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Vlad1618 [11]
3 years ago
15

At what speed, in m/s, would a moving clock lose 4.5 ns in 1.0 day according to experimenters on the ground

Physics
1 answer:
kramer3 years ago
6 0

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:

\Delta t'  = \Delta t \gamma

Making \gamma the subject of the formula, we have:

\gamma= \dfrac{\Delta t'}{\Delta t}

Number of day(s) = 1

1 day in seconds = 24 × 60 × 60

= 84600 seconds

\gamma= \dfrac{86400 \ s}{86400 \ s  - 4.5 \ ns}

\sqrt{1- \dfrac{v^2}{c^2}  }= \dfrac{86400 \ s  - 4.5 \ ns}{86400 \ s}

\sqrt{1- \dfrac{v^2}{c^2}  }= 1 - 5.4 \times 10^{-14}

From the above equation, if we apply binomial expansion to it, we have the following,

\sqrt{1- \dfrac{v^2}{c^2}  }=(1-\dfrac{v^2}{c^2})^{\frac{1}{2}}

⇒1 -\dfrac{1}{2}(\dfrac{v^2}{c^2})

So;

1 -\dfrac{1}{2}(\dfrac{v^2}{c^2}) = 1 - 5.4 \times 10^{-14}

\dfrac{1}{2}(\dfrac{v^2}{c^2}) =  5.4 \times 10^{-14}

\dfrac{v^2}{c^2} =  2 \times 5.4 \times 10^{-14}

\dfrac{v^2}{c^2} =  1.08 \times 10^{-13}

v^2 =  1.08 \times 10^{-13} \times c^2

where ;

c = 3 \times 10^8 \ m/s

v^2 =  1.08 \times 10^{-13} \times (3 \times 10^8  \ m/s)^2

v = \sqrt{ 1.08 \times 10^{-13} \times (3 \times 10^8  \ m/s)^2}

v = \sqrt{ 1.08 \times 10^{-13} } \times (3 \times 10^8  \ m/s)

v =3.286 \times 10^{-7} \times (3 \times 10^8  \ m/s)

\mathbf{v =98.58 \ m/s}

Therefore, the speed at which a moving clock lose 4,5ns in 1.0 day is 98.58 m/s

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