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velikii [3]
4 years ago
6

Using the statistical definition of entropy, what is the entropy of a system where W = 4?

Physics
1 answer:
Oliga [24]4 years ago
5 0
     <span> S = k ln(W)
= 1.38*10^-23 J/K * ln(4)
= 1.91*10^-23 J/K</span>
so it will be c
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3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
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  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

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vova2212 [387]

Answer:

6

Explanation:

0.5(3 × 2^2)

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