Write the nuclear equation for the release of the beta particles by Pb-210.
1 answer:
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If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.
Reasoning:
It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.
Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.
Thus the electric force applied on the charge in my hand due to each other is,
Here k is the Coulomb constant, and r is the distance between the charges.
It is also known that the force on a positive charge due to another positive charge is acted outwards.
Thus, the force on the charge due to the charge on the east is,
And the force on the charge due to the charge on the north is,
As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of from the north or from the north direction.
Thus the net force is acting in the north-east direction.
Learn more about the direction of the force here,
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Explanation:
Given that,
The mean kinetic energy of the emitted electron,
(a) The relation between the kinetic energy and the De Broglie wavelength is given by :
(b) According to Bragg's law,
n = 1
For nickel,
As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.
1) Maximum
2) Maximum
Explanation:
The force acting on a mass on a spring is given by Hooke's law; in magnitude:
where
F is the force
k is the spring constant
x is the displacement
Also we know from Newton's second law that we can write
where
m is the mass
a is the acceleration
So we can write the equation as
(1)
From this relationship, we see that the acceleration is directly proportional to the displacement.
On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by
(2)
where the first term is the elastic potential energy while the second term is the kinetic energy, and where
v is the velocity of the mass
From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.
Therefore this means that:
- When acceleration increases, velocity decreases
- When acceleration decreases, velocity increases
Therefore, the two answers here are:
- When the acceleration of a mass on a spring is zero, the velocity is at a maximum
When the velocity of a mass on a spring is zero, the acceleration is at a maximum