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lawyer [7]
3 years ago
9

Two dimensional motion is not always projectile motion. In two dimensional motion that is not projectile motion, what is not acc

elerating the object
Physics
2 answers:
tigry1 [53]3 years ago
5 0

Two dimensional motion is not always projectile motion. In two dimensional motion that is not projectile motion, gravity is not accelerating the object

MatroZZZ [7]3 years ago
4 0

gravity is the correct answer.

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In the space below, explain why you agree or disagree with the first statement: Each person in a family has the same traits. The
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I disagree.

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When a wave is acted upon by an external damping force what happens to the energy of the wave
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At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen
Misha Larkins [42]

Answer:

  1. The specific mechanical energy of the air in the specific location is 40.5 J/kg.
  2. The power generation potential of the wind turbine at such place is of 2290 kW
  3. The actual electric power generation is 687 kW

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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