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pickupchik [31]
2 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]2 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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A cart of mass m moving right at speed v with respect to the track collides with a cart of mass 0.7m moving left.
Zolol [24]

Answer:

10v / 7

Explanation:

Using the conservation law of momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m v - 0.7 m v₁ =  ( 0.7 m + m) 0 m/s since the cart stuck together after collision. taken right to be positive and left to be negative

m v - 0.7 m v₁ = 0

- 0.7 m v₁ = -m v

v₁ = -m v / - 0.7 m = 10v / 7

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3 years ago
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ludmilkaskok [199]

Answer:

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Explanation:

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3 years ago
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How much heat is absorbed by a 75g iron skillet when it’s temperature rises from 5c to 20c?
alexira [117]

Answer: Q=499.5 J  

Explanation:

The heat (thermal energy) absorbed by the iron skillet can be found using the following equation:

Q=m.C.\Delta T   (1)

Where:

Q is the heat  (absorbed)

m=75 g is the mass of the element (iron in this case)

C is the specific heat capacity of the material. In the case of iron is C=0.444\frac{J}{g\°C}

\Delta T=T_{f}-T_{i}=20\°C - 5\°C= 15\°C is the variation in temperature

Knowing this, lets rewrite (1) with these values:

Q=(75 g)(0.444\frac{J}{g\°C})(15\°C)  (2)

Finally:

Q=499.5 J  

6 0
2 years ago
A comet is traveling through space with speed 3.01 ✕ 104 m/s when it encounters an asteroid that was at rest. The comet and the
Tcecarenko [31]

Answer: 8.493(10)^{-3} m/s

Explanation:

According to the conservation of linear momentum principle, the initial momentum p_{i} (before the collision) must be equal to the final momentum p_{f} (after the collision):

p_{i}=p_{f} (1)

In addition, the initial momentum is:

p_{i}=m_{1}V_{1}+m_{2}V_{2} (2)

Where:

m_{1}=1.71(10)^{14} kg is the mass of the comet

m_{2}=6.06(10)^{20} kg is the mass of the asteroid

V_{1}=3.01(10)^{4} m/s is the velocity of the comet, which is positive

V_{2}=0 m/s is the velocity of the asteroid, since it is at rest

And the final momentum is:

p_{f}=(m_{1}+m_{2})V_{f} (3)

Where:

V_{f} is the final velocity

Then :

m_{1}V_{1}+m_{2}V_{2}=(m_{1}+m_{2})V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{m_{1}V_{1}}{m_{1}+m_{2}} (5)

V_{f}=\frac{(1.71(10)^{14} kg)(3.01(10)^{4} m/s)}{1.71(10)^{14} kg+6.06(10)^{20} kg}

Finally:

V_{f}=8.493(10)^{-3} m/s This is the final velocity, which is also in the positive direction.

8 0
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