Recent content by Earn Success

Wait actually the answer is correct if I just take the 2.94 and the 1.77 and add them together, then the other two diagonals cancel because they are of the same magnitude. Thanks for the help peoples, I live to fight another day of honors physics...

So I found the electrical field for each of the four corners: 2.94 x 106C for the 45microC charge and then 1.77 x 106C for the 27microC charges Now what do I do... I don't understand what the hell and electric field even is so I'm not sure what the directions are. And the answer is...

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Homework Statement Find the electric field for a point in the center of a square, +45microC on one corner and -27microC on the other 3 Homework Equations E = (kQ1)/r2 The Attempt at a Solution I know how to do the Electric field equations in a straight line, but for a square...

i understand it but i don't think it makes a bit of sense that you can subtract invisible things

I thought Nf was a contact force and didn't depend on speed or anything like that

Yes but then it says the change in V-> points towards the center of the circle. That v is is that makes the acceleration happen right? And that v is towards the center because of some sort of vector subtraction which I can never comprehend

A roller coaster car is on a track that forms a circular loop in the vertical plane. If the car is to just maintain contact with the track at the top of the loop, what is the minimum value for its centripetal acceleration at this point? g downward g upward 0.5 g downward 2 g upward...

I think vectors just completely confuse me. What if they used instantaneous velocity, then there would be no resultant vector... As soon as I was about to say, "and no acceleration" I realized that then there would be no acceleration with only 1 vector. Ok so I get that, but after reading...

According to my book, in centripetal acceleration, the direction of the velocity is the same as the acceleration, which points towards the center. There is a picture of a velocity vector pointing out of the circle, and then another point with a second vector pointing out, and the resultant in a...

Ok thanks for all the help. Big test tomorrow, if I get a 90+ all my classes will be above 90 :P

Okay, so the way I'm beginning to see it is that because the FN is just a contact force, and does not depend on the weight of an object, unless that object is at rest. Even if there is an acceleration, the FN is there - as long as there is contact between that object and a surface. When...

"A roller coaster car does a loop-the-loop. When it is upside down at the very top, which of the following is true?" The normal force and the weight are in opposite directions. The normal force and the weight are perpendicular to each other. The weight is zero. The normal force...