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skelet666 [1.2K]
3 years ago
8

Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i

t should be applied to the bar in combination with its surroundings (the lake). Assume that the entropy change of the bar is -1238 J/K . What is the change in total entropy ΔStotal?
Physics
1 answer:
max2010maxim [7]3 years ago
4 0

Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T  

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings

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A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend
Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2 = m·g·h₂ + \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + \dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2

5,061.96 J = 21.5 kg × v₂² + 53.\overline 3 kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0
2 years ago
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Bess [88]

Answer:

Your answer is A

Explanation:

7 0
2 years ago
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Marvin Martian is standing on planet Potatoine.
prisoha [69]

Answer:

W = 113.98 N

Explanation:

Given that,

Radius of PotatoineR=6.4\times 10^6\ m

Mass of Potatoine, M=2\times 10^{24}\ kg

Mass of Marvin, m = 35 kg

We need to find his weight on Potatoine. Weight of an object is given by :

W = mg

g is acceleration due to gravity, g=\dfrac{GM}{R^2}

So,

W=\dfrac{GMm}{R^2}\\\\W=\dfrac{6.67\times 10^{-11}\times 35\times 2\times 10^{24}}{(6.4\times 10^6)^2}\\\\W=113.98\ N

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The masses of blocks A and B are 20kg and 10kg, respectively. The blocks are initially at rest on the floor and are connecte by
Goryan [66]

Answer

mass of the block A = 20 Kg

mass of block B = 10 Kg

acceleration of Block A = ?

acceleration of Block B = ?

Assuming the magnitude of force = 124 N

Applying newton's second law

F = 2 T

T = F/2

now, Tension in the string =

T = 124/2 = 62 N

Weight of the block A

W = m₁ g

W = 20 x 9.8 = 196 N

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W = m₂ g

W = 10 x 9.8

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Weight of blocks is greater than force applied by the pulley so, the blocks will not move.

Hence, acceleration of the block A and B = 0 m/s²

If the Force magnitude is increased to 294 N

T = F/2 = 294/2 = 147 N

Since this Force is less than Weight A  acceleration of the block A = 0 m/s².

For Block B

Tension is more than Weight hence block will move

Net Force = 147 - 98 = 49 N

acceleration =\dfrac{F}{m}

              a = \dfrac{49}{10}

                    a = 4.9 m/s²

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3 years ago
Jeff wrote the following statement:
S_A_V [24]

Answer:

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Explanation:

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