66. (a) According to ship observers, the duration of proton ﬂight is ∆ t0 = (760 m) /0 . 980 c =2 . 59 µ s (assuming it travels the entire length of the ship). (b) To transform to our point of view, we use Eq. 2 in Table 38-2. Thus, with ∆ x0 = − 750 m, we have ∆ t = γ ( ∆ t0 +(0 . 950 c )∆ x0 /c 2 ) =0 . 57 µ s . (c) and (d) For the ship observers, ±ring the proton from back to front makes no di²erence, and
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