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levacccp [35]
3 years ago
10

Where is the location potential energy is converted to kinetic energy? Please answer asap!!!!!

Physics
1 answer:
kolezko [41]3 years ago
7 0

Potential energy is the energy that is possible to be realized, based on position, while  the kinetic energy is the energy that is present and due to motion

The location where the potential energy is converted into kinetic energy is the point (2)

Reason:

The mechanical energy, M.E., of the roller coaster, is given as follows;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy = m·g·h

K.E. = The kinetic energy = \dfrac{1}{2} \cdot m \cdot v^2

Where;

v = The velocity

h = The height

m = The mass

g = Acceleration due to gravity

∴ M.E. = m·g·h + \dfrac{1}{2} \cdot m \cdot v^2

The mechanical energy, M.E., is constant, therefore;

When the height, h = 0, we have, the kinetic energy is maximum, given that the mechanical energy is entirely kinetic energy

Therefore;

The locations where potential energy is converted to kinetic energy are lowest locations, where the height, <em>h</em>, is minimum, such as the point (2)

Learn more about potential and kinetic energy here:

brainly.com/question/18963960

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Answer:

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Explanation:

  1. Under the assumption that the can is a closed system, the conservation law applied to the system would be: E_{in}-E_{out}=E_{change}, where E_{in} is all energy entering the system, E_{out} is the total energy leaving the system and, E_{change} is the change of energy of the system.
  2. As the purpose is to kept the beverage can at constant temperature, the change of energy (E_{change}) would be 0.
  3. The energy  that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4) where \varepsilon is the emissivity of the surface, \sigma=5.67*10^{-8}\frac{W}{m^2K} known as the Stefan–Boltzmann constant, A_S is the total area of the exposed surface, T_S is the temperature of the surface in Kelvin, T_{\infty} is the environment temperature in Kelvin.
  4. For the can the surface area would be ta sum of the top and the sides. The area of the top would be A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2, the area of the sides would be A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2. Then the total area would be A_{total}=A_{top}+A_{sides}=0.01624m^2
  5. Then the radiation heat transferred to the can would be Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W.
  6. The can would lost heat evaporating water, in this case would be Q_{out}=\frac{dm}{dt}*h_{fg}, where \frac{dm}{dt} is the rate of mass of water evaporated and, h_{fg} is the heat of vaporization of the water (2257\frac{J}{g}).
  7. Then in the conservation balance: Q_{in}-Q_{out}=Q_{change}, it would be1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0.
  8. Recall that 1W=1\frac{J}{s}, then solving for \frac{dm}{dt}:\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}
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