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Tresset [83]
3 years ago
15

Compare and contrast potential and kinetic energy

Physics
2 answers:
inysia [295]3 years ago
7 0
Kinetic and Potential Energy HistoryA roller coaster train going down hill represents merely a complex case as a body is descending an inclined plane. Newton's first two laws relate force and acceleration, which are key concepts in roller coaster physics. At amusement parks, Newton's laws can be applied to every ride. These rides range from 'The Swings' to The 'Hammer'. Newton was also one of the developers of calculus which is essential to analyzing falling bodies constrained on more complex paths than inclined planes. A roller coaster rider is in an gravitational field except with the Principle of Equivalence.Potential EnergyPotential energy is the same as stored energy. The "stored" energy is held within the gravitational field. When you lift a heavy object you exert energy which later will become kinetic energy when the object is dropped. A lift motor from a roller coaster exerts potential energy when lifting the train to the top of the hill. The higher the train is lifted by the motor the more potential energy is produced; thus, forming a greater amount if kinetic energy when the train is dropped. At the top of the hills the train has a huge amount of potential energy, but it <span>has very little kinetic energy.Kinetic Energy The word "kinetic" is derived from the Greek word meaning to move, and the word "energy" is the ability to move. Thus, "kinetic energy" is the energy of motion --it's ability to do work. The faster the body moves the more kinetic energy is produced. The greater the mass and speed of an object the more kinetic energy there will be. Hope this helped:))))</span>
mariarad [96]3 years ago
3 0
Kinetic Energy is moving energy. And potential energy is air G that has the potential to be movie or not moving at all
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Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

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Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s   [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

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