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Scientific Theories and Using Formulas

Apply scientific laws, use given formulas with the right units, convert units, and work with scientific notation.

Practice this skill 18 min Scientific theories Leer en español
Key idea

What you'll learn

Scientific theories and laws are powerful because they let us predict. If you know Newton's second law, you can figure out how hard to push a cart. If you know the formula for density, you can tell copper from iron without a chemistry lab. The GED calls this skill scientific theories and using formulas.

In this lesson you will learn to:

  • apply a theory or law (such as conservation of energy, Newton's laws or the rule that less-dense objects float) to a new situation;
  • use a given formula: identify each variable, substitute with units, calculate and check;
  • rearrange a simple formula to find a different variable;
  • do unit conversions with the factor-label method;
  • read, write and calculate with scientific notation.

How it shows up on the GED: Science questions that need a formula usually give it to you right in the passage or question, along with what each letter means. Your job is to use it correctly. Expect fill-in-the-blank questions where you type a number (often with a small rounding allowance), multiple-choice questions where wrong answers come from common slips (dividing upside down, forgetting to convert centimeters to meters), and questions that ask which statement correctly applies a law to an experiment. The on-screen calculator is available on the Science test.

Words to know

Key terms

  • scientific law — a statement, often a formula, that describes what always happens under certain conditions (Newton's second law: F = ma).
  • scientific theory — a well-tested explanation of why things happen, supported by many lines of evidence (cell theory, plate tectonics, evolution). Not a guess.
  • formula — an equation showing how quantities are related (d = m/V).
  • variable — a letter that stands for a quantity (m for mass).
  • substitute — replace a letter with its value (put 54.0 in for m).
  • unit — the measure attached to a number (g, cm³, N, m/s).
  • conversion factor — a fraction equal to 1 that changes units (\dfrac{1\text{ m}}{100\text{ cm}}).
  • density — mass per unit volume (copper: 8.96 g/cm³).
  • speed — distance per unit time (25 m/s).
  • force — a push or pull, measured in newtons (N).
  • work — energy transferred when a force moves an object a distance, measured in joules (J).
  • potential energy — stored energy, such as energy due to height.
  • kinetic energy — energy of motion.
  • scientific notation — a number written as (a number from 1 to less than 10) × (a power of 10) (150{,}000{,}000 = 1.5 \times 10^8).
  • exponent — the small raised number that tells how many times 10 is multiplied (the 8 in 10^8).

All Science words in the glossary

1. Applying theories and laws

A law or theory is a general rule. Applying it means using the rule to explain or predict a specific case.

Law or theoryWhat it saysHow it is applied
Conservation of energyEnergy is not created or destroyed; it changes form.A falling ball's potential energy becomes kinetic energy (plus some heat from air resistance).
Conservation of massIn a chemical reaction, total mass before = total mass after.10 g of reactants make 10 g of products, even if some is a gas that escapes.
Newton's first lawAn object keeps moving (or stays still) unless a net force acts.A passenger lurches forward when a bus stops suddenly.
Newton's second lawF = maSame force on more mass → less acceleration.
Newton's third lawEvery action has an equal and opposite reaction.A rocket pushes gas down; the gas pushes the rocket up.
Density and floatingObjects less dense than a fluid float in it.Oak (0.75 g/cm³) floats in water (1.00 g/cm³).
Cell theoryAll living things are made of cells; cells come from existing cells.A new organism must contain cells.

When results don't match a law. Real measurements often fall a little short of an ideal prediction. A falling ball moves a bit slower than v = \sqrt{2gh} predicts because air resistance changes some energy into heat and sound. The GED's correct answer explains the gap with a condition the ideal case ignored. It does not say the law is wrong or that energy was destroyed.

A ball falling from a height, with an energy bar at three points. At the top all its energy is potential energy (PE). Halfway down it is half potential and half kinetic energy (KE). At the bottom it is mostly kinetic energy plus a small amount changed to heat and sound by air resistance. The total bar length stays the same: energy is conserved. topPEhalfwayPEKEbottomKEheat and soundTotal bar length stays the same

Example. A sealed jar of vinegar and baking soda has a mass of 250 g before mixing. After the fizzing stops, the sealed jar still has a mass of 250 g. If the jar were open, the mass would drop because carbon dioxide gas escapes — but the total mass of all the substances, including the escaped gas, is still conserved.

Conservation of mass with vinegar and baking soda. Before mixing, the sealed jar has a mass of 250 grams. After the fizzing stops, the sealed jar, with carbon dioxide trapped inside, still has a mass of 250 grams. If the jar is open, carbon dioxide escapes and the scale reads less than 250 grams, but the total mass of all substances, including the escaped gas, is conserved. 250 gBefore mixing (sealed)vinegarbaking soda in cup250 gAfter fizzing (sealed)CO₂ trapped insideless than 250 gAfter fizzing (open)CO₂ escapesMass of all substances, including the escaped gas, is conserved.

2. Using a given formula

Follow the same five steps every time:

  1. Write the formula exactly as given.
  2. List the knowns and the unknown, with units. (m = 54.0 g, V = 20.0 cm³, d = ?)
  3. Check the units match what the formula expects. Convert if needed.
  4. Substitute and calculate.
  5. Check that the answer makes sense and has the right units.

Rearranging. Sometimes you know the answer variable and need a different one. For formulas of the form A = BC (like F = ma or W = Fd):

  • A = B \times C
  • B = \dfrac{A}{C}
  • C = \dfrac{A}{B}
Two formula triangles. In the first, F is on top and m and a are side by side below, for F equals m times a; so a equals F divided by m and m equals F divided by a. In the second, m is on top and d and V are below, for density equals mass divided by volume; so m equals d times V and V equals m divided by d. Cover the letter you want: side by side means multiply, top over bottom means divide. FmaF = mamdVd = m ÷ VCover the letter you want: side by side = multiply; top over bottom = divide.a = F ÷ m, m = F ÷ am = dV, V = m ÷ d

Example. A 12 N force acts on a 3.0 kg cart. Find a. From F = ma: a = \dfrac{F}{m} = \dfrac{12}{3.0} = 4.0 m/s².

A cart of mass 3.0 kilograms is pushed by a force of 12 newtons. Its acceleration a is unknown. a equals F divided by m, 12 newtons divided by 3.0 kilograms, which is 4.0 meters per second squared. m = 3.0 kgF = 12 Na = ?a = F ÷ m = 12 N ÷ 3.0 kg = 4.0 m/s²

For d = \dfrac{m}{V}: m = dV and V = \dfrac{m}{d}. A piece of aluminum (2.70 g/cm³) with a mass of 27 g has a volume of V = \dfrac{27}{2.70} = 10 cm³.

Use the units as a guide. "m/s" means divide meters by seconds. "g/cm³" means divide grams by cubic centimeters. "N·m" (a joule) means multiply newtons by meters. If your answer's units don't make sense, the setup is wrong.

3. Unit conversions

Formulas only work when units match. A work formula that expects meters gives a wrong answer — 100 times too big — if you plug in centimeters.

Metric prefixes you should know:

PrefixMeaningExample
kilo- (k)1,0001 km = 1,000 m
centi- (c)\tfrac{1}{100}100 cm = 1 m
milli- (m)\tfrac{1}{1000}1,000 mL = 1 L; 1,000 mg = 1 g
micro- (µ)\tfrac{1}{1{,}000{,}000}1,000,000 µm = 1 m

Time: 60 s = 1 min, 60 min = 1 h, 3,600 s = 1 h.

The factor-label method. Multiply by fractions that equal 1, arranged so unwanted units cancel.

Example. Convert 25 m/s to km/h:

25\ \dfrac{\text{m}}{\text{s}} \times \dfrac{3600\text{ s}}{1\text{ h}} \times \dfrac{1\text{ km}}{1000\text{ m}} = 90\ \text{km/h}

Seconds cancel (one on top, one on bottom), meters cancel, and km/h is left.

Factor-label conversion of 25 meters per second to kilometers per hour: 25 meters over 1 second, times 3600 seconds over 1 hour, times 1 kilometer over 1000 meters, equals 90 kilometers per hour. Meters cancel meters and seconds cancel seconds, leaving kilometers per hour. Each fraction equals 1, so only the units change. 25ms×3600s1h×1km1000m=90 km/hm cancels m · s cancels s · km/h is leftEach fraction equals 1, so only the units change.

Example. Convert 2.5 L to mL: 2.5\text{ L} \times \dfrac{1000\text{ mL}}{1\text{ L}} = 2{,}500 mL.

Sense check: converting to a smaller unit gives a bigger number (2.5 L = 2,500 mL). Converting to a larger unit gives a smaller number (150 cm = 1.5 m).

Metric staircase. Going down from kilometer to meter multiply by 1000; from meter to centimeter multiply by 100; from centimeter to millimeter multiply by 10. Converting down to a smaller unit gives a bigger number, as 2.5 liters equals 2,500 milliliters. Converting up to a larger unit gives a smaller number, as 150 centimeters equals 1.5 meters. kmkilometer× 1000mmeter× 100cmcentimeter× 10mmmillimeterDown to a smaller unit → bigger number (2.5 L = 2,500 mL).Up to a larger unit → smaller number (150 cm = 1.5 m).

4. Scientific notation

Science deals with the very large (distances in space) and the very small (cells, atoms). Scientific notation keeps these numbers manageable.

Form: a \times 10^n, where 1 \le a < 10 and n is a whole number.

Standard formScientific notationHow
150,000,0001.5 \times 10^8decimal moved 8 places left → +8
300,0003.0 \times 10^55 places left → +5
0.00454.5 \times 10^{-3}3 places right → -3
0.000 007 57.5 \times 10^{-6}6 places right → -6
Decimal-point hops for scientific notation. In 150,000,000 the decimal point moves 8 places left, giving 1.5 times 10 to the 8th. In 0.0045 the decimal point moves 3 places right, giving 4.5 times 10 to the minus 3. Moving left gives a positive exponent, a big number; moving right gives a negative exponent, a number between 0 and 1. 150,000,000: decimal point moves 8 places left150000000→ 1.5 × 1080.0045: decimal point moves 3 places right00045→ 4.5 × 10−3Left → positive exponent (big number). Right → negative exponent (between 0 and 1).

A positive exponent means a big number; a negative exponent means a number between 0 and 1.

Multiplying: multiply the front numbers, add the exponents. (2 \times 10^3)(4 \times 10^5) = 8 \times 10^8.

Dividing: divide the front numbers, subtract the exponents. \dfrac{1.5 \times 10^8}{3.0 \times 10^5} = 0.5 \times 10^3 = 5.0 \times 10^2.

Adjusting: if the front number is not between 1 and 10, shift it. 0.5 \times 10^3 = 5.0 \times 10^2 (front number up by ×10, exponent down by 1). 12 \times 10^4 = 1.2 \times 10^5.

Comparing: the number with the larger exponent is larger (when both front numbers are between 1 and 10): 5.6 \times 10^7 < 1.5 \times 10^8.

Formula

Formulas you will often see

GED science questions that need a formula usually state it, with the meaning of each letter. Recognizing these common ones will save time:

QuantityFormulaUnits
Densityd = \dfrac{m}{V}g/cm³
Speedv = \dfrac{d}{t}m/s, km/h
Accelerationa = \dfrac{v_f - v_i}{t}m/s²
Newton's second lawF = maN (newtons)
WorkW = FdJ (joules)
PowerP = \dfrac{W}{t}W (watts)
Kinetic energyKE = \tfrac{1}{2}mv^2J
Gravitational potential energyPE = mgh, g = 9.8 m/s²J

Here v_f is the final speed, v_i the starting speed, and t the time.

Worked example

Worked example 1: Applying a law to a situation

Stimulus. A student burns a 5.0 g piece of steel wool (iron) in an open dish. Afterward, the black product has a mass of 6.4 g.

Question. Which statement best explains the result using the law of conservation of mass?

A. Mass was created when the iron burned. B. The iron combined with oxygen from the air, so the product includes the mass of that oxygen. C. The law of conservation of mass does not apply to burning. D. The scale was broken.

Think-aloud.

  1. The law: mass is neither created nor destroyed in a reaction. So an increase of 6.4 - 5.0 = 1.4 g must have come from somewhere.
  2. Burning iron reacts with oxygen in the air. The 1.4 g is the oxygen that joined the iron. ✓ B
  3. A breaks the law. C — the law applies to all chemical reactions. D — no evidence of error; the result fits the law.
Bar model for burning steel wool. Before: 5.0 grams of iron plus 1.4 grams of oxygen from the air. After: 6.4 grams of black product. 5.0 plus 1.4 equals 6.4, so no mass was created. Beforeiron 5.0 gO₂ 1.4 gfrom the airAfterblack product 6.4 g5.0 g + 1.4 g = 6.4 g: no mass created

Answer: B.

Worked example

Worked example 2: Rearranging a formula

Stimulus. A gold ring has a mass of 9.65 g. Gold's density is 19.3 g/cm³. Formula: d = \dfrac{m}{V}.

Question. What is the volume of the ring?

Think-aloud.

  1. Knowns: m = 9.65 g, d = 19.3 g/cm³. Unknown: V.
  2. Rearrange: V = \dfrac{m}{d}.
  3. Substitute: V = \dfrac{9.65}{19.3} = 0.5 cm³.
  4. Units: g ÷ (g/cm³) = cm³ ✓
  5. Sense check: gold is very dense, so a small volume holds a lot of mass. Half a cubic centimeter for a ring makes sense.
Density triangle for the gold ring: m, 9.65 grams, on top; d, 19.3 grams per cubic centimeter, and V, unknown, on the bottom. Covering V gives V equals m divided by d, 9.65 divided by 19.3, which is 0.5 cubic centimeters. Units: grams divided by grams per cubic centimeter gives cubic centimeters. m9.65 gd19.3 g/cm³V = ?Cover V:V = m ÷ d= 9.65 ÷ 19.3= 0.5 cm³units: g ÷ (g/cm³) = cm³ ✓

Answer: 0.5 cm³.

A common wrong answer is 9.65 \times 19.3 \approx 186 — the units would be g²/cm³, which is not a volume.

Worked example

Worked example 3: Converting units before using a formula

Stimulus. A mover pushes a box with a force of 200 N across a floor for 350 cm. It takes 7.0 seconds. Formulas: W = Fd (with d in meters) and P = \dfrac{W}{t}.

Question. What is the mover's power?

Think-aloud.

  1. Convert distance: 350\text{ cm} \div 100 = 3.5 m.
  2. Work: W = 200 \times 3.5 = 700 J.
  3. Power: P = \dfrac{700\text{ J}}{7.0\text{ s}} = 100 W.
  4. Trap check: using 350 cm directly gives W = 70{,}000 and P = 10{,}000 W — a hundred times too big.
A mover pushes a box with a force of 200 newtons across a floor for a distance of 350 centimeters, which is 3.5 meters, in 7.0 seconds. Work equals 200 newtons times 3.5 meters, 700 joules. Power equals 700 joules divided by 7.0 seconds, 100 watts. boxboxF = 200 Nd = 350 cm = 3.5 m (time: 7.0 s)W = F × d = 200 N × 3.5 m = 700 JP = W ÷ t = 700 J ÷ 7.0 s = 100 W

Answer: 100 W.

Worked example

Worked example 4: Scientific notation

Stimulus. A typical human red blood cell is about 8 \times 10^{-6} m across. A sheet of paper is about 1 \times 10^{-4} m thick.

Question. About how many red blood cells, lined up side by side, would span the thickness of the paper?

Think-aloud.

  1. "How many fit" means divide: \dfrac{1 \times 10^{-4}}{8 \times 10^{-6}}.
  2. Front numbers: 1 \div 8 = 0.125.
  3. Exponents: 10^{-4 - (-6)} = 10^{2}.
  4. Combine: 0.125 \times 10^2 = 12.5.
  5. Sense check: the paper (10^{-4}) is bigger than the cell (10^{-6}), so the answer should be more than 1. ✓
Drawn to scale: a sheet of paper 1 times 10 to the minus 4 meters, or 100 micrometers, thick, with red blood cells 8 times 10 to the minus 6 meters, or 8 micrometers, across lined up beside it. 100 divided by 8 is 12.5, so about 12 or 13 cells span the paper. paper thickness: 1 × 10−4 m = 100 µmone cell: 8 × 10−6 m = 8 µm100 ÷ 8 = 12.5 → about 12 or 13 cells

Answer: about 12 or 13 cells.

Watch the double negative: -4 - (-6) = -4 + 6 = 2.

Calculator tip

Calculator tip (TI-30XS MultiView)

  • For scientific notation, you can type powers of ten with the ^ key: 1.5 \times 10^8 is 1.5 × 10 ^ 8. Press the right arrow to leave each exponent, and put the whole divisor in parentheses: 1.5 × 10 ^ 8 ▶ ÷ ( 3 × 10 ^ 5 ▶ ) enter → 500. (The ×10ⁿ key is a shortcut: 1.5 ×10ⁿ 8.)
  • For a square root, press 2nd then x² (the √ function), type the number, then enter. \sqrt{392} may show as the exact root 14\sqrt{2}; press toggle to see \approx 19.8.
  • For a negative exponent, use the (−) key (not the subtraction key): 10 ^ (−) 6.
  • Do unit conversions as one chain: 25 × 3600 ÷ 1000 enter → 90.
  • Still write the setup on your erasable note board first; the calculator cannot catch an upside-down formula.
Common mistake

Common mistakes

  • Turning the formula upside down. Wrong: speed = 32 s ÷ 800 m = 0.04. Right: v = \dfrac{d}{t} = \dfrac{800}{32} = 25 m/s. Let the units (m/s) guide you.
  • Skipping a unit conversion. Wrong: W = 6.0 \times 150 = 900 J with distance in cm. Right: convert 150 cm to 1.5 m first: W = 9.0 J.
  • Adding exponents when dividing. Wrong: \dfrac{10^8}{10^5} = 10^{13}. Right: subtract: 10^{8-5} = 10^3.
  • Forgetting the last step. Wrong: stopping at v^2 = 392 or at a one-way time when a round trip is asked. Right: take the square root; double for the round trip; convert to the units asked.
  • Saying the law is wrong. Wrong: "Energy was destroyed because the ball was slower than predicted." Right: some energy changed into heat and sound due to air resistance; total energy is conserved.
  • Judging floating by size or weight. Wrong: "The big object floats because it's big." Right: compare density with the fluid's density.
Test strategy

GED strategy for theories and formulas

  1. Find the formula on the screen. It is usually in the stimulus or question. Copy it onto your erasable note board.
  2. Label every number with its letter and unit before you calculate.
  3. Convert first. Compare the units you have with the units the formula needs (meters, kilograms, seconds).
  4. Estimate. Round the numbers and predict roughly what the answer should be. Eliminate choices that are 10 or 100 times off.
  5. Watch the question's final unit. Seconds or minutes? One way or round trip? Mass or density?
  6. For "apply the law" questions, choose the answer that keeps the law true and explains any difference with a real factor (friction, air resistance, escaping gas).
  7. For fill-in answers, follow the rounding instruction exactly and do not type units in the box.

Check yourself

Scientific theoriesFill in
Identifying Mystery Objects by Density

Density is the amount of mass in a given volume. Every pure substance has its own density, so density can help identify what an object is made of. Density is found with this formula:

d = \dfrac{m}{V}

where d is density in grams per cubic centimeter (g/cm³), m is mass in grams (g), and V is volume in cubic centimeters (cm³). An object floats in water if its density is less than the density of water, 1.00 g/cm³, and sinks if its density is greater.

A student measured three solid objects:

ObjectMass (g)Volume (cm³)
A54.020.0
B18.024.0
C89.610.0

Reference densities

SubstanceDensity (g/cm³)
Oak wood0.75
Ice0.92
Water1.00
Aluminum2.70
Iron7.87
Copper8.96

Use the formula d = \dfrac{m}{V} to find the density of Object A.

g/cm³

Type your answer in the box. Fractions like 3/4 are fine.

Hint

Divide the mass by the volume. Mass goes on top.

Check yourself

Scientific theories
Identifying Mystery Objects by Density

Density is the amount of mass in a given volume. Every pure substance has its own density, so density can help identify what an object is made of. Density is found with this formula:

d = \dfrac{m}{V}

where d is density in grams per cubic centimeter (g/cm³), m is mass in grams (g), and V is volume in cubic centimeters (cm³). An object floats in water if its density is less than the density of water, 1.00 g/cm³, and sinks if its density is greater.

A student measured three solid objects:

ObjectMass (g)Volume (cm³)
A54.020.0
B18.024.0
C89.610.0

Reference densities

SubstanceDensity (g/cm³)
Oak wood0.75
Ice0.92
Water1.00
Aluminum2.70
Iron7.87
Copper8.96

The student places each object in a tank of water. Based on the densities, which object or objects will float?

Answer choices
Hint

Find each object's density, then compare it with 1.00 g/cm³.

Summary

  • Laws and theories let you predict: conservation of energy and mass, Newton's laws, density and floating.
  • Use a formula in five steps: write it, list knowns with units, convert, substitute, check.
  • Rearrange A = BC to B = \dfrac{A}{C} or C = \dfrac{A}{B}.
  • Convert with the factor-label method so unwanted units cancel; know kilo-, centi-, milli-.
  • In scientific notation, add exponents to multiply, subtract to divide, and keep the front number between 1 and 10.
  • When data miss an ideal prediction, explain the gap with a real factor — the law still holds.
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