Respuesta :
Answer:
a)
[tex] 5.97\times 10^{7}[/tex] ms⁻¹
b)
[tex]0.0021 [/tex] sec
Explanation:
(a)
[tex]L_{o}[/tex] = Actual length of the vehicle = 50 m
[tex]L[/tex] = length measured by the earthbound observer = 49 m
[tex]v[/tex] = speed of the vehicle
[tex]c[/tex] = speed of light = 3 x 10⁸ ms⁻¹
Length measured by the earthbound observer is given as
[tex]L = L_{o}\sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]49 = 50 \sqrt{1 - \left ( \frac{v}{3\times 10^{8}} \right )^{2}}[/tex]
[tex]0.98 = \sqrt{1 - \left ( \frac{v}{3\times 10^{8}} \right )^{2}}[/tex]
[tex]0.9604 = 1 - \left ( \frac{v}{3\times 10^{8}} \right )^{2}[/tex]
[tex]0.396 = \left ( \frac{v}{3\times 10^{8}} \right )^{2}[/tex]
[tex]v = 5.97\times 10^{7}[/tex] ms⁻¹
b)
[tex]d[/tex] = distance traveled = 6000 km = 6 x 10⁶ m
[tex]t[/tex] = time measured by a clock on the vehicle
[tex]t'[/tex] = time measured by a clock on the earth
Time measured by a clock on the vehicle is given as
[tex]t = \frac{d}{v}[/tex]
[tex]t = \frac{6\times 10^{6}}{5.97\times 10^{7}}[/tex]
[tex]t = 0.1006[/tex] sec
Time measured by a clock on the earth is given as
[tex]t = t'\sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.1006 = t'\sqrt{1 - \left ( \frac{5.97\times 10^{7}}{3\times 10^{8}} \right )^{2}}[/tex]
[tex]t' = 0.1027[/tex] sec
[tex]\Delta t[/tex] = time difference
Time difference is given as
[tex]\Delta t = t' - t [/tex]
[tex]\Delta t = 0.1027 - 0.1006 [/tex]
[tex]\Delta t = 0.0021 [/tex]
[tex]\Delta t = 0.0021 [/tex] sec
