Answer:
(a) a = - 201.8 m/s²
(b) s = 197.77 m
Explanation:
(a)
The acceleration can be found by using 1st equation of motion:
Vf = Vi + at
a = (Vf - Vi)/t
where,
a = acceleration = ?
Vf = Final Velocity = 0 m/s (Since it is finally brought to rest)
Vi = Initial Velocity = (632 mi/h)(1609.34 m/ 1 mi)(1 h/ 3600 s) = 282.53 m/s
t = time = 1.4 s
Therefore,
a = (0 m/s - 282.53 m/s)/1.4 s
<u>a = - 201.8 m/s²</u>
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(b)
For the distance traveled, we can use 2nd equation of motion:
s = Vi t + (0.5)at²
where,
s = distance traveled = ?
Therefore,
s = (282.53 m/s)(1.4 s) + (0.5)(- 201.8 m/s²)(1.4 s)²
s = 395.54 m - 197.77 m
<u>s = 197.77 m</u>
Answer:
The angular momentum of the system is 0.18kgm²/s
Explanation:
L = angular momentum of the particle 1 + angular momentum of the particle 2
L = ml²ω + ml²ω
L = 2ml²ω
Given that,
m = 500g are connected by a massless cord
=0.500kg
l = 30cm
= 0.30m
angular speed ω = 2.0 rev/s
angular momentum of the system =
L = 2ml²ω
L = 2 × 0.500 × (0.30)² × 2
L = 2 × 0.500 × 0.09 × 2
L = 0.18kgm²/s
Thus, the angular momentum of the system is 0.18kgm²/s