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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 = (vvo)/a
t = ((029)/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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This note was uploaded on 07/11/2011 for the course PHYS 225 taught by Professor Bacon during the Summer '09 term at Ferris State.
 Summer '09
 BACON

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