Recent content by babaliaris

That's amazing, that if the local substation is not grounded the current will still flow inside the earth from my home, even though no current will ever travel up to the local station. I think the reason that this confuses me might be that I think of it as But the network system is not a...

Someone once told me "Think of the ground as a large capacitor". So what if I think the circuit like this:

So in other words, if I connect an ampere meter in the cable that connects the neutral and the ground I will see 0A? Even when there is a fault and the neutral has a current flowing through it?

This is a topic that almost no one can actually explain from what I found on the internet. This answer https://physics.stackexchange.com/questions/74625/does-alternating-current-ac-require-a-complete-circuit/74999#74999 is the closest I found so far. The idea is how the grounded neutral...

I will do it tomorrow since I'm exhausted. I'm struggling with that problem for 3 days now. Never in my life, I had to deal with a problem for 3 days continuously. I'm glad that I finally figured it out! How did great scientists try to solve a problem for years and not get crazy?

All the math and physics that I did to reach in the equation ##sinθ' + 2cos(θ') = 2##

So in conclusion, Is this the correct solution? Did I make it!? Thank you for your time and guidance!

Found it ##z = L[1-cosθ']## where ##θ' = θ_{max}## I ended up to ##sinθ' + 2cos(θ') = 2##. Unfortunately, I don't remember how to solve that. I always forget trig formulas... Edited:

##P_{@Θ_{max}} = mgz## where z = the distance from the ground up to the mass m, and z is uknown...

I thought it didn't matter, so I assumed θ=90 where the m mass is indeed L meters from the ground. So this is a mistake? Yes at θmax I don't know the actual height so I can't find the potential energy there. So in order for this to be right, I have to say: ##mgz = \frac{1}{2}mV^2## where z is...

Off Topic:

I thought of something else but I got a math error: V at the lowest point: ##p1+k1 = p2 + k2 <=> mgL + 0 = 0 + \frac{1}{2}mV^2 <=> v = \sqrt{2gL}## So at the highest point I now that: ##V = 0##, ##Θ(t) = Θ' = max{Θ(t)}## and for the lowest point I know that: ##Θ(t) = 0## and ##v =...

My solution is in the pdf I attached. I don't know how to write everything in latex this is why I uploaded the pdf.

But now, how can I use the equation ##|a| = \sqrt{a_{T}^{2} + a_{N}^{2}} = \sqrt{g^{2}sin^{2}(θ(t)) + \frac{|V|^4}{L^{2}}}## and that |a1|=|a2| (1 = highest point in the path, 2 = lowest point)? By the way, since this path is an arc, then ##|V| = ωL##

PS: O MY GOD, I just figured something out... Well if you see the pendulum in a clock, you see it going up and down (oscillating) at the same speed! This means that when the pendulum reaches its maximum height it changes direction instantly! This is why the speed never changes but the direction...