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4 years ago

Fluorine (F) has nine protons and ten neutrons. The atomic mass of fluorine is _____.

1 answer:

wel4 years ago

8 0

The atomic mass is roughly equal to the sum of protons and neutrons of an atom.
9 protons + 10 neutrons = 19
Atomic mass of Flourine is 19.

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For each of the following decays or reactions, name at least. one conservation law that prevents it from occurring.(c) p → π⁺ +

Amanda [17]

(c)p→π⁺₊π⁺₊π

Baryon number is +1 on the left side of the equation, 0 on the

right side. Baryon number is not conserved.

<h3>How do you determine whether a baryon number is conserved?</h3>

According to the law of conservation of baryon number, the sum of the baryon numbers of all incoming particles equals the sum of the baryon numbers of all particles produced by the reaction. Energy, and so on, are conserved even if the incoming proton has sufficient energy and charge.

<h3>What is Baryon Number</h3>

In particle physics, the baryon number denotes which particles are baryons and which particles are not. Each baryon has a baryon number of 1, and each antibaryon has a baryon number of -1. Other non-baryonic particles have a baryon number of 0. Since there are exotic hadrons like pentaquarks and tetraquarks, there is a general definition of baryon number as:

B=1/3( $n_{q} -n_q^{-} }$ )

where $n_{q}$ represents the number of quarks and nq represents the number of antiquarks.

To learn more about Baryon Number refer to

brainly.com/question/10358797

#SPJ4

4 0

2 years ago

A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)