Answer:
The farther star will appear 4 times fainter than the star that is near to the observer.
Explanation:
Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time
Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)
This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by
For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by
Hence we sense it as 4 times fainter than the nearer star.
Answer:
Option C: Current X has a lower potential difference than Current Y.
Explanation:
The chart above only shows the potential difference of difference current.
A careful observation of the chart shows that Current X has a lower potential difference than Current Y.
Answer:
The observer sees the space-probe 9.055m long.
Explanation:
Let 
be the length of the space-probe when measured at rest, and 
be its length as observed by an observer moving at velocity 
, then
Now, we know that 
and 
, and putting these into 
we get:
Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.
1. Because of gravity....
2. No you either feel still ( gravity) or is actually in movement....
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Answer:
B. 80 m/s²
Explanation:
F = ma
a = F/m = (40 N)/(0.5 kg) = 80 m/s²
