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devlian [24]
2 years ago
13

The bar graph shows energy data taken from a roller coaster at a theme park. Analyze the data and assess its validity. In 3–5 se

ntences, record your conclusions.

Physics
1 answer:
alexdok [17]2 years ago
6 0

The data given in the bar graph is valid because it follows the law of conservation of energy, since the GPE at top of 2nd hill plus KE at top of 2nd hill equals KE at bottom of 1st hill.

<h3>What is law of conservation of energy?</h3>

The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.

Based on the law of conservation of energy, kinetic energy of a roller coaster can be converted into potential energy of the roller coaster and vice versa.

ΔK.E = ΔP.E

where;

  • ΔK.E is change in kinetic energy
  • ΔP.E is change in potential energy

The kinetic energy of the coaster is greatest at the bottom of the hill, as the coaster moves upward, the kinetic energy decreases and will be converted into potential energy. The potential energy of the coaster increases as the coaster moves up the hill and will become maximum at the highest point of the hill.

From the given data;

GPE at top of 2nd hill + KE at top of 2nd hill = KE at bottom of 1st hill

Learn more about conservation of energy here: brainly.com/question/166559

#SPJ1

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(a) -1.5\cdot 10^6 N

First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:

v^2 -u ^2 = 2ad

where

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

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Solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2

And the average force on the person is given by

F=ma

with m = 75.0 kg being the mass of the person. Substituting,

F=(75)(-20000)=-1.5\cdot 10^6 N

where the negative sign means the force is opposite to the direction of motion of the person.

b) -1.0\cdot 10^5 N

In this case,

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d  = 15.00 cm = 0.15 m is the displacement of the person with the air bag

So the acceleration is

a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2

So the average force on the person is

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