The astronomical object known as Crab Nebula is thhe remnant of an exploded star. The explosion was seen by the Chinese in 1054
C.E. However, the Crab Nebula is about 3500 LY distance from the Earth In what year did the star actually explode.
2 answers:
The star, Crab Nebula, actually explode in the year 1054.
<h3>Further explanation </h3>The Crab Nebula is the scattered fragments of a supernova, or exploding star, observed by earthly skywatchers in the year 1054. The Crab Nebula is about 6,500 light years from Earth.
The astronomical object known as Crab Nebula is the remnant of an exploded star. The explosion was seen by the Chinese in 1054 C.E. However, the Crab Nebula is about 3500 LY distance from the Earth.
The Crab Nebula is one of the most studied remains of a stellar explosion. The Crab Nebula is widely accepted to be due to a supernova seen in the year 1054 a.d. by Chinese, Japanese, Korean, and Arab astronomers who reported sighting a new bright star in the heavens. The current name "The Crab Nebula" is due to William Parsons, 3rd Earl of Rosse, who observed the object in 1840 using a 36-inch telescope and produced a drawing that looked somewhat like a crab.
<h3>Learn more </h3>- Learn more about Crab Nebula brainly.com/question/11268956
- Learn more about the Chinese brainly.com/question/3913188
- Learn more about The astronomical object brainly.com/question/7725232
<h3>Answer details</h3>
Grade: 9
Subject: physics
Chapter: The astronomical object
Keywords: Crab Nebula, the Chinese, distance, Earth, The astronomical object
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(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².
(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².
(1.c) The relative intensity of the sound as heard by the listener is 103 dB.
(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.
(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.
<h3>Surface area being vibrated</h3>The surface area being vibrated by the time the sound reaches the listener is calculated as follows;
A = 4πr²
A = 4π x (20)²
A = 5,026.55 m²
<h3>Intensity of the sound</h3>The intensity of the sound is calculated as follows;
I = P/A
I = (100) / (5,026.55)
I = 0.02 W/m²
<h3>Relative intensity of the sound</h3>The frequency of the sound heard is determined by applying Doppler effect.
where;
- -v₀ is velocity of the observer moving away from the source
- -vs is the velocity of the source moving towards the observer
- fs is the source frequency
- fo is the observed frequency
- v is speed of sound
Learn more about intensity of sound here: brainly.com/question/17062836