the answer is a!! its pretty simple I just read the graph.
Answer:
According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>
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Explanation:
For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:
v = d / t
where
- v is the speed of the spaceship
- d is the distance
- t is the time
Therefore,
d = v × t
d = (0.999 c)(4 light-years)
d = 3.996 light-years
This distance is<em> very slightly less than 4 light-years. </em>
Answer:
doppler shift's formula for source and receiver moving away from each other:
<em>λ'=λ°√(1+β/1-β)</em>
Explanation:
acceleration of spaceship=α=29.4m/s²
wavelength of sodium lamp=λ°=589nm
as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm
using doppler shift's formula:
<em>λ'=λ°√(1+β/1-β)</em>
putting the values:
700nm=589nm√(1+β/1-β)
after simplifying:
<em>β=0.17</em>
by this we can say that speed at that time is: v=0.17c
to calculate velocity at an acceleration of a=29.4m/s²
we suppose that spaceship started from rest so,
<em>v=v₀+at</em>
where v₀=0
so<em> v=at</em>
as we want to calculate t so:-
<em>t=v/a</em> v=0.17c ,c=3x10⁸ ,a=29.4m/s²
putting values:
=0.17(3x10⁸m/s)/29.4m/s²
<em>t=1.73x10⁶</em>
Answer:
Assessment zone
Explanation:
It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.
