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Its unit symbol is kg."},{"text":"gram (g)","correct":false,"explanation":"The gram is a commonly used SI unit, but it is not the base unit of mass. The SI base unit is the kilogram, with symbol kg."},{"text":"tonne (t)","correct":false,"explanation":"The tonne is accepted for use with SI but is a larger derived unit equal to 1000 kg. The SI base unit of mass is the kilogram."},{"text":"pound (lb)","correct":false,"explanation":"The pound belongs to the imperial system rather than the SI. The SI base unit used to measure mass is the kilogram, with symbol kg."}]},{"id":"af3f7596-808a-4eaa-b6a1-a6debbaccb24","cardType":"basic","points":1,"front":"What is the SI base unit for thermodynamic temperature?","back":"kelvin (K)","mcqOptions":[{"text":"kelvin (K)","correct":true,"explanation":"The kelvin is the SI base unit of thermodynamic temperature. Temperature differences in kelvin have the same numerical size as differences in degrees Celsius."},{"text":"degree Celsius (°C)","correct":false,"explanation":"Degrees Celsius are widely used for temperature, but the kelvin is the SI base unit of thermodynamic temperature. Celsius values are defined relative to the kelvin scale."},{"text":"degree Fahrenheit (°F)","correct":false,"explanation":"The degree Fahrenheit is a non-SI temperature unit and uses a different interval size. The SI base unit of thermodynamic temperature is the kelvin."},{"text":"joule (J)","correct":false,"explanation":"The joule measures energy, including thermal energy, rather than temperature itself. Thermodynamic temperature has the SI base unit kelvin, represented by K."}]},{"id":"5ac27805-814b-47a8-a5b8-bf13afe4833a","cardType":"basic","points":1,"front":"What is the SI base unit for electric current?","back":"ampere (A)","mcqOptions":[{"text":"ampere (A)","correct":true,"explanation":"The ampere is the SI base unit of electric current. It measures the rate of flow of electric charge through a point in a circuit."},{"text":"coulomb (C)","correct":false,"explanation":"The coulomb measures electric charge, not current. Current is charge transferred per unit time, and its SI base unit is the ampere."},{"text":"volt (V)","correct":false,"explanation":"The volt measures potential difference rather than electric current. The SI base unit of current is the ampere, represented by the symbol A."},{"text":"ohm (Ω)","correct":false,"explanation":"The ohm measures electrical resistance, which may be confused with other circuit quantities. Electric current is measured in the SI base unit ampere."}]},{"id":"b4185c0c-e90e-47e6-add9-b36303d47018","cardType":"cloze","points":null,"front":"The SI base unit for amount of substance is [ ] (mol).","back":"The SI base unit for amount of substance is mole (mol).","mcqOptions":[{"text":"mole","correct":true,"explanation":"The mole is the SI base unit for amount of substance. One mole contains exactly $6.02214076\\times10^{23}$ specified elementary entities."},{"text":"kilogram","correct":false,"explanation":"The kilogram is the SI base unit of mass, not amount of substance. Amount of substance is measured in moles, with the unit symbol mol."},{"text":"particle","correct":false,"explanation":"A particle is an individual entity rather than an SI unit. Amount of substance is measured in moles, each representing a specified number of entities."},{"text":"coulomb","correct":false,"explanation":"The coulomb is a derived SI unit of electric charge, not amount of substance. The required SI base unit is the mole, symbol mol."}]},{"id":"0b993346-c54b-47f0-b2ab-5f9c6c46b33d","cardType":"basic","points":2,"front":"Why must a mass in grams be converted to kilograms before using $W=mg$?","back":"The kilogram is the SI base unit; using grams would result in the wrong unit for force.","mcqOptions":[{"text":"The kilogram is the SI base unit; using grams would result in the wrong unit for force.","correct":true,"explanation":"In $W=mg$, gravitational field strength is normally expressed in $N\\,kg^{-1}$, so mass must be in kilograms for the calculated weight to be in newtons."},{"text":"The gram is the SI base unit; using kilograms would result in the wrong unit for force.","correct":false,"explanation":"This reverses the SI convention: the kilogram, not the gram, is the base unit of mass. Using kilograms with $N\\,kg^{-1}$ gives force in newtons."},{"text":"The newton is the SI base unit; using grams would result in the wrong unit for mass.","correct":false,"explanation":"The newton is a derived unit of force, not a base unit. Mass must be converted to kilograms so that $W=mg$ produces weight in newtons."},{"text":"The metre is the SI base unit; using grams would result in the wrong unit for acceleration.","correct":false,"explanation":"The metre is the base unit of length and does not determine the mass conversion. Kilograms are required because mass must use its SI base unit in $W=mg$."}]},{"id":"07e99f29-42a7-4cfe-b57b-ef3a6ce809c5","cardType":"basic","points":2,"front":"Express the newton (N) in terms of SI base units.","back":"$$kg\\,m\\,s^{-2}$","mcqOptions":[{"text":"$$kg\\,m\\,s^{-2}$","correct":true,"explanation":"From $F=ma$, mass has unit kg and acceleration has unit $m\\,s^{-2}$. Therefore one newton is equivalent to $kg\\,m\\,s^{-2}$."},{"text":"$$kg\\,m\\,s^{-1}$","correct":false,"explanation":"Using $s^{-1}$ gives the units of momentum rather than force, reflecting omission of one time factor. Force requires acceleration, so the correct unit is $kg\\,m\\,s^{-2}$."},{"text":"$$kg\\,m^2\\,s^{-2}$","correct":false,"explanation":"The extra factor of metres gives the SI base units of energy or work. A newton measures force, so its base-unit form is $kg\\,m\\,s^{-2}$."},{"text":"$$kg\\,m^{-1}\\,s^{-2}$","correct":false,"explanation":"Placing metres in the denominator does not represent mass multiplied by acceleration. From $F=ma$, the newton must be expressed as $kg\\,m\\,s^{-2}$."}]},{"id":"0d574deb-b8da-446a-a3b7-d53f188d934a","cardType":"basic","points":2,"front":"Express the joule (J) in terms of SI base units.","back":"$$kg\\,m^2\\,s^{-2}$","mcqOptions":[{"text":"$$kg\\,m^2\\,s^{-2}$","correct":true,"explanation":"Work is force multiplied by distance. Combining $N=kg\\,m\\,s^{-2}$ with another factor of metres gives $J=kg\\,m^2\\,s^{-2}$."},{"text":"$$kg\\,m\\,s^{-2}$","correct":false,"explanation":"These are the base units of force, so this omits the distance through which the force acts. Multiplying by metres gives $kg\\,m^2\\,s^{-2}$ for a joule."},{"text":"$$kg\\,m^2\\,s^{-1}$","correct":false,"explanation":"This uses only one inverse power of time, usually from confusing energy with another mechanical quantity. A joule requires $s^{-2}$, giving $kg\\,m^2\\,s^{-2}$."},{"text":"$$kg\\,m^{-2}\\,s^{-2}$","correct":false,"explanation":"Putting squared metres in the denominator reverses the length dependence. Since work equals force multiplied by distance, the correct base units are $kg\\,m^2\\,s^{-2}$."}]},{"id":"7d4adf59-563b-493d-881f-877a681cc78a","cardType":"cloze","points":null,"front":"The watt (W) is equivalent to one [ ] ($J\\,s^{-1}$).","back":"The watt (W) is equivalent to one joule per second ($J\\,s^{-1}$).","mcqOptions":[{"text":"joule per second","correct":true,"explanation":"Power is the rate of energy transfer, so one watt represents one joule of energy transferred each second. Therefore $1\\,W=1\\,J\\,s^{-1}$."},{"text":"joule per metre","correct":false,"explanation":"A joule per metre is equivalent to a newton and therefore represents force, not power. One watt is one joule of energy transferred per second."},{"text":"newton per second","correct":false,"explanation":"A newton per second describes the rate of change of force rather than the rate of energy transfer. One watt is correctly defined as one joule per second."},{"text":"coulomb per second","correct":false,"explanation":"A coulomb per second is an ampere, the unit of electric current. Power is energy transferred per unit time, so one watt is one joule per second."}]},{"id":"78555f7f-af8a-47a6-9776-2f5bec9ff702","cardType":"basic","points":1,"front":"Express the coulomb (C) in terms of SI base units.","back":"$$A\\,s$","mcqOptions":[{"text":"$$A\\,s$","correct":true,"explanation":"Electric charge equals current multiplied by time, so one coulomb is one ampere-second. Its SI base-unit expression is therefore $A\\,s$."},{"text":"$$A\\,s^{-1}$","correct":false,"explanation":"Dividing current by time describes a rate of change of current, not electric charge. Charge is current multiplied by time, giving the base units $A\\,s$."},{"text":"$$A^{-1}\\,s$","correct":false,"explanation":"Reversing the relationship between charge, current and time gives this reciprocal expression. Since $Q=It$, the coulomb must have base units $A\\,s$."},{"text":"$$A\\,s^2$","correct":false,"explanation":"Multiplying current by time squared introduces an extra time factor. Electric charge follows $Q=It$, so its SI base units are $A\\,s$."}]},{"id":"57ee792a-57ca-44bd-9d84-6c4598931b00","cardType":"basic","points":2,"front":"Show the SI base units for potential difference (V).","back":"$$kg\\,m^2\\,s^{-3}\\,A^{-1}$","mcqOptions":[{"text":"$$kg\\,m^2\\,s^{-3}\\,A^{-1}$","correct":true,"explanation":"Potential difference is energy transferred per unit charge. Dividing $kg\\,m^2\\,s^{-2}$ by $A\\,s$ gives $kg\\,m^2\\,s^{-3}\\,A^{-1}$."},{"text":"$$kg\\,m^2\\,s^{-2}\\,A^{-1}$","correct":false,"explanation":"This retains the joule’s time exponent and therefore omits the additional division by seconds contained in a coulomb. Using $V=E/Q$ gives $kg\\,m^2\\,s^{-3}\\,A^{-1}$."},{"text":"$$kg\\,m^2\\,s^{-3}\\,A$","correct":false,"explanation":"Multiplying by charge rather than dividing by it makes the ampere exponent positive. Potential difference is energy per charge, so its base units are $kg\\,m^2\\,s^{-3}\\,A^{-1}$."},{"text":"$$kg\\,m\\,s^{-2}\\,A^{-1}$","correct":false,"explanation":"Using velocity rather than velocity squared when expanding energy loses one factor each of length and inverse time. The correct base units are $kg\\,m^2\\,s^{-3}\\,A^{-1}$."}]},{"id":"671c0817-50ae-4031-8a79-da4bbbfbd4de","cardType":"basic","points":2,"front":"What two conditions must be met for a number to be in standard form ($a \\times 10^n$)?","back":"

$1 \\le a < 10$
$n$ is an integer

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

$1 \\le a < 10$
$n$ is an integer

","correct":true,"explanation":"Standard form requires exactly one non-zero digit before the decimal point, which means $1 \\le a < 10$. The power $n$ must be a whole-number integer."},{"text":"

$0 \\le a < 10$
$n$ is an integer

","correct":false,"explanation":"Allowing $a=0$ is invalid because zero cannot be written with a meaningful power-of-ten exponent in this form. Standard form requires $1 \\le a < 10$ and integer $n$."},{"text":"

$1 \\le a \\le 10$
$n$ is an integer

","correct":false,"explanation":"Including $a=10$ is invalid because $10\\times10^n$ must be rewritten as $1\\times10^{n+1}$. Therefore $1 \\le a < 10$, with integer $n$."},{"text":"

$1 \\le a < 10$
$n$ is a real number

","correct":false,"explanation":"A non-integer exponent would not meet the definition of standard form used here. The coefficient must satisfy $1 \\le a < 10$, and $n$ must be an integer."}]},{"id":"82c37d05-77ed-4000-bbbb-17af95121ba8","cardType":"cloze","points":null,"front":"The prefix mega (M) represents a factor of $\\text{[...]}$.","back":"The prefix mega (M) represents a factor of $\\boxed{10^6}$.","mcqOptions":[{"text":"$$10^6$","correct":true,"explanation":"Mega denotes one million times the base unit. Since one million is $1,000,000=10^6$, the prefix factor is $10^6$."},{"text":"$$10^3$","correct":false,"explanation":"The factor $10^3$ belongs to kilo, not mega. Mega is one million times the base unit, so its prefix factor is $10^6$."},{"text":"$$10^{-6}$","correct":false,"explanation":"The negative exponent describes micro, which is one millionth of a base unit. Mega represents one million base units and therefore has the factor $10^6$."},{"text":"$$10^9$","correct":false,"explanation":"The factor $10^9$ corresponds to giga, which is one thousand times larger than mega. The mega prefix represents the factor $10^6$."}]},{"id":"14e7e4b4-4d8d-48cd-b6c1-63cd9c2df0e1","cardType":"cloze","points":null,"front":"The prefix nano (n) represents a factor of $\\text{[...]}$.","back":"The prefix nano (n) represents a factor of $\\boxed{10^{-9}}$.","mcqOptions":[{"text":"$$10^{-9}$","correct":true,"explanation":"Nano represents one billionth of the base unit. One billionth is $1/10^9=10^{-9}$, so nano has the prefix factor $10^{-9}$."},{"text":"$$10^{-6}$","correct":false,"explanation":"The factor $10^{-6}$ represents micro, not nano. Nano is one thousand times smaller than micro and has the prefix factor $10^{-9}$."},{"text":"$$10^{-12}$","correct":false,"explanation":"The factor $10^{-12}$ represents pico, not nano. Pico is one thousand times smaller than nano, whose correct prefix factor is $10^{-9}$."},{"text":"$$10^9$","correct":false,"explanation":"A positive exponent represents a large multiple rather than a small fraction. Nano means one billionth of the base unit, so its factor is $10^{-9}$."}]},{"id":"e88c4caf-6ca8-4103-aa55-996acf5f55a0","cardType":"basic","points":1,"front":"What is the multiplier for the prefix micro (μ)?","back":"$$10^{-6}$","mcqOptions":[{"text":"$$10^{-6}$","correct":true,"explanation":"Micro means one millionth of the base unit. Since one millionth is $1/10^6$, the multiplier for micro is $10^{-6}$."},{"text":"$$10^{-3}$","correct":false,"explanation":"The multiplier $10^{-3}$ belongs to milli, which is one thousandth. Micro is one thousand times smaller than milli, so its multiplier is $10^{-6}$."},{"text":"$$10^{-9}$","correct":false,"explanation":"The multiplier $10^{-9}$ belongs to nano, which is one thousand times smaller than micro. The correct multiplier for micro is $10^{-6}$."},{"text":"$$10^6$","correct":false,"explanation":"The positive exponent corresponds to mega, meaning one million times the base unit. Micro means one millionth, so its multiplier is $10^{-6}$."}]},{"id":"abbb02e0-43e1-4736-b660-6a00d08449c2","cardType":"basic","points":2,"front":"What is the difference between the prefixes m and M?","back":"

m is milli ($10^{-3}$)
M is mega ($10^6$)

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

m is milli ($10^{-3}$)
M is mega ($10^6$)

","correct":true,"explanation":"SI prefixes are case-sensitive: lowercase m denotes one thousandth, while uppercase M denotes one million. Their factors are therefore $10^{-3}$ and $10^6$, respectively."},{"text":"

m is mega ($10^6$)
M is milli ($10^{-3}$)

","correct":false,"explanation":"This reverses the case-sensitive symbols. Lowercase m is milli with factor $10^{-3}$, whereas uppercase M is mega with factor $10^6$."},{"text":"

m is milli ($10^{-3}$)
M is micro ($10^{-6}$)

","correct":false,"explanation":"Lowercase m does represent milli, but uppercase M is mega rather than micro. The correct factors are $10^{-3}$ for milli and $10^6$ for mega."},{"text":"

m is micro ($10^{-6}$)
M is kilo ($10^3$)

","correct":false,"explanation":"This confuses the symbols for micro and kilo with the case-sensitive pair shown. Lowercase m means milli ($10^{-3}$), while uppercase M means mega ($10^6$)."}]},{"id":"e4736175-bbfe-4714-bc54-9db180531cb9","cardType":"cloze","points":null,"front":"To convert from a base unit to a prefixed unit, you must [ ] by the prefix factor.","back":"To convert from a base unit to a prefixed unit, you must divide by the prefix factor.","mcqOptions":[{"text":"divide","correct":true,"explanation":"A prefixed unit equals its prefix factor multiplied by the base unit. Therefore, converting a base-unit value into the corresponding number of prefixed units requires division by that factor."},{"text":"multiply","correct":false,"explanation":"Multiplication converts a value stated in prefixed units into base units, not the reverse. To convert from a base unit to a prefixed unit, divide by the prefix factor."},{"text":"subtract","correct":false,"explanation":"Prefix conversion is multiplicative rather than additive, so subtracting the factor has no dimensional meaning. A base-unit value must be divided by the prefix factor."},{"text":"add","correct":false,"explanation":"Adding a prefix factor treats it as an offset, but SI prefixes scale quantities multiplicatively. To express a base-unit value in prefixed units, divide by the factor."}]},{"id":"956efa94-1f35-4ae0-a09a-3135f9fff82e","cardType":"basic","points":1,"front":"How is energy converted from electronvolts (eV) to joules (J)?","back":"Multiply by $1.60 \\times 10^{-19}$.","mcqOptions":[{"text":"Multiply by $1.60 \\times 10^{-19}$.","correct":true,"explanation":"One electronvolt is the energy transferred through a potential difference of one volt by one electron, so $1\\,eV = 1.60 \\times 10^{-19}\\,J$."},{"text":"Multiply by $6.24 \\times 10^{18}$.","correct":false,"explanation":"The factor $6.24 \\times 10^{18}$ converts joules to electronvolts, so this reverses the required conversion. To convert electronvolts to joules, multiply by $1.60 \\times 10^{-19}$."},{"text":"Divide by $1.60 \\times 10^{-19}$.","correct":false,"explanation":"Dividing by the joules-per-electronvolt value performs the inverse conversion, from joules to electronvolts. Electronvolts are converted to joules by multiplying by $1.60 \\times 10^{-19}$."},{"text":"Multiply by $1.60 \\times 10^{-16}$.","correct":false,"explanation":"The exponent $-16$ applies when converting kiloelectronvolts to joules, because the kilo prefix contributes $10^3$. For electronvolts, multiply by $1.60 \\times 10^{-19}$."}]},{"id":"85d97f59-04fe-4bbf-a8e0-4e8e44310a5e","cardType":"basic","points":1,"front":"How many joules are in one kilowatt-hour (kW h)?","back":"$$3.60 \\times 10^6\\,J$","mcqOptions":[{"text":"$$3.60 \\times 10^6\\,J$","correct":true,"explanation":"One kilowatt is $1000\\,W$ and one hour is $3600\\,s$, so $1\\,kW\\,h = 1000 \\times 3600 = 3.60 \\times 10^6\\,J$."},{"text":"$$3.60 \\times 10^3\\,J$","correct":false,"explanation":"This converts the hour to $3600\\,s$ but overlooks that one kilowatt equals $1000\\,W$. Including both conversions gives $3.60 \\times 10^6\\,J$."},{"text":"$$3.60 \\times 10^9\\,J$","correct":false,"explanation":"This introduces the kilo factor twice, making the result $1000$ times too large. Applying $1\\,kW = 1000\\,W$ once gives $3.60 \\times 10^6\\,J$."},{"text":"$$1.00 \\times 10^3\\,J$","correct":false,"explanation":"This treats one kilowatt-hour as though the power acts for only one second. Multiplying $1000\\,W$ by $3600\\,s$ gives $3.60 \\times 10^6\\,J$."}]},{"id":"f177d901-82d7-4ba7-aff8-81cc1125c9a8","cardType":"cloze","points":null,"front":"Despite the name, the kilowatt-hour is a unit of [ ] because it is power multiplied by time.","back":"Despite the name, the kilowatt-hour is a unit of energy because it is power multiplied by time.","mcqOptions":[{"text":"energy","correct":true,"explanation":"Power is the rate of energy transfer, so multiplying power by time gives energy. Therefore, the kilowatt-hour is a unit of energy rather than power."},{"text":"power","correct":false,"explanation":"The word “kilowatt” makes power tempting, but multiplying power by time produces energy. A kilowatt-hour is therefore a unit of energy, while the kilowatt alone measures power."},{"text":"potential difference","correct":false,"explanation":"Potential difference is measured in volts and is not obtained by multiplying power by time. Since power multiplied by time equals energy, the missing term is energy."},{"text":"electric charge","correct":false,"explanation":"Electric charge is measured in coulombs and equals current multiplied by time, not power multiplied by time. A kilowatt-hour instead measures energy."}]},{"id":"56d4fdb5-d619-436e-b09f-b649a80da57b","cardType":"basic","points":2,"front":"How should you handle compound units (e.g. $mm\\,ms^{-1}$) during conversion?","back":"Convert the numerator and denominator separately before simplifying.","mcqOptions":[{"text":"Convert the numerator and denominator separately before simplifying.","correct":true,"explanation":"Prefixes attached to numerator and denominator units contribute different factors and powers. Converting each part separately prevents sign and exponent errors before the factors are combined."},{"text":"Convert the numerator only and leave the denominator unchanged before simplifying.","correct":false,"explanation":"This ignores any prefix attached to the denominator, producing an incorrect overall factor. Both the numerator and denominator must be converted separately before the expression is simplified."},{"text":"Convert the denominator only and leave the numerator unchanged before simplifying.","correct":false,"explanation":"This overlooks the conversion factor associated with the numerator’s prefix. Both parts of a compound unit must be converted separately before their factors are simplified."},{"text":"Multiply the numerator and denominator by the same prefix factor before simplifying.","correct":false,"explanation":"Different prefixes, such as milli in $mm$ and $ms$, must be handled according to their positions in the unit. Convert numerator and denominator separately before simplifying."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Use of SI units and their prefixes","title":"Use of SI units and their prefixes Flashcards","topbar":{"backHref":"/as-level/physics/aqa/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"f9d41fef-f7ea-5bc7-b150-ede0ee05b0fe","walled":"$L38"}],"$L39","$L3a"]