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Kitty [74]
3 years ago
6

Which of the following are true about a "simple compressible system"? It cannot be a mixture of different substances (e.g. oxyge

n and nitrogent) It's only relevant reversible work mode is associated with expansion or compression It may not have any heat transfer to or from the system Magnetization is a relevant reversible work mode It can be composed of any phases of a substance: solid, liquid, and/or gas It's state is specified if given two independent, intensive thermodynamic properties
Physics
1 answer:
fomenos3 years ago
4 0

Explanation:

The volume of a simple compressible system is not fixed. At a state of equilibrium, there should be uniformity in the entire system.

From the question we have here, these are the correct options:

1. It cannot be a mixture of different substances (e.g. oxygen and nitrogent)

2. It can be composed of any phases of a substance: solid, liquid, and/or gas

3. It's state is specified if given two independent, intensive thermodynamic properties.

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2 years ago
A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a
pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

120^2=130^2+50^2-2\times130\times50\cos\alpha

\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}

\alpha=\cos^{-1}(0.3846)

\alpha=67.38^{\circ}

We need to calculate the angle β

Using cosine law

50^2=130^2+120^2-2\times130\times120\cos\beta

\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}

\beta=\cos^{-1}(0.923)

\beta=22.63^{\circ}

We need to calculate the force on 130 cm side

Using formula of force

F_{130}=ILB\sin\theta

F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0

F_{130}=0

We need to calculate the force on 120 cm side

Using formula of force

F_{120}=ILB\sin\beta

F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63

F_{120}=0.1385\ N

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

F_{50}=ILB\sin\alpha

F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38

F_{50}=0.1385\ N

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

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Answer:

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3 years ago
Read 2 more answers
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