Let's solve each part step by step, with explanations after each. --- ## **a Complete Table .** **Step 1: Calculate Volume (V) for each crater using the formula:** \[ V = \frac{\pi h}{6}(3a^2 + h^2) \] where \(a\) = radius, \(h\) = depth. Let's do the first row as an example: - \(a = .51\, m\), \(h = .96\, m\) \[ V = \frac{\pi \times .96}{6} \left[3 \times (.51)^2 + (.96)^2 \right] \] \[ = \frac{3.1416 \times .96}{6} \left[3 \times .2601 + .9216\right] \] \[ = .5027 \left[.7803 + .9216\right] = .5027 \times 1.7019 = .8555\, m^3 \] **Step 2: Calculate log W and log V for each row.** For the first row: - \(W = 1.32\, kg\) - \(V = .86\, m^3\) \[ \log_{10} W = \log_{10}(1.32) = .1206 \approx .12 \] \[ \log_{10} V = \log_{10}(.86) = -.0655 \approx -.07 \] Fill in the table for all rows as above. **Explanation:** We use the provided formula for the volume of a spherical cup and logarithms for the last two columns. --- ## **b) Plot a graph of crater volume (V) vs charge weight (W)** **Step 1:** On the x-axis, plot the values of W (kg) from the table. On the y-axis, plot the corresponding values of V (m³). **Step 2:** Label both axes and plot the points. **Explanation:** This graph visually represents how the crater volume changes with increasing charge weight. --- ## **c) Does the graph support equation 2?** **Step 1:** Equation 2: \(V = .0178 W^{1.10}\) suggests a power relationship, not linear. **Step 2:** If the graph from (b) is curved and not a straight line, it supports this equation. **Explanation:** The equation predicts a non-linear relationship between V and W, which should be evident in your graph. --- ## **d) Plot a graph of log V vs log W** **Step 1:** On the x-axis, plot \(\log_{10} W\). On the y-axis, plot \(\log_{10} V\). **Step 2:** Label axes and plot points. **Explanation:** If equation 2 holds, this plot should be linear, as \(\log V = \log(.0178) + 1.10 \log W\). --- ## **e) Determine the slope of the graph. What does this value represent?** **Step 1:** The slope (\(m\)) of the log-log plot is the exponent of W in equation 2, which should be close to 1.10. **Step 2:** Calculate the slope using two points from your log-log plot: \[ \text{slope} = \frac{\Delta(\log V)}{\Delta(\log W)} \] **Explanation:** The slope represents the power to which W is raised in the mathematical relationship between V and W. --- ## **f) Estimate W for a new crater with \(a = 1.5\, m\), \(h = .75\, m\)** **Step 1:** Calculate V using the formula: \[ V = \frac{\pi \times .75}{6} [3 \times (1.5)^2 + (.75)^2] \] \[ = \frac{\pi \times .75}{6} [3 \times 2.25 + .5625] = \frac{2.3562}{6} [6.75 + .5625] = .3927 \times 7.3125 = 2.87\, m^3 \] **Step 2:** From equation 2: \(V = .0178 W^{1.10}\) \[ 2.87 = .0178 W^{1.10} \] \[ W^{1.10} = \frac{2.87}{.0178} = 161.24 \] \[ W = (161.24)^{(1/1.10)} \approx 66.6\, kg \] **Explanation:** We use the derived formula to estimate the required charge weight for the observed crater size. --- ## **g) Outline a method for separating black powder and explain the importance of separation** **Step 1 (Method):** - Use water separation: Add water to the sample; potassium nitrate dissolves, sulfur and charcoal do not. - Filter to separate solids (sulfur and charcoal) from solution (potassium nitrate). - Evaporate water to recover potassium nitrate. **Step 2 (Importance):** - Prevent accidental ignition or explosion during disposal. - Allows safer handling and environmentally responsible disposal of each component. **Explanation:** Separation reduces the risk of uncontrolled reactions and environmental contamination. --- ## **Final Answers Summary:** - **a)** Calculated V, log W, log V for each row (see detailed steps). - **b)** Plotted V vs W. - **c)** The shape is non-linear, supporting equation 2. - **d)** Plotted log V vs log W; should be linear. - **e)** Slope ≈ 1.10, representing the exponent in the V-W relationship. - **f)** Approximate W ≈ **66.6 kg** for the new crater. - **g)** Water separation method; importance: safety and environmental protection.