Recent content by teclo

i have never had to work out a problem as such, but in general for drag force you have F = -Cv or F = -Dv^2 where you use the first for very small speeds and the second for large speeds. C and D are constants that depend on the shape of the object and the viscosity of the material

the problem is that you're using the constants that aren't numbers. i don't think mathematica will do that. use the help browser for plot, i think the command should be something like [Plot, function, {x,a,b}, {c,d}] or something like that, where the things in the fancy brakets give yr...

derive a variation for the nambu-goto action!

it's been a year since I've had e&m, but the first part looks ok. I'm kind of confused on the second part, because you're saying a is a variable. i thought a was a constant? i guess it doesn't really matter, because the general formula on the axis would be a---->x the second one requires...

could you elaborate, please?

the m's do cancel out, though. the equation reads (mu)*m*g = m*a, solving for the acceleration you get a = mu*g. this came up in physics 1 a few years ago and i was super confused. of course if there is a braking force applied (not just coasting to a stop) then it's true the lighter one is...

because acceleration does not equal mass/force (you still forgot the negative)

trust your math skills, not mathematica! (or whatever program you're using). the difference arises if you factor a negative one out before integrating. thus, you are then integrating - 1/p int (1/y-p) using u substitution now requires no extra negative for du. make sense?

not quite, but almost. you forgot the negative when calculating the force.

yes, looks correct to me.

the first thing you'd want to do is get y as a function of t. whoops, i missed your second post. yes, you want to solve for the function y{t} and take the limit as it goes to infinity. you have the right approach, the integration just gets a little messy (but not bad really, assuming that...

the derivative of a constant is zero. what the above equation i posted says is that the velocity (v) is the time derivative of position (x). if you differentiate the initial equation (velocity) you get the acceleration. that's not what you're after.

assuming they have the same coefficient of friction, i don't see why the lighter one would stop any faster than the heavier one. F = (mu)N = ma N = (mg)

doesn't look like it to me. v(t) = dx(t)/dt what is the derivative of option d?

none of the links work for me.