Answer:
minimum number of photon is 4.05 ×
Explanation:
given data
energy = 50 keV = 50 × eV = 50 ×
× 1.602×
J
thickness = 10^-3
contrast = 1%
to find out
number of incident photons
solution
we know here equation that is
E = n × h × ν .......................1
put here all these value
50 × = n × 6.6×
× c/ 1×
50 × × 1.602×
= n × 6.6×
×( 3 ×
/ 1×
)
solve it and find n
n = 4.05 ×
so here minimum number of photon is 4.05 ×
Answer:
a) force between them is attraction, b) F = 1.125 10⁻² N
Explanation:
In this case the electric force is given by Coulomb's law
F =
In the exercise they give us the values of the loads
q1 = - 10 mC = -10 10⁻³ C
q2 = 5 mC = 5 10⁻³ C
d = 20 cm = 0.20 m
let's calculate
F = 9 10⁹
F = 1.125 10⁻² N
To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction