Question.3649 - Homework Assignment for Exam 1 PHYS 100 Chapter 5 Problems due 9/30/24 PROBLEMS Do the following under the Exercise section: Chapter 5 Problems: 1, 2, 5, 8, 10, 11, 13, 14, 15, SP5
Answer Below:
Chapter 5 1) v = 4m/s r = 0.8 m ac = v2/r = 42/0.8 = 20 m/s2. 2) v = 18 m/s r = 40 m ac = v2/r = 324/40 = 8.1 m/s2. 5) m = 0.35 kg ac = 5 m/s2. F = m x a = 1.75 N 8) a) r = 8 m v = 4.5 m/s ac = v2 / r = 20.25/8 = 2.2312 m/s2. b) m = 75 kg F = m x a F = 74 x 2.53 = 189.75 N 10) W1 = 800 N F2 ?= F1 ?(r2 / ?r1??)2 r2 / ?r1 = 3 F2 = 800 x (1/3)2 = 88.89 N 11) F1 = 9.6 N F2 ?= F1 ?(r2 / ?r1??)2 r2 / ?r1 = 4 F2 = 9.6 x (1/ 16) = 0.6 N 13) F1 = 0.28 N r2 / ?r1 = 1/2 F2 ?= F1 ?(r2 / ?r1??)2 = 0.28 x (2/1)2 = 1.12 N 14) we = 270 lb Assuming the gravity on the moon to be 1/6th of that of Earths so Wm = we / 6 = 270/6 = 45lb. 15) we = 150 lb assuming gravity on Jupiter to be 25 m/s2 wj = we x gj / ge = 150 x (25/9.8) = 382.6530 lbSp5) a) Considering mass of Sun to be Ms = 1.99 x 1030 kg Considering mass of Earth to be Me = 5.98 x 1024 kg The distance between Sun and Earth can be r = 1.50 x 1011 m While gravitational constant is 6.674?10?11 Nm2/kg2 F = G (MS?ME/ r2) = 6.674?10?11 x ((1.99 x 1030) x (5.98 x 1024) / (1.50 x 1011)) = 3.54 x 1022 N b) in this case r = 3.82 x 108 m and Mm = 7.36 x 1022 kg F = G (MMME/ r2) = 6.674?10?11 x ((7.36 x 1022) x (5.98 x 1024) / (3.82 x 108)) = 1.98 x 1020 N c) wherein ratio is Fsun/Fmoon = 3.54 x 1022 / 1.98 x 1020 = 179 so F = 6.674?10?11 x ((1.99 x 1030) x (7.36 x 1022) / (1.50 x 1011)) = 4.35 x 1020 NMore Articles From Science
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