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School: Oglethorpe
Chapter 15 Solutions Physics 102 Dr. M. Rulison Wednesday, January 28 1.) A.) U = Q - W = 77 J - 164 J = -87.0 J B.) U env = -U sys = - ( -87.0 J ) = 87.0 J 5.) U env = Qenv - Wenv Wenv = -1.9 107 J Dist = 1.9 10 J 6 105 J/ mi 7 Wenv = Qenv - U
School: Oglethorpe
Chapter 3 Solutions Physics 101 Dr. M. Rulison September 7, 2006 3.) Ar = Ax 2 + Ay 2 = 2^r ^r = 2At At r 2 +r 2 As = At = At 1 4 19600m ? 11200m 2 + 4900m ? 3200m 2 = 8570.3m ^1.5010 11 m = 29863m/s u 29.9km/s 27.8910 6 s 2 1.510 11 m 2r = = 26886
School: Oglethorpe
Chapter 17 Solutions Physics 101 Dr. M. Rulison Monday, February 9 7.) V = v = 343m/s = 5.0m 68.6Hz f For destructive interference: Ax = n V ; n = 1, 3, 5, 7. 2 a Ax = 1 5.0m = 2.5m 2 So, BC = AC + 2.5m = 1.0m + 2.5m = 3.5m BC AC AB Applying the law
School: Oglethorpe
Chapter 16 Solutions Physics 102 Dr. M. Rulison Monday, February 25 11.) t = v net When with the waves: t = v skier wave -v When against the waves: t = v skier wave +v A.) Dividing (1) by (2): 0.60 = 0.50 v wave = 1.091m/s B.) Using this result i
School: Oglethorpe
Chapter 2 Solutions Physics 101 Dr. M. Rulison September 5, 2006 1.) A.) distance = > A x i = 6.9km + 1.8km + 3.7km = 12.4km B.) A x = > A x i = 6.9km ? 1.8km + 3.7km = 8.8km i i 7.) Time for tourist and bear to reach car should be the same. d = d+
School: Oglethorpe
Chapter 15 Solutions Physics 102 Dr. M. Rulison Wednesday, January 28 1.) A.) U = Q - W = 77 J - 164 J = -87.0 J B.) U env = -U sys = - ( -87.0 J ) = 87.0 J 5.) U env = Qenv - Wenv Wenv = -1.9 107 J Dist = 1.9 10 J 6 105 J/ mi 7 Wenv = Qenv - U
School: Oglethorpe
Chapter 3 Solutions Physics 101 Dr. M. Rulison September 7, 2006 3.) Ar = Ax 2 + Ay 2 = 2^r ^r = 2At At r 2 +r 2 As = At = At 1 4 19600m ? 11200m 2 + 4900m ? 3200m 2 = 8570.3m ^1.5010 11 m = 29863m/s u 29.9km/s 27.8910 6 s 2 1.510 11 m 2r = = 26886
School: Oglethorpe
Chapter 17 Solutions Physics 101 Dr. M. Rulison Monday, February 9 7.) V = v = 343m/s = 5.0m 68.6Hz f For destructive interference: Ax = n V ; n = 1, 3, 5, 7. 2 a Ax = 1 5.0m = 2.5m 2 So, BC = AC + 2.5m = 1.0m + 2.5m = 3.5m BC AC AB Applying the law
School: Oglethorpe
Chapter 16 Solutions Physics 102 Dr. M. Rulison Monday, February 25 11.) t = v net When with the waves: t = v skier wave -v When against the waves: t = v skier wave +v A.) Dividing (1) by (2): 0.60 = 0.50 v wave = 1.091m/s B.) Using this result i
School: Oglethorpe
Chapter 2 Solutions Physics 101 Dr. M. Rulison September 5, 2006 1.) A.) distance = > A x i = 6.9km + 1.8km + 3.7km = 12.4km B.) A x = > A x i = 6.9km ? 1.8km + 3.7km = 8.8km i i 7.) Time for tourist and bear to reach car should be the same. d = d+