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College Physics I — Mechanics, Heat and Waves flashcards

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Algebra-based mechanics and thermal physics with worked numbers: unit conversions, kinematics, Newton's laws, circular motion, energy, momentum, statics, rotation, fluids, gas laws, heat, heat engines, oscillations, waves and sound — plus the common homework mistakes.

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Learn deck · part of College Physics I/II (algebra-based) · written against: Standard two-semester algebra-based sequence; OpenStax College Physics 2e, consulted as a reference; no text copied. Updated 2026-10-10.

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  1. What is 72 km/h in meters per second, and what single factor makes km/h-to-m/s conversions quick?
    20 m/s. Since 1 km/h = 1000 m ÷ 3600 s, you divide by 3.6 to go from km/h to m/s and multiply by 3.6 to go back.

    Watch out: mixing km/h with g = 9.8 m/s² in one equation is one of the most common sources of wildly wrong answers.
    Units & Measurement
  2. A displacement of 50 m points 37° south of west. With east and north positive, what are its components?
    x ≈ −40 m, y ≈ −30 m. The angle is measured from the west axis, so the west part uses cosine (50 cos 37° ≈ 40 m) and the south part uses sine (50 sin 37° ≈ 30 m); both components are negative.

    Watch out: cosine goes with the side adjacent to the angle you were given, which is not always the x-axis.
    Vectors & Projectile Motion
  3. A student draws a free-body diagram with an arrow labeled "ma" alongside gravity and the normal force. What is wrong?
    ma is not a force — it is the result of the net force. A free-body diagram shows only real forces from interactions (gravity, normal, tension, friction, applied); their vector sum is then set equal to ma.

    Watch out: "centripetal force" is the same trap — it is the net inward force, not an extra arrow.
    Newton's Laws & Friction
  4. How many joules are in a kilowatt-hour, and how much energy does a 1500 W heater use in 3.0 hours?
    1 kWh = 3.6 × 10⁶ J (1000 W × 3600 s). The heater uses 1.5 kW × 3.0 h = 4.5 kWh, which is 1.6 × 10⁷ J.

    Watch out: the kilowatt-hour is a unit of energy, not power — utilities bill for kWh, and the kW rating is how fast it is used.
    Work & Energy
  5. A 0.045 kg golf ball leaves the club at 60 m/s after a contact time of 0.50 ms. What average force did the club exert?
    The impulse is Δp = 0.045 × 60 = 2.7 kg·m/s, so F = Δp ÷ Δt = 2.7 ÷ (5.0 × 10⁻⁴ s) ≈ 5.4 × 10³ N.

    Watch out: convert milliseconds to seconds; leaving Δt as 0.50 gives an answer a thousand times too small.
    Linear Momentum
  6. A turntable spins at 45 rpm. What is its angular speed in rad/s?
    Multiply by 2π rad per revolution and divide by 60 s per minute: 45 × 2π ÷ 60 ≈ 4.7 rad/s.

    Mnemonic: rpm → rad/s, "times 2π, over 60" (about ÷ 9.55).
    Rotational Motion
  7. Standard atmospheric pressure is how many pascals, millimeters of mercury and pounds per square inch?
    1 atm ≈ 1.013 × 10⁵ Pa = 760 mmHg ≈ 14.7 psi (101.3 kPa).

    Watch out: many formulas, including the ideal gas law in SI form (R = 8.31 J/(mol·K)), need pascals; a pressure in kPa or atm must be converted first.
    Fluids
  8. How much heat turns 0.20 kg of ice at −10 °C into steam at 100 °C? (cicec_{ice} ≈ 2090 J/(kg·°C), cwaterc_{water} ≈ 4186 J/(kg·°C), LfL_f = 3.34 × 10⁵ J/kg, LvL_v = 2.256 × 10⁶ J/kg)
    Add four steps: warm the ice (4.2 kJ) + melt it (66.8 kJ) + warm the water to 100 °C (83.7 kJ) + boil it (451.2 kJ) ≈ 6.1 × 10⁵ J. Vaporization alone is about three quarters of the total.

    Watch out: temperature stays constant during each phase change, so Q = mcΔT cannot span a melting or boiling point.
    Heat & Heat Transfer
  9. OpenStax writes the first law as ΔU = Q − W, while some courses write ΔU = Q + W. How can both be right?
    They define W differently: in ΔU = Q − W, W is work done by the gas; in ΔU = Q + W, W is work done on the gas.

    Both say the same thing — heat in and work done on the gas raise its internal energy.

    Watch out: decide which convention your course uses before plugging in signs, and never mix the two in one problem.
    Thermodynamics
  10. How does the speed of sound in air depend on temperature, and what is it at 30 °C?
    It rises with temperature, roughly v ≈ 331 m/s + (0.6 m/s per °C) × T, giving about 349 m/s at 30 °C. More precisely v is proportional to T\sqrt{T} with T in kelvins.

    Why: warmer molecules move faster and pass the disturbance along more quickly.
    Sound

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  • OpenStax College Physics 2e, consulted as a reference; no text copied.

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