Answer:
Explanation:
From the question we are told that
Nucleus diameter 
a 12C nucleus
Required kinetic energy 
Generally initial speed of proton must be determined,applying the law of conservation of energy we have
where
=initial kinetic energy
=final kinetic energy
=initial electric potential
=final electric potential
mathematically
where
=distance b/w charges
=nucleus charge 
=constant
=proton charge
Generally kinetic energy is know as
Therefore
Generally equation for radius is 
Mathematically solving for radius of nucleus
Generally we can easily solving mathematically substitute into v_1
Therefore the proton must be fired out with a speed of
Answer:
(a)0.0675 J
(b)0.0675 J
(c)0.0675 J
(d)0.0675 J
(e)-0.0675 J
(f)0.459 m
Explanation:
15g = 0.015 kg
(a) Kinetic energy as it leaves the hand
(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J
(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.
(d) The gravitational energy at peak point would also be the same as 0.0675J
(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J
(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:
mgh = 0.0675
0.015*9.81h = 0.0675
h = 0.459 m
The ball would reach a maximum height of 0.459 m
Answers :
1. All points of a conductor are at the same potential. - True
2. Charges prefer to be uniformly distributed throughout the volume of a conductor. - False
3 The electric field inside the conducting material is always zero. -True
4.Just outside the surface of a conductor, the electric field is always zero. - False
Answer:
1. 8437500 N
2. The force between the two charges is attractive.
Explanation:
1. Determination of the force between the two charges.
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
Distance apart (r) = 80 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 2 × 3 / 80²
F = 5.4×10¹⁰ / 6400
F = 8437500 N
Thus, the force of attraction between the two charges is 8437500 N
2. From the question given, the charges are:
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.
Thus the force between them is attractive.
