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:
The solution was to implement an absolute magnitude scale to provide a reference between stars. To do so, astronomers calculate the brightness of stars as they would appear if it were 32.6 light-years, or 10 parsecs from Earth.
Explanation:
Answer:
C
Explanation:
F=ma
given solution
v=12m/s a=v/t
s=6 sec =12m/s÷6sec
=2m/s^2 then we get acceleration now we will find the mass. first derive the the formula of mass by crisis cross then you will get this formula which is m=F/a
=36÷2
= 18
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