 1/r²) and decrease at the same rate with distance—the ratio F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) is independent of r. When comparing gravitational and electric forces: (1) both follow inverse square laws F ∝ 1/r², so distance affects them equally, (2) the ratio of strengths F_elec/F_grav = (kq₁q₂)/(Gm₁m₂) depends on charges and masses but not distance, (3) electric force is intrinsically stronger by a factor of k/G ≈ 10²⁰ when comparing equal numerical values, (4) at atomic scales electric dominates because particles have charge but tiny mass (gravity ≈ 10⁻⁴⁷ N is negligible), and (5) at cosmic scales gravity dominates because objects are massive but electrically neutral (equal + and - charges cancel, so net F_elec ≈ 0). Practical implications: chemistry, molecular biology, material strength, friction, and essentially all everyday phenomena (except falling) are determined by electric forces between atoms and molecules, while planetary orbits, tides, satellite motion, and the large-scale structure of the universe are determined by gravitational forces—the electric force's dominance at small scales is why a charged balloon can lift paper against Earth's entire gravitational pull, yet gravity's dominance at large scales is why planets orbit stars despite any residual electric charges they might have.
To investigate momentum conservation in a 1D collision, you measure masses with a balance and velocities with motion sensors. During analysis, you must assign signs to velocities (e.g., rightward positive, leftward negative). Which calculation correctly represents the total momentum after the collision for two carts?
Let m1,m2 be in kg and v1f,v2f be in m/s.
Explanation: This question tests understanding of experimental design for investigating momentum conservation in collisions by selecting the correct calculation for total momentum after collision. To verify that momentum is conserved (p_before = p_after), an experiment must measure the masses of both colliding objects using a balance, measure their velocities before the collision (v₁ᵢ, v₂ᵢ) and after the collision (v₁f, v₂f) using motion sensors or video analysis, then calculate total momentum before (p_before = m₁v₁ᵢ + m₂v₂ᵢ) and after (p_after = m₁v₁f + m₂v₂f) to verify they are equal within experimental uncertainty. For data analysis: Evidence that momentum is conserved comes from showing that p_before and p_after are approximately equal across multiple trials—for example, if p_before = 1.45 kg⋅m/s and p_after = 1.41 kg⋅m/s, the percent difference is |1.41-1.45|/1.45 × 100% = 2.8%, which is within typical experimental uncertainty and supports conservation; graphing p_after versus p_before for multiple trials should produce a straight line with slope = 1 passing through the origin, further confirming that the momentum after equals the momentum before regardless of initial conditions. Choice A is correct because it properly calculates p_after as the sum of individual momenta m₁v₁f + m₂v₂f, using signs for direction, which is essential for verifying conservation in 1D collisions. Choice B is a tempting distractor but fails because it incorrectly multiplies the total mass by the sum of velocities, which would only apply if velocities were the same (as in perfectly inelastic collisions where they stick), but not generally for all collision types. When designing momentum conservation experiments, remember this checklist: (1) measure masses with a balance—this is non-negotiable since p = mv requires knowing m, (2) measure velocities at two times (immediately before and immediately after collision) using motion sensors or video analysis, (3) calculate both p_before = m₁v₁ᵢ + m₂v₂ᵢ and p_after = m₁v₁f + m₂v₂f with careful attention to direction signs (positive/negative for 1D motion), (4) compare the two values—they should be equal within about 5% for a successful demonstration, and (5) conduct multiple trials to account for random errors. Common mistakes to avoid: (a) forgetting to measure masses (cannot calculate momentum without m), (b) measuring velocities at wrong times (need immediately before and after collision, not minutes later), (c) ignoring direction in 1D collisions (rightward velocity is positive, leftward is negative—this matters for momentum as a vector), (d) comparing individual object momenta instead of system totals (conservation applies to p₁ + p₂, not to p₁ alone), and (e) expecting perfect equality (experimental uncertainty means p_before and p_after will differ by small percentage, typically 2-5% is excellent agreement).
A rectangular coil rotates at constant speed between the poles of a magnet (a simple generator). The magnetic flux through the coil is Φ=BAcos(θ), where θ is the angle between the magnetic field and the coil’s area normal. At which orientation is the magnitude of the induced EMF greatest?
Explanation: This question tests understanding of electromagnetic induction in rotating coils (generators), specifically when induced EMF is maximum during the rotation cycle. For a rotating coil, flux Φ = BA cos(θ) varies sinusoidally with angle θ, and by Faraday's law, induced EMF depends on how fast this flux is changing: |ε| = N|dΦ/dt| = NBA|d(cos θ)/dt| = NBAω|sin θ|, where ω is angular velocity. When θ = 90°, the coil face is parallel to the field (flux Φ = 0), but this is precisely when flux is changing most rapidly because the coil is cutting through field lines at maximum rate—mathematically, |sin(90°)| = 1 is maximum, so |ε| = NBAω is at its peak value. Choice B is correct because it accurately identifies that maximum EMF occurs when θ = 90° where flux is zero but changing most rapidly (|dΦ/dt| is maximum even though Φ = 0). Choice A incorrectly suggests maximum EMF occurs at θ = 0° where flux is maximum, but at this instant the flux is momentarily constant (not changing) as the cosine function peaks—mathematically, |sin(0°)| = 0, so the instantaneous rate of flux change and thus EMF is zero at this orientation. To analyze rotating coil EMF: (1) flux varies as Φ = BA cos(θ), (2) EMF depends on flux rate of change, not flux magnitude, (3) dΦ/dt = -BA sin(θ)·(dθ/dt), (4) EMF is maximum when |sin θ| = 1 (at θ = 90° and 270°), (5) EMF is zero when sin θ = 0 (at θ = 0° and 180°). This explains why AC generators produce sinusoidal voltage—the EMF varies as sin(θ) even though flux varies as cos(θ), with peak voltage occurring as the coil passes through horizontal (parallel to field) not vertical (perpendicular to field) orientation.
