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zvonat [6]
3 years ago
7

If the atomic weight- # of protons = # of neutrons, then how many neutrons does neon have?

Physics
1 answer:
chubhunter [2.5K]3 years ago
3 0

Answer:

atomic weight 20.180

protons 10

neutrons 2,8

Explanation:

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<u>B</u><u>.</u><u> </u><u>Transverse</u><u> </u><u>wave</u><u>.</u>

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Because it has troughs and crests.

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A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed
Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

\int EdS=\frac{Q_{in}}{\epsilon_o}   (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum =  8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

\int EdS=ES=E(4\pi r^2)   (2)

Qin is calculate by using the charge density:

Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho  (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}

Next, you use the results of (3), (2) and (1):

E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}

hence, the electric field is 1580594.95 N/C

7 0
3 years ago
-In the Bohr model, as it is known today, the electron is imagined to move in a circular orbit about a stationary proton. The fo
vampirchik [111]

Answer:

4.3859007196\times 10^{-9}\ m

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

v = Velocity of electron = 2.4\times 10^5\ m/s

q = Charge of electron = 1.6\times 10^{-19}\ C

m = Mass of electron = 9.11\times 10^{-31}\ kg

r = Radius

The electrical and centripetal force will balance each other

\dfrac{kq^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{kq^2}{mv^2}\\\Rightarrow r=\dfrac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times (2.4\times 10^5)^2}\\\Rightarrow r=4.3859007196\times 10^{-9}\ m

The radius of the orbital is 4.3859007196\times 10^{-9}\ m

4 0
3 years ago
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