Answer:
Estimation: year 1110.
Explanation:
We need to know how much time it takes to move 100 arcseconds if it moves at 0.11 arcsecond per year. Similarly to any velocity equation , where in our case the distances are angular, we will obtain the time by doing:
Which, considering from 2019, the explosion ought to have been observed around 1110 (in reality it was observed by Chinese astronomers in 1054).
Answer:Given mass = 4.1kg
Radius = 0.0117m
Velocity V = 8.4m/s
Coefficient of friction = 0.25
Explanation: Below is an attached solution to the problem stated above.
1. The angular acceleration is equal to 524rad/s^2
2. The linear acceleration is equal to 2.45m/s^2
3.the time it takes the ball to begin rolling = 0.98s
4. The distance the ball slides before it begins to roll = 7.05m
Answer:
When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.
Explanation:
As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.
Blueshifts happens when the source of the wave and the observer are moving closer towards one another.
Assume that the wave is of frequency at the source. In other words, the source of the wave sends out a peak after every
.
Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.
The source of this wave sends out a peak after each period of . It would appear to the observer that consecutive peaks arrive every
. That would correspond to a frequency of
.
On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of .
Again, the source of this wave would send out a peak after each period of . However, by the time the source sends out the second peak, the source would have been
closer to the observer then when the source sent out the first peak.
When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than . The observed frequency of this wave would be larger than the original
.