91Ó°ÊÓ

In this problem, we're given: - Time elapsed between the two events for the Earth observer: \(t_e = 50.0 \,\text{ns}\) - Length of the spaceship according to the astronaut: \(L_0 = 12.0 \,\text{m}\)

According to an Earth observer, the spaceship travels \(L_0\) in the time \(t_e\). We can use the formula for speed: \(v = \frac{d}{t}\) Rearranging for the speed of the spaceship: \(v = \frac{L_0}{t_e}\) Plug in the given values: \(v = \frac{12.0 \,\text{m}}{50.0 \times 10^{-9} \,\text{s}} = 240,000,000 \,\text{m/s}\) The speed of the spaceship according to an Earth observer is \(240,000,000 \,\text{m/s}\).

Now, we need to find the elapsed time between light flashes in the astronaut's frame of reference. We can use the time dilation formula: \(t_0 = \frac{t_e}{\sqrt{1 - \frac{v^2}{c^2}}}\) Where \(t_0\) is the elapsed time in the astronaut's frame of reference, \(t_e\) is the elapsed time in the Earth observer's frame of reference, \(v\) is the speed of the spaceship, and \(c\) is the speed of light (\(3\times10^8 \,\text{m/s}\)). Plug in the given values: \(t_0 = \frac{50.0 \times 10^{-9}\,\text{s}}{\sqrt{1 - \frac{(240,000,000\, \text{m/s})^2}{(3\times10^8\, \text{m/s})^2}}}\) Calculate the value of the time: \(t_0 \approx 86.6 \,\text{ns}\) The elapsed time between light flashes in the astronaut's frame of reference is approximately \(86.6\, \text{ns}\).