Aspacecraft with a proper length of 250 m passes by an observer on the earth. according to this observer, it takes 0.820 µs for the spacecraft to pass a fixed point. determine the speed of the spacecraft as measured by the earth-based observer.

0.713c, or 214000 km/s    
And initial check without taking into consideration indicates that the
spacecraft is traveling in excess of 305,000 km/s which is faster than the
speed of light. Obviously, this is impossible, so we need to take into
consideration the feature of Lorentz contraction which will cause the earth
based observe to see the spacecraft as being shorter. The formula for
Lorentz contraction is  
L = L0(sqrt(1-v^2/c^2))  
where  
L = observed length  
L0 = length when relative velocities is 0  
v = speed  
c = speed of light    
Since the observed time will be L/v = t we can construct the equation:  
L/v = t  
L = L0(sqrt(1-v^2/c^2))  
L0(sqrt(1-v^2/c^2))/v = t    
Now solve for v  
L0(sqrt(1-v^2/c^2))/v = t  
L0(sqrt(1-v^2/c^2)) = tv  
sqrt(1-v^2/c^2) = tv/L0  
sqrt(1-v^2/c^2)/v = t/L0  
(1-v^2/c^2)/v^2 = t^2/L0^2  
1/v^2 - 1/c^2 = t^2/L0^2  
1/v^2 = t^2/L0^2 + 1/c^2  
v^2 = 1/(t^2/L0^2 + 1/c^2)  
v = sqrt(1/(t^2/L0^2 + 1/c^2))    
Now substitute the known values and calculate v.  
v = sqrt(1/((0.82x10^-6 s)^2/(250 m)^2 + 1/(299792458 m/s)^2))  
v = sqrt(1/((6.724x10^-13 s^2)/62500 m^2 + 1/8.98755178736818x10^16 m^2/s^2))  
v = sqrt(1/(1.07584x10^-17 s^2/m^2 + 1.11265005605362x10^-17 m^2/s^2))  
v = sqrt(1/(2.18849005605362E-17 m^2/s^2))  
v = sqrt(4.56936049233527x10^16 m^2/s^2)  
v = 213760625.3 m/s    
So we have a calculated velocity of 213760625.3 m/s. Let's see if that's reasonable.  
Due to Lorentz contraction, the observed length will be L0(sqrt(1-v^2/c^2)) meters long. Let's calculate that:  
v/c = 213760625.3/299792458  
v/c = 0.713028696  
L = L0(sqrt(1-v^2/c^2)) 
 L = 250(sqrt(1 - 0.713028696^2))  
L = 250(sqrt(1 - 0.508409921))  
L = 250(0.701134851)  
L = 175.2837127 m    
The speed will be L/t, so  
175.2837127 m / 0.82x10^-6 s = 213760625.3 m/s  
And the result matches the earlier calculated speed. So we have a good cross check on the answer, which when rounded to 3 significant figures is either 0.713c, or 214000 km/s.

May consider a group  g=({0,1},+)g=({0,1},+)where  ++  will work as the xor operator like we work in boolean algebra. note here, that ‘00’ and ‘11’ are not the usual 0 and 1 that we work with everyday. the elements are  called  ‘00’ and ‘11’ and  may not mean  zero (nothing) and one (single). ‘00’ is the identity for the given binary operator +, meaning that any element operated with ‘00’ gets us that element itself.  for a set  ss  in  g′g′  having more elements than just  {0,1}{0,1}  the values for  a+b∈sa+b∈s, such that  a∈s,b∈sa∈s,b∈s, have to be first defined, only then can you determine the value of  1+11+1. the answer could possibly have been  22  if oanswer is25,000