Answer:
The distance is
Explanation:
From the question we are told that
The smallest shift is 
Generally a grid unit is
of an arcsec
This implies that 0.2 grid unit is 
The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

substituting values


Note 
So
The andwer of tye question is 3O2
1) 
The capacitance of a parallel-plate capacitor is given by:

where
is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is

so the area is

While the separation between the plates is

So the capacitance is

And now we can find the energy stored,which is given by:

2) 0.71 J/m^3
The magnitude of the electric field is given by

and the energy density of the electric field is given by

and using
, we find

Hey i dont have an answer but i need the points for finals today. Thank you
Distance = (speed) x (time)
Distance = (20 m/s) x (500 s)
Distance = (20 x 500) (m·s / s)
Distance = 10,000 m