The distance within 0.3s is 0.03m+(0.03m-0.01m) = 0.05m;

When G=2N and m=2kg, the passing distance within 0.4s is 0.025m+(0.025m-0.015m) = 0.035m;

So the answer is: 0.05; ? 0.035; ?

(2) Xiao Ming pulls the car with a hook code through a thin thread without a spring dynamometer, which is convenient to control the pulling force on the car; ?

(3) solar sail area S=(2000m)2=4× 106m2,

So the light pressure on the solar sail is f = 2.5×10-6n/m2× 4×106m2 =10n.

According to the data in the table, when the mass of an object is m= 1kg and the force is F=G= 1N, it moves 0.005m within 0. 1s, and the distance of equal movement increases 0.01m within 0.2s, 0.3s and 0.4s.. On the other hand. When the force acting on an object with a mass of m= 1kg is F=G=2N, it will move 0.005m×2 1=0.0 1m within 0. 1s, and the equal moving distance will increase by 0 in turn within 0.2s, 0.3s and 0.4s.. When the force acting on an object with a mass of m=2kg is F=G=2N, it moves by 0.005m×22=0.005m in 0. 1s, and the equivalent moving distance in 0.2s, 0.3s and 0.4s increases by 0.01m× 22 = 0.0/in turn.

Therefore, the solar sail spacecraft with the mass of m=5t=5× 103kg moves 0.005m× 103 within 0. 1s under the light pressure of F= 10N, and the distance increases in turn within 0.2s, 0.3s and 0.4s..

Therefore, t = 24× 3600s = 86400s = 0. 1 s× 864000 after moving from rest. The total distance of solar sail spacecraft is about 0.005 m×105×103+(0.005 m×105×103+0.01m× 654335438+005 03+0.0 1m× 105× 103×3)+(0.005m× 105× 103+0.0 1m×65438

Answer: The total journey is 7464960m. ..