Difficult
If the speed of light is $3.0 \times 10^{8} \,m s^{-1}$,calculate the distance covered by light in $2.00 \,ns$.
Given:
Speed of light $(v)$ $= 3.0 \times 10^{8} \,m s^{-1}$
Time $(t)$ $= 2.00 \,ns = 2.00 \times 10^{-9} \,s$
Formula:
Distance $(d)$ $= v \times t$
Calculation:
$d = (3.0 \times 10^{8} \,m s^{-1}) \times (2.00 \times 10^{-9} \,s)$
$d = 6.00 \times 10^{-1} \,m$
$d = 0.600 \,m$
Speed of light $(v)$ $= 3.0 \times 10^{8} \,m s^{-1}$
Time $(t)$ $= 2.00 \,ns = 2.00 \times 10^{-9} \,s$
Formula:
Distance $(d)$ $= v \times t$
Calculation:
$d = (3.0 \times 10^{8} \,m s^{-1}) \times (2.00 \times 10^{-9} \,s)$
$d = 6.00 \times 10^{-1} \,m$
$d = 0.600 \,m$
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