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olganol [36]
2 years ago
11

Gas pressure in a closed system is caused by___. A. Heat given off by constant collisions.. B. Compression of the gas. C. Result

ant force of all combined collisions.. D. Rapid diffusion of gas through the system
Physics
2 answers:
jekas [21]2 years ago
8 0
Gas pressure in a closed system is caused by resultant force of all combines collisions. The correct option among all the options that are given in the question is the third option or option "C". The other options can easily be ignored. I hope the answer comes to your help.
Blababa [14]2 years ago
8 0

Answer: Option (A) is the correct answer.

Explanation:

In a closed system, when matter present in the system is hot then there will be number of collisions between the molecules of matter. As a result, molecules will have kinetic energy.

This energy will be given off by the molecules in the form of heat due to constant collisions.

Therefore, we can conclude that gas pressure in a closed system is caused by heat given off by constant collisions.

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A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre
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Answer:

The time taken is  t =  0.356 \ s

Explanation:

From the question we are told that

  The length of steel the wire is  l_1  = 31.0 \ m

   The  length of the  copper wire is  l_2  = 17.0 \ m

    The  diameter of the wire is  d =  1.00 \ m  =  1.0 *10^{-3} \ m

     The  tension is  T  =  122 \ N

     

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              t  =  t_s  +  t_c

Where  t_s is the time taken to transverse the steel wire which is mathematically represented as

         t_s  = l_1 *  [ \sqrt{ \frac{\rho * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_s is the density of steel with a value  \rho_s  =  8920 \ kg/m^3

   So

      t_s  = 31 *  [ \sqrt{ \frac{8920 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_s  = 0.235 \ s

 And

        t_c is the time taken to transverse the copper wire which is mathematically represented as

      t_c  = l_2 *  [ \sqrt{ \frac{\rho_c * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_c is the density of steel with a value  \rho_s  =  7860 \ kg/m^3

 So

      t_c  = 17 *  [ \sqrt{ \frac{7860 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_c  =0.121

So  

   t  = t_c  + t_s

    t =  0.121 + 0.235

    t =  0.356 \ s

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