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(A) The total initial momentum of the system is
(1.30 kg) (27.0 m/s) + (23.0 kg) (0 m/s) = 35.1 kg•m/s
(B) Momentum is conserved, so that the total momentum of the system after the collision is
35.1 kg•m/s = (1.30 kg + 23.0 kg) <em>v</em>
where <em>v</em> is the speed of the combined blocks. Solving for <em>v</em> gives
<em>v</em> = (35.1 kg•m/s) / (24.3 kg) ≈ 1.44 m/s
(C) The kinetic energy of the system after the collision is
1/2 (1.30 kg + 23.0 kg) (1.44 m/s)² ≈ 25.4 J
and before the collision, it is
1/2 (1.30 kg) (27.0 m/s)² ≈ 474 J
so that the change in kinetic energy is
∆<em>K</em> = 25.4 J - 474 J ≈ -449 J
Answer:
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Explanation:
The minimum size copper equipment grounding conductor that is required is ; #4 copper
<h3>Equipment grounding conductor </h3>
An Equipment grounding conductor is a conductive part of ground fault current path which connects the non-conducting metallic parts of the euipment together and also connects the system grounded conductor.
For a copper euipment grounding conductor required to serve a control center connected to a 300 amp over current device the minimum size of copper is #4 copper
Hence we can conclude that The minimum size copper equipment grounding conductor that is required is ; #4 copper.
Learn more about Equipment grounding conductor : brainly.com/question/14124204
Answer:
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