According to the second law of thermodynamics, when energy changes occur, _______ increases. question 2 options:

1 answer:

7 0

The correct answer is entropy.

In fact, second law of thermodynamics states that the total entropy of an isolated system can never decrease. It can remain constant if the system is in equilibrium or in case of a reversible process: however, perfectly reversible process do not exist in the real world, so the entropy of a system always increases whenever there is a process that involves energy changes.

You might be interested in

Which of the following are basic solutions? Check all that apply. A. Laundry detergent B. Vinegar C. Drain cleaner D. Antacids

ratelena [41]

A laundry detergent
B vinegar

4 0

4 years ago

How can you tell there is only one atom of oxygen in H2O?

kow [346]

Answer:

For H2O, there is one atom of oxygen and two atoms of hydrogen. A molecule can be made of only one type of atom. In its stable molecular form, oxygen exists as two atoms and is written O2. to distinguish it from an atom of oxygen O, or ozone, a molecule of three oxygen atoms, O3.

Explanation:

4 0

3 years ago

* Check all of the things that can affect power Time Work Distance Force

gladu [14]

Answer:

fhddmvxmvydlyghclhchc

3 0

3 years ago

How is property belonging to individuals protected under the Fifth Amendment?

kotegsom [21]

Answer:

By preventing the government from taking property without fair payment.

Explanation:

Answer 1

Please mark me as brilliant and thank you

7 0

3 years ago

Read 2 more answers

I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

attashe74 [19]

The angular momentum of a particle in orbit is

l = m v r

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"

m_1 v_1 r_1 = m_2 v_2 r_2

Assuming that the mass did not change, conservation of angular momentum demands that

v_1 r_1 = v_2 r_2

or

v1 = v_2 (r_2/r_1)

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, the orbital speed of this material when it was 40,000 AU from the sun is 4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

3 0

3 years ago