Recent content by Isiudor

Either there's something I'm missing or the above problem cannot be solved using only what you wrote. there's no way to calculate the average velocity since it's unknown how long was spent in each section or the distance traveled in each section... Are you sure there wasn't additional data?

on top of that as far as I understand it now dx/dt is simply a notation saying the derivative of x as a function of t. hence I don't see how is legal. Infact dx/dt is equel to v0 + a*t since it's the derivative of x0 + v0*t + 0.5at^2 and if you substitute that for "v" it solves fine.

I thought d means delta or Δ. Since you say it isn't and since you agree that the instantaneous velocity at t = 3 is 30m/s. explain to me with what values will dx/dt will give you 30m/s. and I don't see how "treating h as a variable" helps, ofcourse it's a variable and given the value h = x his...

you are wrong my friend, that was not your problem. Vfinal = dx/dt is not a statement you can make. Vfinal = dx/dt only when a = 0 which is clearly not the case in a free fall. take for example two situations: a car is cruising at a constant speed over the period of 3 sec it passes 45m, how...

that is when we assume the ball is already rotating in a certain height within the bowl at the same rate as the bowl. what I am asking is why does the bowl exert any force on the ball at all say, we place the ball in the dead center of the bowl and then begin rotating the bowl, there's no...

Homework Statement a ball with a mass of 50 grams is placed inside a circular bowl with a radius of 10cm. the bowl rotates at a rate of 5 hertz what is the height in cms the ball will rise within the bowl? Homework Equations Fr = m*w^2/r \sumF = ma The Attempt at a Solution...

no no no no, well everything the fella above me said is true, but I assume you already accounted for t being the end of the fell from start to finish that is not the problem. dx/dt equels AVERAGE velocity and not the final velocity. quite two different things. assuming constant acceleration...

well doc, thanks for the fast reply, good to see you are still around. anyway, yes the radius is 7m and the question does ask for the total force acting upwards. so you're saying that ontop of the mg force acted back on the skateboarder due to it's weight(597), the ramp exerts additional...

Homework Statement A skateboarder with mass 61kg sliding down a circular ramp reaches a speed of 7.8 m/s at the bottom of the ramp what is the net force acting on the skateboarder at that point? Homework Equations N = mg Fr = m * v^2/r The Attempt at a Solution In my opinion the...

Yes I get it now, the critical speed is the speed I must have to "stay in orbit" due to the mg + N acting on me, although that at the highest point it's just mg. and when I "solve" mg = 1/2mv2 I actually get the critical velocity equation. I guess you are also right regarding the two...

but I WANT the speed to be zero. my goal is not to make a full circle, just to reach the top, if it falls off the highest point, I've reached my goal. since I want to have the minimal starting speed(at bottom of loop) and reach the highest point in the loop, that's my only goal. had my goal...

loop problem, "critical velocity" Homework Statement First I'd like to ask a general question, if a body enters a loop (like a rollercaster loop) and PASSES the heighest point, assuming there's no friction, will it neccasserlly complete the loop or can it's orbit around the loop still...

nothing is wrong, at least not from where I'm sitting. in fact I received back a quiz I had 2 days ago, and got 100(A+ in conventional terms, I guess). Doc Al has been great help in helping me understand it all. thanks again.

well I can definitely understand your solution, I DO understand it... and I could probably now solve any similar question... but what I'm still a bit confused about is the reason why rules seems to change, when A & B are accelerating, and when they are not. in the example of a force of 5N...

you've neglected to mention the kinetic friction between A and the surface, so I'll just calculate it as if kinetic friction is equal to static friction for A, surface. with 5N applied on B, the static friction between A and B is equal to 5N, which is less than max static friction of A, B...