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Without the unit, the number alone does not identify the measured quantity’s scale."},{"text":"A direction.","correct":false,"explanation":"Direction is required for a vector quantity, but not for every physical measurement. A number must be accompanied by a unit to specify its measurement scale."},{"text":"A formula.","correct":false,"explanation":"A formula may describe how a quantity is calculated, but it does not complete a numerical measurement. The number must be accompanied by a unit."},{"text":"A prefix.","correct":false,"explanation":"A prefix changes a unit’s scale, as in kilo- or milli-, but is not always needed. Every complete physical measurement must include a unit."}]},{"id":"c9413361-4b17-4dc2-be27-5f41057cef97","cardType":"cloze","points":null,"front":"Mass uses [ ]; length uses [ ].","back":"Mass uses kilogram (kg); length uses metre (m).","mcqOptions":[{"text":"Mass uses kilogram (kg); length uses metre (m).","correct":true,"explanation":"The kilogram is the SI base unit of mass, while the metre is the SI base unit of length. Their symbols are kg and m respectively."},{"text":"Mass uses gram (g); length uses centimetre (cm).","correct":false,"explanation":"Grams and centimetres are commonly used metric units, but neither is the relevant SI base unit. Mass uses kilogram (kg), and length uses metre (m)."},{"text":"Mass uses metre (m); length uses kilogram (kg).","correct":false,"explanation":"This reverses the two base units. Mass is measured in kilograms (kg), whereas length is measured in metres (m)."},{"text":"Mass uses newton (N); length uses metre (m).","correct":false,"explanation":"A newton measures force or weight, not mass. The SI base unit of mass is kilogram (kg), while the metre (m) correctly measures length."}]},{"id":"5d66c322-31f9-42db-8cc6-242f00ff8466","cardType":"cloze","points":null,"front":"Time uses [ ]; electric current uses [ ].","back":"Time uses second (s); electric current uses ampere (A).","mcqOptions":[{"text":"Time uses second (s); electric current uses ampere (A).","correct":true,"explanation":"The second is the SI base unit of time, and the ampere is the SI base unit of electric current. Their symbols are s and A."},{"text":"Time uses hour (h); electric current uses ampere (A).","correct":false,"explanation":"Hours are accepted for practical time measurements but are not the SI base unit. Time uses the second (s), while electric current uses the ampere (A)."},{"text":"Time uses second (s); electric current uses coulomb (C).","correct":false,"explanation":"The coulomb measures electric charge, not electric current. The SI base unit of current is the ampere (A), while time uses the second (s)."},{"text":"Time uses minute (min); electric current uses volt (V).","correct":false,"explanation":"Minutes are not the SI base unit of time, and volts measure potential difference rather than current. The correct base units are second (s) and ampere (A)."}]},{"id":"f1a03530-18d0-4815-8eec-12469d3e5eee","cardType":"cloze","points":null,"front":"Thermodynamic temperature uses [ ]; amount of substance uses [ ].","back":"Thermodynamic temperature uses kelvin (K); amount of substance uses mole (mol).","mcqOptions":[{"text":"Thermodynamic temperature uses kelvin (K); amount of substance uses mole (mol).","correct":true,"explanation":"The kelvin is the SI base unit of thermodynamic temperature, and the mole is the SI base unit of amount of substance. Their symbols are K and mol."},{"text":"Thermodynamic temperature uses degree Celsius (°C); amount of substance uses mole (mol).","correct":false,"explanation":"Degrees Celsius are widely used, but the SI base unit of thermodynamic temperature is the kelvin (K). Amount of substance is correctly measured in moles (mol)."