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Semenov [28]
3 years ago
14

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.

Physics
2 answers:
nikdorinn [45]3 years ago
7 0

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>
  1. Learn more about Crab Nebula brainly.com/question/11268956
  2. Learn more about the Chinese brainly.com/question/3913188
  3. 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

Finger [1]3 years ago
3 0
Light took 3500 years to reach Earth and reached in 1054 C.E. Therefore, the explosion took place in:
4,554 C.E
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5. If one object has a greater speed than a second object. does the first necessarily have a greater acceleration? Explain, usin
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Explanation:

5. not necessarily so that the first object could have left with initial velocity and the second not, so even if the second has a greater acceleration its velocity is less than that of the first

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     SI System Reductions

     Vo = 80 km / h (1000m / 1km) (1h / 3600s) = 22.2 m / s

     Vf = 90 km / h (1000m / 1km) (1h / 3600s) = 25 m / s

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     am = (Vf-Vo) / t

     am = (25 -22.2) / t = 2.8 / t

      am= 2.8/t

For the bike we have

      Vf = 10 km / h (1000m / 1km) (1h / 3600s) = 2.78 m / s

      Vo = 0

      ab = (Vf -Vo) / t

      ab = (2.78 -0) / t

      ab = 2.8/t

Since time is the same for both of us, if we round to Significant figures the two accelerations are equal

7. If when an object is slowing or slowing down.

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3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
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6 0
3 years ago
Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum
Alina [70]

Answer:

  • Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
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\longrightarrow \:  \:  \sf\Delta x .\Delta p =  \dfrac{h}{4\pi}

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} {4 \times  \frac{22}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} { \frac{88}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34} \times 7} { 8 }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 8  \times 24 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 192 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34}  \times  {10}^{15} } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 2}  \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 21}   } { 192}

\longrightarrow \:  \:  \sf\Delta p = 22.822\times  {10}^{ - 21}

\longrightarrow \:  \:  \sf\Delta p = 2.2822 \times  {10}^{1} \times  {10}^{ - 21}

\longrightarrow \:  \: \underline{ \boxed{ \red{  \bf\Delta p = 2.2822 \times  {10}^{ - 20}  \:  kg/ms}}}

4 0
2 years ago
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