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If the speed of light is \(3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\), calculate the distance covered by light in \(2.00 \mathrm{~ns}\).
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Verified Answer
Distance covered \(=\) Speed \(\times\) Time
\(\begin{aligned}
&=3.0 \times 10^8 \mathrm{~ms}^{-1} \times 2.00 \mathrm{~ns} \\
&=3.0 \times 10^8 \mathrm{~ms}^{-1} \times 2.00 \mathrm{~ns} \times \frac{10^{-9} \mathrm{~s}}{1 \mathrm{~ns}}=6.00 \times 10^{-1} \\
&\mathrm{~m}=0.600 \mathrm{~m}
\end{aligned}\)
\(\begin{aligned}
&=3.0 \times 10^8 \mathrm{~ms}^{-1} \times 2.00 \mathrm{~ns} \\
&=3.0 \times 10^8 \mathrm{~ms}^{-1} \times 2.00 \mathrm{~ns} \times \frac{10^{-9} \mathrm{~s}}{1 \mathrm{~ns}}=6.00 \times 10^{-1} \\
&\mathrm{~m}=0.600 \mathrm{~m}
\end{aligned}\)
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