This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to questions 3 and 4. Question 3: Velocity of canoe relative to still water (v_c) = 4 km/h (North) Velocity of water current (v_w) = 2 km/h (East) a) The magnitude of the resultant velocity of the canoe Step 1: Since the velocities are perpendicular, use the Pythagorean theorem to find the resultant velocity (v_R). v_R = sqrt(v_c^2 + v_w^2) v_R = sqrt((4 km/h))^2 + (2 km/h)^2 Step 2: Calculate the square root. v_R = sqrt(16 km)^2/h^2 + 4 km^2/h^2 v_R = sqrt(20 km)^2/h^2 v_R = 2sqrt(5) km/h v_R ≈ 4.47 km/h The magnitude of the resultant velocity is 4.47 km/h. b) The angle of travel of the canoe relative to North. Step 1: Use the tangent function to find the angle () relative to North. The opposite side is the water current velocity (v_w) and the adjacent side is the canoe's velocity (v_c). () = (v_w)/(v_c) () = 2 km/h4 km/h () = 0.5 Step 2: Calculate the angle. = (0.5) ≈ 26.57^ The angle of travel relative to North is 26.57^ East of North. Question 4: Initial velocity (u) = 0 m/s (starts at rest) Acceleration due to free fall (g) = 10 m/s^2 a) Calculate the velocity of the fruit 1.0 s after it starts to fall. Step 1: Use the kinematic equation v = u + gt. v = 0 m/s + (10 m/s^2)(1.0 s) Step 2: Calculate the final velocity. v = 10 m/s The velocity of the fruit after 1.0 s is 10 m/s. b) The fruit takes 2.0 s to hit the ground. Calculate the distance the fruit falls before hitting the ground. Step 1: Use the kinematic equation s = ut + (1)/(2)gt^2. s = (0 m/s)(2.0 s) + (1)/(2)(10 m/s^2)(2.0 s)^2 Step 2: Calculate the distance. s = 0 + (1)/(2)(10 m/s^2)(4.0 s^2) s = 5 m/s^2 × 4.0 s^2 s = 20 m The distance the fruit falls is 20 m. That's 2 down. 3 left today — send the next one.