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Sauron [17]
2 years ago
11

The bar graph shows energy data taken from a roller coaster at a theme park. analyze the data and assess its validity. 3-5 sente

nces, record your conclusions.

Physics
1 answer:
hichkok12 [17]2 years ago
4 0

This question involves the concepts of the law of conservation of energy, potential energy, and kinetic energy.

The data shown by the bar graph is "valid".

According to the law of conservation of energy, the total energy of the system must remain constant at any given point. Hence, the sum of the kinetic energy and the potential energy of the roller coaster must be constant at any given time.

Considering the bottom of the first hill:

Total Energy at 1st Hill = Kinetic Energy + Potential Energy

From the data given in the bar graph:

Total Energy at 1st Hill = 2,500,000 J + 0 J (since at the bottom potential energy is zero due to zero height)

Total Energy at 1st Hill = 2,500,000 J

Now, considering the top of the second hill:

Total Energy at 2nd Hill = Kinetic Energy + Potential Energy

From the data given in the bar graph:

Total Energy at 2nd Hill = 1,000,000 J + 1,500,000 J

Total Energy at 2nd Hill = 2,500,000 J

Hence,

Total Energy at 1st Hill = Total Energy at 2nd Hill

Therefore, the given bar graph is "valid".

Learn more about the law of conservation of energy here:

brainly.com/question/20971995?referrer=searchResults

The attached picture explains the law of conservation of energy.

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
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(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

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Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

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\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

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