Question

Calculate the velocity watch when it seems to be slows down by 1min in one hour

Answer 1

Answer:

hope it helpful for u......

Answer 2

Concept Introduction:-

The displacement of an item in unit time is referred to velocity.

Given Information:-

We have been given that velocity watch when it seems to be slows down by $1$ min in one hour.

To Find:-

We have to find that the velocity.

Solution:-

According to the problem

Fhom LORENTZ TIME DILATION EQN

$T_{\text {apanas }}=\frac{T_{a}}{\sqrt{1-\frac{y^{2}}{c^{2}}}}$

Where $T_{0}=$ Actual Fiome

Tapbeonat Apparat Itime.

Now. According $\pm A$ the Question

Teppercust $3600 \mathrm{sec}+60 \mathrm{sec}=3.660 \mathrm{sec}$

and $T_{a}=3100 \mathrm{Sec}$

Applying it into the formula.

$T_{\text {efparent }}=\frac{T_{e}}{\sqrt{1-\frac{x^{3}}{c^{2}}}}\\ \Rightarrow 3660=\frac{3600}{\sqrt{\frac{v^{2}}{c t}}}$

$\Rightarrow \sqrt{1-\frac{x^{2}}{c^{2}}}=\frac{36 a^{2}}{346 x}$

$\sqrt{1-\frac{y^{2}}{c^{2}}}=0.98 y$

$1-\frac{x^{2}}{c^{2}} \cdot(\text { ais } 84)^{2}$

$\Rightarrow \frac{y^{2}}{c^{2}}=1-(0.984)^{4}$

$V=0.175 \mathrm{c}=18 \%$

Final Answer:-

The speed of light is $18\%$.