91Ó°ÊÓ

Based on the given information and the steps provided, we can calculate the minimum uncertainty in position for the baseball and proton as follows: Step 1: Calculate the gravitational force acting on the baseball. The mass of the baseball (m_b) is given as 144 g, which we can convert to kg: \(m_b = \frac{144}{1000} \mathrm{kg} = 0.144 \mathrm{kg}\) Step 2: Convert the speed and uncertainty of speed to m/s. \(v_b = 137.32 \frac{1000}{3600} \mathrm{m} / \mathrm{s} \approx 38.144 \mathrm{m/s}\) \(\Delta v_b = 0.10 \frac{1000}{3600} \mathrm{m} / \mathrm{s} \approx 0.0278 \mathrm{m/s}\) Step 3: Calculate the uncertainty in momentum for both the baseball and the proton. For the baseball: \(\Delta p_b = m_b \Delta v_b \approx 0.144 \mathrm{kg} * 0.0278 \mathrm{m/s} \approx 0.0040 \mathrm{kg\cdot m/s}\) For the proton: \(\Delta p_p = m_p \Delta v_b \approx 1.6726219 \times 10^{-27} \mathrm{kg} * 0.0278 \mathrm{m/s} \approx 4.65 \times 10^{-29} \mathrm{kg\cdot m/s}\) Step 4: Calculate the minimum uncertainty in position for both the baseball and the proton. Using the reduced Planck constant, \(\hbar \approx 1.0545718 \times 10^{-34} \mathrm{kg\cdot m^2 / s}\), (a) For the baseball: \(\Delta x_b \geq \frac{1.0545718 \times 10^{-34} \mathrm{kg\cdot m^2 / s}}{2 * 0.0040 \mathrm{kg\cdot m/s}} \approx 1.31 \times 10^{-32} \mathrm{m}\) (b) For the proton: \(\Delta x_p \geq \frac{1.0545718 \times 10^{-34} \mathrm{kg\cdot m^2 / s}}{2 * 4.65 \times 10^{-29} \mathrm{kg\cdot m/s}} \approx 1.13 \times 10^{-6} \mathrm{m}\) So, the minimum uncertainty in position for the baseball is approximately \(1.31 \times 10^{-32} \mathrm{m}\), and the minimum uncertainty in position for the proton is approximately \(1.13 \times 10^{-6} \mathrm{m}\).