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Shkiper50 [21]
2 years ago
11

Which option below best explains why the second law of thermodynamics is

Physics
1 answer:
Llana [10]2 years ago
7 0

''The freezer and room are not an isolated system, since electrical energy flows in.'' is the correct statement.

<h3>What is Second Law of Thermodynamics?</h3>

The Second Law of Thermodynamics says that "in all energy exchanges, if no energy enters or leaves the system, the potential energy of the state will always be less than the energy of the initial state."

So we can conclude that ''The freezer and room are not an isolated system, since electrical energy flows in.'' is the correct statement.

Learn more about law here: brainly.com/question/820417

#SPJ1

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Una tubería con secciones circulares se emplea para el traslado de agua entre el tanque central y un tanque auxiliar. El tanque
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A) 20 L/s

B) 2.55 m/s

C) 10.20 m/s

D) 400.8 kPa

Explanation:

a)

In this problem, we know that the volume of the tank:

V=72 m^3  

is filled in a time of

t = 1 h

We also know that the volume flow rate of water in the pipe is constant in every point of the pipe, so it can be calculated directly from the two data above.

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While the time, in seconds, is

t=1 h = 3600 s

Therefore, the volume flow rate in Liters per second is:

Q=\frac{V}{t}=\frac{72,000}{3600}=20 L/s

b)

The volume flow rate of water through the pipe can be also written as

Q=Av

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A is the cross-sectional area of the pipe

v is the speed of the water

In section B, we have:

r = 0.050 m is the radius of section B

so, the cross-sectional area of section B is:

A=\pi r^2 = \pi (0.050)^2=0.00785 m^2

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c)

As in part B), we know that the volume flow rate must remain constant through the entire pipe.

So, the volume flow rate in section A of the pipe is still

Q=0.02 m^3/s

The radius of the pipe in section A is

r=0.025 m

Therefore, the cross-sectional area in section A of the pipe is

A=\pi r^2 = \pi (0.025)^2=0.00196 m^2

So, since we have

Q=Av

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v=\frac{Q}{A}=\frac{0.02}{0.00196}=10.20 m/s

d)

Here we want to find the gauge pressure in section B.

We know that:

p_A = 2.0 atm = 105,000 Pa is the pressure in section A

h_A=15.0m is the altitude of section A

v_A=10.20 m/s is the speed of water in section A

v_B=2.55 m/s is the speed of water in section B

We can write Bernoulli's equation:

p_A + \rho g h_A + \frac{1}{2}\rho v_A^2 = p_B + \frac{1}{2}\rho v_B^2

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\rho=1000 kg/m^3 is the water density

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And solving for pB, we find:

p_B = p_A + \rho g h_A + \frac{1}{2}\rho (v_A^2-v_B^2)=\\=205,000+(1000)(9.8)(15.0)+\frac{1}{2}(1000)(10.20^2-2.55^2)=400,769 Pa

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p_B = 400.8 kPa

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