標題:
此文章來自奇摩知識+如有不便請留言告知
F6 Phy
發問:
1Two forces p and q have a resultant force rwith a max. value of 60 N and a minimum value of 10NFind r if the angle (feta) between p and q is 40 degree.2 reading of voltmetre 1.30+-0.01V----------------ammeter 0.76+-0.01Alength of the wire 75.4+-0.2cmdiameter of the wire 0.54+-0.02... 顯示更多 1Two forces p and q have a resultant force r with a max. value of 60 N and a minimum value of 10N Find r if the angle (feta) between p and q is 40 degree. 2 reading of voltmetre 1.30+-0.01V ----------------ammeter 0.76+-0.01A length of the wire 75.4+-0.2cm diameter of the wire 0.54+-0.02 mm Cal. with the actual uncertainty, the value of a the resistance of the wire. b the resistivity of the metal of wire . If you have the Qs ,Please answer 3 3 2002 HKALE iiA Q45 MC (Phy)
最佳解答:
1. Max: p + q = 60 ... (1) Min: p - q = 10 ... (2) Solving (1) and (2), p = 35 N, q = 25 N By cosine law, r = sqrt[p2 + q2 - 2pqcos(180* - 40*)] = sqrt[(35)2 + (25)2 - 2(35)(25)cos140*] = 56.5 N 2. Resistance, R = V / I = 1.30 / 0.76 = 1.71 ohms │@R / R│ = │@V / V│+│@I / I│ @R / 1.71 = 0.01 / 1.30 + 0.01 / 0.76 @R = 0.04 ohms So, resistance = 1.71 +- 0.04 ohms b. By R = pl / A = pl /pi(d/2)2 p = pi(d/2)2R / l p = pi(0.54 X 10-3/2)2(1.71) / 0.754 Resistivity, p = 5.19 X 10-7 ohmm │@p / p│ = 2│@d / d│+│@R/R│+│@l / l│ @p / 5.19 X 10-7 = 2(0.02/0.54) + (0.04/1.71) + (0.2/75.4) @p = 5 X 10-8 ohmm Resistivity = 5.2 X 10-7 +- 0.5 X 10-7 ohmm 2002. by Mass = density X volume m = p(pi(d/2)2l) d = 2sqrt(m/plpi) @d / d = @m / m + @l / l @d / d = 4% + 2% @d / d = 6%
其他解答:
留言列表
