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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.

You might be interested in
A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest),  = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0
2 years ago
Pions have a half-life of 1.8 x 10^-8 s. A pion beam leaves an accelerator at a speed of 0.8c. What is the expected distance ove
Nuetrik [128]

Answer:

the expected distance is 4.32 m

Explanation:

given data

half life time = 1.8 × 10^{-8} s

speed = 0.8 c = 0.8 × 3 × 10^{8}

to find out

expected distance over

solution

we know c is speed of light in air is 3 × 10^{8} m/s

we calculate expected distance by given formula that is

expected distance = half life time × speed   .........1

put here all these value

expected distance = half life time × speed

expected distance = 1.8 × 10^{-8} ×  0.8 × 3 × 10^{8}

expected distance = 4.32

so the expected distance is 4.32 m

5 0
2 years ago
A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r
Firlakuza [10]

Answer:

Speed of the this part is given as

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

P_1 + P_2 + P_3 = P_i

here we know that

P_1 = P_2

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

P_1 + P_2 = \sqrt{P^2 + P^2}

P_1 + P_2 = \sqrt2 mv

now from above equation we have

P_3 = -(P_1 + P_2)

mv_3 = -(\sqrt 2 mv)

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0
2 years ago
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person
Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

C=2\pi r    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

L=Lo[1+\alpha \Delta T]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0
2 years ago
Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic
Elena-2011 [213]

Answer:

Answer:

1.1 x 10^9 ohm metre

Explanation:

diameter = 1.5 mm

length, l = 5 cm

Potential difference, V = 9 V

current, i = 230 micro Ampere = 230 x 10^-6 A

radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m

Let the resistivity is ρ.

Area of crossection

A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2

Use Ohm's law to find the value of resistance

V =  i x R

9 = 230 x 10^-6 x R

R = 39130.4 ohm

Use the formula for the resistance

R=\rho \frac{l}{A}

\rho =\frac{RA}{l}

\rho =\frac{39130.4\times 0.05}{1.766\times 10^{-6}}

ρ = 1.1 x 10^9 ohm metre

Explanation:

7 0
2 years ago
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