Answer:
hello your question lacks the required diagram attached below is the complete question with the required diagram
answer : Qtotal = 807.4 Mw
Explanation:
Given Data :
disk properties :
∈ = 0.65
D = 200 mm
Ts = 400⁰c
attached below is the detailed solution
The total rate of Heat transferred from the disk
Qtotal = 807.4 Mw
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
6. Drop to one quarter of its original value
