a) A bus of mass 760 kg requires 120 m to reach certain velocity value Vf. Ignore friction and drag forces and assume the bus en gine exerts a constant forward force F. When the bus is towing a 330-kg small car, how long distance needed to reach same Vf? b) If the Vf of the bus is 28 m/s, what is the tension in the tow cable between bus and small car?
1 answer:
Answer:
Given : A bus of mass 760 kg requires 120 m to reach certain velocity value Vf.
the bus engine exerts a constant forward force F.
To Find : When the bus is towing a 330-kg small car, how long distance needed to reach same Vf?
Solution:
V² - U² = 2aS
V = Vf
U = 0
S = 120 m
=> Vf² - 0 = 2a(120)
=> Vf² = 240a
m = 760 kg
Force = F
F = ma
=> F =760 a
=> a = F/760
Vf² = 240F/760
Case 2 :When the bus is towing a 330-kg small car,
m = 760 + 330 = 1090 kg
a = F/1090
Vf² = 2aS
=> 240F/760 = 2 (F/1090) S
=> S = 120 x 1090 /760
=> S = 172.1 m
172.1 m distance needed to reach same Vf
Explanation:
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Answer:
(a)
(b)
Explanation:
mass, m = 2.3 kg
vx = 40 m/s
vy = 75 m/s
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Answer:
8.61 min
Explanation:
original mass= 12.65
first half life = 12.65/2 = 6.325
second half life = 6.325/2 = 3.1625
Note : 3.1625 is the closest to the value (3.115) given so we work with it
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