Question

1. multiplication with sig figs:

a) 92.45 m · 1.03 m =
b) 0.0024 n · 4.24 s =
c) 4000 kg · 2.001 m/s
e) 0.00698 m² · 100 cm =
f) 2001 kg · 12.6 m/s =
g) 610 n · 4002 s =

1. division with sig figs:

a) 12 m ÷ 31.2 s =
b) 69.4 kg ÷ 38.888 s =
c) 0.012 m² ÷ 0.0002 s =
d) 1800 kg ÷ 410 s ≈
e) 0.102 m ÷ 100 ms =
f) 1001 m³ ÷ 40 ks =

1. addition and subtraction with sig figs:

a) 14 m + 12.2 m =
b) 0.012 kg + 1.0046 kg − 0.0064 kg =
c) 12.46 kg + 9.82 kg − 6.666 kg =
d) 69.45 s + 19.3 s =
e) 200.1 m − 128.28 m =

1. chain calcs with sig figs; round off to the appropriate number of sig figs at the end!

a) (0.045 m · 9.92 kg) ÷ 16.86 s =
b) (9000 m · 4.01 m) − 1.002 m =
c) (0.21 m · 6.23 s) − 1.002 m =
d) (18.01 m · 0.41 m) ÷ (14.62 kg · 12 s) =

Answer:

2) Multiplication with sig figs

a) \( 92.45 \, \text{m} \cdot 1.03 \, \text{m} \)

Step1: Multiply the numbers

\( 92.45 \times 1.03 = 95.2235 \)

Step2: Determine sig figs (92.45 has 4, 1.03 has 3; result takes 3 sig figs)

\( 95.2 \, \text{m}^2 \) (rounded to 3 sig figs)

b) \( 0.0024 \, \text{N} \cdot 4.24 \, \text{s} \)

Step1: Multiply

\( 0.0024 \times 4.24 = 0.010176 \)

Step2: Sig figs (0.0024 has 2, 4.24 has 3; result takes 2 sig figs)

\( 0.010 \, \text{N·s} \) (or \( 1.0 \times 10^{-2} \, \text{N·s} \))

c) \( 4000 \, \text{kg} \cdot 2.001 \, \text{m/s} \)

Step1: Multiply

\( 4000 \times 2.001 = 8004 \)

Step2: Sig figs (4000 has 1 or 4? Assuming 4000 has 4 sig figs (trailing zeros significant), 2.001 has 4; result takes 4)

\( 8004 \, \text{kg·m/s} \) (or \( 8.004 \times 10^3 \))

e) \( 0.00698 \, \text{m}^2 \cdot 100 \, \text{cm} \) (Convert 100 cm to m: \( 100 \, \text{cm} = 1.00 \, \text{m} \))

Step1: Convert units and multiply

\( 0.00698 \times 1.00 = 0.00698 \)

Step2: Sig figs (0.00698 has 3, 1.00 has 3; result takes 3)

\( 0.00698 \, \text{m}^3 \) (or \( 6.98 \times 10^{-3} \, \text{m}^3 \))

f) \( 2001 \, \text{kg} \cdot 12.6 \, \text{m/s} \)

Step1: Multiply

\( 2001 \times 12.6 = 25212.6 \)

Step2: Sig figs (2001 has 4, 12.6 has 3; result takes 3)

\( 25200 \, \text{kg·m/s} \) (or \( 2.52 \times 10^4 \))

g) \( 610 \, \text{N} \cdot 4002 \, \text{s} \)

Step1: Multiply

\( 610 \times 4002 = 2441220 \)

Step2: Sig figs (610 has 2, 4002 has 4; result takes 2)

\( 2.4 \times 10^6 \, \text{N·s} \)

3) Division with sig figs

a) \( 12 \, \text{m} \div 31.2 \, \text{s} \)

Step1: Divide

\( \frac{12}{31.2} \approx 0.3846 \)

Step2: Sig figs (12 has 2, 31.2 has 3; result takes 2)

\( 0.38 \, \text{m/s} \)

b) \( 69.4 \, \text{kg} \div 38.888 \, \text{s} \)

Step1: Divide

\( \frac{69.4}{38.888} \approx 1.7846 \)

