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pickupchik [31]
3 years ago
10

The nucleus of an atom consists of protons and neutrons (no electrons). A nucleus of a carbon‑12 isotope contains six protons an

d six neutrons, while a nitrogen‑14 nucleus comprises seven protons and seven neutrons. A graduate student performs a nuclear physics experiment in which she bombards nitrogen‑14 nuclei with very high speed carbon‑12 nuclei emerging from a particle accelerator. As a result of each such collision, the two nuclei disintegrate completely, and a mix of different particles are emitted, including electrons, protons, antiprotons (with electric charge −???? each), positrons (with charge +???? each), and various neutral particles (including neutrons and neutrinos). For a particular collision, she detects the emitted products and find 17 protons, 4 antiprotons, 7 positrons, and 25 neutral particles. How many electrons are also emitted?
Physics
1 answer:
andriy [413]3 years ago
4 0

Answer:

7 electrons

Explanation:

We can solve the problem by using the law of conservation of electric charge: in fact, the total electric charge before and after the collision must be conserved.

Before the collision, we have:

- A nucleus of carbon-12, consisting of 6 protons (charge +1 each) + 6 neutrons (charge 0 each), so total charge of +6

- A nucleus of nitrogen-14, consisting of 7 protons (charge +1 each) + 7 neutrons (charge 0 each), so total charge of +7

So the total charge before the collision is +6+7=+13 (1)

After the collision, we have:

- 17 protons (charge +1 each): total charge of +17

- 4 antiprotons (charge -1 each): total charge of -4

- 7 positrons (charge +1 each): total charge of +7

- 25 neutral particles (charge 0 each): total charge of 0

- N electrons (charge -1 each): total charge of -N

So the total charge after the collision is +17-4+7+0-N=+20-N (2)

Since the charge must be conserved, we have (1) = (2):

+13 = +20 - N

Solving for N,

N = 20 - 13 = 7

So, there are 7 electrons.

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3 0
3 years ago
Una barra de aluminio que esta a 78 GRADOS CENTIGRADOS entra en contacto con una barra de cobre de la misma longitud y área que
stiks02 [169]

Answer:

Al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

Explanation:

Dado que una barra de aluminio que está a 78 grados centígrados entra en contacto con una barra de cobre de la misma longitud y área que esta a 28 grados centígrados, y posteriormente se lleva acabo la transferencia de energía entre ambas barras llegando a su equilibrio térmico, para determinar la temperatura a la que ambas barras llegarán se debe realizar el siguiente cálculo:

(78 + 28) / 2 = X

106 / 2 = X

53 = X

Por lo tanto, al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

8 0
3 years ago
A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
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B).  No.  The heat in the room goes from the floor to the ceiling
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C).  No.  There's nothing for the heat to soak through between
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D).  No.  Cold water sinks from the surface to the bottom because
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