ch19-p041 - 41. (a) The distribution function gives the...

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() 0 0 2 23 3 3 2 00 0 0 71 4 8. 33 9 v v aa av d v v v v v =− = = ³ Thus, 22 rms rms 0 0 0 0 11 4 1.31 1.31 . 69 v vv v v v =+ = ¡ = 41. (a) The distribution function gives the fraction of particles with speeds between v and v + dv , so its integral over all speeds is unity: ³ P ( v ) dv = 1. Evaluate the integral by calculating the area under the curve in Fig. 19-24. The area of the triangular portion is half the product of the base and altitude, or 1 0 2 av . The area of the rectangular portion is the product of the sides, or av 0 . Thus, 0 13 P v dv av av av = ³ , so 3 0 2 1 av = and av 0 = 2/3=0.67. (b) The average speed is given by avg . P v d v = ³ For the triangular portion of the distribution P ( v ) = av / v 0 , and the contribution of this portion is 0 2 0 0 2 , 9 v av vdv v v == = ³ where 2/3 v 0 was substituted for a . P ( v ) = a in the rectangular portion, and the contribution of this portion is 0 0 2 2 0 0 3 4. v v d v v v = = ³ Therefore, avg avg 0 0 0 0 2 1.22 1.22 9 v v v v = ¡ = . (c) The mean-square speed is given by
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ch19-p041 - 41. (a) The distribution function gives the...

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