At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
<h3>What is the energy of the roller coaster at point E?</h3>
The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.
Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.
At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,
Learn more about potential and kinetic energy at: brainly.com/question/18963960
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Answer:
energy is equal to 1000 J
Explanation:
When the jumper is in the tent, he has a given height, this height gives him a gravitational potential energy, which forms his initial mechanical energy of 1000 J. After jumping, this energy is converted into elastic energy of the rope plus a remainder of potential energy gravitational, it does not reach the ground, but as the friction is negligible the total mechanical energy is conserved, therefore its energy is equal to 1000 J
This is a case of energy transformation, but the total value of mechanical energy does not change
Answer:
-30° C
Explanation:
Data provided in the problem:
The formula for conversion as:
F = (9/5)C + 32
Now,
for the values of F = -22 , C = ?
Substituting the value of F in the above formula, we get
-22 = (9/5)C + 32
or
-22 - 32 = (9/5)C
or
(9/5)C = - 54
or
C = - 54 × (5/9)
or
C = - 30 °
Hence, -22 Fahrenheit equals to -30°C
Answer:
the range or the ball is 48.81 m
Explanation:
given;
Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal.
find:
What is the range of the ball?
solution:
let Ф = 25°
Vo = 25 m/s
<u>consider x-motion using time of fight: x = Vox * t</u>
where x = R = range
t =<u> 2 Voy </u>
g
R =<u> Vo² sin (2Ф)</u>
g
plugin values into the formula:
R = <u>(25)² sin (2*25) </u>
9.81
R = 48.81 m
therefore, the range or the ball is 48.81 m
