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Answer:
(A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Explanation:
Given that,
A nucleus undergoes a nuclear decay.
(A). In alpha decay,
We know that,
When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Fr.
(B). In beta-minus decay,
We know that,
When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Po.
(C). In beta-plus decay,
We know that,
When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Ne.
(D). In gamma decay,
We know that,
When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Tc.
Hence, (A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Answer:
The bullet's initial speed is 243.21 m/s.
Explanation:
Given that,
Mass of the bullet,
Mass of the pendulum,
The center of mass of the pendulum rises a vertical distance of 10 cm.
We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,
V is the speed of the bullet and pendulum at the time of collision
Now using conservation of momentum as :
Put the value of V from equation (1) in above equation as :
So, the bullet's initial speed is 243.21 m/s.