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timama [110]
4 years ago
6

Which scenario describes a transfer of mechanical energy to heat and sound energy? Question 7 options: A) a very high pitched so

und causing a glass to break B) clapping hands C) talking into a microphone. D) a plant using energy from the sun to create food
Chemistry
1 answer:
Likurg_2 [28]4 years ago
8 0

B

<u>Clapping hands</u>  is a best scenario that describes a transfer of mechanical energy to heat and sound energy

Explanation:

The action of clapping your hands is mechanical hence has mechanical energy. When you clap your hands you make a sound because you cause vibrations in the air around the clapping hands. This is a transformation of the mechanical energy to sound energy. You realize that clapping your hands causes your hands to get warmer. Some of the mechanical energy was also transformed into heat energy.

Learn More:

For more on energy transformation check out;

brainly.com/question/11505666

brainly.com/question/11475287

#LearnWithBrainly

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A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/
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Complete Question

The complete question is shown on the first uploaded image  

Answer:

The concentration of  high-pressure side is  C_1 =   0.722 \ kg/m^3

Explanation:

From the question we are told that

    The thickness of the polyethylene is  d  =  1.5 \ mm = 0.0015 \ m

     The  temperature is  T  =  600 \   K

      The flux is  JA  =  2.48 *10^{-5} \  kg/m^2\cdot s

      The concentration on the low-pressure side is  C_2 =  0.5 \ kg/m^3

       The initial diffusivity  is  D_o  =  6.2 *10^{-4} \ m^2 /s

       The activation energy for  diffusion is   Q_d  =  41 \ kJ /mol  =  41*10^3 J /mol

Generally the diffusivity  of the oxygen at 600 K can be  mathematically evaluated  as

        D   = D_o * e^{- \frac{Q_d}{R * T  } }  

substituting values  

         D   = 6.2*10^{-4} * e^{- \frac{41 *10^3}{8.314 * 600  } }  

          D   = 1.671 *10^{-7} \ m^2 /s  

Generally the flux is mathematically represented as

          JA  =  D  *  \frac{C_1 -C_2}{d}

Where C_1 is the concentration of oxygen at the higher side

       So  

             C_1 =    d  * \frac{JA}{D}  + C_2

substituting values  

             C_1 =    0.0015   * \frac{2.48*10^{-5}}{1.671*10^{-7}}  + 0.5

              C_1 =   0.722 \ kg/m^3

 

6 0
3 years ago
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