bw

rank in order of most oh groups to least that will be the same as highest surface tension to lowest

15 years ago

a)3 b)take the derivative of U(x).. U'(x) then take the zero-points U'(x)=0.. one zero is x=0, and it has two zeros symmetric to the y-axis.. take the positive one as answer for x0. c)wolframalpha will do this for you. enter taylor expansion take U'(x) as...

11 years ago

if you use the differential equations from mouse you get after simplification: v0=(F0+F1/2)*t_s/M |a(ts)|=F0/M |a(0)|=(F0+F1)/M s=t_s^2*(F0/2+F1/6)/M W=F0*s

11 years ago

E_heat=0.5*M*v0^2-W

11 years ago

@Shashanoid: do you have x0? make the taylor expansion of U'(x) with x0 K is the factor in the taylor expansion from (x-x0)

11 years ago

@Shashanoid: what is your U'(x) and your x0? @Jack Package: whats wrong with W and E_heat it worked for me

11 years ago

@Shashanoid: the exact value for x0=2*sqrt(5/3) taylor expansion: 3*(x-2*sqrt(5/3))^3+6sqrt(5)(x-2*sqrt(5/3))^2+40*(x-2*sqrt(5/3)) so K=40

11 years ago

@Shashanoid: k=120

11 years ago

T=s*g*(m_1+2*m_2)/d F=-(m_1+m_2)*g+s*g*(m_1+2*m_2)/d

11 years ago

F(t)=beta*t^2 Ff(t)=beta*t^2-m*g*mu a) p(t)=integral F(t) from 0 to t p(t1)=beta/3*t1^3 --------------------------- b) p=m*v, so v(t)=(integral Ff(t) from (the point where it starts to move) to t) /m the point where it starts to move: beta*t^2-m*g*mu=0 t=...

11 years ago

@fem: yes, i would not post it if I had not checked it. @kumar: probably you integrated from 0 to 5? F(t)=0.8*t^2 Ff(t)=0.8*t^2-4 b) time where it starts to move: 0.8*t^2-4=0 ->t=sqrt(4/0.8)=sqrt(5) v(5)=[0.8/3*t^3-4t] integral from sqrt(5) to 5 /2 =9.65...

11 years ago

no, that's the only one I'm not done with, I'll try tomorrow..

11 years ago

merry is just a little conservation of momentum.. m1: mass merry-go-round m2: mass sledgehammer v1: velocity merry-go-round before collision (=0) v2: velocity sledgehammer before collision v1': velocity merry-go-round after collision v2': velocity sledgeh...

11 years ago

I have not yet tried to solve it, maybe you'll find something useful in the thread with id=1386456397 here on jiskha

11 years ago

@Jeff: 44.99 is the solution for kumars values, yours are: c) Ff(t)=1.2*t^2-6 ->t=sqrt(6/1.2)=sqrt(5) v(4)=3.51 (rounded) F(4)=1.2*4^2=19.2 P(4)=67.48 (rounded)

11 years ago

has anyone found F_x?

11 years ago

@fima: how do you get alpha? if i take your values i do not get alpha=50 Probably 0.76 is wrong.. I have a different equation for Fy but I'm not yet sure if it is correct, but it worked for my values..

11 years ago

thx.. now I have it complete :) a) I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*r c) cos...

11 years ago

yes r=L/2

11 years ago

@da: take absolute value b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*|ax| Fy=m*g*cos(theta)-m*w^2*L/2

11 years ago