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3 years ago

A Herbig-Haro (HH) object is:_:_________.

1 answer:

5 0

Given what we know about Herbig-Haro (HH) objects, we can confirm that they are born from the collision between a jet from a star and clouds of interstellar matter.

<h3>What are Herbig-Haro (HH) objects?</h3>

To put it simply, a Herbig-Haro (HH) object is what we observe as a bright spot in distant space. Upon closer observation we can observe that these giant bright patches are not lone objects, but in fact immense groups of interstellar matter.

<h3>How Herbig-Haro (HH) objects are formed.</h3>

The formation of one of these interstellar groups requires an immense amount of energy. This energy come from the speed at which the materials that form these objects collide.
As stated in the question, the jet released from a star in the process of being born smashes into a giant cloud of dust and interstellar material to form the Herbig-Haro (HH) objects.
This collision happens at hundreds of km/s. To put it into perspective, it would be roughly 500,000 miles per hour.

Therefore, we can confirm that option b, which states, "where a jet from a star in the process of being born collides with (and lights up) a nearby cloud of interstellar matter", is the correct choice for the question pertaining to Herbig-Haro (HH) objects.

To learn more about interstellar space visit:

brainly.com/question/7106246

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How molecules react to an increase in temperature.

Slav-nsk [51]

Answer:

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Explanation:

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3 years ago

suppose that during the icy hot lab that 65 kj of energy were transferred to 450 g of water at 20 C. What would have been the fi

balandron [24]

Answer:

The final temperature of water is 54.5 °C.

Explanation:

Given data:

Energy transferred = 65 Kj

Mass of water = 450 g

Initial temperature = T1 = 20 °C

Final temperature= T2 = ?

Solution:

First of all we will convert the heat in Kj to joule.

1 Kj = 1000 j

65× 1000 = 65000 j

specific heat of water is 4.186 J /g. °C

Formula:

q = m × c × ΔT

ΔT = T2 - T1

Now we will put the values in Formula.

65000 j = 450 g × 4.186 J /g. °C × (T2 - 20°C )

65000 j = 1883.7 j /°C × (T2 - 20°C )

65000 j/ 1883.7 j /°C = T2 - 20°C

34.51 °C = T2 - 20°C

34.51 °C + 20 °C = T2

T2 = 54.5 °C

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4 years ago

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Answer:

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Explanation:

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3 years ago

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4 years ago

two friend on roller skate are standing 5m apart facing each other . one of them throw a ball of 2 kg toward the other ,who catc

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This question is about momentum conservation, which belong to physics subject.

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That means that the skater will step back and away from the other skater.

2) When the second skater receives the ball with momentum 2kg*v, he will move backward also and further away.

In conclusion, the two skaters will end standing more than 5 m apart facing each other.

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3 years ago