But to find the volume of the cone, for exemple, the fact that the function isn't parallel to the x-axis doens't influence that "volume is the sum of all areas of all circles"... What changes?
For a given function f(x)=x, if we rotate this function from [0,1] around the x-axis, we'll have a cone... why can't I find its surface area by adding the perimeter of all the circunferences from [0,1] with radius = f(x)?Something like 2*pi∫xdx? The formulas says we have to do 2*pi∫x*√(i+y'²)dx ...
i'm in my freshman year and I'm starting to learn Derivatives in Calculus, and I was wondering, once Ek (kinectic energy) = 1/2mv², then, the derivative of Ek in term of velocity would be mv, which is equal to the linear momentum... I'm finding hard to understand the idea that the variation of...
Why does the ''P2=P1'' (momentum before is equal to momentum after) isn't applied in cases where u throw a ball at wall?
In this case, even is the velocity keeps the same, it will be in another direction, what makes it -v...
Hey guys, in a reaction like 1A + 1B <-> 1C + 1D, which starts with 1 mol of each molecule (A,B,C,D), how can I find out how many mols will be remaining in equilibrium state?
knowing that the Keq is equal to 4
I think I got it... when it comes to friction, the energy becomes heat, so we just cut it off the equation to calculate the Mechanical Energy...
but when it comes to Gravity force, the energy becomes potential energy, so we add it to the equation... is it right?
well, the friction will do a work of -100J (once its againts the movement) and the force will do a 125 work... the work will be 25, as if it were a 5N force only... I learned that way, is it wrong?
I've always seen Work as an extra mechanic energy given to the system... but i really can't figure it out :
If we lift a 2kg ball, that's steady, using g as 10m/s², with 25N, we'll have a 5N up resultant force...
after its 5meters up, the resultant work will be 25J (5N.5m), but the...