A lab tests transmission over the same 1 km cable. For analog voice, intelligibility becomes poor below about 40 dB SNR. A digital voice link with error correction can still work around 15–20 dB SNR, but fails if SNR gets too low to distinguish 0 from 1 reliably. If the measured SNR is 18 dB, which result is most likely?
Explanation: This question tests understanding of how digital and analog transmission methods perform differently under realistic conditions like noise, interference, and long distances. The fundamental difference in transmission performance is that analog signals have noise add directly at every stage (cable, amplifier, relay) with no way to distinguish signal from noise, causing gradual quality degradation proportional to noise level, while digital signals only need to distinguish between two levels (0 and 1), allowing regeneration at repeaters—the receiver detects whether each pulse is closer to 0 or 1 and creates a fresh, clean pulse, effectively removing accumulated noise and maintaining quality over long distances. At a low SNR of 18 dB, analog signals degrade as attenuation reduces amplitude and noise adds to signal, requiring amplification that also amplifies noise, causing signal-to-noise ratio (SNR) to worsen with each stage until signal is buried in hiss/static/snow, while digital signals maintain quality because regeneration detects the 0s and 1s and recreates perfect pulses, and error detection/correction algorithms can identify and fix bit errors that do occur, providing reliable delivery even when channel conditions are poor; for gradual noise increase, analog quality smoothly degrades (slight hiss → loud static as noise increases), while digital maintains perfect quality until noise exceeds the threshold where receiver can't reliably distinguish 0 from 1, then suddenly fails with dropouts or complete loss (cliff effect). Choice B is correct because it recognizes why digital preferred for modern long-distance communication at lower SNR. Choice A reverses the degradation patterns, claiming analog has cliff effect or digital degrades gradually. Practical implications: virtually all modern long-distance communication uses digital (internet, cell phones, satellite, fiber optic cables, digital TV/radio) specifically because regeneration and error correction provide reliable transmission over vast distances despite noise and interference—analog dominated historically when electronics were simpler, but digital's advantages (quality maintenance, error handling, compression, encryption, computer compatibility) led to digital revolution in telecommunications. The trade-off is complexity (digital requires encoding/decoding, analog is direct) but performance benefits overwhelmingly favor digital for any application requiring transmission over distance, multiple copies, or integration with computers, which is why analog transmission is largely obsolete except in legacy systems and niche applications.
In a 1D elastic collision on a frictionless track, two equal-mass carts collide head-on. Cart A has mass mA=1.0kg and initial velocity vAi=+4.0m/s. Cart B has mass mB=1.0kg and initial velocity vBi=−1.0m/s. After the collision, Cart A is observed moving at vAf=−1.0m/s. Using conservation of momentum, what is Cart B's final velocity vBf (in m/s)?
Explanation: This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, total momentum is p_before = m₁v₁ᵢ + m₂v₂ᵢ = (1.0 kg)(4.0 m/s) + (1.0 kg)(-1.0 m/s) = 4 - 1 = 3 kg⋅m/s; after the elastic collision objects separate: using momentum conservation p_after = p_before: (1.0 kg)(-1.0 m/s) + (1.0 kg) v_{Bf} = 3 kg⋅m/s, so -1 + v_{Bf} = 3, v_{Bf} = 4.0 m/s. Choice A is correct because it properly applies momentum conservation with correct signs for directions and properly solves the momentum equation for the unknown velocity. Choice C makes a sign error by treating the initial leftward motion as positive, leading to incorrect total momentum. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. Remember that momentum is conserved in all collisions regardless of whether they're elastic or inelastic—what differs is whether kinetic energy is conserved (elastic only) or lost to other forms like heat and sound (inelastic).
A 0.80kg mass is launched vertically upward by a spring (ignore air resistance). The spring constant is k=300N/m and it is compressed 0.40m. After the mass leaves the spring, how high above the launch point does it rise? (Use g=10m/s2.)
Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. When the spring is compressed by distance x = 0.40 m, it stores elastic potential energy PE_spring = ½kx² = ½(300 N/m)(0.40 m)² = ½(300)(0.16) = 24 J. After the mass leaves the spring, this energy becomes kinetic energy, which then converts to gravitational potential energy as the mass rises: at maximum height h, all energy is gravitational PE, so mgh = 24 J, giving h = 24/(mg) = 24/[(0.80)(10)] = 24/8 = 3.0 m. Choice B is correct because it properly applies conservation of energy, recognizing that all 24 J of spring potential energy ultimately converts to gravitational potential energy at the peak height. Choice A (1.0 m) significantly underestimates the height, possibly from using h = PE/(mg²) incorrectly, while choice C (7.5 m) and choice D (15 m) overestimate, with D appearing to forget the ½ factor in the spring PE formula, calculating as if PE = kx² = 48 J. When solving multi-stage energy problems: (1) identify energy at each stage—here spring PE → KE → gravitational PE, (2) use conservation to equate initial and final energies: ½kx² = mgh, (3) solve for the unknown: h = kx²/(2mg) = (300)(0.16)/(2×0.80×10) = 48/16 = 3.0 m. Key insight: in problems with multiple energy transformations, you can often skip intermediate steps and equate initial energy directly to final energy, as total mechanical energy is conserved throughout when no friction acts.
A point charge Q=+4.0μC is fixed at the origin. Point P is located on the +x-axis at x=0.30m. What is the direction of the electric field E at point P due to Q?
