Fighter jet starting from airbase A flies 300 km east , then 350 km at 30° west of north and then 150km north to arrive finally

Answer:

According to the guideline for kilometers, every three seconds between a lightning strike and the subsequent thunder indicates the distance to the flash in kilometers.

Explanation:

To calculate the speed of sound in meters per second, we need to utilize certain conversion factors. One mile corresponds to 5 seconds after witnessing the lightning. Furthermore, 1 mile comprises 5280 feet, and 1 foot is equivalent to 0.3048 meters. This information is sufficient to solve the issue. The conversion ratios can be set up like this:

Observe how the ratios are organized such that the units cancel out during calculations. One ratio has miles in the numerator while the other has them in the denominator, leading to cancellation. The same applies to the feet.

The question requires us to provide the answer to one significant figure, resulting in the speed of sound rounding to 300m/s.

For the second part, we will again utilize conversions. This time we will set our ratios in reverse and realize that there are 1000 meters in 1 kilometer, leading us to:

This signifies that for every 3.11 seconds, the distance to the lightning strike is 1 kilometer. Since this is a fabric of general knowledge, we round to the nearest whole number for simplicity, establishing the guideline:

According to the rule for kilometers, every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Response:

1) An observer in B 'perceives the two events occurring at the same time

2) Observer B recognizes that the events happen at different times

3)  Δt = Δt₀ /√ (1 + v²/c²)

Clarification:

This scenario illustrates the concept of simultaneity in special relativity. It is important to keep in mind that light's speed remains constant across all inertial frames

1) Since the events are stationary within the frame S ', they propagate at the constant speed of light, resulting in them reaching observation point B'—located equidistantly between both events—simultaneously

Thus, an observer in B 'observes the two events occurring at the same time

2) For an observer B situated within frame S attached to the Earth, both events at A and B appear to take place at the same moment. However, the event at A covers a shorter distance, while the event at B travels a longer distance, since frame S 'is in motion at velocity + v. Hence, with a constant speed, the event covering the lesser distance is perceived first.

Consequently, observer B perceives that the events do not occur simultaneously

3) Let's determine the timing for each event

        Δt = Δt₀ /√ (1 + v²/c²)

where t₀ represents the time in the S' frame, which remains at rest for the events