The captain orders his starship to accelerate from rest at a rate of "1 g" (1 g = 9.8 m/s2). How many days does it take the starship to reach 10% the speed of light? (Light travels at 3.0 × 108 m/s.)

The  days it will take the starship to reach 10% of speed of light is approximately 35 days.

Explanation:

Given;

acceleration of the starship, a = 9.8 m/s²

speed of light is given as, v = 3.0 × 10⁸ m/s

10% of speed light = 0.1 x 3.0 × 10⁸ m/s = 3.0 × 10⁷ m/s

Apply kinematic equation;

acceleration is the time rate of change in velocity.

a = Δv/t

t = Δv/a

Substitute the given values and solve for time

t =  (3.0 × 10⁷)/9.8

t = 3061224.4898 seconds

Therefore, the  days it will take the starship to reach 10% of speed of light is approximately 35 days.

The two space stations are 644.1653 miles far apart.

Explanation:

The distance of Huston and the International Space Station is 323 miles.

The distance of Huston and the Chinese space station Tiangong is 462 miles.

Angle between the two stations = 109⁰

So, applying the law of cosines as;

c² = a² + b² - 2abcosC

From the picture shown below, a = 323 miles and b = 462 miles

So,

c² = 323² + 462² + 2(323)(462)cos109 = 104329 + 213444 - 298452cos109

cos109 = -0.3256

So,

c² = 414948.9712

c = 644.1653 miles