Explanation: This question tests understanding of modeling electric and magnetic fields and predicting how they interact with charges or currents. An electric field E exists in the region around a charge or group of charges and is defined as the force per unit charge that a positive test charge would experience at each point: E = F/q, measured in N/C (Newtons per Coulomb). The field direction is the direction a positive charge would be pushed (away from positive source charges, toward negative source charges), and field magnitude for a point charge is E = kQ/r², decreasing with the square of distance from the source charge Q. For a positive point charge Q = +4.0 μC at the origin, the electric field at point P on the +x-axis at x=0.30 m points radially outward from the charge, which is in the +x direction (away from the origin); a positive test charge at P would experience a force in the +x direction, while a negative test charge would experience force toward the origin in the -x direction. Choice C is correct because it properly identifies electric field direction as radially outward from positive charge. Choice A reverses the electric field direction, showing it pointing inward to positive when actually electric field points away from positive charges (the direction a positive test charge would be pushed) and toward negative charges (the direction a positive test charge would be pulled). When modeling fields and their effects: for electric fields, (1) identify source charges and their signs, (2) remember field points away from positive charges and toward negative charges, (3) calculate magnitude using E = kQ/r² for point charges, (4) for multiple sources add fields as vectors (considering directions), and (5) force on test charge is F = qE in field direction if q positive, opposite if q negative; for magnetic fields, (1) identify source (permanent magnet or current), (2) use right-hand rule for direction (thumb = current, fingers curl = field for wire; or just remember N to S outside magnet), (3) recognize field lines always form closed loops (no monopoles), and (4) force on moving charge is F = qvB perpendicular to both v and B (right-hand rule: fingers = B, thumb = v, palm = F).
In a DC motor, a current-carrying coil sits in a magnetic field and experiences forces that make it rotate. Which energy transformation best describes what happens during steady operation (ignoring losses)?
Explanation: This question tests understanding of how electric and magnetic fields transfer energy without requiring direct physical contact between objects. Magnetic fields can do work on current-carrying coils—electrical energy supplied to the coil creates currents that interact with the magnetic field, producing forces that cause rotation, transferring energy to mechanical (rotational) energy without direct contact. In this scenario, the current in the coil interacts with the magnetic field to produce torque, converting input electrical power to output mechanical power during steady operation. Choice C is correct because it correctly identifies the energy conversion pathway from electrical to mechanical via magnetic forces in the motor. Choice A confuses the process with a generator, reversing the energy flow from mechanical to electrical. Energy transfer via fields: electric fields accelerate charges doing work W = qEd (field energy → kinetic energy), magnetic forces on currents do work W = Fd with F = BIL (electrical → mechanical in motors), changing magnetic fields induce currents transferring energy between circuits (Faraday's law: electromagnetic induction), and electromagnetic waves carry energy through space at light speed (radiation energy). The key insight is fields serve as energy carriers, storing energy when created and releasing it to objects within the field, enabling energy transfer without material contact—this is fundamentally different from conduction (needs contact) or convection (needs fluid motion), and explains wireless charging (induction), motors (magnetic force work), particle accelerators (electric field acceleration), and solar panels (EM wave absorption).
A rectangular wire loop is placed in a uniform magnetic field B directed to the right (→). The loop is oriented so that the left vertical side carries current upward (↑) and the right vertical side carries current downward (↓). What best describes the forces on the two vertical sides due to the magnetic field?
Explanation: This question tests understanding of the relationship between electric current and magnetic fields, specifically how magnetic fields exert forces on current-carrying wires in a loop. A current-carrying wire placed in an external magnetic field experiences a force perpendicular to both the current direction and the field direction, with magnitude F = BIL sin(θ) where B is field strength, I is current, L is wire length in field, and θ is angle between current and field (maximum force when perpendicular); the force direction is given by the right-hand rule: point your fingers along the magnetic field B, point your thumb along the current I, and your palm faces the direction of force F—this is the motor principle that makes electric motors work. For the left side with current upward and field to the right, using the force right-hand rule: point your fingers to the right (B), point your thumb upward (I), and your palm faces into the page—meaning force into the page; for the right side with current downward, thumb downward reverses the force to out of the page. Choice C is correct because it correctly uses the force right-hand rule for each side: left into the page, right out of the page, accurately predicting opposite forces that could cause rotation. Choice B misapplies the right-hand rule by pointing thumb in the wrong direction for each side, leading to reversed predictions for the force directions. To solve current-magnetic field problems, identify whether you're dealing with (1) current creating a magnetic field or (2) current in an external field experiencing a force, then apply the appropriate right-hand rule: for force on current, fingers point along external field B, thumb points along current I, and palm faces force F direction. Common errors to avoid: (a) using left hand instead of right (gives opposite direction), (b) confusing which right-hand rule applies (field-from-current vs force-on-current), (c) thinking force is parallel to current or field (it's perpendicular to both), (d) forgetting that field lines circle around wire not radiate outward, and (e) assuming the compass points toward the wire (it aligns tangent to field circles).
A rectangular coil rotates at constant speed between the poles of a magnet (a simple generator). The magnetic field in the gap is approximately uniform. As the coil rotates, which change is directly responsible for producing an alternating induced EMF in the coil?
