Answer:
The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is
8.00 x 10-13J
Explanation:
In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).
Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)
Although, the kinetic energy is converted to potential energy in Coulomb's law equation.
That is,
1/2(mv^2) = (K* q1q2)/r
Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter
Answer with Explanation:
We are given that
Area of loop=

Resistance, R=
B=
We know that magnetic flux

Emf ,
Current, 
Current, 
Substitute t=0 s
Then, I=
=1.6 A
Substitute t=1 s
Then, I=
=0
Substitute
t=2 s
Current, I=
=1.6 A
The answer is : 35 to 52 minutes.
Hope this helps!
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