Step2: Sig figs (69.4 has 3, 38.888 has 5; result takes 3)

\( 1.78 \, \text{kg/s} \)

c) \( 0.012 \, \text{m}^2 \div 0.0002 \, \text{s} \)

Step1: Divide

\( \frac{0.012}{0.0002} = 60 \)

Step2: Sig figs (0.012 has 2, 0.0002 has 1; result takes 1? Wait, 0.0002 is 1 sig fig, 0.012 is 2. Ambiguity: if 0.0002 is 1 sig fig, result is 60 (1 sig fig: \( 6 \times 10^1 \)); if 0.0002 is 2 (e.g., 0.00020), but here it’s 0.0002. Assume 1 sig fig: \( 6 \times 10^1 \, \text{m}^2/\text{s} \)

d) \( 1800 \, \text{kg} \div 410 \, \text{s} \)

Step1: Divide

\( \frac{1800}{410} \approx 4.390 \)

Step2: Sig figs (1800 has 2, 410 has 2; result takes 2)

\( 4.4 \, \text{kg/s} \)

e) \( 0.102 \, \text{m} \div 100 \, \text{ms} \) (Convert 100 ms to s: \( 100 \, \text{ms} = 0.100 \, \text{s} \))

Step1: Convert and divide

\( \frac{0.102}{0.100} = 1.02 \)

Step2: Sig figs (0.102 has 3, 0.100 has 3; result takes 3)

\( 1.02 \, \text{m/s} \)

f) \( 1001 \, \text{m}^3 \div 40 \, \text{ks} \) (Convert 40 ks to s: \( 40 \, \text{ks} = 40000 \, \text{s} \))

Step1: Convert and divide

\( \frac{1001}{40000} \approx 0.025025 \)

Step2: Sig figs (1001 has 4, 40 has 1 or 2? If 40 has 2, result takes 2: \( 0.025 \, \text{m}^3/\text{s} \) (or \( 2.5 \times 10^{-2} \)))

4) Addition and subtraction with sig figs

a) \( 14 \, \text{m} + 12.2 \, \text{m} \)

Step1: Add

\( 14 + 12.2 = 26.2 \)

Step2: Sig figs (14 has 2 decimal places? No, 14 has uncertainty in the ones place, 12.2 in the tenths. Result rounded to ones place: \( 26 \, \text{m} \) (wait, 14 is two sig figs, 12.2 is three. For addition, decimal places: 14 has 0, 12.2 has 1. Result should have 0 decimal places: \( 26 \, \text{m} \))

b) \( 0.012 \, \text{kg} + 1.0046 \, \text{kg} - 0.0064 \, \text{kg} \)

Step1: Add/Subtract step by step

\( 0.012 + 1.0046 = 1.0166 \); \( 1.0166 - 0.0064 = 1.0102 \)

Step2: Sig figs (0.012 has 2 decimal places? No, 0.012 has 2 sig figs (1 and 2), 1.0046 has 5, 0.0064 has 2. For addition/subtraction, decimal places: 0.012 (thousandths place: 0.012 = 0.0120), 1.0046 (ten-thousandths), 0.0064 (ten-thousandths). Result rounded to thousandths? Wait, 0.012 is 0.012 (3 decimal places? No, 0.012 is two sig figs, so uncertainty in the thousandths place. So result: \( 1.01 \, \text{kg} \) (rounded to three decimal places? Wait, better: 0.012 (2 sig figs), 1.0046 (5), 0.0064 (2). The least precise is 0.012 (uncertainty ±0.001). So result: \( 1.01 \, \text{kg} \) (three sig figs? Wait, 1.0102 rounded to three sig figs: 1.01)

c) \( 12.46 \, \text{kg} + 9.82 \, \text{kg} - 6.666 \, \text{kg} \)

Step1: Add/Subtract

\( 12.46 + 9.82 = 22.28 \); \( 22.28 - 6.666 = 15.614 \)

