Let's solve the problem step by step: --- ## Step 1: Find the Minimum Uncertainty in the Block's Velocity Given: - Uncertainty in position, \(\Delta x = 1. \times 10^{-10}\) m - Mass of block, \(m = 8.4\) kg - Planck's constant, \(h = 6.626 \times 10^{-34}\) J·s (use \(\hbar = \frac{h}{2\pi}\)) Using Heisenberg's Uncertainty Principle: \[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \] \[ \Delta p = m \Delta v \] So, \[ \Delta x \cdot m \Delta v \geq \frac{\hbar}{2} \] \[ \Delta v \geq \frac{\hbar}{2 m \Delta x} \] Plug in the values: - \(\hbar = \frac{6.626 \times 10^{-34}}{2\pi} \approx 1.055 \times 10^{-34}\) J·s \[ \Delta v \geq \frac{1.055 \times 10^{-34}}{2 \times 8.4 \times 1. \times 10^{-10}} \] \[ = \frac{1.055 \times 10^{-34}}{16.8 \times 10^{-10}} \] \[ = \frac{1.055}{16.8} \times 10^{-24} \] \[ \approx .0628 \times 10^{-24} \] \[ = 6.28 \times 10^{-26} \text{ m/s} \] **Explanation:** We used Heisenberg's uncertainty principle to find the minimum uncertainty in the block's velocity. --- ## Step 2: How Far Might the Block Move in 6 Years Due to This Velocity? Time, \(t = 6\) years \(= 6 \times 365 \times 24 \times 360\) s \(= 1.89 \times 10^8\) s Distance = \(\Delta x_{\text{move}} = \Delta v \times t\) \[ = 6.28 \times 10^{-26} \text{ m/s} \times 1.89 \times 10^8 \text{ s} \] \[ = 1.187 \times 10^{-17} \text{ m} \] **Explanation:** We multiplied the uncertainty in velocity by the total time to estimate how far the block might move due to this velocity uncertainty. --- ## Final Answers **(a) Minimum uncertainty in the block's velocity:** \[ \boxed{6.28 \times 10^{-26} \text{ m/s}} \] **(b) How far the block might move in 6 years due to this velocity:** \[ \boxed{1.19 \times 10^{-17} \text{ m}} \]