Explanation: This question tests understanding of electromagnetic induction in generators, specifically how rotation changes flux to produce alternating EMF. Faraday's law states that changing magnetic flux induces EMF (ε = -N(ΔΦ/Δt)), where flux Φ = BA cos(θ) depends on the angle θ between the magnetic field and the normal to the coil—as a coil rotates, this angle changes continuously, causing flux to vary sinusoidally. As the rectangular coil rotates in the uniform magnetic field, the angle θ between the field direction and the normal to the coil's plane changes continuously from 0° to 360° with each rotation—this causes the flux Φ = BA cos(θ) to vary from maximum (BA when θ = 0°, coil face perpendicular to field) to zero (when θ = 90°, coil face parallel to field) to negative maximum (−BA when θ = 180°) and back. This continuous flux variation means ΔΦ/Δt is always non-zero (except at instantaneous extrema), inducing a continuously varying EMF that alternates in direction as cos(θ) goes positive and negative—this produces alternating current (AC). Choice B is correct because it accurately identifies that the changing angle θ in the flux formula Φ = BA cos(θ) is what causes the flux to change with time, properly citing the flux formula and recognizing rotation changes θ. Choice C incorrectly suggests the coil's area A changes during rotation, but a rigid rectangular coil maintains constant area—it's the effective area (A cos(θ), the projection perpendicular to the field) that changes, which is captured by the changing angle θ in the flux formula, not by changing A itself. To understand generator operation: (1) recognize that flux Φ = BA cos(θ) depends on coil orientation via angle θ, (2) rotation causes θ to change continuously with time, (3) changing θ means changing cos(θ), thus changing flux, (4) by Faraday's law, changing flux induces EMF proportional to rotation speed. The induced EMF follows ε = −NBA(d/dt)[cos(ωt)] = NBAω sin(ωt), producing sinusoidal AC voltage with frequency equal to rotation frequency—this is how power plants generate electricity by spinning coils in magnetic fields.
A car bumper insert is being chosen for a low-speed crash (parking collision). The design goal is to maximize energy absorption (reduce rebound) while keeping repair costs low, and it only needs to work once in a severe impact. Which material behavior is most appropriate for reducing forces on occupants by dissipating kinetic energy rather than returning it?
Which choice best matches the physics goal?
Explanation: This question tests understanding of applying physics principles—specifically work-energy relationships—to design collision protection systems. The kinetic energy of a moving object (KE = ½mv²) must be absorbed during a collision through work done by stopping forces: W = Fd, so for a fixed amount of kinetic energy to absorb, increasing the deformation distance d allows smaller forces F to do the necessary work—this is why crushable materials, compressible padding, and extendable restraints improve safety by spreading energy absorption over longer distances. For vehicle crumple zones: In a vehicle collision, the front of the car is designed to crumple and deform over a distance of 0.5-1 m during impact, which extends the collision time from perhaps 0.01 s (rigid car) to 0.1 s (with crumple zone)—using F_avg = Δp/Δt, this 10-fold increase in collision time reduces the average force on the vehicle (and indirectly on occupants) by a factor of 10, from potentially unsurvivable levels to forces the passenger compartment structure can withstand. Simultaneously, the deformation distance of 0.5-1 m allows the kinetic energy to be absorbed through work (W = Fd) with smaller peak forces, as the crumpling material does work against the collision force. Choice C is correct because it uses work-energy relationship showing that increasing deformation distance reduces force for the same energy absorption, with plastic deformation dissipating energy through permanent crushing rather than elastic rebound. Choice D incorrectly suggests using stiffer, more rigid materials, when actually rigid materials cause shorter collision times and higher peak forces—safety requires materials that deform or compress to extend Δt and increase deformation distance d, which is why crumple zones crumple and padding compresses rather than staying rigid. When designing collision protection systems, apply two key physics principles: (1) Impulse-momentum theorem F_avg = Δp/Δt shows that for a given momentum change (stopping an object), extending the collision time Δt reduces the average force F_avg—achieve this through crumple zones, padding compression, airbag deflation, or seat belt stretching; (2) Work-energy principle W = Fd shows that for a fixed kinetic energy to absorb (KE = ½mv²), increasing the deformation distance d reduces the required force F—achieve this through crushable materials, thick padding, or structures designed to fold/compress over longer distances. Remember that effective collision protection typically involves: (a) materials that deform plastically (crush permanently) rather than elastically (bounce back), because plastic deformation maximizes energy absorption, (b) progressive resistance that increases gradually with deformation rather than sudden stiffening (avoids peak force spikes), (c) distribution of forces over large surface areas to reduce local pressure (airbags spread over chest, helmets over skull), and (d) multi-stage systems where soft initial padding handles minor impacts comfortably while firmer secondary layers engage for severe impacts—the goal is always to extend Δt and increase d while keeping F below injury thresholds at all times during the collision.
A student is designing a hand-crank generator that converts mechanical rotation → electrical energy (electromagnetic induction) to charge a USB power bank during a camping trip. The device must provide an electrical output power Pout≥5.0 W at 5.0 V for at least 10 minutes. The generator is expected to be η=70% efficient (electrical output divided by mechanical input). What minimum mechanical input power Pin (from cranking) is required to meet the electrical power requirement?
Use: η=PinPout.
Explanation: This question tests understanding of designing energy conversion devices to meet output requirements while respecting efficiency and constraints. When designing energy conversion devices, the relationship between input, output, and efficiency is η = (output/input), which can be rearranged to determine required input: input = output/η, or expected output: output = η × input—for example, if a device must provide 100 W output and operates at 80% efficiency (η = 0.80), it requires input = 100/0.80 = 125 W, with 25 W lost as waste heat. For this design challenge requiring 5.0 W of electrical output with efficiency η = 0.70, the required input is P_in = P_out/η = 5.0/0.70 = 7.14 W, which means the design must have sufficient mechanical input from cranking to provide this power level. Choice C is correct because it correctly calculates required input using η = output/input rearranged to P_in = P_out/η = 5.0/0.70 = 7.1 W. Choice A (3.5 W) makes error in efficiency calculation, using inverted formula P_in = P_out × η instead of P_in = P_out/η, while Choice B (5.0 W) ignores efficiency entirely, and Choice D (14 W) doubles the correct answer. Design strategy: (1) identify required output (5.0 W electrical), (2) determine efficiency η (0.70 given), (3) calculate required input = output/η = 7.1 W, (4) select components providing this input capacity, (5) verify constraints satisfied (hand-cranking feasible), (6) account for waste energy = 7.1 - 5.0 = 2.1 W (becomes heat in generator). Remember that efficiency is always <100%, so input must exceed output; waste energy usually becomes thermal, requiring heat management in design.
