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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/as-level/physics/edexcel/as-si-units-and-physical-quantities/flashcards","label":"SI units and physical quantities","subjectCode":"as_level_physics","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"1337c051-6cc8-4861-b4c3-1bea05d3054e","title":"SI units and physical quantities Flashcards","topicId":"10b680bb-890c-5de4-a875-eb3d9fac1322","topicSlug":"as-si-units-and-physical-quantities","cards":[{"id":"1a2b5366-ee65-474b-86ec-cb1eccce28fb","cardType":"basic","points":2,"front":"What two components are required to define a physical quantity?","back":"

Numerical value
Unit

","mcqOptions":[{"text":"

Numerical value
Unit

","correct":true,"explanation":"A physical quantity requires a numerical value to specify its magnitude and a unit to establish the measurement standard used for comparison."},{"text":"

Numerical value
Dimension

","correct":false,"explanation":"A dimension identifies the type of quantity, such as mass or length, but does not provide a measurement standard. A physical quantity requires a numerical value and a unit."},{"text":"

Magnitude
Direction

","correct":false,"explanation":"Magnitude and direction define a vector quantity, not every physical quantity. Any measured physical quantity must be expressed using a numerical value and a unit."},{"text":"

Symbol
Unit

","correct":false,"explanation":"A symbol is convenient notation but is not required to specify a measured quantity. The two essential components are a numerical value and a unit."}]},{"id":"697b3441-170b-451c-9bf7-9629ff4d4ef8","cardType":"cloze","points":null,"front":"The SI base unit for electric current is the [ ], and for thermodynamic temperature it is the kelvin (K).","back":"The SI base unit for electric current is the ampere (A), and for thermodynamic temperature it is the kelvin (K).","mcqOptions":[{"text":"ampere (A)","correct":true,"explanation":"The ampere, symbol A, is one of the seven SI base units and measures electric current directly rather than being constructed from other SI units."},{"text":"coulomb (C)","correct":false,"explanation":"The coulomb measures electric charge and is a derived unit equal to an ampere second. The SI base unit for electric current is the ampere (A)."},{"text":"volt (V)","correct":false,"explanation":"The volt measures potential difference, not electric current, and is an SI derived unit. Electric current has the SI base unit ampere (A)."},{"text":"ohm (Ω)","correct":false,"explanation":"The ohm measures electrical resistance rather than current and is derived from volts per ampere. The SI base unit for electric current is the ampere (A)."}]},{"id":"c78c714d-390c-40f3-9737-2551e6568022","cardType":"cloze","points":null,"front":"The SI base unit for electric current is the ampere (A), and for thermodynamic temperature it is the [ ].","back":"The SI base unit for electric current is the ampere (A), and for thermodynamic temperature it is the kelvin (K).","mcqOptions":[{"text":"kelvin (K)","correct":true,"explanation":"The kelvin, symbol K, is the SI base unit of thermodynamic temperature. Temperature differences and absolute temperatures can therefore be stated directly in kelvins."},{"text":"degree Celsius (°C)","correct":false,"explanation":"Degrees Celsius are commonly used and have the same interval size as kelvins, but Celsius is not the SI base unit. The base unit is the kelvin (K)."},{"text":"joule (J)","correct":false,"explanation":"The joule measures energy, including thermal energy, which may cause confusion with temperature. The SI base unit for thermodynamic temperature is the kelvin (K)."},{"text":"fahrenheit (°F)","correct":false,"explanation":"Fahrenheit is a non-SI temperature scale and does not measure thermodynamic temperature in SI base units. The required base unit is the kelvin (K)."}]},{"id":"d6d5dad1-7fed-42ff-871f-a2d7a9933e0c","cardType":"cloze","points":null,"front":"The SI base unit for amount of substance is the [ ], and for luminous intensity it is the candela (cd).","back":"The SI base unit for amount of substance is the mole (mol), and for luminous intensity it is the candela (cd).","mcqOptions":[{"text":"mole (mol)","correct":true,"explanation":"The mole, symbol mol, is the SI base unit for amount of substance and relates particle number to the fixed Avogadro constant."},{"text":"kilogram (kg)","correct":false,"explanation":"The kilogram measures mass, which is not the same as amount of substance even though both describe samples of matter. Amount of substance is measured in moles (mol)."},{"text":"gram (g)","correct":false,"explanation":"The gram is a unit of mass rather than amount of substance, and it is not itself an SI base unit. The required unit is the mole (mol)."},{"text":"particle (pt)","correct":false,"explanation":"Particles may be counted when determining amount of substance, but “particle” is not an SI unit. The SI base unit is the mole (mol)."}]},{"id":"45b7eb57-be19-41df-a7aa-103c05546a6a","cardType":"cloze","points":null,"front":"The SI base unit for amount of substance is the mole (mol), and for luminous intensity it is the [ ].","back":"The SI base unit for amount of substance is the mole (mol), and for luminous intensity it is the candela (cd).","mcqOptions":[{"text":"candela (cd)","correct":true,"explanation":"The candela, symbol cd, is the SI base unit of luminous intensity, describing the strength of visible light emitted in a particular direction."},{"text":"lumen (lm)","correct":false,"explanation":"The lumen measures luminous flux, the total perceived light output, rather than luminous intensity in a direction. The SI base unit for luminous intensity is the candela (cd)."},{"text":"lux (lx)","correct":false,"explanation":"The lux measures illuminance, or luminous flux received per unit area, rather than luminous intensity. Luminous intensity has the SI base unit candela (cd)."},{"text":"watt (W)","correct":false,"explanation":"The watt measures power and can describe radiant energy output, but it does not account specifically for perceived luminous intensity. The required unit is the candela (cd)."}]},{"id":"5c96269b-314b-4e56-9a5a-6bf3b7bf40cb","cardType":"basic","points":1,"front":"What is the SI base unit for mass?","back":"kilogram (kg)","mcqOptions":[{"text":"kilogram (kg)","correct":true,"explanation":"The kilogram, symbol kg, is the SI base unit for mass. Unusually among base units, its standard name already contains the prefix kilo."},{"text":"gram (g)","correct":false,"explanation":"The gram is a commonly used mass unit, but it is one-thousandth of the SI base unit. The SI base unit for mass is the kilogram (kg)."},{"text":"newton (N)","correct":false,"explanation":"The newton measures force, not mass, and is derived as $\\mathrm{kg\\,m\\,s^{-2}}$. The SI base unit for mass is the kilogram (kg)."},{"text":"tonne (t)","correct":false,"explanation":"The tonne is an accepted unit equal to $1000\\,\\mathrm{kg}$, but it is not an SI base unit. Mass has the base unit kilogram (kg)."}]},{"id":"396d33b3-364b-4cde-af75-94e33868a3e5","cardType":"basic","points":3,"front":"List the seven SI base quantities.","back":"

Length
Mass
Time
Electric current
Temperature
Amount of substance
Luminous intensity

","mcqOptions":[{"text":"

Length
Mass
Time
Electric current
Temperature
Amount of substance
Luminous intensity

","correct":true,"explanation":"These are the seven independent SI base quantities. All other SI quantities, including force, energy and electric charge, can be expressed in terms of these quantities."},{"text":"

Length
Mass
Time
Electric charge
Temperature
Amount of substance
Luminous intensity

","correct":false,"explanation":"Electric charge is derived from electric current multiplied by time, so it is not a base quantity. The correct fourth quantity is electric current."},{"text":"

Length
Mass
Time
Electric current
Energy
Amount of substance
Luminous intensity

","correct":false,"explanation":"Energy is derived from mass, length and time, so it cannot replace a base quantity. Temperature is the missing SI base quantity in this list."},{"text":"

Length
Mass
Time
Electric current
Temperature
Volume
Luminous intensity

","correct":false,"explanation":"Volume is derived from length cubed and is therefore not an independent base quantity. Amount of substance is the SI base quantity that should appear instead."}]},{"id":"2b9d7def-7975-47a8-b4b6-bc96c9bbd2c7","cardType":"basic","points":2,"front":"State the multiplier for the prefix giga (G) and tera (T).","back":"

Giga: $10^9$
Tera: $10^{12}$

","mcqOptions":[{"text":"

Giga: $10^9$
Tera: $10^{12}$

","correct":true,"explanation":"The prefix giga represents one billion, or $10^9$, while tera represents one trillion, or $10^{12}$. Each successive listed prefix differs by a factor of $10^3$."},{"text":"

Giga: $10^6$
Tera: $10^9$

","correct":false,"explanation":"These values are shifted down by one prefix level: $10^6$ is mega and $10^9$ is giga. Giga is $10^9$ and tera is $10^{12}$."},{"text":"

Giga: $10^9$
Tera: $10^{15}$

","correct":false,"explanation":"The giga value is correct, but $10^{15}$ corresponds to peta, skipping the next prefix. Tera is $10^{12}$, three powers of ten above giga."},{"text":"

Giga: $10^{12}$
Tera: $10^9$

","correct":false,"explanation":"This reverses the two prefix multipliers. Giga is the smaller prefix at $10^9$, while tera is the larger prefix at $10^{12}$."}]},{"id":"32682ba5-4cdf-4ccc-a477-2ccbcdf92447","cardType":"basic","points":2,"front":"State the multiplier for the prefix nano (n) and pico (p).","back":"

Nano: $10^{-9}$
Pico: $10^{-12}$

","mcqOptions":[{"text":"

Nano: $10^{-9}$
Pico: $10^{-12}$

","correct":true,"explanation":"The prefix nano represents one billionth, $10^{-9}$, while pico represents one trillionth, $10^{-12}$. Each successive SI prefix here differs by a factor of $10^3$."},{"text":"

Nano: $10^{-6}$
Pico: $10^{-9}$

","correct":false,"explanation":"These multipliers are shifted upwards by one prefix step: $10^{-6}$ is micro and $10^{-9}$ is nano. Nano is $10^{-9}$ and pico is $10^{-12}$."},{"text":"

Nano: $10^{-12}$
Pico: $10^{-9}$

","correct":false,"explanation":"This reverses the two multipliers. Nano is the larger multiplier, $10^{-9}$, while pico represents the smaller multiplier, $10^{-12}$."},{"text":"

Nano: $10^{9}$
Pico: $10^{12}$

","correct":false,"explanation":"The positive exponents belong to large multiples rather than small fractions. Nano and pico use negative exponents: $10^{-9}$ and $10^{-12}$ respectively."}]},{"id":"a18a0297-0e2d-4109-ad4d-5dac2b57c67f","cardType":"basic","points":1,"front":"State the symbol and multiplier for the prefix micro.","back":"$$μ$ and $10^{-6}$","mcqOptions":[{"text":"$$μ$ and $10^{-6}$","correct":true,"explanation":"The Greek letter $μ$ is the SI symbol for micro, and the prefix multiplies a unit by $10^{-6}$, meaning one millionth."},{"text":"$$m$ and $10^{-3}$","correct":false,"explanation":"The symbol $m$ with multiplier $10^{-3}$ denotes milli, not micro. Micro uses the Greek letter $μ$ and has multiplier $10^{-6}$."},{"text":"$$n$ and $10^{-9}$","correct":false,"explanation":"The symbol $n$ and multiplier $10^{-9}$ denote nano, which is one thousand times smaller than micro. Micro is $μ$ with multiplier $10^{-6}$."},{"text":"$$M$ and $10^{6}$","correct":false,"explanation":"The uppercase symbol $M$ and positive multiplier $10^6$ denote mega. Micro instead uses $μ$ and the negative power $10^{-6}$."}]},{"id":"50fd332a-f3fb-46f7-98e4-522cb2b0c74f","cardType":"cloze","points":null,"front":"Unit symbols are case-sensitive; for example, [ ] stands for milli, whereas M stands for mega.","back":"Unit symbols are case-sensitive; for example, m stands for milli, whereas M stands for mega.","mcqOptions":[{"text":"m","correct":true,"explanation":"The lowercase prefix symbol m denotes milli, meaning a factor of $10^{-3}$. Prefix symbols are case-sensitive, so it must not be confused with uppercase M."},{"text":"μ","correct":false,"explanation":"The symbol μ denotes micro, a factor of $10^{-6}$, not milli. The correct symbol for milli is lowercase m, representing $10^{-3}$."},{"text":"k","correct":false,"explanation":"The lowercase symbol k denotes kilo, a factor of $10^{3}$, not milli. Lowercase m is the correct prefix symbol for $10^{-3}$."},{"text":"c","correct":false,"explanation":"The lowercase symbol c denotes centi, a factor of $10^{-2}$, not milli. Milli is represented by lowercase m and means $10^{-3}$."}]},{"id":"5647394b-2c9d-4a46-b725-918b4d8097c0","cardType":"cloze","points":null,"front":"Unit symbols are case-sensitive; for example, m stands for milli, whereas [ ] stands for mega.","back":"Unit symbols are case-sensitive; for example, m stands for milli, whereas M stands for mega.","mcqOptions":[{"text":"M","correct":true,"explanation":"The SI prefix for a factor of $10^6$ is mega, represented by an uppercase symbol. SI prefixes are case-sensitive, so this must not be confused with the lowercase symbol for milli."},{"text":"k","correct":false,"explanation":"The lowercase prefix represents kilo, a factor of $10^3$, not mega. Confusing the powers of ten would make a measurement one thousand times larger rather than one million times larger."},{"text":"μ","correct":false,"explanation":"The prefix represents micro, a factor of $10^{-6}$, so it indicates a very small quantity rather than the millionfold increase associated with mega."},{"text":"G","correct":false,"explanation":"The prefix represents giga, a factor of $10^9$. It is tempting because it is another large-scale prefix, but it is one thousand times larger than mega."}]},{"id":"eb1dac90-019e-41ab-af72-43cdc67df320","cardType":"cloze","points":null,"front":"When a prefix is placed before a unit, it acts as a [ ] for a power of ten.","back":"When a prefix is placed before a unit, it acts as a multiplier for a power of ten.","mcqOptions":[{"text":"multiplier","correct":true,"explanation":"An SI prefix multiplies the unit by a specified power of ten. For example, kilo applies a multiplier of $10^3$ and milli applies $10^{-3}$."},{"text":"divisor","correct":false,"explanation":"Treating every prefix as a divisor fails for prefixes such as kilo and mega, which increase the unit. A prefix acts as a multiplier given by a power of ten."},{"text":"dimension","correct":false,"explanation":"A prefix does not change a quantity’s physical dimension; kilometres and metres both measure length. It acts as a multiplier corresponding to a power of ten."},{"text":"base quantity","correct":false,"explanation":"Base quantities include length, mass and time, whereas a prefix only rescales a unit. It acts as a multiplier specified by a power of ten."}]},{"id":"33db6722-9a2c-4554-a1ef-eabed9ea6a66","cardType":"cloze","points":null,"front":"When a prefix is placed before a unit, it acts as a multiplier for a [ ].","back":"When a prefix is placed before a unit, it acts as a multiplier for a power of ten.","mcqOptions":[{"text":"power of ten","correct":true,"explanation":"Each SI prefix corresponds to a power of ten, such as $10^3$ for kilo or $10^{-6}$ for micro, which multiplies the associated unit."},{"text":"base unit","correct":false,"explanation":"A prefix may be attached to suitable units, but it is not itself a multiplier for a base unit. It represents a multiplier that is a power of ten."},{"text":"physical quantity","correct":false,"explanation":"A physical quantity is what is measured, such as length or mass, and is not a numerical multiplier. An SI prefix specifies a power of ten."},{"text":"unit symbol","correct":false,"explanation":"A unit symbol identifies the unit, while a prefix determines its scale. The multiplier represented by an SI prefix is a power of ten."}]},{"id":"5e9ad96b-874b-49b3-b4df-b32d6e857dd0","cardType":"basic","points":1,"front":"Express the newton (N) in SI base units.","back":"$$kg m s^{-2}$","mcqOptions":[{"text":"$$kg m s^{-2}$","correct":true,"explanation":"From Newton’s second law, $F=ma$. Mass has unit $kg$ and acceleration has unit $m s^{-2}$, giving the newton as $kg m s^{-2}$."},{"text":"$$kg m s^{-1}$","correct":false,"explanation":"This is the SI base-unit expression for momentum, obtained from mass times velocity. Force is mass times acceleration, so a newton is $kg m s^{-2}$."},{"text":"$$kg m^{-1} s^{-2}$","correct":false,"explanation":"This is the base-unit expression for pressure, found by dividing force by area. A newton is a unit of force and equals $kg m s^{-2}$."},{"text":"$$kg m^2 s^{-2}$","correct":false,"explanation":"This is the base-unit expression for energy because it includes an additional factor of distance. Force alone is measured in newtons, equal to $kg m s^{-2}$."}]},{"id":"c78fa23b-644a-4208-956a-15b4d4d9fa31","cardType":"basic","points":1,"front":"Express the joule (J) in SI base units.","back":"$$kg m^2 s^{-2}$","mcqOptions":[{"text":"$$kg m^2 s^{-2}$","correct":true,"explanation":"A joule is a newton metre. Multiplying $kg m s^{-2}$ by another metre gives the SI base-unit expression $kg m^2 s^{-2}$."},{"text":"$$kg m s^{-2}$","correct":false,"explanation":"This is the SI base-unit expression for force in newtons. Energy equals force multiplied by distance, adding another factor of metres to give $kg m^2 s^{-2}$."},{"text":"$$kg m^2 s^{-3}$","correct":false,"explanation":"This is the SI base-unit expression for power, or energy transferred per second. A joule measures energy itself, so its time exponent is $-2$, giving $kg m^2 s^{-2}$."},{"text":"$$kg m s^{-1}$","correct":false,"explanation":"This is the SI base-unit expression for momentum, calculated as mass times velocity. A joule is force times distance and equals $kg m^2 s^{-2}$."}]},{"id":"9245c05b-9fc5-452a-a30d-9d0ae5ba5cf8","cardType":"basic","points":1,"front":"Express the watt (W) in SI base units.","back":"$$kg m^2 s^{-3}$","mcqOptions":[{"text":"$$kg m^2 s^{-3}$","correct":true,"explanation":"A watt is one joule per second. Since $1\\text{ J}=1\\text{ kg m}^2\\text{ s}^{-2}$, dividing by seconds gives $kg m^2 s^{-3}$."},{"text":"$$kg m^2 s^{-2}$","correct":false,"explanation":"This is the base-unit form of a joule, which measures energy. Power is energy divided by time, so a further factor of $s^{-1}$ gives $kg m^2 s^{-3}$."},{"text":"$$kg m s^{-3}$","correct":false,"explanation":"This results from omitting one factor of length when converting energy into base units. A watt is a joule per second, giving the correct form $kg m^2 s^{-3}$."},{"text":"$$kg m^2 s^{-1}$","correct":false,"explanation":"This uses the wrong power of time by failing to include both the $s^{-2}$ from energy and division by time. The correct unit is $kg m^2 s^{-3}$."}]},{"id":"4f7a888e-af23-453c-811c-6cb1562ec12f","cardType":"basic","points":2,"front":"Express the volt (V) in SI base units.","back":"$$kg m^2 s^{-3} A^{-1}$","mcqOptions":[{"text":"$$kg m^2 s^{-3} A^{-1}$","correct":true,"explanation":"Potential difference is energy transferred per unit charge. Using $J C^{-1}$, with $J=kg m^2 s^{-2}$ and $C=A s$, gives $kg m^2 s^{-3} A^{-1}$."},{"text":"$$kg m^2 s^{-3} A^{-2}$","correct":false,"explanation":"This is the base-unit form of an ohm because it divides voltage by current again. A volt is power per ampere, so its unit is $kg m^2 s^{-3} A^{-1}$."},{"text":"$$kg m^2 s^{-3}$","correct":false,"explanation":"This is the base-unit form of a watt and omits the current factor. Since voltage is power divided by current, the correct unit includes $A^{-1}$."},{"text":"$$kg m^2 s^{-2} A^{-1}$","correct":false,"explanation":"This divides energy by current rather than by charge, missing the extra time factor in $C=A s$. The base-unit form of a volt is $kg m^2 s^{-3} A^{-1}$."}]},{"id":"21e03095-54dd-4606-8600-ae972417d6be","cardType":"basic","points":1,"front":"What defines an equation as dimensionally consistent?","back":"Every term in the equation has the same units.","mcqOptions":[{"text":"Every term in the equation has the same units.","correct":true,"explanation":"Terms can only be added, subtracted or equated when they represent the same type of physical quantity. Therefore, every term in a dimensionally consistent equation has matching units."},{"text":"Every variable in the equation has the same units.","correct":false,"explanation":"Different variables may represent different physical quantities and therefore have different units. Dimensional consistency instead requires every complete term that is added or equated to have the same units."},{"text":"Every term in the equation has the same numerical value.","correct":false,"explanation":"Dimensional consistency concerns units, not whether separate terms have identical numerical values. Every term must have the same units so that addition and equality are physically meaningful."},{"text":"Every term in the equation has different units.","correct":false,"explanation":"Terms with different units cannot validly be added, subtracted or equated. A dimensionally consistent equation requires every term to have the same units, even if variables within terms differ."}]},{"id":"48f44ac4-35a6-4f5b-b255-5bd0134c4740","cardType":"basic","points":2,"front":"Why does dimensional consistency not prove an equation is correct?","back":"It cannot account for dimensionless constants (e.g., numbers like 2 or $1/2$) that have no units.","mcqOptions":[{"text":"It cannot account for dimensionless constants (e.g., numbers like 2 or $1/2$) that have no units.","correct":true,"explanation":"A missing or incorrect dimensionless factor does not change either side’s units. Consequently, an equation may be dimensionally consistent while still containing an incorrect numerical constant."},{"text":"It cannot account for quantities with derived units (e.g., force in newtons or energy in joules) within an equation.","correct":false,"explanation":"Derived units can be replaced by their SI base-unit equivalents and checked dimensionally. The genuine limitation is that dimensionless numerical constants such as 2 or $1/2$ cannot be detected."},{"text":"It requires all quantities to be written in SI base units (e.g., metres, kilograms and seconds) before comparison.","correct":false,"explanation":"Dimensions may be compared using consistent derived units or their base-unit equivalents, so this is not the limitation. Dimensional checks cannot reveal incorrect dimensionless constants such as 2."},{"text":"It cannot be applied to equations containing more than two terms because only two sets of units can be compared.","correct":false,"explanation":"Any number of terms can be checked, provided every term has matching dimensions. The actual limitation is that dimensionless numerical factors do not affect units and therefore escape detection."}]},{"id":"50f42355-5d0c-4933-b7ab-ea984003fe9f","cardType":"basic","points":1,"front":"How is the unit of a graph gradient calculated?","back":"$$unit on vertical axis / unit on horizontal axis$","mcqOptions":[{"text":"$$unit on vertical axis / unit on horizontal axis$","correct":true,"explanation":"A graph’s gradient is calculated as change in the vertical quantity divided by change in the horizontal quantity, so its unit follows the same vertical-unit over horizontal-unit ratio."},{"text":"$$unit on vertical axis × unit on horizontal axis$","correct":false,"explanation":"Multiplying the axis units would give the unit of a product, not a gradient. Since gradient is change in $y$ divided by change in $x$, the units must also be divided."},{"text":"$$unit on horizontal axis / unit on vertical axis$","correct":false,"explanation":"This reverses the rise-over-run calculation and gives the units of the reciprocal gradient. The vertical-axis unit must be divided by the horizontal-axis unit."},{"text":"$$unit on vertical axis / unit on horizontal axis^2$","correct":false,"explanation":"Squaring the horizontal-axis unit is not part of the ordinary gradient calculation. Gradient is change in the vertical quantity divided once by change in the horizontal quantity."}]},{"id":"03290fd7-d0cc-498e-b88f-78178ad011e6","cardType":"cloze","points":null,"front":"In a table of results, units should be placed in the [ ] rather than in every individual cell.","back":"In a table of results, units should be placed in the column headings rather than in every individual cell.","mcqOptions":[{"text":"column headings","correct":true,"explanation":"Each column heading should state the measured quantity and its unit. This keeps the table concise while making the unit applying to every value in that column unambiguous."},{"text":"table caption","correct":false,"explanation":"A caption identifies or describes the whole table but does not label each measured quantity. Units belong in the column headings so every value’s unit is clear without repetition."},{"text":"first data row","correct":false,"explanation":"Placing units in the first data row could make them look like measurements and leaves the column label incomplete. Units should instead appear in the column headings."},{"text":"row headings","correct":false,"explanation":"Row headings usually identify individual trials or categories, rather than the units of measured variables. Units should be written in the relevant column headings above the data."}]},{"id":"61db1980-8063-4cb6-a7f0-8ce41723664d","cardType":"basic","points":1,"front":"Convert $750\\text{ }μ\\text{s}$ into seconds using scientific notation.","back":"$$7.5 × 10^{-4}\\text{ s}$","mcqOptions":[{"text":"$$7.5 × 10^{-4}\\text{ s}$","correct":true,"explanation":"The prefix micro means $10^{-6}$. Therefore, $750\\,μ\\text{s}=750×10^{-6}\\text{ s}$, which in standard scientific notation is $7.5×10^{-4}\\text{ s}$."},{"text":"$$7.5 × 10^{-3}\\text{ s}$","correct":false,"explanation":"This treats the micro prefix as though it produced a factor ten larger after conversion. Since $μ=10^{-6}$, $750\\,μ\\text{s}=750×10^{-6}\\text{ s}=7.5×10^{-4}\\text{ s}$."},{"text":"$$7.5 × 10^{-5}\\text{ s}$","correct":false,"explanation":"This moves the decimal one place too far when rewriting $750×10^{-6}$. Changing 750 to 7.5 increases the power by two, giving $7.5×10^{-4}\\text{ s}$."},{"text":"$$7.5 × 10^{4}\\text{ s}$","correct":false,"explanation":"This reverses the sign of the power and makes a microscopic time extremely large. Micro means $10^{-6}$, so the correct conversion is $7.5×10^{-4}\\text{ s}$."}]},{"id":"98070268-36d4-4050-9916-d183e684cd12","cardType":"basic","points":2,"front":"Identify the units for the gradient of a graph with potential difference (V) on the y-axis and current (A) on the x-axis.","back":"$$V A^{-1}$ (or ohms, $Ω$)","mcqOptions":[{"text":"$$V A^{-1}$ (or ohms, $Ω$)","correct":true,"explanation":"Gradient is change in the y-axis quantity divided by change in the x-axis quantity. Thus the unit is volts per ampere, $V A^{-1}$, equivalent to ohms."},{"text":"$$A V^{-1}$ (or siemens, $S$)","correct":false,"explanation":"This is current divided by potential difference, the reciprocal of the required gradient and the unit of conductance. Calculating vertical change over horizontal change gives $V A^{-1}$, or ohms."},{"text":"$$V A$ (or watts, $W$)","correct":false,"explanation":"Multiplying potential difference by current gives electrical power, measured in watts. A graph gradient requires division of the y-axis unit by the x-axis unit, giving $V A^{-1}$."},{"text":"$$V A^{-2}$ (or volts per ampere squared)","correct":false,"explanation":"This incorrectly divides by the current unit twice. The gradient calculation divides change in potential difference by change in current once, producing $V A^{-1}$, equivalent to ohms."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"SI units and physical quantities","title":"SI units and physical quantities Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"10b680bb-890c-5de4-a875-eb3d9fac1322","walled":"$L39"}],"$L3a","$L3b"]