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Zinaida [17]
2 years ago
12

Which transition represents a time when water molecules are moving closer together?

Physics
2 answers:
oee [108]2 years ago
4 0
The best transition between the four options presented to represent a time when water molecules are moving closer together would be A. Frost forms on a window pane.

The closest distance that the water molecules can do is when the water is in the state of being solid. It is known that the solid state of matter has the closest distance from molecule to molecule that when a molecule tries to move, the others move as well creating a vibration and thus producing heat in the process. When they are in a liquid state, they are quite far from each other. In a gas state, they really are far from each. This explains the difference in their characteristics.
Marta_Voda [28]2 years ago
4 0

A. Frost forms on a window pane.



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Correct answer choices are :

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D) Nitrogen


Explanation:


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2 years ago
Given the diagram below determine the net force acting on the object
crimeas [40]

A) 0

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3 years ago
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When a scientist contrasts two or more objects what is he or she looking for?
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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
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