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antiseptic1488 [7]
3 years ago
9

phase, color, and ductility are all examples of what type of property? chemical changeable physical external

Physics
1 answer:
svet-max [94.6K]3 years ago
3 0
<span>Phase, color, and ductility are all examples of physical external properties</span>
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tilte in the earth

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Which atomic model proposed that electrons move in specific orbits around the nucleus of an atom
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A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res
Luden [163]
The first thing you should know for this case is the definition of distance.
 d = v * t
 Where,
 v = speed
 t = time
 We have then:
 d = v * t
 d = 9 * 12 = 108 m
 The kinetic energy is:
 K = ½mv²
 Where,
 m: mass
 v: speed
 K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
 The work due to friction is
 w = F * d
 Where,
 F = Force
 d = distance:
 w = 400 * 108 = 4.32 * 10 ^ 4
 The power will be:
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 answer:
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3 years ago
When drinking at a private even, you should assume that drinks will ___.
allochka39001 [22]

When drinking at a private event, you should assume that drinks will be STRONGER THAN NORMAL.

At private events, some hosts have the habit of mixing different drinks together in order to increase the intoxicating power of the drinks. This does not normally happen when one is buying from restaurants or other commercial places. Thus, to be on the safe side, one should always assume that drinks will be stronger when one is attending a private event, this will caution one to drink responsibly in order to avoid intoxication.

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3 years ago
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
6 0
3 years ago
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