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The unit identifies what has been measured and gives the number physical meaning."},{"text":"It lacks a direction.","correct":false,"explanation":"Direction is required only for vector quantities, whereas many physical quantities are scalars. The number 5 is incomplete because it lacks a unit specifying what it measures."},{"text":"It lacks a dimension.","correct":false,"explanation":"Dimensions classify quantities, but they are not normally written beside every measured value. The number 5 is incomplete because no unit has been given."},{"text":"It lacks a prefix.","correct":false,"explanation":"A prefix merely indicates a unit multiplier, such as kilo or milli, and is not always needed. What 5 lacks is a unit identifying the physical quantity."}]},{"id":"aae0326b-26bb-4bdc-8bf2-cb0f446ecda6","cardType":"basic","points":2,"front":"Why convert quantities to SI units before using physics equations?","back":"Equations assume SI units, avoiding wrong scales.","mcqOptions":[{"text":"Equations assume SI units, avoiding wrong scales.","correct":true,"explanation":"Standard physics equations generally produce the stated SI result only when input quantities use compatible SI units. Conversion prevents errors caused by prefixes and non-SI scale factors."},{"text":"Equations assume integer values, avoiding decimal answers.","correct":false,"explanation":"Physics equations can use decimal values without difficulty; quantities do not need to be integers. They should instead be converted to SI units to avoid scale errors."},{"text":"Equations assume vector values, avoiding missing directions.","correct":false,"explanation":"Equations may involve scalar or vector quantities depending on the context, so vector form is not universally required. SI conversion is needed to keep numerical scales consistent."},{"text":"Equations assume base quantities, avoiding derived units.","correct":false,"explanation":"Derived SI units such as joules and newtons can be used directly in appropriate equations. The essential requirement is compatible SI units, which prevent incorrect scale factors."}]},{"id":"d8e4cc24-c564-4eba-b4a6-a1867b78d399","cardType":"cloze","points":null,"front":"Scientific notation writes a number as [ ] multiplied by [ ].","back":"Scientific notation writes a number as a coefficient from 1 to 10 multiplied by a power of ten.","mcqOptions":[{"text":"Scientific notation writes a number as a coefficient from 1 to 10 multiplied by a power of ten.","correct":true,"explanation":"In scientific notation, the coefficient has one non-zero digit before the decimal point, so it lies from 1 up to but not including 10, and multiplies an integer power of ten."},{"text":"Scientific notation writes a number as a coefficient from 0 to 1 multiplied by a power of ten.","correct":false,"explanation":"This places the coefficient below 1, whereas standard scientific notation requires it to be at least 1 and less than 10. It must multiply a power of ten."},{"text":"Scientific notation writes a number as a coefficient from 1 to 10 multiplied by a power of one hundred.","correct":false,"explanation":"Using powers of one hundred skips alternate decimal place values and is not standard scientific notation. The coefficient from 1 to 10 must multiply a power of ten."},{"text":"Scientific notation writes a number as a coefficient from 10 to 100 multiplied by a power of ten.","correct":false,"explanation":"This leaves two digits before the decimal point, a common normalisation error. Standard scientific notation uses a coefficient from 1 to 10 multiplied by a power of ten."}]},{"id":"27890c78-3c58-4ffd-bcd5-aceb33b68927","cardType":"basic","points":1,"front":"Convert 5 cm into metres for an SI calculation.","back":"5 × 10^-2 m","mcqOptions":[{"text":"5 × 10^-2 m","correct":true,"explanation":"The prefix centi means 10^-2, so 1 cm equals 10^-2 m. Therefore, 5 cm equals 5 × 10^-2 m."},{"text":"5 × 10^-3 m","correct":false,"explanation":"This treats centimetres as millimetres, using 10^-3 instead of 10^-2. Since centi means one hundredth, 5 cm is 5 × 10^-2 m."},{"text":"5 × 10^-1 m","correct":false,"explanation":"This divides by 10 rather than by 100, making the result ten times too large. Since 100 cm equals 1 m, 5 cm is 5 × 10^-2 m."},{"text":"5 × 10² m","correct":false,"explanation":"This reverses the conversion and multiplies by 100 rather than dividing by 100. Converting centimetres to metres gives 5 × 10^-2 m."}]},{"id":"32c0587c-a5b3-4dc3-bcc6-ddee6e1e7494","cardType":"cloze","points":null,"front":"One order of magnitude means a factor of [ ]; three orders means a factor of [ ].","back":"One order of magnitude means a factor of 10; three orders means a factor of 10³.","mcqOptions":[{"text":"One order of magnitude means a factor of 10; three orders means a factor of 10³.","correct":true,"explanation":"Each order of magnitude corresponds to multiplication or division by 10. Three successive orders therefore give a factor of 10 × 10 × 10 = 10³."},{"text":"One order of magnitude means a factor of 1; three orders means a factor of 3.","correct":false,"explanation":"This confuses the number of orders with an ordinary numerical difference. One order is a factor of 10, so three orders correspond to a factor of 10³."},{"text":"One order of magnitude means a factor of 10; three orders means a factor of 30.","correct":false,"explanation":"This multiplies 10 by the number of orders instead of applying the factor repeatedly. Three orders give 10 × 10 × 10, which is 10³."},{"text":"One order of magnitude means a factor of 10²; three orders means a factor of 10⁶.","correct":false,"explanation":"This incorrectly assigns two powers of ten to each order, doubling every exponent. One order is a factor of 10, and three orders are a factor of 10³."}]},{"id":"6a7cb563-54b3-4e1c-bccc-b227dfc79786","cardType":"basic","points":1,"front":"What is the order of magnitude of 3 × 10⁶ J?","back":"10⁶ J","mcqOptions":[{"text":"10⁶ J","correct":true,"explanation":"The value 3 × 10⁶ J is on the million-joule scale, so its order of magnitude is represented by the power 10⁶ J."},{"text":"10³ J","correct":false,"explanation":"This mistakes the coefficient 3 for the exponent and gives the thousand-joule scale. The exponent in 3 × 10⁶ identifies the order as 10⁶ J."},{"text":"10⁷ J","correct":false,"explanation":"This rounds the exponent upward even though the value remains on the 10⁶ J scale. The stated order of magnitude is therefore 10⁶ J."},{"text":"10^-6 J","correct":false,"explanation":"This reverses the exponent’s sign, changing millions of joules into millionths of a joule. The positive exponent gives an order of magnitude of 10⁶ J."}]},{"id":"dc492c72-3761-45e3-8452-c65f5dda9b63","cardType":"basic","points":2,"front":"A hair is about 10^-4 m wide and an atom about 10^-10 m. How many orders of magnitude wider is the hair?","back":"Six orders of magnitude.","mcqOptions":[{"text":"Six orders of magnitude.","correct":true,"explanation":"The exponent difference is -4 - (-10) = 6. Equivalently, 10^-4 divided by 10^-10 is 10⁶, so the hair is six orders wider."},{"text":"Four orders of magnitude.","correct":false,"explanation":"This uses only the hair’s exponent and ignores the atom’s scale. Subtracting the exponents gives -4 - (-10) = 6, so the difference is six orders."},{"text":"Ten orders of magnitude.","correct":false,"explanation":"This takes the atom’s exponent magnitude as the answer instead of comparing the two widths. Their exponent difference is 6, so the hair is six orders wider."},{"text":"Fourteen orders of magnitude.","correct":false,"explanation":"This adds the magnitudes of the exponents rather than subtracting them. For a ratio, 10^-4 ÷ 10^-10 = 10⁶, giving six orders."}]},{"id":"76f7ade7-4773-41f9-bc02-1c685c5cde25","cardType":"basic","points":1,"front":"If asked for the nearest order of magnitude, what form should the final answer take?","back":"A single power of ten with the unit.","mcqOptions":[{"text":"A single power of ten with the unit.","correct":true,"explanation":"An order of magnitude specifies the scale of a quantity rather than a precise coefficient. The final estimate should therefore be written as one power of ten with its unit."},{"text":"A decimal coefficient with the unit.","correct":false,"explanation":"A decimal coefficient gives more precision but does not by itself state an order of magnitude. The requested form is a single power of ten accompanied by the unit."},{"text":"A percentage uncertainty with the unit.","correct":false,"explanation":"Percentage uncertainty describes measurement precision and is dimensionless, so attaching the quantity’s unit is inappropriate. A nearest-order answer should be a single power of ten with the unit."},{"text":"A range of values with the unit.","correct":false,"explanation":"A range may communicate uncertainty but does not provide the requested nearest order of magnitude. The final answer should instead be one power of ten with the appropriate unit."}]},{"id":"2ccd4a66-273a-4576-b39b-e71ffddc0d0a","cardType":"cloze","points":null,"front":"For x = c × 10ⁿ, if c < √10 round to [ ]; if c > √10 round to [ ].","back":"For x = c × 10ⁿ, if c < √10 round to 10ⁿ; if c > √10 round to 10ⁿ⁺¹.","mcqOptions":[{"text":"For x = c × 10ⁿ, if c < √10 round to 10ⁿ; if c > √10 round to 10ⁿ⁺¹.","correct":true,"explanation":"√10 is the halfway point on a logarithmic scale between $10^n$ and $10^{n+1}$. Below it, $x$ is nearer $10^n$; above it, it is nearer $10^{n+1}$."},{"text":"For x = c × 10ⁿ, if c < √10 round to 10ⁿ⁺¹; if c > √10 round to 10ⁿ.","correct":false,"explanation":"This reverses the rounding direction. A coefficient below √10 is closer to $10^n$, while a coefficient above √10 is closer to the next power, $10^{n+1}$."},{"text":"For x = c × 10ⁿ, if c < √10 round to 10ⁿ; if c > √10 round to 10^n-1.","correct":false,"explanation":"The second term uses the wrong exponent direction. If the coefficient is larger than √10, the number is nearer the next larger power, so it rounds to $10^{n+1}$."},{"text":"For x = c × 10ⁿ, if c < √10 round to 10^n-1; if c > √10 round to 10ⁿ.","correct":false,"explanation":"This shifts both powers one step too low. The nearest-order choice is between $10^n$ and $10^{n+1}$, with √10 as the boundary between them."}]},{"id":"8cb381d9-af91-4401-8f69-0dfc45eaed43","cardType":"basic","points":2,"front":"What is the quick coefficient rule for rounding to the nearest order of magnitude?","back":"

1, 2, 3 usually round down
4 to 9 usually round up

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

1, 2, 3 usually round down
4 to 9 usually round up

","correct":true,"explanation":"For order-of-magnitude rounding, the exact boundary is √10 ≈ 3.16. As a quick rule, coefficients 1, 2, 3 round down, while 4 to 9 round up."},{"text":"

1, 2, 3, 4 usually round down
5 to 9 usually round up

","correct":false,"explanation":"This uses the ordinary decimal rounding threshold of 5. For nearest order of magnitude the boundary is √10 ≈ 3.16, so 4 should usually round up."},{"text":"

1, 2 usually round down
3 to 9 usually round up

","correct":false,"explanation":"This puts the boundary too low. Since √10 is about 3.16, a coefficient of 3 is still closer to the lower power of ten and usually rounds down."},{"text":"

1 to 5 usually round down
6 to 9 usually round up

","correct":false,"explanation":"This treats the coefficients as if the halfway point were about 5.5. On a logarithmic order-of-magnitude scale the boundary is √10, so 4 and above usually round up."}]},{"id":"342035a5-3e12-4598-8d29-b8e1cf7dd0a7","cardType":"basic","points":1,"front":"Round 0.0048 m to the nearest order of magnitude.","back":"10^-2 m","mcqOptions":[{"text":"10^-2 m","correct":true,"explanation":"0.0048 m is $4.8 \\times 10^{-3}$ m. Since 4.8 is greater than √10, it rounds up to the next power of ten, $10^{-2}$ m."},{"text":"10^-3 m","correct":false,"explanation":"This rounds the coefficient 4.8 down as if only the exponent mattered. Because 4.8 is greater than √10, the nearest order of magnitude is $10^{-2}$ m."},{"text":"10^-4 m","correct":false,"explanation":"This is a decimal-place error, making the value ten times smaller than the base power in $4.8 \\times 10^{-3}$ m. The coefficient is above √10, so the answer is $10^{-2}$ m."},{"text":"10² m","correct":false,"explanation":"This loses the negative exponent from a number much smaller than 1 m. Since 0.0048 m is $4.8 \\times 10^{-3}$ m, it rounds to $10^{-2}$ m."}]},{"id":"c003e67a-115f-4810-8282-7add9c1b2bf4","cardType":"cloze","points":null,"front":"An answer to one significant figure keeps a [ ]; an answer to nearest order of magnitude is just a [ ].","back":"An answer to one significant figure keeps a coefficient; an answer to nearest order of magnitude is just a power of ten.","mcqOptions":[{"text":"An answer to one significant figure keeps a coefficient; an answer to nearest order of magnitude is just a power of ten.","correct":true,"explanation":"One significant figure keeps one non-zero digit as the coefficient, such as $3 \\times 10^4$. Nearest order of magnitude removes that coefficient and gives only a power of ten."},{"text":"An answer to one significant figure keeps a power of ten; an answer to nearest order of magnitude is just a coefficient.","correct":false,"explanation":"This reverses the two ideas. One significant figure still includes a coefficient, while an order-of-magnitude answer is reduced to a single power of ten."},{"text":"An answer to one significant figure keeps a unit; an answer to nearest order of magnitude is just a power of ten.","correct":false,"explanation":"Units are not what distinguishes one significant figure from order-of-magnitude rounding. One significant figure keeps a coefficient, while nearest order of magnitude gives only a power of ten."},{"text":"An answer to one significant figure keeps a coefficient; an answer to nearest order of magnitude is just a significant figure.","correct":false,"explanation":"This confuses the two levels of rounding. A nearest order-of-magnitude answer is not a significant figure with a coefficient; it is just a power of ten."}]},{"id":"979ab9b5-3d65-4a8c-a857-c354403896ab","cardType":"basic","points":1,"front":"For nearest-order estimates, being within what factor is often the main expectation?","back":"Within a factor of 10.","mcqOptions":[{"text":"Within a factor of 10.","correct":true,"explanation":"Order-of-magnitude estimates are intended to get the correct power of ten. Being within a factor of 10 is usually enough to show the scale is right."},{"text":"Within a factor of 2.","correct":false,"explanation":"A factor of 2 is a much tighter expectation than an order-of-magnitude estimate. For nearest-order estimates, the main aim is usually to be within a factor of 10."},{"text":"Within a factor of 100.","correct":false,"explanation":"A factor of 100 allows an error of two orders of magnitude, which is too loose for a nearest-order estimate. The usual expectation is within a factor of 10."},{"text":"Within a factor of 1.","correct":false,"explanation":"A factor of 1 would mean an exact answer, not an estimate. Order-of-magnitude work mainly checks the scale, so within a factor of 10 is the usual target."}]},{"id":"6912a67c-8bd1-447a-bbff-643cbe329425","cardType":"cloze","points":null,"front":"Benchmark masses: a typical human is about [ ], while a car is about [ ].","back":"Benchmark masses: a typical human is about 10² kg, while a car is about 10³ kg.","mcqOptions":[{"text":"Benchmark masses: a typical human is about 10² kg, while a car is about 10³ kg.","correct":true,"explanation":"A typical adult mass is roughly 50-100 kg, which is order $10^2$ kg. A car is roughly 1000 kg, so its benchmark mass is order $10^3$ kg."},{"text":"Benchmark masses: a typical human is about 10¹ kg, while a car is about 10³ kg.","correct":false,"explanation":"The human mass is underestimated by one order of magnitude, closer to a small child or heavy object. A typical human is about $10^2$ kg, while a car is about $10^3$ kg."},{"text":"Benchmark masses: a typical human is about 10² kg, while a car is about 10⁴ kg.","correct":false,"explanation":"The car mass is overestimated by one order of magnitude, more like a heavy lorry. For estimates, a typical car is about $10^3$ kg and a human about $10^2$ kg."},{"text":"Benchmark masses: a typical human is about 10³ kg, while a car is about 10² kg.","correct":false,"explanation":"This effectively swaps the scale of the two masses. Humans are around $10^2$ kg, whereas cars are heavier, typically around $10^3$ kg."}]},{"id":"580bfb41-d23c-4baf-af8b-4e89dd9855fd","cardType":"basic","points":1,"front":"What benchmark value is used for gravitational field strength near Earth?","back":"g ≈ 10 N kg^-1","mcqOptions":[{"text":"g ≈ 10 N kg^-1","correct":true,"explanation":"Near Earth, gravitational field strength is about 9.8 N kg^-1, which is rounded to 10 N kg^-1 for estimation. This is the standard benchmark value."},{"text":"g ≈ 1 N kg^-1","correct":false,"explanation":"This is too small by a factor of about 10 and would seriously underestimate weights. Near Earth the benchmark gravitational field strength is $g \\approx 10\\,\\text{N kg}^{-1}$."},{"text":"g ≈ 100 N kg^-1","correct":false,"explanation":"This is too large by a factor of about 10 and would overestimate weights. Near Earth, $g$ is approximately 9.8 N kg^-1, rounded to 10 N kg^-1."},{"text":"g ≈ 10 N kg^-2","correct":false,"explanation":"The numerical value is tempting, but the unit exponent is wrong. Gravitational field strength is force per unit mass, so its benchmark unit is N kg^-1, not N kg^-2."}]},{"id":"26d53637-0f0f-4250-90bf-1ea6b384d8ba","cardType":"cloze","points":null,"front":"Benchmark values: density of water ≈ [ ]; atmospheric pressure ≈ [ ].","back":"Benchmark values: density of water ≈ 10³ kg m^-3; atmospheric pressure ≈ 10⁵ Pa.","mcqOptions":[{"text":"Benchmark values: density of water ≈ 10³ kg m^-3; atmospheric pressure ≈ 10⁵ Pa.","correct":true,"explanation":"Water has density about 1000 kg m^-3, giving $10^3$ kg m^-3. Atmospheric pressure is about 100000 Pa, giving the benchmark $10^5$ Pa."},{"text":"Benchmark values: density of water ≈ 10⁰ kg m^-3; atmospheric pressure ≈ 10⁵ Pa.","correct":false,"explanation":"This confuses 1 g cm^-3 with SI units. Converting to kg m^-3 gives water a density of $10^3$ kg m^-3, while atmospheric pressure is about $10^5$ Pa."},{"text":"Benchmark values: density of water ≈ 10³ kg m^-3; atmospheric pressure ≈ 10⁴ Pa.","correct":false,"explanation":"The density value is right, but the pressure is one order of magnitude too low. Standard atmospheric pressure is about 100000 Pa, so the benchmark is $10^5$ Pa."},{"text":"Benchmark values: density of water ≈ 10⁵ kg m^-3; atmospheric pressure ≈ 10³ Pa.","correct":false,"explanation":"This mixes up the common powers of ten and assigns unrealistic scales. Water is about $10^3$ kg m^-3, and atmospheric pressure is about $10^5$ Pa."}]},{"id":"6e909e13-78c3-4f4d-b798-1e493c3e87ef","cardType":"basic","points":1,"front":"Which equation estimates kinetic energy from mass and speed?","back":"E_k = ½mv²","mcqOptions":[{"text":"E_k = ½mv²","correct":true,"explanation":"Kinetic energy equals half the object’s mass multiplied by the square of its speed. The squared speed means changes in speed have a particularly large effect on kinetic energy."},{"text":"E_k = mv²","correct":false,"explanation":"Omitting the factor of ½ makes the estimated kinetic energy twice its correct value. The correct equation is E_k = ½mv²."},{"text":"E_k = ½mv","correct":false,"explanation":"Failing to square the speed gives incorrect dimensions and underestimates the effect of speed. The correct equation is E_k = ½mv²."},{"text":"E_k = ½m²v","correct":false,"explanation":"This incorrectly squares the mass rather than the speed, so it does not represent kinetic energy. The correct equation is E_k = ½mv²."}]},{"id":"014ab89f-a8c4-44f5-99c8-a9770f97ff96","cardType":"basic","points":2,"front":"Using m ≈ 10³ kg and v ≈ 30 m s^-1, what is a car’s kinetic energy to nearest order?","back":"10⁶ J","mcqOptions":[{"text":"10⁶ J","correct":true,"explanation":"Using E_k = ½mv² gives approximately ½ × 10³ × 30² = 4.5 × 10⁵ J, whose nearest order of magnitude is 10⁶ J."},{"text":"10⁵ J","correct":false,"explanation":"This rounds 4.5 × 10⁵ J down based only on its written exponent. It is closer to 10⁶ J on a logarithmic scale, so that is the nearest order."},{"text":"10⁷ J","correct":false,"explanation":"This is one order of magnitude too large, often caused by over-rounding the squared speed. The calculation gives 4.5 × 10⁵ J, whose nearest order is 10⁶ J."},{"text":"10³ J","correct":false,"explanation":"This effectively neglects the factor of v², which contributes roughly 10³. Including it gives about 4.5 × 10⁵ J, with nearest order 10⁶ J."}]},{"id":"f3fd4d8e-5086-4dcc-bca2-b2300a64b78c","cardType":"basic","points":2,"front":"In E_k = ½mv², what happens to kinetic energy if speed doubles?","back":"It becomes four times larger.","mcqOptions":[{"text":"It becomes four times larger.","correct":true,"explanation":"Kinetic energy is proportional to v². Doubling speed therefore multiplies kinetic energy by 2² = 4, while the mass remains unchanged."},{"text":"It becomes twice as large.","correct":false,"explanation":"This treats kinetic energy as directly proportional to speed and overlooks the square. Since E_k ∝ v², doubling speed makes the kinetic energy four times larger."},{"text":"It becomes eight times larger.","correct":false,"explanation":"This applies an unnecessary cube to the speed change. Kinetic energy depends on the square of speed, so doubling speed gives a factor of 2² = 4."},{"text":"It becomes half as large.","correct":false,"explanation":"This reverses the relationship between speed and kinetic energy. A greater speed increases kinetic energy, and doubling speed makes it four times larger because speed is squared."}]},{"id":"042ad79f-ff8c-494b-b4b1-8c014ae8f38a","cardType":"cloze","points":null,"front":"When multiplying powers of ten, [ ] the indices; when dividing powers of ten, [ ] the indices.","back":"When multiplying powers of ten, add the indices; when dividing powers of ten, subtract the indices.","mcqOptions":[{"text":"When multiplying powers of ten, add the indices; when dividing powers of ten, subtract the indices.","correct":true,"explanation":"For the same base, multiplication combines powers by adding indices, while division combines them by subtracting indices: 10^a × 10^b = 10^a+b and 10^a/10^b = 10^a-b."},{"text":"When multiplying powers of ten, multiply the indices; when dividing powers of ten, subtract the indices.","correct":false,"explanation":"Multiplying the indices is the rule for raising a power to another power, not multiplying powers with the same base. Here the indices are added for multiplication and subtracted for division."},{"text":"When multiplying powers of ten, add the indices; when dividing powers of ten, add the indices.","correct":false,"explanation":"Adding indices is correct for multiplication but not division. Dividing by 10^b removes that power, so the denominator’s index must be subtracted."},{"text":"When multiplying powers of ten, subtract the indices; when dividing powers of ten, add the indices.","correct":false,"explanation":"This reverses both index laws. Multiplication adds the indices because the factors combine, whereas division subtracts the denominator’s index from the numerator’s index."}]},{"id":"5479c769-fa25-4278-83ce-c6758d5dc8c5","cardType":"basic","points":1,"front":"When raising a power of ten to a power, what happens to the indices?","back":"Multiply the indices.","mcqOptions":[{"text":"Multiply the indices.","correct":true,"explanation":"When a power is raised to another power, the indices multiply. For example, (10³)² = 10^{3 × 2} = 10⁶."},{"text":"Add the indices.","correct":false,"explanation":"Adding indices applies when multiplying separate powers with the same base. When one power is raised to another, the correct rule is to multiply the indices."},{"text":"Subtract the indices.","correct":false,"explanation":"Subtracting indices applies when dividing powers with the same base. Raising a power of ten to another power requires the two indices to be multiplied."},{"text":"Divide the indices.","correct":false,"explanation":"Dividing the indices is not the rule for a power raised to a power. The outer power repeats the inner power, so the indices must be multiplied."}]},{"id":"e93ce3e6-9dc0-46ee-9188-35355a2c7f75","cardType":"basic","points":1,"front":"Which equation estimates pressure from force and area?","back":"p = F/A","mcqOptions":[{"text":"p = F/A","correct":true,"explanation":"Pressure is the normal force distributed over an area, so it equals force divided by area. This gives the unit N m^-2, equivalent to pascals."},{"text":"p = FA","correct":false,"explanation":"Multiplying force by area gives incorrect dimensions and predicts that spreading a force increases pressure. Pressure is force divided by area: p = F/A."},{"text":"p = A/F","correct":false,"explanation":"This reverses the pressure relationship and gives units of m² N^-1, not pascals. Pressure is calculated by dividing force by area."},{"text":"p = F/A²","correct":false,"explanation":"Squaring the area gives the wrong dimensions and makes pressure decrease too rapidly as contact area increases. The area appears only to the first power in p = F/A."}]},{"id":"84e1d0be-6d87-4cd9-8a49-071e8d42cdb6","cardType":"basic","points":2,"front":"What is the order of magnitude of the pressure beneath a standing person in the notes?","back":"10⁴ Pa","mcqOptions":[{"text":"10⁴ Pa","correct":true,"explanation":"A person’s weight is roughly 10³ N and the combined contact area of their feet is roughly 10^-1 m². Dividing gives pressure of order 10⁴ Pa."},{"text":"10³ Pa","correct":false,"explanation":"This is one order too small, typically from using an unrealistically large contact area of about 1 m². Using roughly 10^-1 m² gives 10⁴ Pa."},{"text":"10⁵ Pa","correct":false,"explanation":"This is one order too large, often from estimating the foot contact area as only 10^-2 m². The notes use estimates that give 10⁴ Pa."},{"text":"10⁶ Pa","correct":false,"explanation":"This overestimates the pressure by two orders of magnitude, implying an extremely small contact area. A weight near 10³ N over about 10^-1 m² gives 10⁴ Pa."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Estimation of physical quantities","title":"Estimation of physical quantities Flashcards","topbar":{"backHref":"/as-level/physics/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"13e13249-b495-5ff4-8304-512b71f51b93","walled":"$L39"}],"$L3a","$L3b"]