A 2.0 kg pendulum bob is released from rest at a point 5.0m above its lowest point (ignore air resistance, g=10m/s2). What is the bob’s speed at the lowest point?
Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. At the highest point of the swing, the pendulum has maximum gravitational PE = mgh = (2.0 kg)(10 m/s²)(5.0 m) = 100 J and zero kinetic energy (starting from rest), while at the lowest point it has zero PE (using lowest point as reference) and maximum KE = ½mv². Energy conservation gives mgh_top = ½mv²_bottom, so 100 J = ½(2.0 kg)v², which gives v² = 100 m²/s², therefore v = 10 m/s. Choice B (10 m/s) is correct because it properly applies conservation of energy, recognizing that all 100 J of initial gravitational potential energy converts to kinetic energy at the lowest point. Choice C (14 m/s) would require KE = ½(2.0)(14)² = 196 J, which exceeds the available 100 J of initial PE, violating conservation of energy. When solving pendulum problems: (1) identify the reference level (usually the lowest point where PE = 0), (2) calculate initial energy (all PE at release point if starting from rest), (3) at the lowest point all energy is KE, so PE_initial = KE_final, (4) solve using mgh = ½mv², which simplifies to v = √(2gh). Remember that in a frictionless pendulum, mechanical energy oscillates between PE at the extremes and KE at the bottom, but the total mechanical energy remains constant throughout the swing.
For this experiment comparing elastic vs perfectly inelastic collisions, you will use two dynamics carts on a low-friction track. You can swap magnetic bumpers (elastic) and velcro bumpers (perfectly inelastic). Which set of measurements is essential to verify pbefore=pafter for each collision type?
Explanation: This question tests understanding of experimental design for investigating momentum conservation in collisions, specifically the essential measurements needed for different collision types. To verify that momentum is conserved (p_before = p_after), an experiment must measure the masses of both colliding objects using a balance, measure their velocities before the collision (v₁ᵢ, v₂ᵢ) and after the collision (v₁f, v₂f) using motion sensors or video analysis, then calculate total momentum before (p_before = m₁v₁ᵢ + m₂v₂ᵢ) and after (p_after = m₁v₁f + m₂v₂f) to verify they are equal within experimental uncertainty. For essential measurements in comparing elastic and inelastic collisions, the procedure requires determining masses m₁ and m₂ with a balance (as they cannot be assumed) and velocities before and after with precise tools like motion sensors or video to enable momentum calculations for both types, while controlling factors like track friction to ensure validity. Choice A is correct because it identifies both essential measurements: masses with a balance and velocities with motion sensors or video analysis, which are necessary to calculate p = mv for the system before and after. Choice B is a tempting distractor but fails because it omits initial velocities—momentum conservation requires comparing p_before (which needs v₁ᵢ and v₂ᵢ) to p_after, so both sets are essential, and low friction alone does not eliminate the need for initial data. When designing momentum conservation experiments, remember this checklist: (1) measure masses with a balance—this is non-negotiable since p = mv requires knowing m, (2) measure velocities at two times (immediately before and immediately after collision) using motion sensors or video analysis, (3) calculate both p_before = m₁v₁ᵢ + m₂v₂ᵢ and p_after = m₁v₁f + m₂v₂f with careful attention to direction signs (positive/negative for 1D motion), (4) compare the two values—they should be equal within about 5% for a successful demonstration, and (5) conduct multiple trials to account for random errors. To improve experimental quality: use a low-friction track or air track to minimize external forces that would violate conservation, ensure the track is level so gravity doesn't add a constant force, use precise velocity measurement tools (motion sensors better than stopwatch/meterstick), take multiple trials and average to reduce random error, and always include uncertainty analysis showing that p_before and p_after agree within the combined measurement uncertainties.
A head-on elastic collision occurs between two carts on a low-friction track. Cart A has mass mA=1.0 kg with initial velocity vAi=+4.0 m/s (right). Cart B has mass mB=1.0 kg with initial velocity vBi=−2.0 m/s (left). After the collision, Cart B is observed moving right at vBf=+4.0 m/s. Using conservation of momentum only, what must Cart A’s final velocity vAf be (in m/s) for momentum to be conserved?
Explanation: This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, total momentum is p_before = m_A v_Ai + m_B v_Bi = (1.0 kg)(+4.0 m/s) + (1.0 kg)(-2.0 m/s) = +4 - 2 = +2 kg⋅m/s. After the collision, using momentum conservation p_after = p_before: (1.0 kg) v_Af + (1.0 kg)(+4.0 m/s) = +2 kg⋅m/s, solving for the unknown velocity gives v_Af + 4 = +2, v_Af = +2 - 4 = -2.0 m/s. Choice B is correct because it properly applies momentum conservation with correct signs for directions and accurately solves the momentum equation for the unknown velocity. Choice D incorrectly assumes kinetic energy conservation alone or doubles a value without basis, leading to an overstated negative velocity. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. Remember that momentum is conserved in all collisions regardless of whether they're elastic or inelastic—what differs is whether kinetic energy is conserved (elastic only) or lost to other forms like heat and sound (inelastic).
On a low-friction track (rightward positive), Cart A has mass mA=4.0kg and is initially at rest (vAi=0). Cart B has mass mB=1.0kg and moves right at vBi=+12m/s. They collide and stick together (perfectly inelastic). What is the common final velocity vf?
