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Question: How big shall the area of a parachute be to let a 80 kilogram human land in 10 metres/second in an atmosphere of 4200 millibar and affected by 0.5 G of gravity?
To determine the size of the parachute needed, we can use the drag equation: Fd = 0.5 * Cd * A * rho * V^2 Where, Fd is the drag force (N) Cd is the drag coefficient (dimensionless, a common value for a parachute is 1.5) A is the area of the parachute (m^2) rho is the air density (kg/m^3) V is the desired landing speed (m/s) We can first calculate the air density using the given atmospheric pressure, using the ideal gas law: rho = P / (R * T) where P is the atmospheric pressure (Pa), R is the specific gas constant for air (287 J/kg·K), and T is the temperature (K). For this problem, we aren't provided with a temperature, so we will assume standard conditions at sea level, which is 15°C or 288.15 K. Keep in mind that changes in temperature will affect the parachute size. Given: P = 4200 mbar = 4200 * 100 (to convert to Pa) = 420000 Pa R = 287 J/kg·K T = 288.15 K rho = 420000 Pa / (287 J/kg·K * 288.15 K) ≈ 5.08 kg/m^3 Now we can calculate the force required to achieve 0.5 G deceleration for an 80 kg person: F = m * a F = 80 kg * (0.5 * 9.81 m/s^2) = 392.4 N Now we can solve for the parachute area: A = Fd / (0.5 * Cd * rho * V^2) A = 392.4 N / (0.5 * 1.5 * 5.08 kg/m^3 * (10 m/s)^2) ≈ 6.87 m^2 So, the parachute needs to have an area of approximately 6.87 square meters to allow an 80 kg person to land at 10 meters/second in an atmosphere of 4200 millibars and under 0.5 G of gravity. Please note that this is a simplified calculation and there are many factors that can influence the actual parachute size required.
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