anonymous
  • anonymous
In the figure, a stuntman drives a car (without negative lift) over the top of a hill, the cross section of which can be approximated by a circle of radius R = 262 m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill?
Physics
schrodinger
  • schrodinger
See more answers at brainly.com
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
1 Attachment
anonymous
  • anonymous
is FBD necessary?
anonymous
  • anonymous
I just need the FBD thank you!!

Looking for something else?

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

More answers

IrishBoy123
  • IrishBoy123
|dw:1442581163215:dw|
anonymous
  • anonymous
I got 50.67 answer but I feel so discontent to my work. It is like Ineed to know something
anonymous
  • anonymous
hmm okay
anonymous
  • anonymous
my prof just gave a formula of mg=mv^2/r but I want to know where did he derive it
anonymous
  • anonymous
I know mg is always down sooo mv^2r <- kinda confuse
IrishBoy123
  • IrishBoy123
just imagine you are the car or in the car
IrishBoy123
  • IrishBoy123
for derivation:
IrishBoy123
  • IrishBoy123
|dw:1442656237102:dw| for a particle accelerating in a circle, radius r \[\vec s = r < \cos \theta, \sin \theta>\] \[\vec { \dot s } = \vec v = r<-\sin \theta, \cos \theta> \dot \theta\] \[\vec a = \vec{ \dot v} = r<-cos \theta, -\sin \theta>\dot \theta ^2 + r<-\sin \theta, \cos \theta> \ddot \theta\] \[= r<-\cos \theta \, \dot \theta ^2 - \sin \theta \, \ddot \theta, -\sin \theta \, \dot \theta ^2+ \cos \theta \, \ddot \theta>\] projecting that along the radial vector \(\vec r = <\cos \theta, \sin \theta>\) \[a_r = r<-\cos \theta \, \dot \theta ^2 - \sin \theta \, \ddot \theta, -\sin \theta \, \dot \theta ^2+ \cos \theta \, \ddot \theta> \bullet <\cos \theta, \sin \theta>\] \[= -r \cos^2 \theta \, \dot \theta ^2 - r \sin \theta \cos \theta \ddot \theta - r \sin^2 \theta \, \dot \theta^2 + r \sin \theta \cos \theta \, \ddot \theta\] \[= - r \dot \theta ^2 = -\omega^2 r \] the particle is accelerating inwardly with \(a = \omega^2 r = \frac{v^2}{r}\) the car going over the hump must do the same or else it will continue in a straight line along the tangent to the hump. gravity must provide the force to accelerate the car in a circle

Looking for something else?

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