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pickupchik [31]
2 years ago
8

An electron is emitted from a nucleus- State the effect this has on the charge of the nucleus

Physics
2 answers:
abruzzese [7]2 years ago
8 0

The charge of the nucleus becomes positive.

The effect of an electron being removed has on the charge of the nucleus is actually quite simple.

Think about it like this. An atom has an equal amount of electrons and protons.

Say, Hydrogen.

Protons:1

Electrons:1

The charge is equal. Now, take away an electron.

Protons:1

Electrons:0

There is no negativity now, that an electron is absent.

This makes the charge of the nucleus <em>positive</em>, due to the absence of an electron.

Hope this helps and please consider marking branliest.

Likurg_2 [28]2 years ago
5 0

Question

An electron is emitted from a nucleus- State the effect this has on the charge of the nucleus

Answer:

an electron orbiting around the nucleus combines with a nuclear proton to produce a neutron, which remains in the nucleus, and a neutrino, which is emitted.  As in positron emission, the nuclear positive charge and hence the atomic number decreases by one unit, and the mass number remains the same.

hope this helps!!!

-xXxAnimexXx-

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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m
marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

4 0
3 years ago
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
2 years ago
Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
yaroslaw [1]

Answer:

 K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

          p₀ = 0

final moment

          p_f = m_a v_a + m_b v_b

          p₀ = p_f

          0 = m_a v_a + m_b v_b

tells us that

          m_a = 8 m_b

         

           0 = 8 m_b v_a + m_b v_b

           v_b = - 8 v_a                    (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

           Em₀ = K₀ = 73 J

final point. After separating the body

          Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

           K₀ = K_f

           73 = ½ m_a (v_a² + v_b² / 8)

           

we substitute equation 1

           73 = ½ m_a (v_a² + 8² v_a² / 8)

           73 = ½ m_a (9 v_a²)

           73/9 = ½ m_a (v_a²) = K_a

            K_a = 8,111 J

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3 years ago
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ikadub [295]
The correct answer is False
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earnstyle [38]
Divide 14 by 6 and there is your answer with the unit of m
8 0
3 years ago
Read 2 more answers
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