A 0.91 kg ball and a 2 kg ball are connected by a 0.95 m long rigid, massless rod. The rod is rotating clockwise about its cente
1 answer:
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Explanation:
The given data is as follows.
radius (r) = 3.25 cm,
Now, we will calculate the tangential acceleration as follows.
Putting the given values into the above formula as follows.
=
= 37.7
Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 .
The initial speed of car A is 15.18 m/s.
Momentum is defined as mass in motion. If there are two objects (the two objects in motion or only one object in motion and the other in stationary) that collide and no other forces work in the system, the law of momentum conservation applies in the system.
p=p'
pa+pb = pa'+pb'
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
- ma = mass of object A (kg) = 1,783 kg
- mb = mass of object B (kg) = 1,600 kg
- va = speed of object A before collides (m/s)
- va' = speed of object A after collides (m/s) = 8 m/s
- vb = speed of object B before collides (m/s) = 0 m/s
- vb' = speed of object B after collides (m/s) = 8 m/s
- p = momentum before collision (Ns)
- p' = momentum after collision (Ns)
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)
(1,783×va) + 0 = 14,264+12,800
(1,783×va) = 27,064
va = 15.18 m/s
Learn more about The law of momentum conservation here: brainly.com/question/7538238
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