Some workers do 15000 J of work on a 3500 N crate to push it up a ramp. The ramp is 4 m high. What is the efficiency of the ramp?

93.3%

Explanation:

Efficiency of ramp is given by

Where n is the efficiency, W is work and subscripts o and i represent output and input respectively.

The work input is what the workers do and that is given as 15 kJ

The work output

Since work is product of force and perpendicular distance and substituting force with 3500 N and distance with 4 m we get that the work output is 3500*4=14000 J= 14 kJ

Finally, substituting 14 kJ and 15 kJ for work output and input respectively we deduce that

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: a. same unit

explanation: