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4 years ago

5

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

r at 101. K. The rod and reservoirs make up a closed system. Of 7190. J are conducted from one end of the rod to the other uniformly (no change in temperature along the rod) what is the change in entropy of a. each reservoirb. the rodc. the system

1 answer:

Masteriza [31]4 years ago

3 0

Answer:

a) $S_1=-9.65}\ J/K$

b) $S_2=71.18\ J/K$

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

Negative sign because heat is leaving.

$S_1=-9.65}\ J/K$

b)

The entropy change of reservoir 101 K

$S_2=\dfrac{7190}{101}\ J/K$

$S_2=71.18\ J/K$

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

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Explanation:

It is given that,

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3 years ago

An object glides on a horizontal tabletop with a coefficient of kinetic friction of 0.5. If its initial velocity is 4.3 m/s, how

Shkiper50 [21]

Answer:

Time, t = 0.87 seconds

Explanation:

Given that,

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Using first equation of motion to find it as :

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3 years ago

Ray creates an energy transfer diagram for a hair dryer. However, the diagram contains an error that could be corrected in sever

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2 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

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Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

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Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

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The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

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$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

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$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

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Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

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3 years ago

there are two slides at the park between which you are deciding. Both start at the height of 6 meter. One is short and Steve whi

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It doesn't matter. If the slides are truly frictionless, then
your kinetic energy at the bottom will be equal to the
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The only way I can think of that it would make a difference
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and you didn't have anything to eat all the way down.
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top. Then, in order to have the same kinetic energy at the
bottom, you'd need to be going a little bit faster.

But if it takes less than, say, two or three days, to go down the
long, shallow slide, then this effect would probably be too small
to make any difference.

6 0

3 years ago