Recent content by spoon

Okay, so I solving m du/dt = -(u^2 + 10u): t/500 + C1 = 1/10 ln(u+10) - 1/10 ln(u) t/50 + C2 = ln((u+10)/u) where C2 = 10*C1 e^(t/50 + C2) = Ce^(t/50)-1 = 10/u u = 10(Ce^(t/50))^-1 v^2 = 10(Ce^(t/50))^(-1) -----> v = +/-[10(Ce^(t/50))^-1]^(1/2) But v should just be the negative right?

Sorry, looking back that was pretty clear... So then it would be the integral of: dz/(1-z)z = ln(1-z) + ln(z) Then set this equal to: C + Integral( r(t)) and just solve for z?

Thanks for the point about r(t), so I fixed that part, but I'm still confused as how to go about the rest... If I substitute y with kz and y' with ?, then how do I integrate the following: ?/[(1-z)kz] Because I could integrate it if the "?" were a "dz" right?

Following that... since m = 1000 (500)u d/dt + u^2 + u = 0 Is it a legal operation to factor a "u" out of this equation?

I'm also attempting another problem... A population has a periodic growth rate r(t) = A[1 + sin(t/(2*pi))], but otherwise follows the logistic population model with carrying capacity K. There is no threshold and the initial population is Yo = Y knot = K/2. a. Modify the basic logistic...

So integrating that I got t/m = .1 ln|v| - .05 ln|v^2 + 10| I wasn't really sure how to integrate the right hand side on paper, so I used my calculator Solving for v, using m = 1000: e^(t/50) = v/((v^2) + 10) I used a variable g to = e^(t/50) which led to: gv^2 +10g = v I tried using the...

Alright, so I starting with Newton's equation...I'm thinking: F = m*V = 1000*V then I'm assuming F = Stopping force which is: F = V^3 +10V = 1000V Integrating both sides... (V^4)/4 + 5V^2 = 500V^2 Then solving for V, I got 13.416 But this totally seems wrong to me...

So far? Nothing...I'm really looking for a place to start =(

Driving your 1000 kg car on I-5 at 100 km/h, you suddenly noticed a large group of animals. You slam on a break and this provides a stopping force equal to v^3 plus the friction force of 10v. Set up and solve a differential equation for velocity. You need to find a suitable change of variable.

Actually now I got part A...I just need part B... Thanks for the help by the way

I kind of get what you're saying but I'm still sort of lost?

An 8.90-cm-diameter, 310 g sphere is released from rest at the top of a 2.00-m-long, 17 degree incline. It rolls, without slipping, to the bottom. a) What is the sphere's angular velocity at the bottom of the incline? b) What fraction of its kinetic energy is rotational? If someone could...