},{"text":"Thermodynamic temperature uses kelvin (K); amount of substance uses kilogram (kg).","correct":false,"explanation":"A kilogram measures mass, not amount of substance. Amount of substance uses the mole (mol), while thermodynamic temperature correctly uses the kelvin (K)."},{"text":"Thermodynamic temperature uses joule (J); amount of substance uses gram (g).","correct":false,"explanation":"Joules measure energy and grams measure mass, so neither is the required base unit here. The correct units are kelvin (K) and mole (mol)."}]},{"id":"0ab3f155-e40c-406c-bc0a-232f837ce492","cardType":"basic","points":1,"front":"What is the SI base unit for luminous intensity?","back":"Candela (cd).","mcqOptions":[{"text":"Candela (cd).","correct":true,"explanation":"The candela, symbol cd, is the SI base unit of luminous intensity. It describes the perceived power of visible light emitted in a particular direction."},{"text":"Lumen (lm).","correct":false,"explanation":"The lumen is a derived unit of luminous flux, representing the total perceived light output. Luminous intensity instead uses the SI base unit candela (cd)."},{"text":"Lux (lx).","correct":false,"explanation":"The lux measures illuminance, or luminous flux received per unit area. The SI base unit for luminous intensity is the candela (cd)."},{"text":"Watt (W).","correct":false,"explanation":"The watt measures power and does not account for the eye’s sensitivity to different wavelengths. Luminous intensity is measured in candela (cd)."}]},{"id":"4dc02300-8e0d-40cd-bcaa-304157ae0b5c","cardType":"basic","points":1,"front":"What is wrong with writing 5 kgs?","back":"Unit symbols are not pluralised: write 5 kg.","mcqOptions":[{"text":"Unit symbols are not pluralised: write 5 kg.","correct":true,"explanation":"SI unit symbols remain unchanged after plural numerical values, so kg never gains an s. A space separates the value from the symbol, giving 5 kg."},{"text":"Unit symbols are always pluralised: write 5 kgs.","correct":false,"explanation":"This confuses unit symbols with written unit names, which may take plural forms. SI symbols never take a plural s, so the correct notation is 5 kg."},{"text":"Unit names are not pluralised: write 5 kilogram.","correct":false,"explanation":"When a unit name is written out after a value other than one, it is normally pluralised as kilograms. Using the symbol, the correct notation is 5 kg."},{"text":"Unit symbols are written without spaces: write 5kg.","correct":false,"explanation":"SI notation requires a space between a numerical value and its unit symbol. The symbol must also remain unpluralised, so the correct form is 5 kg."}]},{"id":"74895569-5c6d-4771-8c11-1c9a167b998f","cardType":"basic","points":1,"front":"What is wrong with writing 300 °K?","back":"Kelvin uses K, not °K.","mcqOptions":[{"text":"Kelvin uses K, not °K.","correct":true,"explanation":"The kelvin is an SI base unit whose symbol is K, written without a degree sign. Degree symbols belong with Celsius and other angular measurements."},{"text":"Kelvin uses °C, not °K.","correct":false,"explanation":"The degree sign is not used with kelvin, but Celsius is not the appropriate replacement here. The SI symbol for kelvin is K without a degree sign."},{"text":"Kelvin uses °K, not K.","correct":false,"explanation":"Writing °K incorrectly adds a degree sign to the kelvin symbol. The correct convention is K, with no degree sign, even when stating an absolute temperature."},{"text":"Kelvin uses K, not °C.","correct":false,"explanation":"Celsius is also written with a degree sign, so it does not demonstrate the error in the expression given. Kelvin must be written K, without °."}]},{"id":"3ad99735-5687-4e06-810e-aac8b58345a4","cardType":"basic","points":1,"front":"What are the SI units of speed?","back":"m s^-1.","mcqOptions":[{"text":"m s^-1.","correct":true,"explanation":"Speed is distance divided by time. Using the SI base units metre and second therefore gives metres per second, written m s^-1."},{"text":"m s^-2.","correct":false,"explanation":"Metres per second squared measures acceleration, which is the rate of change of velocity. Speed is distance divided by time, so its units are m s^-1."},{"text":"m s.","correct":false,"explanation":"Multiplying distance by time gives m s, which does not represent speed. Speed requires distance divided by time, producing the SI units m s^-1."},{"text":"kg m s^-1.","correct":false,"explanation":"Including kilograms gives the units of momentum because momentum equals mass multiplied by velocity. Speed contains no mass term and has units m s^-1."}]},{"id":"71f73b12-3934-4703-b8ed-41099289129b","cardType":"basic","points":1,"front":"What are the SI units of acceleration?","back":"m s^-2.","mcqOptions":[{"text":"m s^-2.","correct":true,"explanation":"Acceleration is velocity change per unit time. Since velocity has units m s^-1, dividing by seconds gives m s^-2."},{"text":"m s^-1.","correct":false,"explanation":"These are the units of velocity or speed, because they contain only one factor of time in the denominator. Acceleration requires a second division by time."},{"text":"kg m s^-2.","correct":false,"explanation":"This expression has the dimensions of acceleration multiplied by mass, corresponding to force rather than acceleration. Acceleration alone is measured in m s^-2."},{"text":"m² s^-2.","correct":false,"explanation":"Squaring the distance unit is a common dimensional slip. Acceleration involves distance divided by time squared, not area divided by time squared."}]},{"id":"d2b8b385-a91c-48ce-bc0c-7c5f80a40687","cardType":"basic","points":1,"front":"From $p=mv$, what are the SI units of momentum?","back":"kg m s^-1.","mcqOptions":[{"text":"kg m s^-1.","correct":true,"explanation":"Substituting mass in kilograms, velocity in metres per second, and multiplying gives momentum in kg m s^-1."},{"text":"kg m s^-2.","correct":false,"explanation":"These are the units of force, obtained when velocity is incorrectly treated as acceleration. Momentum uses one power of time in the denominator, not two."},{"text":"kg m² s^-1.","correct":false,"explanation":"The extra metre comes from confusing momentum with energy or work-related dimensions. Multiplying mass by velocity gives only one metre factor."},{"text":"kg m s.","correct":false,"explanation":"This reverses the time factor and represents a different quantity. Momentum is mass multiplied by velocity, so seconds must remain in the denominator."}]},{"id":"cdd1cb0d-2db0-41fa-8ca1-d519c636f9aa","cardType":"basic","points":1,"front":"From $ρ=m/V$, what are the SI units of density?","back":"kg m^-3.","mcqOptions":[{"text":"kg m^-3.","correct":true,"explanation":"Density is mass divided by volume. A kilogram divided by cubic metres therefore has units kg m^-3."},{"text":"kg m^-2","correct":false,"explanation":"This uses area rather than volume in the denominator, giving mass per unit area. Density specifically requires mass divided by cubic metres."},{"text":"kg^-1 m³","correct":false,"explanation":"This is the inverse dimensional relationship, corresponding to volume per unit mass or specific volume. Density is mass per unit volume."},{"text":"kg m^-2 s^-1","correct":false,"explanation":"The extra time factor suggests a rate of mass flow through an area, not density. Density contains mass and volume only, with no time unit."}]},{"id":"74d28023-4199-4659-bfef-f57860a268c7","cardType":"cloze","points":null,"front":"For frequency, 1 Hz = [ ]; for force, 1 N = [ ].","back":"For frequency, 1 Hz = s^-1; for force, 1 N = kg m s^-2.","mcqOptions":[{"text":"For frequency, 1 Hz = s^-1; for force, 1 N = kg m s^-2.","correct":true,"explanation":"Frequency counts cycles per second, so its unit is the reciprocal of the second. Force is mass multiplied by acceleration, giving kilograms metres per second squared."},{"text":"For frequency, 1 Hz = s^-1; for force, 1 N = kg m s^-1.","correct":false,"explanation":"The frequency relationship is correct, but the force expression has only one inverse second and therefore represents momentum rather than force. Acceleration contributes s^-2."},{"text":"For frequency, 1 Hz = s; for force, 1 N = kg m s^-2.","correct":false,"explanation":"The force expression is correct, but frequency has been confused with time period. A hertz means cycles per second, so it is s^-1, not s."},{"text":"For frequency, 1 Hz = s; for force, 1 N = kg m s^-1.","correct":false,"explanation":"Both relationships contain dimensional errors: frequency is not measured in seconds, and force needs mass as well as acceleration. The correct forms are s^-1 and kg m s^-2."}]},{"id":"241baea6-187c-43b1-92e5-b165e2085456","cardType":"cloze","points":null,"front":"For energy or work done, 1 J = [ ]; for power, 1 W = [ ].","back":"For energy or work done, 1 J = kg m² s^-2; for power, 1 W = kg m² s^-3.","mcqOptions":[{"text":"For energy or work done, 1 J = kg m² s^-2; for power, 1 W = kg m² s^-3.","correct":true,"explanation":"Work is force multiplied by distance, giving kg m² s^-2. Power is energy transferred per second, adding another inverse second to give kg m² s^-3."},{"text":"For energy or work done, 1 J = kg m² s^-2; for power, 1 W = kg m² s^-2.","correct":false,"explanation":"The energy expression is correct, but power has been given the same time dependence as energy. Dividing energy by time introduces one additional factor of s^-1."},{"text":"For energy or work done, 1 J = kg m² s^-3; for power, 1 W = kg m² s^-3.","correct":false,"explanation":"The power expression is correct, but energy has been assigned an extra inverse second. Energy is force times distance, so its time factor is s^-2."},{"text":"For energy or work done, 1 J = kg m s^-2; for power, 1 W = kg m s^-3.","correct":false,"explanation":"The distance factor has been reduced incorrectly in both expressions. Work and energy involve force through a distance, producing m²; power then adds one inverse second."}]},{"id":"dba55ca5-4559-4df9-a5f2-86cee2220570","cardType":"cloze","points":null,"front":"For pressure, 1 Pa = [ ]; for charge, 1 C = [ ].","back":"For pressure, 1 Pa = kg m^-1 s^-2; for charge, 1 C = A s.","mcqOptions":[{"text":"For pressure, 1 Pa = kg m^-1 s^-2; for charge, 1 C = A s.","correct":true,"explanation":"Pressure is force divided by area, so its dimensions are kg m^-1 s^-2. Charge is current multiplied by time, giving A s."},{"text":"For pressure, 1 Pa = kg m^-1 s^-2; for charge, 1 C = A.","correct":false,"explanation":"The pressure expression is correct, but charge has been confused with current. Charge accumulates when current flows for a time, so its unit is A s."},{"text":"For pressure, 1 Pa = kg m s^-2; for charge, 1 C = A s.","correct":false,"explanation":"The charge expression is correct, but pressure has been given the dimensions of force rather than force per area. Dividing by area adds an inverse metre."},{"text":"For pressure, 1 Pa = kg m s^-2; for charge, 1 C = A s^-1.","correct":false,"explanation":"This reverses the area factor and gives neither pressure nor charge correctly. Pressure divides force by area, while charge multiplies current by time."}]},{"id":"485323e8-17a1-4482-b27a-8635884cc310","cardType":"cloze","points":null,"front":"For potential difference, 1 V = [ ]; for resistance, 1 Ω = [ ].","back":"For potential difference, 1 V = kg m² s^-3 A^-1; for resistance, 1 Ω = kg m² s^-3 A^-2.","mcqOptions":[{"text":"For potential difference, 1 V = kg m² s^-3 A^-1; for resistance, 1 Ω = kg m² s^-3 A^-2.","correct":true,"explanation":"Potential difference is energy per charge, producing kg m² s^-3 A^-1. Resistance is potential difference per current, adding another A^-1."},{"text":"For potential difference, 1 V = kg m² s^-3 A^-1; for resistance, 1 Ω = kg m² s^-3 A^-1.","correct":false,"explanation":"The potential-difference expression is correct, but resistance has been confused with potential difference. Ohm’s law divides voltage by current, introducing a second inverse ampere overall."},{"text":"For potential difference, 1 V = kg m² s^-3 A^-2; for resistance, 1 Ω = kg m² s^-3 A^-2.","correct":false,"explanation":"The resistance expression is correct, but potential difference has been assigned resistance’s extra inverse ampere. Voltage contains A^-1, whereas resistance contains A^-2."},{"text":"For potential difference, 1 V = kg m² s^-2 A^-1; for resistance, 1 Ω = kg m² s^-2 A^-2.","correct":false,"explanation":"The metre and second powers here match energy rather than electrical potential difference. Dividing energy by charge introduces s^-1 and A^-1."}]},{"id":"a98caea3-f4cd-4a33-bacb-694c8229fe7b","cardType":"cloze","points":null,"front":"For capacitance, 1 F = [ ]; for magnetic flux density, 1 T = [ ].","back":"For capacitance, 1 F = kg^-1 m^-2 s⁴ A²; for magnetic flux density, 1 T = kg s^-2 A^-1.","mcqOptions":[{"text":"For capacitance, 1 F = kg^-1 m^-2 s⁴ A²; for magnetic flux density, 1 T = kg s^-2 A^-1.","correct":true,"explanation":"Capacitance is charge divided by potential difference, giving kg^-1 m^-2 s⁴ A². From $F=BIl$, magnetic flux density has base units kg s^-2 A^-1."},{"text":"For capacitance, 1 F = kg^-1 m^-2 s³ A²; for magnetic flux density, 1 T = kg s^-2 A^-1.","correct":false,"explanation":"The capacitance expression loses one factor of time, commonly from treating charge as current rather than current multiplied by time. The correct power is s⁴, while the tesla expression is correct."},{"text":"For capacitance, 1 F = kg^-1 m^-2 s⁴ A²; for magnetic flux density, 1 T = kg s^-3 A^-1.","correct":false,"explanation":"The extra inverse second in the tesla expression results from confusing magnetic flux density with its rate of change. From $B=F/(Il)$, the correct time factor is s^-2."},{"text":"For capacitance, 1 F = kg m² s^-4 A^-2; for magnetic flux density, 1 T = kg s^-2 A.","correct":false,"explanation":"The capacitance dimensions have been inverted, as if using voltage divided by charge, and the current exponent for tesla has the wrong sign. The correct expressions use A² and A^-1, respectively."}]},{"id":"43b69fdc-932b-4f8a-85e7-6a74132e3e62","cardType":"basic","points":1,"front":"What is one weber in SI base units?","back":"kg m² s^-2 A^-1.","mcqOptions":[{"text":"kg m² s^-2 A^-1.","correct":true,"explanation":"Magnetic flux equals voltage multiplied by time. Since a volt is kg m² s^-3 A^-1, multiplying by seconds gives kg m² s^-2 A^-1."},{"text":"kg s^-2 A^-1.","correct":false,"explanation":"These are the base units of a tesla, not a weber. A weber measures magnetic flux and has SI base units kg m² s^-2 A^-1."},{"text":"kg m² s^-3 A^-1.","correct":false,"explanation":"These are the base units of a volt, which measures potential difference. Multiplying voltage by time gives a weber, so its time exponent is -2 rather than -3."},{"text":"kg m² s^-2 A^-2.","correct":false,"explanation":"These are the base units of a henry, introducing an extra inverse ampere. A weber is magnetic flux, with base units kg m² s^-2 A^-1."}]},{"id":"6f3dd944-cb18-4f48-9a2f-c4f572ba6798","cardType":"basic","points":1,"front":"When T is used before a unit, as in TW, what multiplier does it mean?","back":"Tera, $10^{12}$.","mcqOptions":[{"text":"Tera, $10^{12}$.","correct":true,"explanation":"The uppercase prefix T denotes tera, which multiplies the following unit by $10^{12}$. Therefore, one terawatt is $10^{12}$ watts."},{"text":"Giga, $10^{9}$.","correct":false,"explanation":"Giga uses the uppercase prefix G and represents $10^9$, not the uppercase prefix T. T denotes tera, which has the multiplier $10^{12}$."},{"text":"Mega, $10^{6}$.","correct":false,"explanation":"The prefix for $10^6$ is mega, written M. An uppercase T instead denotes tera, so TW means $10^{12}$ watts."},{"text":"Pico, $10^{-12}$.","correct":false,"explanation":"The prefix for $10^{-12}$ is pico, written p; confusing it with tera reverses the sign of the exponent. Uppercase T means tera, or $10^{12}$."}]},{"id":"ea754df6-e50d-4e9d-bc20-f7f70a11e00d","cardType":"basic","points":1,"front":"Convert 4.7 kΩ into SI units.","back":"4.7 × 10³ Ω.","mcqOptions":[{"text":"4.7 × 10³ Ω.","correct":true,"explanation":"The prefix kilo means $10^3$, so multiplying 4.7 by $10^3$ converts kilohms into the SI unit of resistance, ohms."},{"text":"4.7 × 10^-3 Ω.","correct":false,"explanation":"Using $10^{-3}$ confuses kilo with milli. Kilo represents $10^3$, so 4.7 kΩ is 4.7 × 10³ Ω."},{"text":"4.7 × 10⁶ Ω.","correct":false,"explanation":"Using $10^6$ treats kilo as mega. Kilo contributes a factor of $10^3$, so the correct conversion is 4.7 × 10³ Ω."},{"text":"4.7 × 10² Ω.","correct":false,"explanation":"Using $10^2$ moves the decimal point only twice. The kilo prefix requires three places, giving 4.7 × 10³ Ω."}]},{"id":"bce1fe43-e7b4-41eb-8934-2b29cb1f36e6","cardType":"basic","points":1,"front":"Convert 3.2 mA into SI units.","back":"3.2 × 10^-3 A.","mcqOptions":[{"text":"3.2 × 10^-3 A.","correct":true,"explanation":"The prefix milli means $10^{-3}$. Therefore, converting milliamperes into amperes gives 3.2 × 10^-3 A."},{"text":"3.2 × 10³ A.","correct":false,"explanation":"A positive exponent confuses milli with kilo and makes the value larger rather than smaller. Milli means $10^{-3}$, giving 3.2 × 10^-3 A."},{"text":"3.2 × 10^-6 A.","correct":false,"explanation":"The exponent -6 belongs to the micro prefix, not milli. Since milli means $10^{-3}$, the current is 3.2 × 10^-3 A."},{"text":"3.2 × 10^-2 A.","correct":false,"explanation":"Using $10^{-2}$ moves the decimal point only twice. Converting milli to the base unit requires three places, producing 3.2 × 10^-3 A."}]},{"id":"f5ef8b51-9e2e-4212-b9cd-1aa83bbb7703","cardType":"basic","points":1,"front":"Convert 1 cm³ into m³.","back":"1 × 10^-6 m³.","mcqOptions":[{"text":"1 × 10^-6 m³.","correct":true,"explanation":"Since 1 cm equals $10^{-2}$ m, cubing the length conversion gives $(10^{-2})^3 = 10^{-6}$. Thus 1 cm³ equals 1 × 10^-6 m³."},{"text":"1 × 10^-2 m³.","correct":false,"explanation":"This applies the centimetre-to-metre conversion without cubing it. Volume requires $(10^{-2})^3$, so 1 cm³ equals 1 × 10^-6 m³."},{"text":"1 × 10^-4 m³.","correct":false,"explanation":"This squares the conversion factor as though converting area. Cubic centimetres require cubing $10^{-2}$, giving 1 × 10^-6 m³."},{"text":"1 × 10^-3 m³.","correct":false,"explanation":"This incorrectly associates cubic units with an exponent of -3. The entire linear conversion $10^{-2}$ must be cubed, producing $10^{-6}$ m³."}]},{"id":"0566f874-40cd-4413-a0da-6df2e1ae9dbd","cardType":"basic","points":1,"front":"For a homogeneous equation, what must be true of the units on both sides?","back":"The LHS and RHS units match.","mcqOptions":[{"text":"The LHS and RHS units match.","correct":true,"explanation":"A physically valid homogeneous equation must have identical units on its left-hand and right-hand sides. A mismatch shows that the proposed relationship cannot be dimensionally correct."},{"text":"The LHS and RHS numbers match.","correct":false,"explanation":"Homogeneity concerns units, not whether both sides happen to have equal numerical coefficients. The LHS and RHS must have matching units for the equation to be valid."