Answer:

Explanation:
To develop this exercise we proceed to use the kinetic energy equations,
In the end we replace


Here
meaning the 4 wheels,
So replacing

So,




It’s just E because ethe positiv and negative current are supposed to flow thorough the bulb in opppsote sides at a equel level.In some them negerive/postive is absent and some of them are connected to the same side
Answer:
A
Explanation:
because thats what I put and got it right
Answer: It's hard to say without characterizing the collision. But it will be either A if the collision is totally in-elastic, or B if the collision is totally elastic. It could be anywhere in between for partially elastic collisions.
Explanation:
momentum is conserved, so initial system momentum will be left to right.
The velocity of the center of mass is 50(5) / 550 = 0.4545... m/s
In an elastic collision, the lead ball will move off at twice that speed or 0.91 m/s to the right.
The steel ball will bounce back and move away at 0.91 - 5 = -4.1 m/s . The negative sign indicates the steel ball has reversed course and has negative momentum
In a totally in-elastic collision, both balls would move to the right at 0.45 m/s. The steel ball will still have positive momentum.