Question

Part (a) Homework. Unanswered We have a rather small solar sail, which has a mass of 30.0 kg, and a sail area of 60000 m. At a particular distance from the Sun, the Sun's gravitational field is 0.004 N/kg, and the intensity of sunlight is 1000 W/m². Assuming the plane of the sail is perpendicular to the incident light, and the light is 100% reflected, what is the magnitude of the force exerted on the sail by the light? N Numeric Answer: Unanswered Submit Part (b) Homework. Unanswered We have a rather small solar sail, which has a mass of 30.0 kg, and a sail area of 60000 m. At a particular distance from the Sun, the Sun's gravitational field is 0.004 N/kg, and the intensity of sunlight is 1000 W/m². Calculate the magnitude of the solar sail's acceleration. (Don't forget to account for gravity.) m/s2 Numeric Answer: Unanswered 5 attempts left Submit Part (c) Homework. Unanswered We have a rather small solar sail, which has a mass of 30.0 kg, and a sail area of 60000 m2. At a particular distance from the Sun, the Sun's gravitational field is 0.004 N/kg, and the intensity of sunlight is 1000 W/m². Assuming the solar sail starts from rest and it has a constant acceleration, how many seconds would it take the solar sail to cover a distance of 3 x 108 m? Note that this is comparable to the distance from the Earth to the Moon, and that light covers this distance in 1 second. Numeric

Answer 1

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ai ven ma Given me 30.0kg A= 60,000m2 and c(speed of light) = 3x108 m/s I = 1000 W/m2 (a) FORCE , F= 2 - X A = 2 x 1000 - X60000 = 4 x 10 3x108 = 0.4N (6) Fnet = 0.4 - Fg =ma Fq = 0.004 x 30 = 0.12N 0.4-0.12=ma → a= 0-2% = 0.0093 m 192 30 (C) We know & ut + + 6at² a=0.0093 m/82 8= 3x108 3 x 100 = 0x t t. X 0.0093 x + 2 #2 = 645-16 x108bec t = 25.4 x10 sec