Answer:
(a): 
(b): 
(c): 
Explanation:
Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053
m.
Part (a):
According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges 
and 
respectively is given by
where,
= Coulomb's constant = 
= distance of separation between the charges.
For the given system,
The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, 
The charge on the electron, 
These two are separated by the distance, 
Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by
Part (b):
The gravitational force of attraction between two objects of masses 
and respectively is given by
where,
= Universal Gravitational constant = 
- = distance of separation between the masses.
For the given system,
The mass of proton, 
The mass of the electron, 
Distance between the two,
Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by
The ratio 
:
Answer:
See explanation
Explanation:
We have to convert to angular velocity in rads-1 as follows;
Angular velocity in rad/s = 2π/60 × 1900 rpm = 199 rad/s
Given that
angular velocity =angle turned /time taken
Time taken = angle turned/angular velocity
Converting 35° to radians we have;
35 × π/180 = 0.61 radians
Time taken = 0.61 radians/199 rad/s
Time taken = 0.0031 seconds
For the part a) we need only the momentum of the box and we have the data to find it.
Momentum is given by,
where clearly, p is the momentum, m the mass of the box and v is the velocity.
Substituting,
For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,
<em>It is the same that part a)</em>
Answer:
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Explanation:
