1) A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force? 2) The same spring is stretched 5 m. What was the force?
3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?
4) How far will a 50 N force stretch the same spring?
5) A spring is stretched 0.15 m and the force was 1500 N. What is the spring constant?
6) A spring is stretched 0.75 m by using a force of 30 N. What is the spring constant?
7) How far will a hanging 0.400 kg object stretch a spring if its spring constant is 300 N/m?

8) A hanging 7.5 kg object stretches a spring 1.1 m. What is the spring constant?
Find the force by multiplying the mass (7.5 kg) x gravity (9.8 m/s2)
Use that to calculate the spring constant
9) A spring has a constant of 875 N/m. What hanging mass will cause this spring to stretch 4.5 m?
10) A spring with a spring constant of 25 N/m is stretched 65 cm. What was the force used?
11) A 25 N force stretches a spring 280 cm. What was the spring constant?
12) A 75 N force stretches a spring 175 cm. What was the proportionality constant?

b. latent learning (the word "latent" itself means something that exists but isn't obvious, only under certain conditions! )

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).

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr