Какое минимальное время должен длиться полет корабля, чтобы использование экономичного

The change in velocity (Δv) of a spacecraft under constant acceleration can be calculated using the formula Δv = a * t, where a is the acceleration and t is the time. For the first engine, Δv1 = a1 * t1 = 4g * 10min = 40g * 10min, where g is the acceleration due to gravity (9.8m/s^2). To convert minutes to seconds, we can multiply by 60, so Δv1 = 40 * 9.8 * 10 * 60 = 237600 m/s. For the second engine, Δv2 = a2 * t2 = 0.5g * 2 hours = 0.5g * 2 * 60 minutes = 60g * 2 minutes = 120g * 2min, where 1 hour = 60 minutes. We want to find the minimum time the spacecraft must fly to make the second engine more economical. We know that Δv = a1 * t1 = a2 * t2, so t2 = Δv1 / a2 = 237600 / 0.5g = 237600 / (0.5 * 9.8) = 485333.33 seconds. To convert seconds to minutes, we can divide by 60, so t2 = 485333.33 / 60 = 8090.56 minutes. Rounding to the nearest integer, the minimum time the spacecraft must fly is 8090 minutes, or approximately 134 hours. More details...shorturl.at/lE129