Answer:
25.52 seconds
Explanation:
Speed of space ship = 0.92 c = v
c = Speed of light
Time observed from asteroid = Δt = 10 seconds
Time dilation
[tex]\Delta t'=\frac{\Delta t}{\sqrt{1-\frac {v^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{10}{\sqrt{1-\frac {0.92^2c^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{10}{\sqrt{1-0.92^2}}\\\Rightarrow \Delta t'=25.52\ s[/tex]
∴ Time interval be if measured by an observer on the Enterprise would be 25.52 seconds
