Answer:
the speed at which a moving clock lose 4,5ns in 1.0 day is 98.58 m/s
Explanation:
Let's first find the relation for the time dilation by using the formula:
[tex]\Delta t' = \Delta t \gamma[/tex]
Making [tex]\gamma[/tex] the subject of the formula, we have:
[tex]\gamma= \dfrac{\Delta t'}{\Delta t}[/tex]
Number of day(s) = 1
1 day in seconds = 24 × 60 × 60
= 84600 seconds
[tex]\gamma= \dfrac{86400 \ s}{86400 \ s - 4.5 \ ns}[/tex]
[tex]\sqrt{1- \dfrac{v^2}{c^2} }= \dfrac{86400 \ s - 4.5 \ ns}{86400 \ s}[/tex]
[tex]\sqrt{1- \dfrac{v^2}{c^2} }= 1 - 5.4 \times 10^{-14}[/tex]
From the above equation, if we apply binomial expansion to it, we have the following,
[tex]\sqrt{1- \dfrac{v^2}{c^2} }=(1-\dfrac{v^2}{c^2})^{\frac{1}{2}}[/tex]
⇒[tex]1 -\dfrac{1}{2}(\dfrac{v^2}{c^2})[/tex]
So;
[tex]1 -\dfrac{1}{2}(\dfrac{v^2}{c^2}) = 1 - 5.4 \times 10^{-14}[/tex]
[tex]\dfrac{1}{2}(\dfrac{v^2}{c^2}) = 5.4 \times 10^{-14}[/tex]
[tex]\dfrac{v^2}{c^2} = 2 \times 5.4 \times 10^{-14}[/tex]
[tex]\dfrac{v^2}{c^2} = 1.08 \times 10^{-13}[/tex]
[tex]v^2 = 1.08 \times 10^{-13} \times c^2[/tex]
where ;
[tex]c = 3 \times 10^8 \ m/s[/tex]
[tex]v^2 = 1.08 \times 10^{-13} \times (3 \times 10^8 \ m/s)^2[/tex]
[tex]v = \sqrt{ 1.08 \times 10^{-13} \times (3 \times 10^8 \ m/s)^2}[/tex]
[tex]v = \sqrt{ 1.08 \times 10^{-13} } \times (3 \times 10^8 \ m/s)[/tex]
[tex]v =3.286 \times 10^{-7} \times (3 \times 10^8 \ m/s)[/tex]
[tex]\mathbf{v =98.58 \ m/s}[/tex]
Therefore, the speed at which a moving clock lose 4,5ns in 1.0 day is 98.58 m/s