Explanation: This question tests understanding of the conservation of momentum in collisions. The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision equals the total momentum after the collision: p_before = p_after, or m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f, where momentum is the product of mass and velocity and must account for direction. Before the collision, Cart A has momentum p₁ = m_A v_{Ai} = (4.0 kg)(0) = 0 kg⋅m/s, and Cart B has momentum p₂ = (1.0 kg)(+12 m/s) = +12 kg⋅m/s, giving total initial momentum p_before = 0 + 12 = +12 kg⋅m/s. After the perfectly inelastic collision, the objects stick together with combined mass (4.0 + 1.0 = 5.0 kg), so v_f = p_before / (m_A + m_B) = +12 / 5.0 = +2.4 m/s. Choice A (+2.4 m/s) is correct because it properly applies momentum conservation and correctly calculates the common final velocity for the combined mass. Choice C (+12 m/s) incorrectly assumes the moving cart maintains its velocity after collision, ignoring that momentum must be shared among the larger combined mass—this would violate conservation of momentum. When solving momentum conservation problems: (1) define a positive direction (typically right or forward), (2) assign signs to all velocities based on direction, (3) calculate p_before = m₁v₁ᵢ + m₂v₂ᵢ (with signs), (4) set equal to p_after = m₁v₁f + m₂v₂f, then (5) solve algebraically for the unknown. A key distinction: in perfectly inelastic collisions, objects stick together and move with a common final velocity v_f = (m₁v₁ᵢ + m₂v₂ᵢ)/(m₁ + m₂), while in other collisions objects bounce apart with different final velocities that must be determined by both momentum conservation and additional information about the collision type.
Two sound waves travel through the same room air (so their speed is the same, v≈343 m/s). Wave 1 has frequency f1=250 Hz and Wave 2 has frequency f2=1000 Hz. How does the wavelength of Wave 1 compare to the wavelength of Wave 2?
Explanation: This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. For waves traveling at constant speed v, the equation v = fλ shows that frequency and wavelength are inversely proportional: if frequency increases by a factor of 4 (from 250 Hz to 1000 Hz), wavelength must decrease by the same factor to keep their product (wave speed) constant. Mathematically, since v₁ = v₂, we have f₁λ₁ = f₂λ₂, so λ₁/λ₂ = f₂/f₁ = 1000/250 = 4, meaning λ₁ is 4 times λ₂. Choice A is correct because it accurately describes the inverse relationship between f and λ, recognizing that the lower frequency wave (250 Hz) has 4 times the wavelength of the higher frequency wave (1000 Hz). Choice D incorrectly claims wavelength and frequency are independent in the same medium, when actually they're inversely proportional: as frequency increases, wavelength must decrease to maintain constant wave speed. The key insight is that at constant wave speed, frequency and wavelength are inversely related—quadruple the frequency means one-quarter the wavelength, which is why low-pitched sounds have long wavelengths and high-pitched sounds have short wavelengths.
A single rectangular loop is rotated at constant speed between the poles of a strong magnet, so the angle θ between the magnetic field and the loop’s area normal changes with time. The loop is connected to a load, forming a closed circuit. Which statement correctly explains why an alternating voltage is produced?
Explanation: This question tests understanding of electromagnetic induction in rotating coils, specifically how continuous rotation produces alternating voltage through changing magnetic flux. Faraday's law states that induced EMF depends on the rate of flux change, where flux Φ = BA cos(θ) varies as the angle θ between the field and the loop's area normal changes—as the loop rotates at constant speed, θ changes continuously, causing cos(θ) to vary from +1 to -1 and back, making the flux oscillate between maximum positive and maximum negative values. As the loop rotates in the magnetic field, the angle θ changes continuously, causing flux Φ = BA cos(θ) to change from maximum (Φ = BA when θ = 0°, loop perpendicular to field) through zero (Φ = 0 when θ = 90°, loop parallel to field) to negative maximum (Φ = -BA when θ = 180°) and back—this continuous flux change induces a continuously varying EMF that alternates in direction, producing alternating current (AC). Choice A is correct because it accurately identifies that the changing angle during rotation causes the flux Φ = BA cos(θ) to change continuously, and by Faraday's law this changing flux induces an EMF that changes sign as cos(θ) goes from positive to negative. Choice C incorrectly claims the field must reverse direction each half-turn to produce AC, but the field remains constant—it's the changing orientation (and thus changing flux) that produces the alternating voltage, not any change in the field itself. To understand AC generation: (1) as the loop rotates, angle θ between field and area normal changes continuously, (2) flux Φ = BA cos(θ) varies sinusoidally with rotation, (3) by Faraday's law, EMF = -N(dΦ/dt) = NBA ω sin(ωt) where ω is angular velocity, (4) the sinusoidal EMF alternates between positive and negative, driving alternating current. This is the fundamental principle of all AC generators: mechanical rotation in a magnetic field produces electrical power through continuous flux change, explaining how power plants convert mechanical energy (turbine rotation) to electrical energy, with the rotation frequency determining the AC frequency (60 Hz in the US means 60 rotations per second).
In the Earth–Moon system, the mass of Earth is ME=6.0×1024 kg and the mass of the Moon is MMoon=7.3×1022 kg. The distance between their centers is r=3.8×108 m. Using Newton’s Law of Universal Gravitation, F=Gr2m1m2 with G=6.67×10−11 N⋅m2/kg2, what is the magnitude of the gravitational force between Earth and the Moon?
