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storchak [24]
3 years ago
12

Suppose we have two identical boxes

Physics
1 answer:
Marianna [84]3 years ago
8 0

Answer:

(C) The number of microstates of system 2 is twice as high as those of system 1.

Explanation:

In thermal contact, energy is exchanged between the two systems until thermal equilibrium is reached.

In thermal equilibrium, the temperature of system one is always equal to temperature of system two.

Also from second law of thermodynamic;

The entropy of an isolated system never decreases: in equilibrium, the entropy stays the same; otherwise the entropy increases until equilibrium is reached.

Since the two boxes are in thermal equilibrium, the entropy of system 1 is equal to the entropy of system 2.

However, a microstate is a specific microscopic configuration of a thermodynamic system that the system may occupy with a certain probability in the course of its thermal fluctuations.

Since system 2 = twice system 1

Then, microscopic configuration of system 2 is equal to twice microscopic configuration of system 1.

P₂ (Box(A+B))= P₁(2A), since the two boxes are identical.

Therefore, the number of microstates of system 2 is twice as high as those of system 1.

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A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth
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Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

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