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insens350 [35]
3 years ago
15

Based on the second law of thermodynamics, how would you expect a system to change over time?

Physics
2 answers:
deff fn [24]3 years ago
5 0

B. It's randomness would increase

Because the Second Law of Thermodynamics states that as energy is transferred or transformed, more and more of it is wasted. It also states that there is a natural tendency of any isolated system to degenerate into a more disordered state.

jenyasd209 [6]3 years ago
5 0

Answer:

B. Its randomness would increase.

Explanation:

Thermodynamics is referred as the study of heat and energy. It explains the movement of energy within and outside any system.

The second law of thermodynamics explains about the inefficiency and decay of a system. Amount of work done will increase the amount of energy released which is irreversible hence the feasibility and efficiency will keep on increasing.

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To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
Lesechka [4]

Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

3 0
2 years ago
Electric Circuits Homework
nirvana33 [79]
<h2>Sorry, But I don't know!!</h2>
3 0
2 years ago
Giving brainly, please help
brilliants [131]

Answer:

when the mug is heated thus its temperature rises increasing the kinetic energy of the molecules , the oscillations around the rest position in the mug(solid) increases which increase the spaces between molecules and the mug expands. what cause the cracking is that outside of the mug expands before the inside and the mug cracks

3 0
2 years ago
When you and a friend move a couch to another room you exert a force of 75 N over 5m how much work will you do
sesenic [268]

The work done is 375 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this problem,

F = 75 N is the force applied to the couch

d = 5 m is the displacement

Assuming the force applied to the couch is parallel to the motion, \theta=0

And so, the work done is

W=(75)(5)(cos 0)=375 J

Learn more about work:

brainly.com/question/6763771

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6 0
3 years ago
Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th
choli [55]

Answer:

1.368\times 10^{-20}\ J

Explanation:

q = Charge in the potassium ion = 19e-18e

e = Charge of electron = 1.6\times 10^{-19}\ C

V_2-V_1 = Change in potential = 0-(-85.5\times 10^{-3})

Change in electric potential is given by

E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J

The energy is 1.368\times 10^{-20}\ J

3 0
3 years ago
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