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PSYCHO15rus [73]
3 years ago
5

Resistors and reactors, for use over 600 volts, shall not be installed in close enough proximity to combustible materials to con

stitute a fire hazard and shall have a clearance of not less than ___ from combustible materials.
Physics
1 answer:
suter [353]3 years ago
4 0

Resistors and reactors, for use over 600 volts, shall not be installed in close enough proximity to combustible materials to constitute a fire hazard and shall have a clearance of not less than<u> 300 mm </u>from combustible materials.

Explanation:

  • The hazards associated with high power industrial resistors are primarily due to their open construction, which is necessary for cooling.
  • The exposed conductors which make up the resistors can be not only a shock hazard but also a thermal burn hazard.
  • When a resistor fails, it either goes open or the resistance increases. When the resistance increases, it can burn the board, or burn itself up.
  • Avoid touching non-flammable resistors in operation; the surface temperature ranges from approximately 350 °C to 400°C when utilized at the full rated value. Maintaining a surface temperature of 200°C or less will extend resistors service life.
  • Do not apply power to a circuit while measuring resistance. When you are finished using an ohmmeter, switch it to the OFF position if one is provided and remove the leads from the meter.
  • Always adjust the ohmmeter for 0 (or in shunt ohmmeter) after you change ranges before making the resistance measurement.

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g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 × 1026 W and assume that th
vagabundo [1.1K]

Answer:

The value is N  =  1.107 *10^{45 }  \ photons    

Explanation:

From the question we are told that

   The  power output from the sun is  P_o =  4 * 10^{26} \  W

   The average wavelength of each photon is  \lambda  = 550 \  nm  =  550 *10^{-9} \  m

Generally the energy of each photon emitted is mathematically represented as

        E_c =  \frac{h * c  }{ \lambda }

Here  h is the Plank's constant with value  h  =  6.62607015 * 10^{-34} J \cdot s

          c is the speed of light with value  c =  3.0 *10^{8} \  m/s

So

       E_c =  \frac{6.62607015 * 10^{-34}  * 3.0 *10^{8}  }{ 550 *10^{-9} }          

=>   E_c =  3.614 *10^{-19} \  J          

Generally the  number of photons emitted by the Sun in a second is mathematically represented as

         N  =  \frac{P }{E_c}

=>      N  =  \frac{4 * 10^{26} }{3.614 *10^{-19}}

=>      N  =  1.107 *10^{45 }  \ photons    

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