Answers by visitors named: grey

is this a conservation problem?

i don't know if should use , 1/2mv^2=kx to solve it or not

what mass do i use?

sin(30) = sin(a) (.5)(1.6) (1.2)(.5) sin(30)(.6) = sin(a)(.8) sin(a)= (sin(30)(.6))/.8 sin(a)= .375 a= 22.024º Therefore, b= equals: 180º- 22.024º- 30º = 127.976 sin(30) = sin(127.976) (.5)(1.6) (.5)(v) sin(30)(.5)(v) = sin(127.976)(.5)(1.6) v = (sin(127.976)(.5)(1.6))/ (sin(30)(.5)) v= .6306148577 / .25 v= 2.522m/s The cue ball original speed is 2.5 m/s.

try kx=Ffmgh , then solve for h , (its an idea)

do i solve for vf in the vf=vi+at equation , then put thr average velocity ?

Its A.

What is the product? negative 9 x (5 minus 2x) 18 x squared minus 45 x negative 18 x squared minus 45 x negative 18 x minus 45 x 18 x minus 45 x

thanks seen it done that, I had to check my answers and C A B is correct.