To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.
The energy fluency is given by the equation

Where
The energy fluency
c = Activity of the source
r = distance
E = electric field
In the other hand we have the equation for current in materials, which is given by

Then replacing our values we have that


We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.
Now the energy fluency would be,



The uncollided flux density at the outer surface of the tank nearest the source is 
Answer:
a) 39.6 m/s b) 4123 N
Explanation:
a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).
Fnet=ma
ma=m(v^2/R) (centripetal acceleration)
mg=m(v^2/R)
m cancels out (this is why pilot feels weightless) so,
g=(v^2/R)
9.8 m/s^2 = v^2/160 m
v^2=1568 m^2/s^2
v=39.6 m/s
b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.
Convert 300 km/hr to m/s
300 km/hr=83.3 m/s
Convert pilot's weight into mass:
760 N = 77.55 kg
Fnet=ma
n-mg=m(v^2/R)
n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)
n=3363.2 N+760 N=4123 N
Explanation:
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Answer:
(a) 0.3778 eV
(b) Ratio = 0.0278
Explanation:
The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

(a) The energy of the electron in n= 6 excited state is:


Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV
(b) For first orbit energy is:




Ratio = 0.0278