Calculating the Radius of Curvature for a Car on a Hill

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To calculate the radius of curvature for a car moving over a hill at 85.7 km/h, the speed must be converted to meters per second, resulting in approximately 2.38 m/s. The only force acting on a package in the car is gravity, which allows for the use of the centripetal acceleration formula. By applying F = ma, the equation simplifies to mg = mv^2/r, leading to the cancellation of mass. This results in the formula r = v^2/g, allowing for the calculation of the radius of curvature without needing the mass of the package. The discussion emphasizes understanding the relationship between speed, gravity, and curvature in this context.
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An automobile moves at a constant speed over the crest of a hill traveling at a speed of 85.7km.h. At the top of the hill a package on a seat in the rear of the car barely remains in contact with the seat. What is the radius of curvature (m) of the hill?

85.7km/h=2.38m.s v=2.38m/s ac=v^2/r r=v^2/a

I know that the only force acting on the package is gravity. Does this mean that the mass of the package is 9.80? mg=m*ac

I am not sure where to go from here
 
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Error in "85.7km/h=2.38m.s"
In 85.7 km/h you must replace the km with 1000 m and the h with 3600 s.

You are so close when you say "only force acting on the package is gravity"!
Put this into F = ma and realize that the acceleration is centripetal so put that formula in for a. Solve for R.
 
So...
F=m*v^2/r

9.80=m*23.8^2/r

How do I solve for r if I don't know the mass of the package?
 
F = ma
mg = m*v^2/r
and the m's cancel.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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