solve the following using guess method
Answer (1)
Answer:Section 1: (3 points each) 1. Momentum of the Car: - Given: mass (m) = 1463 kg, velocity (v) = 27 m/s - Formula: Momentum (p) = m * v - Calculation: p = 1463 kg * 27 m/s = 39461 kg·m/s - Answer: The momentum of the car is 39461 kg·m/s. 2. Kinetic Energy of the Coaster Car: - Given: mass (m) = 1345 kg, velocity (v) = 22.0 m/s - Formula: Kinetic Energy (KE) = 1/2 * m * v² - Calculation: KE = 0.5 * 1345 kg * (22.0 m/s)² = 326,290 J - Answer: The kinetic energy of the coaster car is 326,290 J. 3. Missy Diwater's Speed: - Given: kinetic energy (KE) = 14,000 J, mass (m) = 53 kg - Formula: KE = 1/2 * m * v² => v = √(2 * KE / m) - Calculation: v = √(2 * 14000 J / 53 kg) ≈ 22.9 m/s - Answer: Missy Diwater's speed was approximately 22.9 m/s. 4. Height of the Cannon Ball: - Given: potential energy (PE) = 14065 J, mass (m) = 45 kg, acceleration due to gravity (g) ≈ 9.81 m/s² - Formula: PE = m * g * h => h = PE / (m * g) - Calculation: h = 14065 J / (45 kg * 9.81 m/s²) ≈ 31.9 m - Answer: The cannon ball was approximately 31.9 m high. 5. Potential Energy of the Object: - Given: mass (m) = 12 kg, height (h) = 15 m, acceleration due to gravity (g) ≈ 9.81 m/s² - Formula: PE = m * g * h - Calculation: PE = 12 kg * 9.81 m/s² * 15 m = 1765.8 J - Answer: The potential energy of the object before it fell was 1765.8 J. Section 2: (5 points each) 1. Marble Thrown Horizontally: - Given: horizontal velocity (vₓ) = 1.75 m/s, horizontal distance (x) = 0.85 m - Assumptions: We'll assume negligible air resistance. The time it takes for the marble to fall vertically is the same as the time it takes to travel horizontally. - Calculations: - Find time of flight (t): We need to find the vertical velocity (vᵧ) first. Since the marble is thrown horizontally, its initial vertical velocity is 0. Using the equation of motion: y = vᵧt + 1/2gt², where y is the vertical distance (height of the table), vᵧ = 0, and g = 9.81 m/s². We need to solve for 't' and 'y'. We can use the horizontal motion to find 't': x = vₓt => t = x/vₓ = 0.85m / 1.75 m/s ≈ 0.486 s. - Find height (y): Now that we have 't', we can use the vertical motion equation: y = 1/2gt² = 0.5 * 9.81 m/s² * (0.486 s)² ≈ 1.15 m - Final velocity (v): To find the final velocity, we need to find the final vertical velocity (vᵧf) using vᵧf = vᵧi + gt = 0 + 9.81 m/s² * 0.486 s ≈ 4.77 m/s. The final velocity is the vector sum of the horizontal and vertical velocities: v = √(vₓ² + vᵧf²) = √(1.75² + 4.77²) ≈ 5.08 m/s. The angle of impact can be determined using trigonometry (arctan(4.77/1.75)). - Answer: The lab table is approximately 1.15 m high. The marble's velocity just before hitting the floor is approximately 5.08 m/s. 2. Girl Jumping into a Raft: - Given: mass of girl (m₁) = 53.0 kg, mass of raft (m₂) = 100 kg, initial velocity of girl (v₁) = 4.50 m/s, initial velocity of raft (v₂) = 0 m/s - Principle: Conservation of momentum: m₁v₁ + m₂v₂ = (m₁ + m₂)v_f, where v_f is the final velocity of the girl and raft. - Calculation: (53.0 kg * 4.50 m/s) + (100 kg * 0 m/s) = (53.0 kg + 100 kg) * v_f => v_f = (53.0 kg * 4.50 m/s) / (153 kg) ≈ 1.56 m/s - Answer: The final velocity of the girl and the raft is approximately 1.56 m/s. 3. Colliding Carts: - Given: mass of cart 1 (m₁) = 17.0 kg, velocity of cart 1 (v₁) = 5.0 m/s, mass of cart 2 (m₂) = 7.5 kg, velocity of cart 2 (v₂) = -2.25 m/s (negative since it's moving to the left). - Principle: Conservation of momentum: m₁v₁ + m₂v₂ = (m₁ + m₂)v_f - Calculation: (17.0 kg * 5.0 m/s) + (7.5 kg * -2.25 m/s) = (17.0 kg + 7.5 kg) * v_f => v_f = (85 - 16.875) / 24.5 ≈ 2.78 m/s - Answer: The final speed of the two carts is approximately 2.78 m/s. 4. Depth of the Well: - Given: time (t) = 5.16 s, acceleration due to gravity (g) ≈ 9.81 m/s² - Formula: Assuming negligible air resistance, we can use the equation of motion: d = 1/2 * g * t² where d is the depth of the well. - Calculation: d = 0.5 * 9.81 m/s² * (5.16 s)² ≈ 130 m - Answer: The depth of the well is approximately 130 m. Remember that these calculations use approximations for g (acceleration due to gravity). More precise values could slightly alter the results.
