Answer:
The distance is [tex]2.58\times10^{7}\ m[/tex].
Explanation:
Given that,
Time [tex]\Delta t = 0.2\ s[/tex]
The velocity is no more than a 14 % error in the speed of light.
So,
Velocity [tex]v= c\times 86\%[/tex]
We need to calculate the distance
Using formula of speed
[tex]v = \dfrac{d}{t}[/tex]
[tex]d = v\times t[/tex]
Where, v = speed
d = distance
t = time
Put the value into the formula
[tex]d = 3\times10^{8}\times\dfrac{86}{100}\times0.2[/tex]
[tex]d=516\times10^{5}\ m[/tex]
We know that,
The one side distance d' is
[tex]d'=\dfrac{d}{2}[/tex]
[tex]d'=\dfrac{516\times10^{5}}{2}[/tex]
[tex]d'=2.58\times10^{7}\ m[/tex]
Hence, The distance is [tex]2.58\times10^{7}\ m[/tex].
The distance in this scenario is 2.58 × 10⁷ m.
This is defined as the amount of space between two points or lines.
Δt = 0.2s
Velocity = no more than a 14 % error in the speed of light = c × 86%
Distance = velocity/time
3× 10⁸ × 86/100 × 0.2
= 516 × 10⁵ m
Since only one side is involved
516 × 10⁵ /2 = 2.58 × 10⁷ m.
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