},{"text":"The LHS units always cancel.","correct":false,"explanation":"Units do not need to cancel on the left-hand side unless that side is dimensionless. Homogeneity requires the complete LHS and RHS to have matching units."},{"text":"The RHS is always dimensionless.","correct":false,"explanation":"The right-hand side need not be dimensionless; it must instead have the same units as the left-hand side. Homogeneity requires matching units across the equation."}]},{"id":"004e6909-ad66-410e-aede-9f4ef03671fc","cardType":"basic","points":1,"front":"What unit rule applies to terms being added or subtracted?","back":"They must have the same units.","mcqOptions":[{"text":"They must have the same units.","correct":true,"explanation":"Addition and subtraction are meaningful only between quantities of the same physical type and units. For example, a time cannot be directly added to a distance."},{"text":"They must have the different units.","correct":false,"explanation":"Quantities with different units cannot be directly added or subtracted, even if conversion factors are available. They must first be expressed using the same units."},{"text":"They must have the opposite units.","correct":false,"explanation":"There is no concept of opposite units that permits addition or subtraction. Each term must represent the same kind of quantity and have the same units."},{"text":"They must have the reciprocal units.","correct":false,"explanation":"Reciprocal units can arise through division, but they do not permit terms to be added or subtracted. Added or subtracted terms must have the same units."}]},{"id":"d6c6819f-03c7-4dcb-8c7c-f3bc537fe2d9","cardType":"basic","points":1,"front":"In unit checking, what units do constants like $1/2$ and functions like $cosθ$ have?","back":"No units - they are dimensionless.","mcqOptions":[{"text":"No units - they are dimensionless.","correct":true,"explanation":"Pure numerical factors such as $1/2$ carry no units, and trigonometric functions return dimensionless ratios. Their arguments must also be dimensionless, with angles conventionally expressed in radians."},{"text":"Radian units - they measure angles.","correct":false,"explanation":"Radians are used to express angles, but a trigonometric function such as $cosθ$ evaluates to a dimensionless ratio. Pure numbers such as $1/2$ are also dimensionless."},{"text":"Matching units - they copy the variable.","correct":false,"explanation":"Constants such as $1/2$ do not acquire the units of nearby quantities. They are pure numbers, while functions such as $cosθ$ also produce dimensionless values."},{"text":"Inverse units - they divide the variable.","correct":false,"explanation":"Inverse units arise when one dimensional quantity is divided by another, not from pure numerical constants or trigonometric functions. Both $1/2$ and $cosθ$ are dimensionless."}]},{"id":"c37bca0d-f06c-40f2-8fb2-acb49b977a98","cardType":"basic","points":2,"front":"Why can a homogeneous equation still be incorrect?","back":"It may be missing a dimensionless constant.","mcqOptions":[{"text":"It may be missing a dimensionless constant.","correct":true,"explanation":"Dimensional homogeneity only confirms that both sides have matching dimensions. It cannot determine pure numerical factors, such as 2 or π, so a homogeneous equation may still be incomplete."},{"text":"It may be using an incorrect base unit.","correct":false,"explanation":"Changing the base unit does not make a dimensionally valid physical relationship incorrect if conversions are handled consistently. The real limitation is that dimensional analysis cannot reveal a missing dimensionless constant."},{"text":"It may have an unbalanced physical dimension.","correct":false,"explanation":"An equation with unbalanced dimensions is not homogeneous, so this contradicts the premise. A homogeneous equation can instead be wrong because dimensional analysis cannot detect a missing dimensionless constant."},{"text":"It may contain a derived S.I. unit.","correct":false,"explanation":"Derived S.I. units are valid combinations of base units and do not make an equation incorrect. The undetectable error may instead be a missing dimensionless constant, such as ½."}]},{"id":"5903f0e8-e423-4567-9b5b-847cc63d98f0","cardType":"basic","points":1,"front":"What format should graph axes and table headings use?","back":"quantity / unit.","mcqOptions":[{"text":"quantity / unit.","correct":true,"explanation":"A graph axis or table heading should show the measured quantity followed by a slash and its unit. This keeps numerical entries free from repeated units and clearly identifies both."},{"text":"unit / quantity.","correct":false,"explanation":"This reverses the standard order and can make the label ambiguous. The accepted format places the physical quantity first and its unit second: quantity / unit."},{"text":"quantity × unit.","correct":false,"explanation":"The multiplication sign incorrectly suggests that the quantity is being multiplied by its unit. Standard axis and heading notation separates them using the format quantity / unit."},{"text":"quantity = unit.","correct":false,"explanation":"An equals sign falsely states that a physical quantity and its unit are identical. Graph axes and table headings conventionally use quantity / unit instead."}]},{"id":"aed79f87-1b25-43bb-8d1d-4807ec29c0df","cardType":"basic","points":2,"front":"How are the units of a graph gradient found?","back":"Vertical-axis units ÷ horizontal-axis units.","mcqOptions":[{"text":"Vertical-axis units ÷ horizontal-axis units.","correct":true,"explanation":"Gradient is change in the vertical quantity divided by change in the horizontal quantity. Its units therefore equal the vertical-axis units divided by the horizontal-axis units."},{"text":"Horizontal-axis units ÷ vertical-axis units.","correct":false,"explanation":"This calculates the units of the reciprocal gradient by reversing rise and run. Gradient units are vertical-axis units divided by horizontal-axis units."},{"text":"Vertical-axis units × horizontal-axis units.","correct":false,"explanation":"Multiplying the axis units confuses gradient with a product of quantities. Because gradient is change in vertical quantity divided by change in horizontal quantity, the units must also be divided."},{"text":"Vertical-axis units ÷ horizontal-axis values.","correct":false,"explanation":"Units must be combined with units, not with numerical axis values. The gradient’s units are found by dividing the vertical-axis units by the horizontal-axis units."}]},{"id":"8aba5534-ffca-4618-8cf9-dd7604488f0c","cardType":"basic","points":1,"front":"What are the units of the gradient of a velocity-time graph?","back":"m s^-2.","mcqOptions":[{"text":"m s^-2.","correct":true,"explanation":"The gradient of a velocity-time graph is change in velocity divided by change in time. Dividing m s^-1 by s gives m s^-2, the unit of acceleration."},{"text":"m s^-1.","correct":false,"explanation":"These are the units of velocity itself, not its gradient against time. Dividing the velocity change by a time interval gives m s^-2."},{"text":"m² s^-2.","correct":false,"explanation":"This results from multiplying velocity units rather than dividing them by time. A velocity-time gradient has units m s^-1 ÷ s, which simplifies to m s^-2."},{"text":"m s^-3.","correct":false,"explanation":"This divides by time twice and corresponds to the rate of change of acceleration rather than velocity. A velocity-time gradient divides by time once, giving m s^-2."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"S.I. units","title":"S.I. units Flashcards","topbar":{"backHref":"/a-level/physics/ocr/a/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"741ec95f-565b-5182-9f82-a82cf4beac8c","walled":"$L39"}],"$L3a","$L3b"]