Answer:
a

b

Explanation:
From the question we are told that
The speed of the spaceship is 
Here c is the speed of light with value 
The length is 
The distance of the star for earth is 
The speed is 
Generally the from the length contraction equation we have that
![l = l_o \sqrt{1 -[\frac{v}{c } ]}](https://tex.z-dn.net/?f=l%20%20%3D%20%20l_o%20%20%5Csqrt%7B1%20-%5B%5Cfrac%7Bv%7D%7Bc%20%7D%20%5D%7D)
Now the when at rest the length is 
So



Considering b
Applying above equation
![l =l_o \sqrt{1 - [\frac{v}{c } ]}](https://tex.z-dn.net/?f=l%20%20%3Dl_o%20%5Csqrt%7B1%20-%20%20%5B%5Cfrac%7Bv%7D%7Bc%20%7D%20%5D%7D)
Here 
So



Answer:
3 m/s
Explanation:
Average Speed = 
Plug in the numbers, it will be (6m + 3m) divided by (2s + 1s), which is 9m/3s, which equals to 3m/s.
If it were possible for an object to fall freely near the surface of the Earth,
-- The direction of its velocity would always be "down"; that is, toward the center of the Earth.
-- The size of its velocity would continually increase, at the rate of 9.8 meters per second for every second it falls.
Answer:
yes
Explanation:
this is simple
the horizontal line is adjacent
the vertical line is opposite
recall that cos x=adj/hyp
adj=hyp(cos x)
while opp=hyp(sin x)