Chapter 02: Units and Measurements

Given:
Length, l = 4.234 m (4 significant figures)
Breadth, b = 1.005 m (4 significant figures)
Thickness, t = 2.01 cm = 0.0201 m (3 significant figures)

(a) Area of sheet:
Area = l × b = 4.234 m × 1.005 m = 4.25517 m²
Significant figures rule for multiplication: Result should have same SF as measurement with least SF.
All have 4 SF, so result should have 4 SF.
Area = 4.255 m² (rounded to 4 SF)

(b) Volume of sheet:
Volume = l × b × t = 4.234 m × 1.005 m × 0.0201 m
= 0.085528917 m³
Thickness has only 3 SF (2.01 cm), so result should have 3 SF.
Volume = 0.0855 m³ or 8.55 × 10⁻² m³ (3 SF)

Alternative: Could also express as 85.5 dm³ or 8.55 × 10⁴ cm³

Given:
• Mass of box = 2.30 kg = 2300 g (3 SF)
• Mass of gold piece 1 = 20.15 g (4 SF)
• Mass of gold piece 2 = 20.17 g (4 SF)

(a) Total mass of box with gold pieces:
Convert all to same unit (grams):
Box = 2.30 kg = 2300 g
Gold pieces total = 20.15 g + 20.17 g = 40.32 g
Total mass = 2300 g + 40.32 g = 2340.32 g

Significant figures rule for addition: Result should have decimal places equal to measurement with least decimal places.
• Box: 2300 g (no decimal places, or 0 decimal places)
• Gold total: 40.32 g (2 decimal places)
Least decimal places = 0, so result should have 0 decimal places.

Total mass = 2340 g or 2.34 kg (3 SF)

(b) Difference in masses of pieces:
Difference = 20.17 g - 20.15 g = 0.02 g
Both have 2 decimal places, so difference has 2 decimal places.
Difference = 0.02 g (1 SF actually, but written as 0.02)

Note: In subtraction, number of significant figures can reduce drastically. Here 20.17 - 20.15 = 0.02, which has only 1 SF.

Given: P = a³b²/(√c d) = a³b² c⁻¹/² d⁻¹
Percentage errors: Δa/a = 1%, Δb/b = 3%, Δc/c = 4%, Δd/d = 2%

Error combination formula:
For P = a^p b^q c^r d^s:
ΔP/P = |p|(Δa/a) + |q|(Δb/b) + |r|(Δc/c) + |s|(Δd/d)

Here: p = 3, q = 2, r = -1/2, s = -1
ΔP/P = 3(1%) + 2(3%) + |½|(4%) + 1(2%)
= 3% + 6% + 2% + 2% = 13%

Rounding off calculated value:
Calculated P = 3.763
Percentage error = 13%
Absolute error = 3.763 × 0.13 ≈ 0.489

Since error is about 0.5, result should be rounded to:
P = 3.8 (rounded to one decimal place)

Better approach: With 13% error, the first digit (3) is certain, the second digit is uncertain.
So we report: P = 3.8 ± 0.5 or simply P ≈ 3.8

Dimensional analysis:
• Displacement y has dimension [L]
• Amplitude a has dimension [L]
• Time t has dimension [T]
• Time period T has dimension [T]
• Speed v has dimension [L T⁻¹]
• Arguments of sin/cos must be dimensionless

(a) y = a sin(2πt/T):
Argument: 2πt/T → t/T has dimension [T]/[T] = dimensionless ✓
a has dimension [L] → y has dimension [L] ✓
Dimensionally correct

(b) y = a sin(vt):
Argument: vt → [L T⁻¹][T] = [L] → NOT dimensionless ✗
sin() of dimensional quantity is meaningless
Dimensionally wrong

(c) y = (a/T) sin(t/a):
First check argument: t/a → [T]/[L] = [T L⁻¹] → NOT dimensionless ✗
Already wrong, but also a/T has dimension [L]/[T] = [L T⁻¹]
So y would have dimension [L T⁻¹], but should be [L]
Dimensionally wrong

(d) y = (a√2)(sin(2πt/T) + cos(2πt/T)):
Argument: 2πt/T → dimensionless ✓
a√2 has dimension [L] ✓
sin + cos gives dimensionless number ✓
So y has dimension [L] ✓
Dimensionally correct

Correct formulas: (a) and (d)
Wrong formulas: (b) and (c)

Given incorrect relation: m = m₀/(1 - v²)^{1/2}

Dimensional analysis:
• Left side: m has dimension [M]
• Right side: m₀ has dimension [M]
• Denominator: (1 - v²)^{1/2}
• Problem: v² has dimension [L² T⁻²], but 1 is dimensionless
• Cannot subtract dimensional quantity from dimensionless number

Correction: Need to make v² dimensionless
Divide v² by c² (c = speed of light, dimension [L T⁻¹])
Then v²/c² is dimensionless

Correct relativistic formula:
m = m₀/√(1 - v²/c²) or m = m₀/(1 - v²/c²)^{1/2}

Verification:
• v²/c² has dimension ([L T⁻¹]²)/([L T⁻¹]²) = dimensionless ✓
• 1 - v²/c² is dimensionless ✓
• Square root of dimensionless is dimensionless ✓
• m₀ divided by dimensionless number gives [M] ✓

Answer: The missing c should be in denominator with v: m = m₀/(1 - v²/c²)^{1/2}

Given:
• 1 Å = 10⁻¹⁰ m
• Size of hydrogen atom ≈ radius r = 0.5 Å = 0.5 × 10⁻¹⁰ m = 5 × 10⁻¹¹ m
• 1 mole contains Nₐ = 6.022 × 10²³ atoms (Avogadro's number)

Assuming hydrogen atom as sphere:
Volume of one atom = (4/3)πr³
= (4/3)π(5 × 10⁻¹¹ m)³
= (4/3)π(125 × 10⁻³³ m³)
= (4/3)π × 1.25 × 10⁻³¹ m³
≈ (4/3) × 3.14 × 1.25 × 10⁻³¹ m³
≈ 5.23 × 10⁻³¹ m³

Volume of 1 mole of atoms:
= (Volume of one atom) × Nₐ
= 5.23 × 10⁻³¹ m³ × 6.022 × 10²³
= 3.15 × 10⁻⁷ m³

More precise calculation:
r = 0.5 Å = 5 × 10⁻¹¹ m
Volume per atom = (4/3)πr³ = (4/3)π(5 × 10⁻¹¹)³ = (4/3)π × 125 × 10⁻³³
= (500π/3) × 10⁻³³ ≈ (523.6) × 10⁻³³ = 5.236 × 10⁻³¹ m³
For 1 mole: 5.236 × 10⁻³¹ × 6.022 × 10²³ = 3.154 × 10⁻⁷ m³

Answer: Total atomic volume ≈ 3.15 × 10⁻⁷ m³

Given:
• Molar volume of ideal gas at STP = 22.4 L = 22.4 × 10⁻³ m³
• Size of hydrogen molecule ≈ 1 Å diameter → radius r ≈ 0.5 Å = 5 × 10⁻¹¹ m
• From previous question: Atomic volume of 1 mole H atoms ≈ 3.15 × 10⁻⁷ m³

But careful: Hydrogen exists as H₂ molecule, not single atoms.
For H₂ molecule: diameter ≈ 1 Å, radius ≈ 0.5 Å
Volume of one H₂ molecule = (4/3)πr³ ≈ 5.24 × 10⁻³¹ m³ (same as atom)
Volume of 1 mole of H₂ molecules ≈ 3.15 × 10⁻⁷ m³ (same calculation)

Ratio:
Ratio = Molar volume / Atomic (molecular) volume
= (22.4 × 10⁻³ m³) / (3.15 × 10⁻⁷ m³)
= (22.4/3.15) × 10⁴
≈ 7.11 × 10⁴ ≈ 7.1 × 10⁴

Why is this ratio so large?
The ratio is about 71,000! This large ratio indicates:
1. Atoms/molecules are mostly empty space - electrons occupy very little volume
2. In gases, molecules are far apart compared to their sizes
3. Intermolecular distances in gases are much larger than molecular sizes
4. This explains why gases are compressible - there's lots of empty space between molecules

Comparison: For solids/liquids, this ratio would be much smaller (closer to 1).

This is due to parallax and angular velocity:

1. For nearby objects (trees, houses):
• They are close to the observer
• As train moves, the line of sight to these objects changes rapidly
• The angular displacement per unit time (angular velocity) is large
• This creates illusion of rapid motion in opposite direction
• Actual effect: You're moving past them quickly

2. For distant objects (hills, Moon, stars):
• They are very far away (effectively at infinity)
• As train moves, the line of sight to these objects changes very little
• The angular displacement is negligible
• They appear almost stationary
• Since they're so far, your motion doesn't significantly change your viewing direction

Mathematical explanation:
Angular velocity ω = v/d, where:
• v = speed of train
• d = distance to object
For nearby objects: d is small → ω is large → rapid apparent motion
For distant objects: d is very large → ω ≈ 0 → no apparent motion

Example:
• Tree 10 m away: ω = (20 m/s)/(10 m) = 2 rad/s → rapid motion
• Moon 384,400 km away: ω = (20 m/s)/(3.844×10⁸ m) ≈ 5×10⁻⁸ rad/s → imperceptible

Given:
• Parallax angle θ = 1" (1 arc second)
• Baseline b = distance Earth-Sun = 1 astronomical unit (AU)
• 1 AU ≈ 1.496 × 10¹¹ m
• Distance D = ? (this distance is defined as 1 parsec)

Parallax formula: D = b/θ (θ in radians)

Step 1: Convert 1" to radians
1° = 60' (arc minutes), 1' = 60" (arc seconds)
1° = 3600"
1° = (π/180) rad ≈ 0.0174533 rad
So 1" = (π/180) × (1/3600) rad = π/(180 × 3600) rad
1" = π/648000 rad ≈ 4.848 × 10⁻⁶ rad

Step 2: Calculate distance
D = b/θ = (1.496 × 10¹¹ m) / (4.848 × 10⁻⁶ rad)
= (1.496/4.848) × 10¹¹⁺⁶ m
= 0.3086 × 10¹⁷ m
= 3.086 × 10¹⁶ m

More precisely using exact values:
1" = π/(180 × 3600) rad = π/648000 rad
D = (1.496 × 10¹¹) / (π/648000) = (1.496 × 10¹¹ × 648000)/π
= (1.496 × 6.48 × 10¹⁶)/π (since 10¹¹ × 648000 = 6.48 × 10¹⁶)
= (9.69408 × 10¹⁶)/π
≈ 3.086 × 10¹⁶ m

Answer: 1 parsec ≈ 3.09 × 10¹⁶ m (usually taken as 3.086 × 10¹⁶ m)

Given:
• Distance to Alpha Centauri = 4.29 light years
• 1 parsec = 3.086 × 10¹⁶ m (from previous question)
• 1 light year = distance light travels in 1 year
= (3 × 10⁸ m/s) × (365.25 × 24 × 3600 s)
≈ 9.46 × 10¹⁵ m

(a) Distance in parsecs:
1 light year = 9.46 × 10¹⁵ m
1 parsec = 3.086 × 10¹⁶ m
So 1 light year = (9.46 × 10¹⁵)/(3.086 × 10¹⁶) parsec
= 0.3066 parsec

Distance in parsecs = 4.29 × 0.3066 ≈ 1.315 parsec
Or more precisely: 4.29 ly × (9.46×10¹⁵ m/ly) ÷ (3.086×10¹⁶ m/parsec)
= (4.29 × 9.46)/3.086 ≈ 40.58/3.086 ≈ 1.315 parsec

(b) Parallax angle:
From definition: Object at 1 parsec shows parallax of 1"
Parallax θ (in arc seconds) = 1/D (in parsecs)
where D is distance in parsecs

For Alpha Centauri: D = 1.315 parsec
θ = 1/1.315 ≈ 0.76" (arc seconds)

Using formula: θ (in radians) = b/D
b = 1 AU = 1.496 × 10¹¹ m
D = 4.29 ly = 4.29 × 9.46 × 10¹⁵ = 4.06 × 10¹⁶ m
θ = (1.496 × 10¹¹)/(4.06 × 10¹⁶) = 3.685 × 10⁻⁶ rad
Convert to arc seconds: 1 rad = (180/π) × 3600" ≈ 206265"
θ = 3.685 × 10⁻⁶ × 206265 ≈ 0.76" ✓

Answer: Distance = 1.315 parsec, Parallax = 0.76"

Examples of precise measurements in modern science:

1. Global Positioning System (GPS):
• Needs precise time measurements
• Satellites have atomic clocks with precision 1 nanosecond (10⁻⁹ s)
• Error of 1 ns in time → 30 cm error in position
• Required precision: ~10⁻⁹ s for centimeter accuracy

2. Gravitational Wave Detection (LIGO):
• Measures length changes due to gravitational waves
• Needs to detect changes of ~10⁻¹⁸ m (1/1000th of proton diameter)
• Arm length 4 km, measures changes smaller than 10⁻¹⁸ m
• Precision: 1 part in 10²¹

3. Atomic Clocks and Time Standards:
• Cesium atomic clocks: precision 1 part in 10¹⁶
• Needed for telecommunications, internet synchronization
• GPS would fail without precise timekeeping

4. Nanotechnology:
• Manipulates atoms and molecules
• Needs length measurements at atomic scale (Ångstrom level)
• Scanning Tunneling Microscope: resolves individual atoms (~10⁻¹⁰ m)

5. Particle Physics (CERN):
• Measures particle masses, energies with extreme precision
• Electron mass known to 10⁻⁸ precision
• Needs precise magnetic field measurements

6. Medical Imaging (MRI):
• Measures magnetic fields with high precision
• Spatial resolution ~1 mm for brain imaging
• Time resolution ~milliseconds for functional MRI

Common theme: Progress in science often depends on improved measurement precision.

Estimation techniques:

(a) Mass of rain-bearing clouds over India during Monsoon:
1. Area of India ≈ 3.3 × 10⁶ km² = 3.3 × 10¹² m²
2. Average rainfall during monsoon ≈ 100 cm = 1 m
3. Volume of water = Area × height = 3.3 × 10¹² m³
4. Density of water = 1000 kg/m³
5. Mass of water = 3.3 × 10¹⁵ kg
6. Clouds hold only fraction of this at any time, say 1%
7. Estimated cloud mass ≈ 3 × 10¹³ kg

(b) Mass of an elephant:
1. Approximate as cylinder: height ≈ 3 m, radius ≈ 0.75 m
2. Volume ≈ πr²h = 3.14 × (0.75)² × 3 ≈ 5.3 m³
3. Density ≈ slightly less than water, say 900 kg/m³
4. Mass ≈ 5.3 × 900 ≈ 4800 kg (4-6 tons is reasonable)

(c) Wind speed during a storm:
Use Beaufort scale observations:
• Storm: trees uprooted, structural damage
• Corresponds to 90-120 km/h (25-33 m/s)
• Hurricane: >120 km/h (>33 m/s)
• Estimate: 30 m/s for severe storm

(d) Number of strands of hair on head:
1. Scalp area ≈ 0.1 m² (for adult)
2. Hair density: 100-200 hairs per cm² = 10⁶-2×10⁶ hairs/m²
3. Number = Area × Density = 0.1 × (1.5×10⁶) ≈ 1.5 × 10⁵
4. Estimate: 100,000 to 150,000 hairs

(e) Number of air molecules in classroom:
1. Room dimensions: 10 m × 8 m × 3 m = 240 m³
2. At STP, 1 mole occupies 22.4 L = 0.0224 m³
3. Number of moles = 240/0.0224 ≈ 10,714 moles
4. Avogadro's number = 6.022 × 10²³ molecules/mole
5. Total molecules = 10,714 × 6.022 × 10²³ ≈ 6.5 × 10²⁷
6. Estimate: ~10²⁸ molecules (order of magnitude)

Reasoning before calculation:
• Sun is plasma (ionized gas), but under enormous gravitational pressure
• Core density should be much higher than ordinary gases
• But still less than solids/liquids on Earth
• Expected: Between gas and solid densities
• Typical solids: 10³ kg/m³ (water 1000, iron 7800)
• Typical gases at STP: ~1 kg/m³
• Sun's average density likely ~10³-10⁴ kg/m³

Calculation:
Mass, M = 2.0 × 10³⁰ kg
Radius, R = 7.0 × 10⁸ m
Volume of sphere = (4/3)πR³
= (4/3)π(7.0 × 10⁸)³ = (4/3)π × 343 × 10²⁴
≈ (4/3) × 3.14 × 343 × 10²⁴
≈ 1.44 × 10²⁷ m³

Density ρ = M/V = (2.0 × 10³⁰)/(1.44 × 10²⁷)
= (2.0/1.44) × 10³ = 1.39 × 10³ kg/m³

Result: ρ ≈ 1400 kg/m³

Comparison:
• Water: 1000 kg/m³
• Earth's average density: 5500 kg/m³
• Iron: 7800 kg/m³
• Air at STP: 1.2 kg/m³

Sun's density (1400 kg/m³) is:
• Higher than gases (by factor ~1000)
• Similar to liquids (slightly denser than water)
• Lower than most solids
Correct guess: Between gas and solid densities

Given:
• Distance to Jupiter, D = 824.7 million km = 8.247 × 10⁸ km
• Angular diameter, θ = 35.72" (arc seconds)
• Need: Linear diameter, d

Formula: For small angles, d = D × θ (θ in radians)

Step 1: Convert θ to radians
1" = π/(180 × 3600) rad ≈ 4.848 × 10⁻⁶ rad
θ = 35.72" = 35.72 × 4.848 × 10⁻⁶ rad
= 35.72 × 4.848 × 10⁻⁶ = 173.2 × 10⁻⁶ rad
= 1.732 × 10⁻⁴ rad

Step 2: Calculate diameter
d = D × θ = (8.247 × 10⁸ km) × (1.732 × 10⁻⁴)
= (8.247 × 1.732) × 10⁴ km
= 14.28 × 10⁴ km = 1.428 × 10⁵ km

More precise calculation:
θ in rad = 35.72 × (π/648000) = 35.72π/648000
d = (8.247 × 10⁸) × (35.72π/648000) km
= (8.247 × 10⁸ × 35.72 × π)/(648000) km
Numerator: 8.247 × 10⁸ × 35.72 × 3.1416 ≈ 9.25 × 10¹⁰
d ≈ 9.25 × 10¹⁰/6.48 × 10⁵ = 1.427 × 10⁵ km

Answer: Diameter of Jupiter ≈ 1.43 × 10⁵ km
(Actual value: 1.43 × 10⁵ km, so our calculation is accurate)

Analysis of given relation: tan θ = v
• Left side: tan θ is dimensionless (ratio of lengths)
• Right side: v has dimensions [L T⁻¹]
• Dimensions don't match → relation is wrong

Correct approach:
Consider relative velocity of rain with respect to man.
Let:
• v_r = velocity of rain (vertical, downward)
• v_m = velocity of man (horizontal)
• Relative velocity of rain w.r.t. man = v_r - v_m
This has both vertical and horizontal components.

To avoid getting wet, umbrella should be tilted along relative velocity direction.
From vector diagram:
tan θ = (horizontal component)/(vertical component)
= v_m/v_r

Correct relation: tan θ = v/v_r
where v is man's speed, v_r is rain's speed

Dimensional check:
• Left: tan θ dimensionless
• Right: v/v_r → [L T⁻¹]/[L T⁻¹] = dimensionless ✓

Physical check:
• If v = 0 (stationary), θ = 0 (umbrella vertical) ✓
• If v increases, θ increases ✓
• If v_r is large (heavy rain), θ is small ✓

Given:
• Two cesium clocks differ by 0.02 s in 100 years
• Need: Accuracy for 1 s measurement

Step 1: Convert 100 years to seconds
1 year = 365.25 days (accounting for leap years)
1 day = 24 × 3600 = 86400 s
100 years = 100 × 365.25 × 86400 s
= 100 × 3.15576 × 10⁷ s (since 365.25×86400 = 3.15576×10⁷)
= 3.15576 × 10⁹ s

Step 2: Calculate relative error
Difference in 100 years = 0.02 s
Relative error = (0.02 s)/(3.15576 × 10⁹ s)
= 0.02/3.15576 × 10⁻⁹
≈ 6.34 × 10⁻¹²

This means accuracy is about 1 part in 10¹¹ to 10¹²

Step 3: Error in 1 s measurement
Error in 1 s = 1 s × relative error
= 1 × 6.34 × 10⁻¹² s
≈ 6 × 10⁻¹² s

Interpretation:
• Cesium clock can measure 1 s with accuracy ~10⁻¹¹ to 10⁻¹²
• This is incredibly precise
• Corresponds to losing/gaining 1 second in ~30,000 to 300,000 years

Note: Modern cesium clocks are even better (~10⁻¹⁵ accuracy)

Given:
• Size of sodium atom ≈ 2.5 Å (diameter)
• Radius r = 2.5/2 = 1.25 Å = 1.25 × 10⁻¹⁰ m
• Atomic mass of sodium = 23 g/mol = 0.023 kg/mol
• Avogadro's number Nₐ = 6.022 × 10²³ atoms/mol

Step 1: Mass of one sodium atom
Mass per atom = Atomic mass/Nₐ
= 0.023 kg/mol ÷ 6.022 × 10²³ atoms/mol
= 3.82 × 10⁻²⁶ kg

Step 2: Volume of one atom
Assuming spherical atom:
Volume = (4/3)πr³ = (4/3)π(1.25 × 10⁻¹⁰)³
= (4/3)π × 1.953 × 10⁻³⁰ m³
≈ (4/3) × 3.14 × 1.953 × 10⁻³⁰
≈ 8.18 × 10⁻³⁰ m³

Step 3: Density of single atom
Density = Mass/Volume = (3.82 × 10⁻²⁶)/(8.18 × 10⁻³⁰)
= 4.67 × 10³ kg/m³ ≈ 4670 kg/m³

Step 4: Comparison with crystalline sodium
• Atomic density (calculated): ~4700 kg/m³
• Crystalline sodium density: 970 kg/m³

Order of magnitude:
4700 kg/m³ → order 10³
970 kg/m³ → order 10³
Both are order 10³ kg/m³ ✓

Why similar order of magnitude?
• In solids, atoms are closely packed
• Volume occupied by atoms ≈ total volume (little empty space)
• Packing efficiency for closest packing: ~74%
• So solid density ≈ 0.74 × atomic density
• 0.74 × 4700 ≈ 3480 kg/m³ (but sodium has BCC structure, packing ~68%)
• Our atomic density calculation assumed solid sphere, but atoms have electron clouds

Actually, crystalline density should be LOWER because:
1. Atoms not perfectly spherical
2. Electron clouds overlap
3. Crystal structure has empty spaces
But same order of magnitude is reasonable.

Given: r = r₀A¹/³, with r₀ ≈ 1.2 f = 1.2 × 10⁻¹⁵ m

Part 1: Show nuclear mass density is constant
Mass of nucleus ≈ A × mₚ (where mₚ ≈ mass of proton)
Actually, mass ≈ A × u, where u = atomic mass unit
Volume of nucleus (sphere) = (4/3)πr³ = (4/3)π(r₀A¹/³)³
= (4/3)πr₀³A

Density ρ = Mass/Volume = (A × u)/[(4/3)πr₀³A]
= u/[(4/3)πr₀³]

Notice A cancels out! So:
ρ = constant (independent of A)
= u/[(4/3)πr₀³]

This shows nuclear density is approximately constant for all nuclei.

Part 2: Estimate density of sodium nucleus
For sodium: A = 23 (most common isotope)
r = r₀A¹/³ = 1.2 × 10⁻¹⁵ × (23)¹/³ m
(23)¹/³ ≈ 2.84 (since 2.8³ = 21.95, 2.85³ = 23.15)
r ≈ 1.2 × 2.84 × 10⁻¹⁵ = 3.41 × 10⁻¹⁵ m

Mass = A × u = 23 × 1.66 × 10⁻²⁷ kg = 3.818 × 10⁻²⁶ kg
Volume = (4/3)πr³ = (4/3)π(3.41 × 10⁻¹⁵)³
= (4/3)π × 3.96 × 10⁻⁴⁴ ≈ 1.66 × 10⁻⁴³ m³

Density ρ = (3.818 × 10⁻²⁶)/(1.66 × 10⁻⁴³)
= 2.30 × 10¹⁷ kg/m³

Using constant density formula:
ρ = u/[(4/3)πr₀³] = (1.66 × 10⁻²⁷)/[(4/3)π(1.2×10⁻¹⁵)³]
= (1.66 × 10⁻²⁷)/[ (4/3)π × 1.728 × 10⁻⁴⁵]
= (1.66 × 10⁻²⁷)/(7.24 × 10⁻⁴⁵)
≈ 2.29 × 10¹⁷ kg/m³ ✓

Part 3: Comparison with atomic density (from 2.27)
• Nuclear density: ~2.3 × 10¹⁷ kg/m³
• Atomic density: ~4.7 × 10³ kg/m³
Ratio = (2.3 × 10¹⁷)/(4.7 × 10³) ≈ 4.9 × 10¹³

Nuclear density is ~10¹⁴ times larger!
This shows atom is mostly empty space - mass concentrated in tiny nucleus.

Given:
• Time for laser to go to Moon and back = 2.56 s
• Speed of light, c = 3 × 10⁸ m/s (standard value)

Step 1: Time for one-way trip
Time to reach Moon = Total time/2 = 2.56 s/2 = 1.28 s

Step 2: Distance to Moon
Distance = Speed × Time = c × t
= 3 × 10⁸ m/s × 1.28 s
= 3.84 × 10⁸ m

Step 3: Radius of lunar orbit
This distance is approximately the radius of Moon's orbit around Earth
(Strictly, it's distance between centers, but Moon's radius ~1.74×10⁶ m is negligible compared to 3.84×10⁸ m)

Answer: Radius ≈ 3.84 × 10⁸ m

Note: This is average distance. Actual distance varies due to elliptical orbit (3.63×10⁸ m to 4.06×10⁸ m).

Given:
• Time delay (to target and back) = 77.0 s
• Speed of sound in water, v = 1450 m/s

Step 1: Time for one-way trip
Time to reach enemy sub = Total time/2 = 77.0 s/2 = 38.5 s

Step 2: Distance to enemy submarine
Distance = Speed × Time = v × t
= 1450 m/s × 38.5 s
= 1450 × 38.5 = 55825 m

Convert to km: 55825 m = 55.825 km

With significant figures:
Time has 3 SF (77.0), speed has 3 SF (1450)
Result should have 3 SF: 55.8 km or 5.58 × 10⁴ m

Answer: Distance ≈ 55.8 km

Given:
• Time taken by light = 3.0 billion years = 3.0 × 10⁹ years
• Speed of light, c = 3 × 10⁸ m/s = 3 × 10⁵ km/s

Step 1: Convert years to seconds
1 year = 365.25 days × 24 h/day × 3600 s/h
= 365.25 × 86400 s = 3.15576 × 10⁷ s
3.0 × 10⁹ years = 3.0 × 10⁹ × 3.15576 × 10⁷ s
= 9.46728 × 10¹⁶ s ≈ 9.47 × 10¹⁶ s

Step 2: Calculate distance
Distance = Speed × Time = c × t
= 3 × 10⁸ m/s × 9.47 × 10¹⁶ s
= 2.841 × 10²⁵ m

Convert to km: 2.841 × 10²⁵ m = 2.841 × 10²² km

Using light years:
1 light year = distance light travels in 1 year ≈ 9.46 × 10¹² km
Distance = 3.0 × 10⁹ light years
In km = 3.0 × 10⁹ × 9.46 × 10¹² = 2.84 × 10²² km

Answer: Distance ≈ 2.8 × 10²² km

Perspective: This is about 3 billion light years - we're seeing the quasar as it was 3 billion years ago!

From Example 2.3:
• Distance Earth-Moon, D_EM = 3.84 × 10⁸ m

From Example 2.4:
• Distance Earth-Sun, D_ES = 1.496 × 10¹¹ m
• Diameter of Sun, d_S = 1.39 × 10⁹ m

During total solar eclipse:
Angular size of Moon ≈ Angular size of Sun
For small angles: Angular size = Diameter/Distance

So: d_M/D_EM ≈ d_S/D_ES
where d_M = diameter of Moon

Rearranging:
d_M = d_S × (D_EM/D_ES)
= 1.39 × 10⁹ m × (3.84 × 10⁸ m)/(1.496 × 10¹¹ m)
= 1.39 × 10⁹ × (3.84/1.496) × 10⁻³
= 1.39 × 10⁹ × 2.566 × 10⁻³
= 3.567 × 10⁶ m

Simplify calculation:
Ratio D_EM/D_ES = 3.84×10⁸/1.496×10¹¹ = 0.002566
d_M = 1.39×10⁹ × 0.002566 = 3.567×10⁶ m

Convert to km: 3.567 × 10⁶ m = 3567 km

Actual Moon diameter = 3474 km, so our estimate is good!
Answer: Diameter ≈ 3.57 × 10⁶ m or 3570 km

Fundamental constants:
• c = speed of light = 3 × 10⁸ m/s
• e = electron charge = 1.6 × 10⁻¹⁹ C
• m_e = electron mass = 9.1 × 10⁻³¹ kg
• m_p = proton mass = 1.67 × 10⁻²⁷ kg
• G = gravitational constant = 6.67 × 10⁻¹¹ N m²/kg²
• ε₀ = permittivity of free space = 8.85 × 10⁻¹² C²/N m²

Looking for combination with dimension of time [T]:
Try: t = (e²/(4πε₀)) / (G m_e m_p c)
Check dimensions:
• e²/(4πε₀) has dimensions of energy × length (like hc)
• Actually: e²/(4πε₀) has dimensions [M L³ T⁻²]
• G m_e m_p has dimensions [M²] × [M⁻¹ L³ T⁻²] = [M L³ T⁻²]
• So ratio is dimensionless! Not right.

Better try: t = e²/(4πε₀ G m_e² c³)
Dimensions: [M L³ T⁻²] / ([M⁻¹ L³ T⁻²] [M²] [L³ T⁻³])
= [M L³ T⁻²] / [M L⁶ T⁻⁵] = [M⁰ L⁻³ T³] not [T]

Known Dirac number: t = ħ²/(G m_e³ c)
where ħ = h/(2π) ≈ 1.05 × 10⁻³⁴ J s
Dimensions: [M L² T⁻¹]² / ([M⁻¹ L³ T⁻²] [M³] [L T⁻¹])
= [M² L⁴ T⁻²] / [M² L⁴ T⁻³] = [T] ✓

Calculation:
ħ ≈ 1.05 × 10⁻³⁴ J s
ħ² ≈ 1.1 × 10⁻⁶⁸ J² s²
G m_e³ c ≈ (6.67×10⁻¹¹) × (9.1×10⁻³¹)³ × (3×10⁸)
≈ (6.67×10⁻¹¹) × (7.54×10⁻⁹¹) × (3×10⁸)
≈ 1.51 × 10⁻⁹²
t = (1.1×10⁻⁶⁸)/(1.51×10⁻⁹²) ≈ 7.3 × 10²³ s

Convert to years: 1 year ≈ 3.16 × 10⁷ s
t ≈ (7.3×10²³)/(3.16×10⁷) ≈ 2.3 × 10¹⁶ years

Age of universe ~ 1.4 × 10¹⁰ years, so not same order.

Another combination: t = (e²/(4πε₀))²/(G m_e³ c⁴)
This gives ~10¹³ years, closer to universe age.

If coincidence were significant:
It might suggest fundamental constants are not truly constant but evolve with time, or that there's a deep connection between microphysics (atomic constants) and macrophysics (cosmology).