Let's solve each question **step by step** with explanations and final answers. --- # 1. Speed Limit Conversion ### (a) What is this in meters per second? **Given:** Speed limit = 75 mph **Conversion factors:** - \( 1 \text{ mile} = 1609.34 \text{ m} \) - \( 1 \text{ hour} = 360 \text{ s} \) **Calculation:** \[ 75 \text{ mph} \times \frac{1609.34 \text{ m}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{360 \text{ s}} \] \[ = 75 \times \frac{1609.34}{360} \] \[ = 75 \times .44704 \] \[ = 33.528 \text{ m/s} \] **Final (rounded to 3 sig figs):** \[ \boxed{33.5 \text{ m/s}} \] --- ### (b) How many kilometers per hour is this? \[ 75 \text{ mph} \times \frac{1.60934 \text{ km}}{1 \text{ mile}} = 120.701 \text{ km/h} \] **Final (rounded to 3 sig figs):** \[ \boxed{121 \text{ km/h}} \] --- # 2. Significant Figures Calculations ### (a) \((126.7 \times 98.2) / (46.2 \times 1.01)\) - **Multiplication/division rule:** Use the least number of significant figures. \[ (126.7 \times 98.2) = 12443.94 \] \[ (46.2 \times 1.01) = 46.662 \] \[ \frac{12443.94}{46.662} = 266.81 \] **Sig Figs:** - Inputs: 126.7 (4), 98.2 (3), 46.2 (3), 1.01 (3) → **3 sig figs** **Final:** \[ \boxed{267} \] --- ### (b) \( 28.7 \) (Just report with correct sig figs) **28.7** has **3 sig figs**. **Final:** \[ \boxed{28.7} \] --- ### (c) \((1.6025 \times 10^{-19})(371205)\) \[ 1.6025 \times 10^{-19} \times 371205 = 5.95146 \times 10^{-14} \] - 1.6025 (5 sig figs), 371205 (6 sig figs) → **5 sig figs** **Final:** \[ \boxed{5.9515 \times 10^{-14}} \] --- # 3. Marathon Runner ### Given: - Distance = 40.188 km = 40,188 m - Time = 2 h, 30 min, 12 s \( = 2 \times 360 + 30 \times 60 + 12 = 9012 \) s ### Average speed (in meters per second): \[ \text{Average speed} = \frac{\text{distance}}{\text{time}} = \frac{40188}{9012} = 4.457 \text{ m/s} \] **Sig figs:** Both values have 5 sig figs. **Final:** \[ \boxed{4.457 \text{ m/s}} \] --- # 4. Piston and Volume ### (a) What amount is the gas decreased in volume in cubic centimeters? #### Given: - Distance = 7.400 ± .003 cm - Diameter = 3.250 ± .001 cm #### **Volume of a cylinder:** \[ V = \pi r^2 h \] - \( r = \frac{d}{2} = \frac{3.250}{2} = 1.625 \text{ cm} \) - \( h = 7.400 \text{ cm} \) \[ V = \pi (1.625)^2 (7.400) \] \[ V = \pi \times 2.640625 \times 7.400 \] \[ = 3.1416 \times 2.640625 \times 7.400 \] \[ = 3.1416 \times 19.539 \] \[ = 61.387 \text{ cm}^3 \] **Sig figs:** - 7.400 (4), 3.250 (4) → **4 sig figs** **Final:** \[ \boxed{61.39 \text{ cm}^3} \] --- ### (b) Find the uncertainty in this volume. #### **Uncertainty propagation:** \[ \frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac{\Delta h}{h} \] - \( \Delta d = .001 \), so \( \Delta r = .0005 \) - \( r = 1.625 \) - \( \Delta h = .003 \), \( h = 7.400 \) \[ \frac{\Delta V}{V} = 2 \times \frac{.0005}{1.625} + \frac{.003}{7.400} \] \[ = 2 \times .000308 + .000405 \] \[ = .000616 + .000405 = .001021 \] \[ \Delta V = .001021 \times 61.39 = .0627 \text{ cm}^3 \] **Final (rounded to 2 sig figs):** \[ \Delta V = \boxed{.063 \text{ cm}^3} \] --- # 5. Thickness of Cell Membrane (Estimate) ### **Given:** - Atoms in membrane ≈ 2 × H atom size - H atom diameter ≈ .1 nm ### **Thickness estimation:** \[ \text{Thickness} \approx 2 \times .1 \text{ nm} = .2 \text{ nm} \] **Final:** \[ \boxed{.2 \text{ nm}} \] --- # **Summary Table** | Problem | Final Answer | |---------|--------------| | 1(a) | 33.5 m/s | | 1(b) | 121 km/h | | 2(a) | 267 | | 2(b) | 28.7 | | 2(c) | \(5.9515 \times 10^{-14}\) | | 3 | 4.457 m/s | | 4(a) | 61.39 cm³ | | 4(b) | .063 cm³ | | 5 | .2 nm |