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If the speed of a roller coaster at the bottom of a hill is 20.0 m/s , what is the tallest hill the coaster can climb, assuming no friction losses ?

Respuesta :

Answer:

20.4 meters

Explanation:

KE = 1/2 m v^2 = mgh = PE (convert all kinetic energy to potential energy, because at the highest point the coaster can climb, it will be stopped!)

=> h = (1/2 m v^2) / (mg) = v^2/(2*g) = (20 m/s)^2 / (2*9.8 m/s^2) = 20.4 m

The tallest height that the roller coaster can climb is 20 m.

A roller coaster is a device in which energy conversions from potential to kinetic energy takes place. Hence the roller coaster can easily be used to demonstrate the conversion of mechanical energy from one form to another.

Given that the velocity of the roller coaster at the bottom of the hill is  20.0 m/s, potential energy is converted into kinetic energy.

Hence;

mgh = 1/2mv^2

gh =1/2v^2

h = 1/2v^2/g

h = 0.5 × (20)^2/10

h = 20 m

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