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Lynna [10]
3 years ago
15

Which part of the diagram represents the proton

Physics
2 answers:
blsea [12.9K]3 years ago
7 0

Answer:

C

Explanation:

An atom consists essentially of three types of particles:

- Protons: these are contained in the nucleus, and they are positively charged

- Neutrons: this are also contained in the nucleus, but they have no electric charge

- Electrons: they orbit around the nucleus, and they have negative charge

Based on this diagram, we see that the particle labeled by C is a proton, since it is in the nucleus and it is positively charged. The particle labeled by B is a neutron, while the particle labeled by A is an electron.

Alexandra [31]3 years ago
6 0
Protons have a positive charge which is indicated by a + sign.
Hence, the answer is C.
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State coulomb's law mathematically
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Coulomb's law is express as:

\begin{gathered} F=k\frac{q_1q_2}{r^2} \\ \text{ where} \\ k\text{ is Coulomb's constant} \\ q_1\text{ and }q_2\text{ are the charges} \\ r\text{ is the distance between charges} \end{gathered}

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
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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
  • V_1 = The first volume of air inside the tire
  • 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
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • n = number of moles of air

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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