Explanation: This question tests understanding of Newton's Law of Universal Gravitation and the ability to calculate gravitational force between two masses. Newton's Law of Universal Gravitation states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers: F = G(m₁m₂)/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, m₁ and m₂ are the masses in kilograms, r is the center-to-center distance in meters, and F is the attractive force in Newtons. To find the gravitational force between Earth with mass m₁ = 6.0 × 10²⁴ kg and the Moon with mass m₂ = 7.3 × 10²² kg separated by distance r = 3.8 × 10⁸ m, we substitute into F = G(m₁m₂)/r²: F = (6.67 × 10⁻¹¹ N·m²/kg²)(6.0 × 10²⁴ kg)(7.3 × 10²² kg)/(3.8 × 10⁸ m)² = (6.67 × 10⁻¹¹)(4.38 × 10⁴⁷)/(1.444 × 10¹⁷) = (2.92 × 10³⁷)/(1.444 × 10¹⁷) = 2.02 × 10²⁰ N. Choice B is correct because it properly applies F = G(m₁m₂)/r² with correct values and units, handling scientific notation accurately to yield 2.0 × 10²⁰ N. Choice A uses an incorrect power of 10 in scientific notation (11 instead of 20), likely from an error in subtracting exponents when dividing powers of 10. When calculating gravitational force: (1) identify both masses in kg and distance between centers in m, (2) for objects on a planet's surface use r = radius, for objects in orbit use r = radius + altitude, (3) substitute carefully into F = G(m₁m₂)/r² keeping track of scientific notation, (4) check that the result makes sense: gravitational forces between everyday objects are tiny (~10⁻⁸ N), between person and Earth are hundreds of Newtons (matching weight), and between astronomical objects are enormous (~10²⁰ N or larger). The key insight of the inverse square law is that gravitational force decreases rapidly with distance: double the distance and force drops to 1/4, triple the distance and force drops to 1/9—this r² in the denominator makes gravity a short-range force in practical terms, though it technically extends infinitely.
A battery charging system receives Ein=4000 J of electrical energy. Measurements show Elost=600 J is dissipated as heat. How much useful chemical energy is stored in the battery, Eout?
Use: Ein=Eout+Elost
Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this battery charging system, the input energy is E_in = 4000 J and the waste energy is E_lost = 600 J, so using E_in = E_out + E_lost, we get E_out = E_in - E_lost = 4000 - 600 = 3400 J. The remaining energy E_waste = 600 J is lost as thermal energy, which is characteristic of battery chargers where electrical resistance and chemical reactions produce heat. Choice B is correct because it properly calculates useful energy stored as E_out = E_in - E_lost = 4000 - 600 = 3400 J. Choice A (4600 J) incorrectly adds energies instead of subtracting; Choice C (2400 J) makes an arithmetic error; Choice D (0.85 J) appears to calculate efficiency as a decimal but gives energy units. To calculate efficiency: (1) identify total input energy, (2) identify waste energy, (3) compute useful output = input - waste, (4) verify output is positive and less than input. Higher efficiency means more input converts to useful chemical energy, less wasted as heat.
Two masses are connected by a light rope over a frictionless pulley. Mass A (mA=4.0kg) falls 2.0m while mass B (mB=2.0kg) rises 2.0m. The system starts from rest and, after moving 2.0m, both masses have speed v=3.0m/s. Take g=9.8m/s2. How much energy is converted to thermal energy (e.g., due to axle friction/air resistance) during this motion?
Explanation: This question tests understanding of how energy transfers between objects through physical interactions like collisions or forces doing work. When objects interact, energy can transfer from one to another through work done by forces during the interaction: in collisions, contact forces during impact transfer kinetic energy between objects (some may convert to thermal in inelastic collisions), and when one object pushes or pulls another, the applied force does work W = Fd transferring energy from the source (person, falling object, spring) to the recipient object as kinetic or potential energy. Mass A loses gravitational PE: ΔPE_A = m_A g h = (4.0)(9.8)(2.0) = 78.4 J. Mass B gains gravitational PE: ΔPE_B = m_B g h = (2.0)(9.8)(2.0) = 39.2 J. Both masses gain kinetic energy: KE_total = ½(m_A + m_B)v² = ½(6.0)(3.0)² = 27 J. Energy from A's PE loss goes to: B's PE gain (39.2 J) + total KE gain (27 J) + thermal energy. Therefore: E_thermal = 78.4 - 39.2 - 27 = 12.2 J, demonstrating energy transfer through rope tension with partial dissipation. Choice C is correct because it accurately calculates thermal energy as A's PE loss minus B's PE gain minus total KE gain: 78.4 - 39.2 - 27 = 12.2 J. Choice A (0.8 J) makes a calculation error; choice B (5.0 J) appears arbitrary; choice D (39.2 J) incorrectly uses B's PE gain as thermal energy. To analyze energy transfer through interactions: (1) calculate initial energy source (A's PE loss = 78.4 J), (2) identify all energy gains (B's PE = 39.2 J, total KE = 27 J), (3) thermal energy = source - all gains = 78.4 - 39.2 - 27 = 12.2 J, (4) verify conservation: 78.4 J (lost) = 39.2 J (B's PE) + 27 J (KE) + 12.2 J (thermal), (5) mechanism: rope tension transfers energy from A to B while some dissipates. Key insight: in real pulley systems, not all of falling mass's PE transfers to rising mass and kinetic energy; some converts to thermal due to friction/air resistance.
A 4kg cart on a horizontal track experiences an applied force Fapp=25N to the right and a friction force Ff=5N to the left. Using g=10m/s2, what is the cart's acceleration (magnitude and direction)?
Explanation: This question tests understanding of net force and acceleration using Newton's Second Law. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when forces don't balance, the object accelerates in the direction of the net force with magnitude a = F_net / m. The forces acting on the cart horizontally are: applied force F_app = 25 N to the right and friction force F_f = 5 N to the left. The net horizontal force is F_net = 25 N - 5 N = 20 N to the right. Using Newton's Second Law: a = F_net / m = 20 N / 4 kg = 5 m/s² to the right, in the direction of the net force. Choice A is correct because it correctly calculates net force as the vector sum with directions and applies Newton's Second Law to find acceleration. Choice B incorrectly divides the applied force by mass without accounting for friction (a = 25 N / 4 kg = 6.25 m/s²)—acceleration depends on net force, not just the applied force. When calculating acceleration from forces: (1) find the net force by vector addition of all forces, (2) apply F_net = ma to find acceleration magnitude, (3) remember acceleration direction always matches net force direction, and (4) include all forces when finding net force—don't forget opposing forces like friction.
