A97 kg man holding a 0.365 kg ball stands on a frozen pond next to a wall. he throws the ball at the wall with a speed of 11.3 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. how fast is he moving after he catches the ball? ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall.

Explanation:

First when man throws the ball then the speed of the man is given as

now ball will rebound after collision with the wall

so speed of the ball will be same as initial speed

now again by momentum conservation we will have

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).

answer: strength i guess

explanation: