How is energy transferred throughout the roller coaster

Physics

2 answers:

Angelina_Jolie [31]3 years ago

5 0

Answer:

On a roller coaster, energy changes from potential to kinetic energy and back again many times over the course of a ride. Kinetic energy is energy that an object has as a result of its motion. ... Traditionally, the coaster cars are pulled up the first hill by a chain; as the cars climb, they gain potential energy.

Explanation:

solmaris [256]3 years ago

3 0

Answer:

On a roller coaster, energy changes from potential to kinetic and back again many times over and over the course of the ride. Kinetic energy is energy that an object has because of its motion. All moving objects possess kinetic energy, which is determined by the mass and speed of the object.

Explanation:

You might be interested in

The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in

bezimeni [28]

1) Potential difference: 1 V

2) $V_b-V_a = -1 V$

Explanation:

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

$\Delta U=q\Delta V$

where

q is the charge's magnitude

$\Delta V$ is the potential difference between the initial and final position

In this problem, we have:

$q=4.80\cdot 10^{-19}C$ is the magnitude of the charge

$\Delta U = 4.80\cdot 10^{-19}J$ is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

$\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V$

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

$V_b-V_a = - 1V$