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Fundamentals of Physics I最新文献

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Waves I 波我
Pub Date : 1991-11-01 DOI: 10.2307/j.ctvmd85m7.22
Jeff Duntemann
Page 1 of 3 http://edugen.wileyplus.com/edugen/courses/crs7165/halliday9781118230...Xk5NzgxMTE4MjMwNzI1YzE2LXNlYy0wMDQ0Lnhmb3Jt.enc?course=crs7165&id=ref Print this page Review & Summary Transverse and Longitudinal Waves Mechanical waves can exist only in material media and are governed by Newton's laws. Transverse mechanical waves, like those on a stretched string, are waves in which the particles of the medium oscillate perpendicular to the wave's direction of travel. Waves in which the particles of the medium oscillate parallel to the wave's direction of travel are longitudinal waves.
第1页/ 3页http://edugen.wileyplus.com/edugen/courses/crs7165/halliday9781118230...Xk5NzgxMTE4MjMwNzI1YzE2LXNlYy0wMDQ0Lnhmb3Jt.enc?course=crs7165&id=ref打印本页回顾与总结横波和纵波机械波只能存在于物质介质中,并受牛顿定律支配。横向机械波,就像拉伸的弦上的那些,是介质的粒子垂直于波的传播方向振荡的波。介质中的粒子与波的传播方向平行振荡的波称为纵波。
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引用次数: 12
Thermodynamics II
Pub Date : 1900-01-01 DOI: 10.2307/j.ctvmd85m7.27
Mandeep Dalal
CHAPTER 6 Thermodynamics – II  Clausius-Clapeyron Equation The Clausius-Clapeyron equation was initially proposed by a German physics Rudolf Clausius in 1834 and then further developed by French physicist Benoît Clapeyron in 1850. This equation is extremely useful in characterizing a discontinuous phase transition between two phases of a single constituent.  Derivation of Clausius-Clapeyron Equation In order to derive the Clausius-Clapeyron equation, consider a system at equilibrium i.e. the free energy change for the ongoing process is zero (ΔG = 0). However, we know from the principles of thermodynamics that the variation of free energy with temperature and pressure can be formulated by the following differential equation.
Clausius-Clapeyron方程Clausius-Clapeyron方程最初是由德国物理学家Rudolf Clausius在1834年提出的,然后由法国物理学家beno t Clapeyron在1850年进一步发展。这个方程在描述单一组分的两相之间的不连续相变时非常有用。为了推导Clausius-Clapeyron方程,考虑一个处于平衡状态的系统,即正在进行的过程的自由能变化为零(ΔG = 0)。然而,我们从热力学原理知道,自由能随温度和压力的变化可以用以下微分方程表示。
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引用次数: 0
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Fundamentals of Physics I
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