anonymous
  • anonymous
Please help :) http://screencast.com/t/nCUCq22y
Physics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Draw FBD
ghazi
  • ghazi
@Yahoo! solve this...it is for you...i will assume your kinematics is prepared
anonymous
  • anonymous
u sure the information is enough

Looking for something else?

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

More answers

anonymous
  • anonymous
decelleration = 0.5m/s^2
anonymous
  • anonymous
@zaphod Do u have answer key
anonymous
  • anonymous
Now, find alpha
anonymous
  • anonymous
i knw how to do the next one...i think we have to resolve it a = -g ( sin alpha)
anonymous
  • anonymous
mg sin x = ma
anonymous
  • anonymous
sin x = a/g
anonymous
  • anonymous
x = sin^-1 a/g
anonymous
  • anonymous
|dw:1346508274972:dw|
anonymous
  • anonymous
@ajprincess wat do u think?
anonymous
  • anonymous
anonymous
  • anonymous
for a) assuming constant acceleration, v^2 = u^2 + 2as
anonymous
  • anonymous
Yes.....we want 2 nd question
anonymous
  • anonymous
6.25 = 2.25 + 8a a = 1/2
anonymous
  • anonymous
now \[mgsin \alpha = ma\] |dw:1346511156184:dw|
anonymous
  • anonymous
so \[gsin \alpha = \frac{1}{2}\] \[\sin \alpha = \frac{1}{2g}\]
anonymous
  • anonymous
now inverse sine gets you alpha
anonymous
  • anonymous
Yup..thxx.....and i was correct

Looking for something else?

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