Recent content by ItsImpulse

A sphere (with density less than that of water) is placed into 2 different cylinders. One with a spinning column of water and the other with a still column of water. Why does the sphere in the spinning column of water take much longer to float to the top of the cylinder? I was thinking about...

Hey there PF, I'm currently a high school student and I'm planning to do physics in college when I get there. However, I want to get my math up to scratch before I go deeper into it. I do have undergraduate level texts for physics but not for mathematics. I have managed to borrow 2 books...

Homework Statement How many blocks can you stack on top of each other before the tower falls? Each block has a mass m and a length l. The first block is on the table. The second block is stacked on top of the first but moved by 0.15L to the right. This pattern continues Homework...

so in other words the one that rotates more will go down the ramp slower? it would be mgh = 0.5mv^2 + 0.5Iw^2 right?

1. yes starting with 0 speed. 2. it's rolling down a ramp but yes at same height. 3. vertical acceleration is just gsin(theta) am I right?

suppose you had a ball rolling down a ramp, without slipping and compare it to a ball that starts with a velocity u that is horizontally to the side. how would the time taken be different to reach the bottom?

But then assuming the current flows from A to B and CD and EF has no potential difference, how does the current get into the black square in the first place?

I would not think so since current does not flow in CD of EF?

I meant that consider that between node C and node E is a parallel of circuit, by which one branch has a resistance of R while the other branch has a resistance of sqrt(2)R/2. I'm sorry, by inner square do you mean the second largest square? or the largest square in black. If its the earlier...

As I have mentioned above, probably not clearly, I took into account that the resistance of CE and DF is simply sqrt(2)R/2 and R in parallel. This gives an effective resistance of sqrt(2)/(sqrt(2) + 2) across CE and DF

Firstly I apologize for the bad image as I don't know how to properly photo edit things. Now, I started off by taking using A and B as start and end points, respectively. Upon entering A, the current will split into the two branches. At C, the current would have gone through a resistance of R/2...

Sorry I forgot. Assume the wires to be of a constant material thus length of the largest square to be of resistance R. Yes they are all connected such that the corner of each square bisects the length of the larger square.

Homework Statement Find the effective resistance across this circuit from two diagonally opposite points on the outer squarehttp://aiminghigh.aimssec.org/wp-content/uploads/2012/01/7squares.gif (I took the image off google it isn't exactly a circuit but its the correct orientation) given that...

It is true that the block does not slide. Since The maximum static friction is given by μs(mg), plugging the values in gives 0.33(10)(9.81)= 32.373N hence you would need a force greater than 32.373N to be able to move it in the first place. Only after movement will the kinetic friction be used.

Oh I see that's how you go about doing it. Would be accurate to say that the max range is at 45 degrees but with the cave inhibiting motion, it's the greatest angle possible?