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Helen [10]
2 years ago
11

You drop a rock into a pond, and water waves spread out in circles. (a) The waves carry water outward, away from where the rock

hit. That moving water carries energy outward. (b) The waves only make the water move up and down. No energy is carried outward from where the rock hit. (c) The waves only make the water move up and down, but the waves do carry energy outward, away from where the rock hit.
Physics
1 answer:
coldgirl [10]2 years ago
4 0

Answer:

Explanation:

wave motion is defined as the propagation of the disturbance due to continous vibration of the molecules of the medium. Waves oduced due to disturbance carry only energy but not the matter.

The waves move water up and down and carry the energy.

Therefoe, the statement "(a) The waves carry water outward away from where the rock hit. The moving water carries energy outward"is incorrect.

The waves move water up and down and carry the energy, but not matter.

Therefore, the statement "(c) The waves ony make the water move u and down. No energy is carried outward from where the rock hit" is incorrect.

When a rock is droed ito a pd, a disturbance starts from the point where the stone hits the water. The waves produced carry the energy from the starting point toother locatons over a large distance which is not possible by the single molecules of water.

Therefore, the statement "(c) The waves only make the water move up and down, but the waves do carry energy outward away from where the rock hit" is correct.

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  2. The power generation potential of the wind turbine at such place is of 2290 kW
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Explanation:

  1. The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: \frac{1}{2} V^2 where V is the velocity of the air.
  2. The specific kinetic energy would be: \frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}.
  3. The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
  4. To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: \frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}
  5. Then the mass flow is obtain from the volumetric flow times the density of the air: m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}
  6. Then, the Power generation potential is: 40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW
  7. The actual electric power generation is calculated using the definition of efficiency:\eta=\frac{E_P}{E_I}}, where η is the efficiency, E_P is the energy actually produced and, E_I is the energy input. Then solving for the energy produced: E_P=\eta*E_I=0.30*2290kW=687kW
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