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baherus [9]
2 years ago
11

Examine the roller coaster track above. Assume there is negligible friction as the roller coaster moves from position A to posit

ion F. Also assume that the roller coaster has minimal kinetic energy at point A. Create three different models below to describe the kinetic energy, the potential energy, and the total energy of the roller coaster at point E. Examples of models include, but are not limited to: pie charts, bar graphs, equations, textual descriptions, and analogies.

Physics
1 answer:
anyanavicka [17]2 years ago
5 0

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

#SPJ1

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Two vehicles A and B accelerate uniformly from rest.
spayn [35]

Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, a_A = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - a_A·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, V_{18A} = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, a_B = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ v_{18B} = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, v_{18B} = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle S_A = A₁ + A₂ + A₃

∴ S_A = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle S_A = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, S_B = 440 m

The distance of the two vehicles apart at the 18t second, S_{AB} = S_B - S_A

∴ S_{AB} = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, S_{AB} = 60 m.

5 0
3 years ago
Randomization in an experiment means that the experimental units or subjects are selected as a simple random sample from the who
vampirchik [111]
True
The sample of the experiment is randomized in randomization.
4 0
3 years ago
What does Newtons first law of motion say about objects at rest and objects in motion?
Alborosie

Answer:

Explanation:

an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

6 0
3 years ago
Which statement is correct?
Sedbober [7]

Answer:

The answer is "Choice C ".

Explanation:

The relationship between the E and V can be defined as follows:

\to E= -\Delta V

Let,

\to E= \frac{\delta V}{\delta x}

When E=0

\to   \frac{\delta V}{\delta x}=0

v is a constant value

Therefore, In the electric potential in a region is a constant value then the electric-field must be into zero that is everywhere in the given region, that's why in this question the "choice c" is correct.

7 0
2 years ago
How much heat is released when 30 g of water at 96°C cools to 25°C?The specific heat of water is 1 cal/gºC.
Blababa [14]
2130 cal is how much is released qhen 30 g of water at
3 0
3 years ago
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