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fomenos
3 years ago
9

Which type of circuit is becoming more common in residential electrical design and construction?

Physics
1 answer:
jekas [21]3 years ago
3 0

Communication circuit  <em>(D)</em> is becoming more common in residential electrical design and construction.  

LAN Ethernet cables, outlets, and even hubs and bridges, are being built into the walls of new homes, along with the usual electrical outlet wiring, to give the owner the networking infrastructure and internet access that everybody needs now ... without stringing a mess of cables on the floor and through doors all over the house.

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Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
  • T_3 = The third temperature of air inside the tire = 85^\circ C =(273+85)\ K=358 \ K
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  • V_2 = The second volume of air inside the tire = 30\% V_1 = 0.3V_1
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
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<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
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Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)

Part (a):

Using the above equation for this part of compression in the air, we have

\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa

Hence, the pressure in the tire after the compression is 3.81\times 10^5\ Pa.

Part (b):

Again using the equation for this part for the air, we have

\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

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