Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance

Where, d = distance
A = amplitude
Put the value into the formula


Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Answer:
The answer is 0.5 Hz
Explanation:
Its pretty easy to get the answer. One hertz (Hz) is equal to one cycle or period per second. So, just divide the period by the number of seconds.
1 period/2 secs = 1/2 Hz or 0.5 Hz
Answer:
(D) energy from one place to another
It is talking about how from a different perspective things look different.
That picture should help.
Answer:
- The separation will be spacelike.
- The first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.
Explanation:
We can define the separation between two events (using the + - - - signature) as :

where the separation will be lightlike if is equal to zero, timelike if is positive and spacelike if is negative.
For our problem








So the separation will be spacelike, and the first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.