Answer:
a) 0, = -0.33 us
b) 140m
c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.
Explanation:
a)
the lorentz factor expression is written as;
y = 1₀ / √(1 - (v²/c²))
where v is the relative speed of an observer and c is the speed of light
so we were given that relative speed to be o.7c
therefore
y = 1 / √(1 - ((0.7c)² / c²))
y = 1 / √(1 - (0.49c² / c²))
y = 1 / √(1 - 0.49)
y = 1 / 0.7141
y = 1.4
1 - the coordinates of the first event, the s' frame of reference is,
x1 ' = y(x1 - vt1) = 0
y1 ' = y1, z1' = z1 and
t1 ' = y [t1 - v/c²x1]
= 0
2 - the coordinates of the second event, the s ' frame of reference is'
x2 ' = y(x2-vt2)
= 1.4(100m - 0)
= 140m
y2 ' = y2, z2 ' = z2
t2 ' = y [ t2 - v/c²x2 ]
= 1.4 [ 0 - 0.7c/c²(100) ]
using speed of light c as 3*10^8
1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]
= -0.33 us
b)
distance between
delltaX' = X2' - X1'
= 140m - 0
= 140m
c)
No, The event are not simultaneous i.e they did not occur at the same time.
the second even (-0.33 us) occurs 0.33 us earlier than the first event.
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