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oee [108]
3 years ago
14

Venn diagrams are used for comparing and contrasting topics. The overlapping sections show characteristics that the topics have

in common, and the sections that are unique to each topic show the characteristics that apply to only that topic. This Venn diagram is comparing mechanical and electromagnetic waves. 2 circles with an overlap. The left circled is labeled Mechanical Wave with further labels Carry Energy Through Matter and Examples: Ocean Waves and Sound Waves. The right circle is labeled Electromagnetic Wave with further labels Carry Energy Through Space and Examples: Light and X-ray. What could be put into the center section of this Venn diagram? “Carry energy through space” and “Follow a pattern” “Carry energy” and “Follow a pattern” “Carry energy through space” and “Can be heard” “Carry energy” and “Can be heard”
Physics
2 answers:
Veronika [31]3 years ago
6 0

Answer:

carry energy follow pattern

Explanation:

tangare [24]3 years ago
5 0

Answer:

Carry energy and follow a pattern

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Suatu gas memiliki volume 2 m³ dipanaskan dengan kondisi isobarik sehingga volumenya menjadi 5,5 m³ jika tekanan 4 atmosfer hitu
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3 years ago
Which of the following involve an increase in the entropy of the system under consideration? a. melting of a solid b. evaporatio
Mama L [17]

Explanation:

Since, entropy is the measure of degree of randomness. So, more randomly the molecules of a substance are moving more will be its entropy.

  • For example, when a solid melts then it means heat is absorbed by it due to which its molecules have gained energy. As a result, they collide with each other and hence, entropy will increase.
  • Evaporation of a liquid will also cause the liquid to change its state from liquid to gas. This means molecules will go far away from each other leading to an increase in the entropy.
  • Sublimation is a process of conversion of a solid into gaseous phase without going through liquid phase. So, in this case also entropy will increase due to gain in energy by the molecules of a solid.
  • In freezing, molecules of a substance come closer to each other and acquire less energy. Hence, entropy decreases.
  • Mixing is a process of combining two or more substances physically with each other. This leads to increase in entropy of a substance.
  • In separation molecules are separated from each other leading to a decrease in energy. Hence, entropy will also decrease.
  • Diffusion is a process in which molecules are able to rapidly move from one place to another. Hence, entropy increases when diffusion takes place.

Thus, we can conclude that melting of a solid, evaporation of a liquid, sublimation, mixing and diffusion involve an increase in the entropy of the system under consideration.

8 0
3 years ago
A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
Stells [14]

Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

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Explanation:

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