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sergij07 [2.7K]
3 years ago
14

A special vehicle of length 50 m is designed to take passengers at extremely high speeds between different places on earth.

Physics
1 answer:
olga nikolaevna [1]3 years ago
6 0

Answer:

a)

5.97\times 10^{7} ms⁻¹

b)

0.0021  sec

Explanation:

(a)

L_{o} = Actual length of the vehicle = 50 m

L = length measured by the earthbound observer = 49 m

v = speed of the vehicle

c = speed of light = 3 x 10⁸ ms⁻¹

Length measured by the earthbound observer is given as

L = L_{o}\sqrt{1 - \left ( \frac{v}{c} \right )^{2}}

49 = 50 \sqrt{1 - \left ( \frac{v}{3\times 10^{8}} \right )^{2}}

0.98 = \sqrt{1 - \left ( \frac{v}{3\times 10^{8}} \right )^{2}}

0.9604 = 1 - \left ( \frac{v}{3\times 10^{8}} \right )^{2}

0.396 = \left ( \frac{v}{3\times 10^{8}} \right )^{2}

v = 5.97\times 10^{7} ms⁻¹

b)

d = distance traveled = 6000 km = 6 x 10⁶ m

t = time measured by a clock on the vehicle

t' = time measured by a clock on the earth

Time measured by a clock on the vehicle  is given as

t = \frac{d}{v}

t = \frac{6\times 10^{6}}{5.97\times 10^{7}}

t = 0.1006 sec

Time measured by a clock on the earth is given as

t = t'\sqrt{1 - \left ( \frac{v}{c} \right )^{2}}

0.1006 = t'\sqrt{1 - \left ( \frac{5.97\times 10^{7}}{3\times 10^{8}} \right )^{2}}

t' = 0.1027 sec

\Delta t = time difference

Time difference is given as

\Delta t = t' - t

\Delta t = 0.1027 - 0.1006

\Delta t = 0.0021

\Delta t = 0.0021  sec

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