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Anna11 [10]
3 years ago
12

Which statement describes the relationship between energy and entropy in the universe?

Physics
2 answers:
rusak2 [61]3 years ago
8 0

In fact, entropy of an isolated system never decreases (2nd law of thermodynamics), unless some external energy is provided in order to "restore" order in the system and decrease its entropy.

(note that when external energy is added to the system, it is no longer "isolated").


*This is only true if the question is referring to a certain system within the universe. If we are considering the universe itself as the system, then this option is no longer correct, because no external energy can be provided to the universe, and since the universe is an isolated system, its entropy can never decrease. If we are considering the universe itself as the system, none of the options is true.




netineya [11]3 years ago
8 0

Answer:

<h2>Energy must be created to increase entropy.</h2>

Explanation:

The choices for this quetion are

  • Energy must be destroyed for entropy to be constant.
  • Energy must be decreased for entropy to be constant.
  • Energy must be added to decrease entropy.
  • Energy must be created to increase entropy.

According to physics,<em> entropy can be defined as the measurement of randomness in a system. </em>It depends of the observations at the moment, when we talk about entropy, the time past before to reach the current state, doesn't matter. With entropy, all that matters is the now.

Now, energy is the potential to do some work, that is, to get movement in a system. That meas, to molecules move, we need to add energy, that's how it works. So, in thermodynamics, when we increase the temperature of a system, the randomness also increse, that is, the entropy increase, because the increase of temperature is related to adding energy. So, basically, more energy increases entropy, that's why entropy is always in one direction, from a current state to another more random state, more disorder.

So, when if we say that we add energy to decrease entropy, that's wrong, because the energy increases the degree of random activity of the particles involved.

Therefore, the right answer is the last choice: Energy must be created to increase entropy.

However, this true statement must be rewritten, because according to thermodynamics, energy cannot be created or destroyed, just can be transformed. So, we are talking the last choice as energy transformation.

It's important to say that other options aren't true because, as we said before, entropy doesn't decrease, and it doesn't remain constant.

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The moon is 3.85 x 10 to the 8 m from earth and has a diameter of 3.48 x 10 to the 6 m. You have a pea (diameter = 0.50 cm) and
polet [3.4K]

Answer:

(a) dime

Explanation:

Convert all to metric unit:

0.5 cm = 0.005 m

1.8 cm = 0.018 m

71 cm = 0.71 m

In order to find out we would need to calculate the ratio R between the object diameter d and their distance s to our eyes:

R_m = \frac{d_m}{s_m} = \frac{3.48*10^6}{3.85*10^8} \approx 0.009

R_p = \frac{d_p}{s_p} = \frac{0.005}{0.71} \approx 0.007

R_d = \frac{d_d}{s_d} = \frac{0.018}{0.71} \approx 0.0253

Since the ratio of the dime is larger than the ratio of the moon, and the ratio of the pea is smaller than the ratio of the moon, only the (a) dime can cover your view of the moon.

3 0
3 years ago
3 small identical balls have charges -3×10^-12 ,8×10^-12 and 4×10^-12 find the charge​
Norma-Jean [14]

Answer:

  3×10^-12 C  

Explanation:

The total of the three charges is ...

   (-3 +8 +4)×10^-12 C = 9×10^-12 C

Assuming the charge is equally distributed between the balls when they are brought in contact, the charge on each ball will be ...

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c

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As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

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