Let's tackle each part step by step, as requested. --- ## **3(a) Volume of a box with uncertainties** #### **Step 1: Calculate the volume** Given: - Height (\(h\)) = 3.1 ± .6 in - Width (\(w\)) = 9.2 ± .9 in - Depth (\(d\)) = 25.3 ± 3.9 in Volume, \( V = h \cdot w \cdot d \) \[ V = 3.1 \times 9.2 \times 25.3 = 722.876 \approx 723 \, \text{in}^3 \] **Explanation:** We multiply the nominal values to get the central value of the volume. --- #### **Step 2: Calculate the uncertainty in the volume** For multiplication, relative uncertainties add: \[ \left(\frac{\Delta V}{V}\right) = \left(\frac{\Delta h}{h}\right) + \left(\frac{\Delta w}{w}\right) + \left(\frac{\Delta d}{d}\right) \] \[ \left(\frac{\Delta V}{V}\right) = \frac{.6}{3.1} + \frac{.9}{9.2} + \frac{3.9}{25.3} = .1935 + .0978 + .1542 = .4455 \] \[ \Delta V = .4455 \times 723 \approx 322 \, \text{in}^3 \] **Explanation:** We add the relative uncertainties and multiply by the calculated volume to get the absolute uncertainty. --- **Final Answer (3a):** \[ \boxed{V = 723 \pm 322 \ \text{in}^3} \] --- ## **3(b) Volume and average density of the Earth with uncertainties** #### **Step 1: Calculate Earth's volume as a sphere** Given: - \( r_E = 6367 \pm 11 \) km - \( M_E = (5.9737 \pm .08) \times 10^{24} \) kg Volume of a sphere: \[ V = \frac{4}{3}\pi r^3 \] \[ V = \frac{4}{3}\pi (6367 \times 10^3 \text{ m})^3 = 1.083 \times 10^{21} \text{ m}^3 \] **Explanation:** Convert radius to meters and use the formula for a sphere's volume. --- #### **Step 2: Calculate density and its uncertainty** Density: \[ \rho = \frac{M_E}{V} \] \[ \rho = \frac{5.9737 \times 10^{24}}{1.083 \times 10^{21}} = 5516 \text{ kg/m}^3 \] Relative uncertainties: \[ \left(\frac{\Delta \rho}{\rho}\right) = \left(\frac{\Delta M_E}{M_E}\right) + \left(\frac{\Delta V}{V}\right) \] \[ \frac{\Delta M_E}{M_E} = \frac{.08 \times 10^{24}}{5.9737 \times 10^{24}} = .0134 \] For \( V \), \[ \frac{\Delta V}{V} = 3 \frac{\Delta r}{r} = 3 \frac{11}{6367} = .00518 \] Total relative uncertainty: \[ .0134 + .00518 = .0186 \] \[ \Delta \rho = .0186 \times 5516 \approx 103 \text{ kg/m}^3 \] **Explanation:** Relative uncertainty in volume for a sphere is \(3 \Delta r / r\). Add mass and volume uncertainties, then multiply by density. --- **Final Answer (3b):** \[ \text{Volume} = (1.083 \pm .017) \times 10^{21} \text{ m}^3 \] \[ \text{Density} = 5516 \pm 103 \text{ kg/m}^3 \] --- ## **3(c) Volume of an oblate ellipsoid Earth and density accuracy** #### **Step 1: Calculate ellipsoid volume** Given: - \( a_E = 6378 \) km, \( c_E = 6357 \) km \[ V = \frac{4}{3}\pi a^2 c \] \[ V = \frac{4}{3}\pi (6378 \times 10^3)^2 (6357 \times 10^3) \] \[ V = 1.083 \times 10^{21} \text{ m}^3 \ (\text{compare with sphere: } 1.083 \times 10^{21} \text{ m}^3) \] **Explanation:** Plug in the actual equatorial and polar radii to get the ellipsoid volume. --- #### **Step 2: Check if the difference is within previous uncertainty** \[ \Delta V = |V_{\text{ellipsoid}} - V_{\text{sphere}}| \approx |1.083 \times 10^{21} - 1.083 \times 10^{21}| = \text{ (very small difference)} \] **Explanation:** The difference between the sphere and ellipsoid volume is much smaller than the calculated uncertainty in (b), so the uncertainty covers this. --- **Final Answer (3c):** The change in volume from using the ellipsoid model is much smaller than the uncertainty from (b), so the density calculation is still accurate within the given uncertainty. --- If you want to proceed with question 4 as well, let me know!