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Nastasia [14]
2 years ago
14

The quark composition of the proton and neutron are, respectively, uud and udd, where u is an up quark (charge +23e) and d is a

down quark (charge −13e). There are also anti-up u¯ (charge −23e) and anti-down d¯ (charge +13) quarks. The combination of a quark and an antiquark is called a meson. The mesons known as pions have the composition π+=ud¯ and π−=u¯d. Suppose a proton collides with an antineutron. During such collisions, the various quarks and antiquarks annihilate whenever possible.When the remaining quarks combine to form a single particle, it is a:_________.A) ProtonB) NeutronC) π+D) π−
Physics
1 answer:
ludmilkaskok [199]2 years ago
8 0

Answer:

Option C=> π+.

Explanation:

Just as it is given in the question above, we can see that the addition or combination of proton and neutron gives what is known as QUARKS.

Quarks are not easily measured because one can not see and study a quark independently; quarks move in multiples together.

The study of Quarks is very important in physics because they relate very well with electromagnetic force, strong force, weak force and Gravitational force.

"...When the remaining quarks combine to form a single particle, it is a π+".

π+ is a meson or a pion and it contains quarks and anti-quarks too

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There are areas in California that use the heat from deep in the Earth to generate electricity.

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It comes from the earth itself and we use a lot of things that comes from the earth and deep within it.

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compared to an object that does not roll, but instead slides without friction, should a rolling object be released from the same
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from a lesser hight

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Why is the sky blue?
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Elsie is finishing second grade. If she goes to school 147 day per year and she have 10 years of school left, how many days of s
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6 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
3 years ago
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