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grigory [225]
3 years ago
5

The car on the roller coaster is released from the position shown and allowed to roll freely.

Physics
2 answers:
xxTIMURxx [149]3 years ago
8 0

Answer:

Gravitational potential energy,Kinetic energy.     when goes down potential goes into kinetic.    Cant reach x because not enough potential energy

Explanation:

shepuryov [24]3 years ago
5 0

Answer:

The energy changes from gravitational potential energy to kinetic energy to potential energy (static) again.

It does not have sufficient potential energy to overcome the 30 m elevation.  

Explanation:

The car at its highest point possesses gravitational potential energy which is given by the following:

E_{p} = mgh

This energy upon rolling is converted to kinetic energy given by:\

E_{k} = \frac{1}{2}mv^{2}

The kinetic energy is then changed to static potential energy after the 10 m hump.

The car cannot react point X because from the 10 m hump it does not have sufficient potential energy to overcome the 30 m elevation.  

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Answer:

819.78 m

Explanation:

<u>Given:</u>

  • OA = range of initial position of the airplane from the point of observation = 375 m
  • OB = range of the final position of the airplane from the point of observation = 797 m
  • \theta = angle of the initial position vector from the observation point = 43^\circ
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A diagram has been attached with the solution in order to clearly show the position of the plane.

\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} =  (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} =  (-708.34\ \hat{i}+412.67\ \hat{j})\ m

The magnitude of the displacement vector = \sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

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