Friday

November 28, 2014

November 28, 2014

Posted by **Leslie** on Friday, November 9, 2012 at 7:01am.

For this problem I made a = (0, -g) and used the equation Vyo= 1.5 - g(t^2)/2. And made Vyo=0 so, solving for t, I got sqrt(3/g).

Then v=2pir/T,so (22x2pi)/44s = pi rev/s

so w= pi rev/s

Then, using the equation v=Rw , v = (.5m)(pi rev/s) = pi/2 m/s

After that I didn't know what to do, so if I made any errors above, please let me know. I think for finding the distance I would use the equation D = sqrt(R^2 = Xmax^2) but I don't know how I could incorporate that into the problem

Thank you

- Physics -
**Elena**, Friday, November 9, 2012 at 10:42amThe drops begin to move with horizontal speed

v(x)=2•π•n•R=2•π•N•R/t=

=2•π•22•0.5/44=1.57 m/s

The time for falling down is

t=sqrt(2h/g) = sqrt(2•1.5/9.8)=0.55 s.

L=v(x) •t =1.57•0.55 =0.87 m

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