anonymous
  • anonymous
hey, sorry about going off subject, i would like to ask a question on physics. Since there isn't anyone in that section i am helpless. here is my question: A cylinder of radius 20cm and mass 100g rolls down an inclined plane of height 60cm without slipping. the linear velocity of the cylinder at the bottom is?? (g=980cm/s/s)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
mgh = KE
anonymous
  • anonymous
1/2 mv^2+1/2 I w^2= Mgh
anonymous
  • anonymous
i did that but i m not gettin the answer.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
KE = KE(r) + KE(t) 1/2(mr^2/2)w^2 + 1/2 x m x v^2
anonymous
  • anonymous
rw=v w=v/r substitute
anonymous
  • anonymous
3/4 x mv^2 = mgh
anonymous
  • anonymous
i went through all that buddy. can you solve ang post your anser.
anonymous
  • anonymous
*and *answer
anonymous
  • anonymous
v=2.8 m/s
anonymous
  • anonymous
ow rats! i made a calculation mistake! neway thnx buddy..
anonymous
  • anonymous
no prob
anonymous
  • anonymous
wat is the radius of curveture if a particle moves in a trajectory given by y=f(x) at any point(x,y)
anonymous
  • anonymous
parameterize the eqn in terms of time....r =v^2/a(radial)
anonymous
  • anonymous
it is given as \[R=[1+(dy/dx)^{2}\left[ \right]^{3/2}\]
anonymous
  • anonymous
divied by d^2y\dx^2
anonymous
  • anonymous
if u know the answer then y do u ask..i dont memorize formulae..its wiser to go through the basic approach...
anonymous
  • anonymous
i want to kno how it came

Looking for something else?

Not the answer you are looking for? Search for more explanations.