Step2: Sig figs (12.46: 4, 9.82: 3, 6.666: 4. Least decimal places: 2 (12.46 and 9.82 have 2 decimal places, 6.666 has 3). Result rounded to 2 decimal places: \( 15.61 \, \text{kg} \) (or 15.6 if considering sig figs? Wait, 12.46 (4), 9.82 (3), 6.666 (4). The sum/subtraction: 22.28 (2 decimal) - 6.666 (3 decimal) = 15.614. Round to 2 decimal places: 15.61, which is 4 sig figs? Wait, 12.46 (4), 9.82 (3) → sum is 22.28 (4 sig figs? 12.46 + 9.82 = 22.28 (exact here). Then 22.28 - 6.666 = 15.614. 6.666 has 4 sig figs. So result: 15.61 (4 sig figs) or 15.6 (3)? Wait, 9.82 has 3 sig figs, so the sum 22.28 is limited by 9.82 (3 sig figs? No, addition is about decimal places. 12.46 (2 decimals), 9.82 (2 decimals) → sum has 2 decimals (22.28). Then subtract 6.666 (3 decimals) → result has 2 decimals: 15.61.

d) \( 69.45 \, \text{s} + 19.3 \, \text{s} \)

Step1: Add

\( 69.45 + 19.3 = 88.75 \)

Step2: Sig figs (69.45 has 2 decimal places, 19.3 has 1. Result rounded to 1 decimal place: \( 88.8 \, \text{s} \)

e) \( 200.1 \, \text{m} - 128.28 \, \text{m} \)

Step1: Subtract

\( 200.1 - 128.28 = 71.82 \)

Step2: Sig figs (200.1 has 1 decimal place, 128.28 has 2. Result rounded to 1 decimal place: \( 71.8 \, \text{m} \)

5) Chain calcs with sig figs (round at end)

a) \( (0.045 \, \text{m} \cdot 9.92 \, \text{kg}) \div 16.86 \, \text{s} \)

Step1: Multiply first: \( 0.045 \times 9.92 = 0.4464 \)

Step2: Divide: \( \frac{0.4464}{16.86} \approx 0.02647 \)

Step3: Sig figs (0.045: 2, 9.92: 3, 16.86: 4. Least sig figs: 2. Round to 2 sig figs: \( 0.026 \, \text{m·kg/s} \) (or \( 2.6 \times 10^{-2} \))

b) \( (9000 \, \text{m} \cdot 4.01 \, \text{m}) - 1.002 \, \text{m} \)

Step1: Multiply: \( 9000 \times 4.01 = 36090 \)

Step2: Subtract: \( 36090 - 1.002 = 36088.998 \)

Step3: Sig figs (9000: 1 or 4? Assume 9000 has 4 (trailing zeros significant), 4.01: 3, 1.002: 4. Multiplication: 9000×4.01 = 36090 (4 sig figs from 9000, 3 from 4.01 → 3 sig figs? Wait, 9000 is ambiguous, but if it’s 9000. (4 sig figs), 4.01 (3) → product is 36090 (3 sig figs: 3.61×10⁴). Then subtract 1.002: 36090 - 1.002 ≈ 36089. Rounded to 3 sig figs: \( 3.61 \times 10^4 \, \text{m}^2 \) (wait, units: 9000 m × 4.01 m = 36090 m², then -1.002 m? Wait, units mismatch! 9000 m × 4.01 m is m², 1.002 m is m. This is a unit error, but assuming it’s a typo (maybe 1.002 m²), proceed. If units are correct, the problem is invalid, but assuming m²: 36090 m² - 1.002 m² = 36088.998 m² ≈ 3.61×10⁴ m² (3 sig figs).

c) \( (0.21 \, \text{m} \cdot 6.23 \, \text{s}) - 1.002 \, \text{m} \)

Step1: Multiply: \( 0.21 \times 6.23 = 1.3083 \)

Step2: Subtract: \( 1.3083 - 1.002 = 0.3063 \)

Step3: Sig figs (0.21: 2, 6.23: 3, 1.002: 4. Multiplication: 2 sig figs (0.21) → 1.3 (2 sig figs). Then subtract 1.002: 1.3 - 1.002 = 0.298 ≈ 0.30 (2 sig figs)

d) \( (18.01 \, \text{m} \cdot 0.41 \, \text{m}) \div (14.62 \, \text{kg} \cdot 12 \, \text{s}) \)

Step1: Multiply numerator