Physics 211 Exam 1 typed

Physics 211 Exam 1 typed - Physics 211 Exam 1 1 Part A...

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Unformatted text preview: Physics 211: Exam 1 1. Part A: Answer: 125.4 m Part B: Answer: 2. Part A: Answer : 47.2 inches 3. Part A: F= Answer: Part B: = = Answer: 4. Part A: = Answer: Lost about $53, 131.26 Part B: 1 slug = 14.59 kg = 3.2808398 m 15.0 N = 15.0 = 15.0 x 1kg x 1m x 15.0 x 0.068540096 slug x 3.280839895 ft x Answer: 3.37 lb 5. Part A: a = (Vf – Vi) / time a = (4.1 m/s – 0)/0.11s = 3.373036218 lb a = 37.3 Answer: 37.3 Part B: (2 x 37.3m/ ) x = 16.81/74.6 x = 0.225m Answer: 0.23m 6. Part A: = 2a = (0-(20m/ ) / (2(-4.0m/ )) = 50.0m Answer: 50.0m 7. Part A: ½ b x h + L x w ½(8)(2) + 2 x 4 = 16m Answer: 16m Part B: ½ bh ½ (3)(12) = 18m Answer: 18m Part C: ½ bh ½ (1)(-4) = -2m Answer: -2m Part D: Answer: Yes Part E: Answer: At a time of 3seconds. 8. Part A: Distance traveled during time to react: x = vt x = (16m/s)(0.5s) x = 8m Time it takes to stop: Vf = Vi + at 0 = 16 + (-12)t t = 1.33s Distance traveled when stopping: x = xo + vt + ½ x = 8 + (16)(1.33) + ½ (-12)( x = 18.67 m 35m – 18.67m = 16.33m Answer: 16.33m between you and the deer. Part B: = + 2ax 0= + 2(-12)(35) Vi = 28.98m/s Answer: 28.98m/s 9. Part A: Answer: 33.7m/s Part B: V = Vo + at t = (v-vo)/a t = ((0-29)/-9.8) + ((33.7 -0)/9.8) t = 2.96 + 3.44 t = 6.40s Answer: 6.4s 10. Work for both parts A and B combined: X = Vot + ½ x = (0)15 + ½ (1.40)( x = 157.5m Final speed of train: = V(2(1.40)(157.5)) = 441 V = 21m/s Distance traveled: t = x/v t = 1100/21 11. 12. 13. 14. t = 52.38s x = 2ax x=( / (2)(2.20) x = 100.227m Vf = Vi + at 0 = 21 + -2.2t t = 9.545s Part A: Added up all distances found to get the answer 157.5m + 1100m + 100.227m = 1357.727 Answer: 1357.73m Part B: Added up all the time found. 15s + 52.38s + 9.545s = 76.925s Answer: 76.93s Part A: Acceleration is equal to g. Any object in freefall will always have an acceleration with a magnitude that is equal to gravity(g). Part B: Acceleration is equal to g and always will be. The only way that this might change will be if terminal velocity is reached. Part A: t^2 = 2x/a t^2 = 2(62)/9.8 t^2 = 12.65 t = 3.56s Answer: 3.56s Part B: vt = x x = (31)3.56 x = 110.36m Answer: 110.36m Part A: V cos(degree) 100 cos(25) Velocity = 90.6m/s Part A: 9.8 sin 29* = 4.751m Time: 15. 16. 17. 18. 4.751 = ½ 4.751 = ½(4.751) =2 t = 1.41s Answer: 1.41s Part A: x=½ = 2x/a = 2(0.025)/9.8 t = 0.0714s Answer: 0.0714s Part B: x = vt 56 = v(0.0714) V = 784.3m/s Answer: 784.3m/s Part A: X = 1/2 110 = ½ (9.8) t = 4.74s x = vt x = (4.74)(100) x = 474m Answer: 474m Part A: magnitude = √ + 10 = √ + + 36 = 100 = 64 y = 8m Answer: 8m Part A: E = 2A + 3B Ex = 2Ax + 3Bx = 2(5) + 3(-3) = 1 Answer: 1x Part B: Ey = 2Ay + 3By = 2(2) +3(-5) = -11 Answer: -11y Part C: E = √( + ) = √ (( + Answer: 11.05 Part D: = 11.05 Tan ɵ = (Ey/Ex) Tan ɵ = -11/1 =-11 ɵ = -84.81 degrees Answer: -84.81 degrees ...
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