Recent content by xiaoB

Try using this way: Let y=ex :smile:

Oh!thanks.I not think so...

Yes ! the derive should be this wA*cos(wt) . But if v(t)= - wAsin(wt + ∅) the sin(wt + ∅) for the phase ∅ is 900 so the sin(wt + 90)=cos(wt).:smile: Therefore,v(t)= - wAsin(wt + ∅) nothings going wrong.:smile:

You just derive this x = 0.20m*sin(0.44 rad/s * t) respected for t : dx/dt=V.

Hi,eczeno!:smile: Actually he correct because the sine has the phase ∅.

(0.44rad/pi)x1800=25.120

For easy, (1)First find up the degree respect for horizontal-x. (2)Find the components-xy for each forces: (i)Rx=∑Fx=F4cos(90-42)+F3cos(35)+... (ii)Ry=∑Fy=F4sin(90-42)+F3sin(35)+... (3)Finally, find the resultant R=√(Rx2+Ry2) And sketch the diagram for resultant R to find up its angle...

As i know, the calculation of the book was using radian(0.44) changed to degree(25.120).So,the answer would get -0.01655 m/s. But for me, normally we usage radian.:smile:

This: Here got a mistake please change it ∅=tan-1(e tan θ).

Your calculation got problem because the answers should be : -3.675+9.8t - 4.9t2=0 => t=0.5s / t=1.5s :smile:

∫(dT/dt)dt=∫(6.4-0.4T) dt T=6.4t-0.4Tt+c T+0.4Tt=6.4t+c T(1+0.4t)=6.4t+c You should have answer!

First can you tell me is this answer you want when t=2.3,T=52.460 ? :smile:

As i see if not going wrong, i would chose b) because the question has been said that is elastic collision. Except the question say is inelastic collision then i will chose a).:smile:

Let say the maximum height for Y2 go through the window is equal to 6 m meanwhile the final vertical velocity Vyf=0 for Y2. Therefore, apply V2yf=V2yi-2gY2. Because Viy =V1sinθ.So,you can get the V1 after that you can get the V0.:smile:

Normally, the Range of X2=V1cosθt1 .So, Range of Velocity is V1cosθ= X2/t1 => VX2=X2/t1. And same things to Range of X1.Apply this θ=tan-1(e tan 600). Please check it out the answer. Y2 is too low. Very sorry to you.:redface: