someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (the block of wood is sitting on a frictionless surface, so it moves freely when the bullet hits it). the wood block is initially at rest. the bullet is traveling 300 m/s when it hits the wood block and sticks inside it. now the bullet and the wood block move together as one object. how fast are they traveling?

22.2 m/s

Explanation:

We can answer this question by using the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. Therefore, we can write:

where is the total momentum before the collision and the total momentum after the collision. Re-writing the equation,

where:

m = 0.04 kg is the mass of the bullet

u = 300 m/s is the initial velocity of the bullet

M = 0.5 kg is the mass of the block of wood

U = 0 is the initial velocity of the block (it is at rest)

v  is the final velocity of the bullet+block combined, after the collision

Solving for u, we find the final velocity of the bullet+block after the collision:

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

b would be the answer