91Ó°ÊÓ

First, we need to find the initial momentum of the ship. Momentum (p) is given by the product of mass (m) and velocity (v): \[p = m \times v\] Using the given mass and velocity: \[p = (1.00 \times 10^{7} \mathrm{kg}) \times (0.750 \mathrm{m}/\mathrm{s})\] Calculating the value of p: \[p = 7.50 \times 10^6 \mathrm{kg} \cdot \mathrm{m}/\mathrm{s}\]

As the cruise ship comes to a stop, its final velocity is 0. Therefore, the final momentum (pf) is: \[pf = m \times v_f\] \[pf = (1.00 \times 10^{7} \mathrm{kg}) \times (0 \mathrm{m}/\mathrm{s})\] \[pf = 0 \mathrm{kg} \cdot \mathrm{m}/\mathrm{s}\]

Now, let's calculate the change in momentum (Δp) by subtracting the final momentum from the initial momentum: \[Δp = p_f - p_i\] Substitute the values of the initial and final momentum: \[Δp = 0 \mathrm{kg} \cdot \mathrm{m}/\mathrm{s} - 7.50 \times 10^6 \mathrm{kg} \cdot \mathrm{m}/\mathrm{s}\] \[Δp = - 7.50 \times 10^6 \mathrm{kg} \cdot \mathrm{m}/\mathrm{s}\]

To find the time it took to stop the ship, we can use the equation for acceleration. Since we know the initial velocity (v_i), the final velocity (v_f), and the distance (d) the ship traveled: \[v_f^2 = v_i^2 + 2ad\] We can find the acceleration(a) and then use it to find the time(t): \[a = \frac{v_f^2 - v_i^2}{2d}\] Substitute the values: \[a = \frac{(0 \mathrm{m}/\mathrm{s})^2 - (0.750 \mathrm{m}/\mathrm{s})^2}{2 \times 6.00 \mathrm{m}}\] Calculating the value of acceleration(a): \[a \approx - 0.09375 \mathrm{m}/\mathrm{s}^2\] Now, using the acceleration and initial velocity, we can find the time (t): \[t = \frac{v_f - v_i}{a}\] Substitute the values: \[t = \frac{0 \mathrm{m}/\mathrm{s} - 0.750 \mathrm{m}/\mathrm{s}}{- 0.09375 \mathrm{m}/\mathrm{s}^2}\] Calculating the value of time(t): \[t \approx 8.00 \mathrm{s}\]

Now that we know the change in momentum and the time, we can calculate the average force exerted on the pier using the concept of impulse (I = Δp = F × t): \[F = \frac{Δp}{t}\] Substitute the values of the change in momentum and time: \[F = \frac{- 7.50 \times 10^6 \mathrm{kg} \cdot \mathrm{m}/\mathrm{s}}{8.00 \mathrm{s}}\] Calculating the average force: \[F \approx - 9.375 \times 10^5 \mathrm{N}\] Since force is a vector, the negative sign indicates that the force exerted on the pier is in the opposite direction to the initial velocity of the cruise ship. So, the average force exerted on the pier is approximately \(9.375 \times 10^5 \mathrm{N}\) in the opposite direction to the initial motion of the ship.