A car at v=25 m/s must stop with an average deceleration magnitude less than amax=30 m/s2 to reduce injury risk. Assume constant deceleration and use v2=2ad to estimate the minimum stopping distance required.
Front-end designs (maximum controlled deformation distance available):
Vehicle constraint: d≤10.5 m. Which design meets both the deceleration limit and the vehicle constraint?
Explanation: This question tests understanding of evaluating collision protection designs using kinematics to meet deceleration and space constraints. Using v² = 2ad, we can find the minimum stopping distance needed: d = v²/(2a) = 25²/(2×30) = 625/60 = 10.42 m to keep deceleration below 30 m/s². Design C provides exactly 10.5 m, which exceeds the minimum 10.42 m needed and fits within the vehicle constraint of d ≤ 10.5 m. Checking designs against the minimum: Design A (8 m) gives a = 625/16 = 39.1 m/s² (exceeds limit), Design B (12 m) would give a = 625/24 = 26.0 m/s² but exceeds vehicle constraint, Design C (10.5 m) gives a = 625/21 = 29.8 m/s² (meets limit), Design D (6 m) gives a = 625/12 = 52.1 m/s² (exceeds limit). Choice B is correct because Design C provides at least the minimum stopping distance (10.42 m) with its 10.5 m design and fits exactly within the 10.5 m vehicle constraint. Choice D incorrectly suggests Design B despite it violating the size constraint at 12 m > 10.5 m maximum. When evaluating crumple zone designs: (1) calculate minimum distance d = v²/(2a_max), (2) check each design provides d ≥ minimum, (3) verify design fits within vehicle constraints, (4) eliminate designs failing either requirement, and (5) select the feasible design meeting both criteria. Physical constraints often limit ideal safety designs.
A bicycle helmet is being redesigned to reduce concussion risk. In a fall, a rider’s head (modeled as 5.0 kg effective mass) hits the ground at 6.0 m/s and comes to rest. Without a helmet, the collision time is about 0.010 s; with a helmet liner that compresses, the collision time can be increased to 0.060 s. Using the impulse-momentum theorem J=Δp=FavgΔt, which statement best explains why the compressible liner improves safety?
Explanation: This question tests understanding of applying physics principles—specifically impulse-momentum and work-energy relationships—to design collision protection systems. The impulse-momentum theorem (J = Δp = F_avg × Δt) shows that for a given momentum change (stopping an object: Δp = mv), the average force experienced is inversely proportional to the collision time: F_avg = Δp/Δt, which means extending the collision time Δt reduces the average force F_avg—this is the fundamental principle behind crumple zones, airbags, padding, and other safety features that increase the time over which an object comes to rest. For padding/helmets: Protective padding like in helmets or sports equipment extends the collision time by compressing during impact—for example, if a 5 kg helmet and head decelerate from 5 m/s to 0 in a fall, the momentum change is Δp = (5 kg)(5 m/s) = 25 kg⋅m/s. Without padding (Δt ≈ 0.005 s), F_avg = 25/0.005 = 5000 N, but with 2 cm of foam padding that compresses during impact (Δt ≈ 0.02 s), F_avg = 25/0.02 = 1250 N—a factor of 4 reduction that could prevent skull fracture. Choice A is correct because it applies impulse-momentum theorem showing that extending collision time reduces force for the same momentum change. Choice C confuses elastic bouncing with energy absorption, suggesting materials that bounce back are best for protection, when actually plastic deformation (permanent crushing that doesn't bounce) absorbs the most energy and is ideal for one-time protection like vehicle crashes—elastic materials store and release energy, which can cause secondary impacts. When designing collision protection systems, apply two key physics principles: (1) Impulse-momentum theorem F_avg = Δp/Δt shows that for a given momentum change (stopping an object), extending the collision time Δt reduces the average force F_avg—achieve this through crumple zones, padding compression, airbag deflation, or seat belt stretching; (2) Work-energy principle W = Fd shows that for a fixed kinetic energy to absorb (KE = ½mv²), increasing the deformation distance d reduces the required force F—achieve this through crushable materials, thick padding, or structures designed to fold/compress over longer distances. Remember that effective collision protection typically involves: (a) materials that deform plastically (crush permanently) rather than elastically (bounce back), because plastic deformation maximizes energy absorption, (b) progressive resistance that increases gradually with deformation rather than sudden stiffening (avoids peak force spikes), (c) distribution of forces over large surface areas to reduce local pressure (airbags spread over chest, helmets over skull), and (d) multi-stage systems where soft initial padding handles minor impacts comfortably while firmer secondary layers engage for severe impacts—the goal is always to extend Δt and increase d while keeping F below injury thresholds at all times during the collision.
An electric motor has efficiency η=80% and delivers useful mechanical power Pout=320 W. What electrical input power Pin is required? (Use η=(Pout/Pin)×100%.)
Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this motor with efficiency η = 80% and output power P_out = 320 W, we need to find input power: η = (P_out/P_in) × 100%, so 80 = (320/P_in) × 100, which gives P_in = 320/(0.80) = 400 W. The remaining power P_waste = P_in - P_out = 80 W is lost as thermal energy, which is characteristic of electric motors where electrical resistance and friction cause heating. Choice B is correct because it properly calculates input power from η = (P_out/P_in) × 100%, rearranging to P_in = P_out/(η/100) = 320/0.80 = 400 W. Choice A incorrectly multiplies 320 × 0.80 = 256 W; Choice C inverts the efficiency calculation; Choice D makes an arithmetic error. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted.