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Problem 1.1 – Calculate the number of seconds in (a) day (b) week (c) month and state answers using SI prefixes.

Given:

• 1 minute = 60 seconds

• 1 hour = 60 minutes

• 1 day = 24 hours

• 1 week = 7 days

• 1 month = 30 days

(a) Seconds in 1 Day

Solution:

1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds

Seconds in 1 day = 24 × 60 × 60
= 24 × 3600
= 86,400 seconds

Using SI prefixes:
86,400 s = 86.4 × 10³ s = 86.4 ks

✅ Answer: 86.4 ks

(b) Seconds in 1 Week

Solution:

1 week = 7 days
1 day = 86,400 seconds

Seconds in 1 week = 7 × 86,400
= 604,800 seconds

Using SI prefixes:
604,800 s = 604.8 × 10³ s = 604.8 ks

✅ Answer: 604.8 ks

(c) Seconds in 1 Month

Solution:

1 month = 30 days
1 day = 86,400 seconds

Seconds in 1 month = 30 × 86,400
= 2,592,000 seconds

Using SI prefixes:
2,592,000 s = 2.592 × 10⁶ s = 2.592 Ms

✅ Answer: 2.592 Ms

Problem 1.2 – State the answers of problem 1.1 in scientific notation.

Solution:

✅ Answer:

• (a) 8.64 × 10⁴ s

• (b) 6.048 × 10⁵ s

• (c) 2.592 × 10⁶ s

Problem 1.3 – Solve the following addition or subtraction. State answers in scientific notation.

(a) 4 × 10³ kg + 3 × 10² kg

Solution:

Step 1: Convert both to same power of 10
4 × 10³ kg = 4 × 10³ kg
3 × 10² kg = 0.3 × 10³ kg

Step 2: Add
(4 × 10³) + (0.3 × 10³) = (4 + 0.3) × 10³
= 4.3 × 10³ kg

✅ Answer: 4.3 × 10³ kg

(b) 5.4 × 10⁴ m - 3.2 × 10⁴ m

Solution:

Step 1: Both are already same power of 10 (10⁴)
5.4 × 10⁴ - 3.2 × 10⁴ = (5.4 - 3.2) × 10⁴
= 2.2 × 10⁴ m

✅ Answer: 2.2 × 10⁴ m

Problem 1.4 – Solve the following multiplication or division. State answers in scientific notation.

(a) (5 × 10² m) × (3 × 10² m)

Solution:

= (5 × 3) × (10² × 10²)
= 15 × 10⁴
= 1.5 × 10⁵ m²

✅ Answer: 1.5 × 10⁵ m²

(b) (3 × 10² kg) / (1.5 × 10⁴ m³)

Solution:

= (3/1.5) × (10²/10⁴)
= 2 × 10²⁻⁴
= 2 × 10⁻² kg m⁻³

✅ Answer: 2 × 10⁻² kg m⁻³

Problem 1.5 – Calculate the following and state your answer in scientific notation.

(3 × 10² kg) × (4.0 × 10¹ m) / (5 × 10⁰ s²)

Note: 5 × 10⁰ = 5 × 1 = 5

Solution:

Step 1: Multiply numerator
(3 × 4.0) × (10² × 10¹) = 12 × 10³ = 1.2 × 10⁴

Step 2: Divide by denominator
= (1.2 × 10⁴) / (5 × 10⁰)
= (1.2/5) × 10⁴
= 0.24 × 10⁴
= 2.4 × 10³

Step 3: Write units
kg × m / s² = kg m s⁻²

✅ Answer: 2.4 × 10³ kg m s⁻²

Problem 1.6 – State the number of significant digits in each measurement.

(a) 0.0045 m

Rule Used: Leading zeros are not significant

Solution:

• 0.0045 has digits: 4 and 5 (both are significant)

• Leading zeros (0.00) are NOT significant

✅ Answer: 2 significant figures

(b) 2.047

Rule Used: All non-zero digits and zeros between non-zero digits are significant

Solution:

• 2, 0, 4, 7 → All are significant

• Zero between 2 and 4 is significant

✅ Answer: 4 significant figures

(c) 3.40 m

Rule Used: Trailing zeros after decimal are significant

Solution:

• 3, 4, 0 → All are significant

• Zero after decimal is significant

✅ Answer: 3 significant figures

(d) 3.420 × 10⁻⁴ m

Rule Used: All digits in coefficient (before × 10ⁿ) are significant

Solution:

• In 3.420, all digits (3, 4, 2, 0) are significant

✅ Answer: 4 significant figures

Problem 1.7 – Write in scientific notation.

(a) 0.00035 m

Solution:

Step 1: Move decimal point 4 places to the right
0.00035 = 3.5 × 10⁻⁴

Step 2: Check: 3.5 × 10⁻⁴ = 0.00035 ✓

✅ Answer: 3.5 × 10⁻⁴ m

(b) 206.4 × 10² m

Solution:

Step 1: 206.4 × 10² = 206.4 × 100 = 20,640

Step 2: Convert to scientific notation
20,640 = 2.064 × 10⁴

OR directly:
206.4 × 10² = 2.064 × 10² × 10²
= 2.064 × 10⁴

✅ Answer: 2.064 × 10⁴ m

Problem 1.8 – Write using correct prefixes.

(a) 5.0 × 10² cm

Solution:

Step 1: 5.0 × 10² cm = 500 cm
Step 2: Convert cm to m: 500 cm = 5.0 m
Step 3: Write as 0.5 × 10¹ m = 0.5 km

OR simply:
5.0 × 10² cm = 0.5 km

✅ Answer: 0.5 km

(b) 580 × 10² g

Solution:

Step 1: 580 × 10² g = 58,000 g
Step 2: Convert to kg: 58,000 g = 58 kg

✅ Answer: 58 kg

(c) 45 × 10⁻³ s

Solution:

Step 1: 45 × 10⁻³ s = 0.045 s
Step 2: Convert to ms (milliseconds): 0.045 s = 45 ms

OR:
45 × 10⁻³ s = 45 ms

✅ Answer: 45 ms

Problem 1.9 – Light year is a unit of distance used in Astronomy. It is the distance covered by light in one year. Taking the speed of light as 3.0 × 10⁸ m/s, calculate the distance.

Given:

• Speed of light = 3.0 × 10⁸ m/s

• Time = 1 year

Solution:

Step 1: Convert 1 year to seconds
1 year = 365.25 days (including leap year)
1 year = 365.25 × 24 × 60 × 60 seconds
= 31,557,600 seconds
= 3.15576 × 10⁷ s

Step 2: Distance = Speed × Time
Distance = (3.0 × 10⁸) × (3.15576 × 10⁷)
= 9.46728 × 10¹⁵ m
= 9.46 × 10¹⁵ m (rounded to 3 SF)

✅ Answer: 9.46 × 10¹⁵ m

Problem 1.10 – Express the density of mercury given as 13.6 g/cm³ in kg/m³.

Given:

• Density = 13.6 g/cm³

Solution:

Step 1: Convert grams to kilograms
1 g = 10⁻³ kg
13.6 g = 13.6 × 10⁻³ kg

Step 2: Convert cm³ to m³
1 cm = 10⁻² m
1 cm³ = (10⁻²)³ m³ = 10⁻⁶ m³

Step 3: Calculate
Density = 13.6 g/cm³
= 13.6 × (10⁻³ kg) / (10⁻⁶ m³)
= 13.6 × 10³ kg/m³
= 1.36 × 10⁴ kg/m³

✅ Answer: 1.36 × 10⁴ kg/m³