Let's solve **Question 1a** step by step as requested. --- ## Question 1a ### **Step 1: Estimate Power Consumption for Laptops** - **Assume** each laptop uses **50 W** (typical for a laptop). - **Boeing 747-800** typical passenger count: **~410** passengers. - **Flight distance**: Vancouver to London ≈ **7,600 km**. - **Flight time**: Assume average speed of 900 km/h → \( t = \frac{760}{900} \approx 8.44 \) hours. **Total energy required:** \[ \text{Energy} = \text{Power per laptop} \times \text{Number of laptops} \times \text{Flight time} \] \[ = 50\,W \times 410 \times 8.44\,h = 50 \times 410 \times 8.44 \times 360\,J \] \[ = 50 \times 410 \times 30384\,J = 50 \times 12,457,440\,J = 622,872,000\,J \] **Explanation:** We calculated the total energy needed to power all laptops for the flight duration. --- ### **Step 2: Convert Jet Fuel to Electricity (Include Efficiency)** - **Jet fuel to electricity efficiency**: \( \eta = .38 \) - **Energy from fuel needed:** \( \text{Energy needed} = \frac{\text{Electricity required}}{\eta} \) - **Jet fuel energy density:** \( 1\,L = 34\,MJ \) \[ \text{Total jet fuel energy required} = \frac{622,872,000\,J}{.38} \approx 1,638,610,526\,J \] \[ \text{Fuel needed (L)} = \frac{1,638,610,526\,J}{34,000,000\,J/L} \approx 48.2\,L \] **Explanation:** We divided the total electrical energy by the conversion efficiency, then calculated the equivalent jet fuel volume needed. --- ### **Final Answer** **A Boeing 747-800 must carry approximately 48 liters of additional jet fuel to power every passenger's laptop for the flight from Vancouver to London.**

# Question 1: Flight Analysis ## Question 1a ### Step 1: Estimate Power Consumption for Laptops - **Assume** each laptop uses **50 W**. - **Boeing 747-800** typical passenger count: **~410**. - **Flight distance**: Vancouver to London ≈ **7,600 km**. - **Average speed**: **900 km/h** → \( t = \frac{7600}{900} \approx 8.44 \) hours. **Total energy required:** \[ \text{Energy} = \text{Power per laptop} \times \text{Number of laptops} \times \text{Flight time} \] \[ = 50\,W \times 410 \times 8.44\,h = 50 \times 410 \times 30384\,J = 622,872,000\,J \] **Explanation:** We calculated the total energy needed to power all laptops for the entire flight duration. ### Step 2: Convert Jet Fuel to Electricity (Include Efficiency) - **Jet fuel to electricity efficiency**: \( \eta = 0.38 \) - **Energy from fuel needed:** \[ \text{Energy needed} = \frac{622,872,000\,J}{0.38} \approx 1,638,610,526\,J \] - **Jet fuel energy density:** \( 1\,L = 34\,MJ \) \[ \text{Fuel needed (L)} = \frac{1,638,610,526\,J}{34,000,000\,J/L} \approx 48.2\,L \] **Explanation:** The total electrical energy required is divided by the conversion efficiency, and then converted to the equivalent jet fuel volume. ### Final Answer **A Boeing 747-800 must carry approximately 48 liters of additional jet fuel to power every passenger's laptop for the flight from Vancouver to London.** --- ## Question 1b ### Step 1: Calculate Fuel Consumption for Boeing 747-300 - **Assume** fuel consumption: **12 L/km** (average). - **Flight distance**: 7,600 km. **Total fuel consumption:** \[ \text{Total Fuel} = \text{Fuel consumption rate} \times \text{Distance} \] \[ = 12\,L/km \times 7600\,km = 91,200\,L \] ### Step 2: Calculate Fuel Consumption for Toyota Prius - **Fuel economy**: **4.7 L/100 km**. - **Average number of passengers**: **1.5**. **Fuel consumption per passenger (1/100 km)**: \[ \text{Fuel per passenger} = \frac{4.7\,L}{100\,km \times 1.5} \approx 0.0313\,L/100\,km \] **Comparison**: - **Boeing 747-300 (per passenger)**: \[ \text{Fuel per passenger} = \frac{91,200\,L}{410} \approx 222.0\,L \] ### Final Answer **The Boeing 747-300 consumes significantly more fuel per passenger per 100 km than a 2010 Toyota Prius, with approximately 222.0 L compared to 0.0313 L.** --- ## Question 1c ### Step 1: Calculate Potential Energy to Altitude - **Cruising altitude**: **10,000 m** (typical). - **Mass of the plane**: Assume **400,000 kg**. **Potential Energy**: \[ PE = mgh = 400,000\,kg \times 9.81\,m/s^2 \times 10,000\,m = 39,240,000,000\,J \] ### Step 2: Calculate Kinetic Energy to Speed - **Cruising speed**: \(300\,m/s\) (approx). **Kinetic Energy**: \[ KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 400,000\,kg \times (300\,m/s)^2 = 18,000,000,000\,J \] **Total Energy**: \[ \text{Total} = PE + KE = 39,240,000,000\,J + 18,000,000,000\,J = 57,240,000,000\,J \] **Explanation:** Total energy required includes both potential and kinetic energy to reach cruising altitude and speed. ### Final Answer **The plane requires approximately 57.24 billion joules of energy to reach cruising altitude and speed.** --- ## Question 1d ### Step 1: Fuel Cost for 20 Minutes at Cruising Altitude - **Cruising speed**: **300 m/s**. - **Time**: **20 min = 1200 s**. **Distance flown**: \[ \text{Distance} = \text{Speed} \times \text{Time} = 300\,m/s \times 1200\,s = 360,000\,m = 360\,km \] **Fuel consumption for 20 min**: \[ \text{Fuel} = \text{Fuel rate} \times \text{Distance} = 12\,L/km \times 360\,km = 4,320\,L \] ### Step 2: Compare Fuel Costs - **Fuel cost from part c)**: \(91,200\,L\) for the flight. **Comparison**: - Fuel cost for ascending to cruising speed and altitude vs. cruising fuel cost shows that cruising consumes less fuel over time. ### Final Answer **The plane consumes approximately 4,320 liters of fuel in 20 minutes at cruising altitude, which is much less than the energy required to take off and reach cruising speed. Planes should fly at higher altitudes to optimize fuel efficiency.**